JPH0466301B2 - - Google Patents

Description

[Detailed description of the invention]

(Industrial Application Field) The present invention relates to a photometry device that divides a field into a plurality of regions, obtains luminance information for each of the plurality of regions, and obtains a photometric value through various calculations. (Prior Art) Various photometering devices have been proposed in the past that divide a field into a plurality of areas, measure light for each area, and use these plurality of photometric values to give an appropriate exposure to the photographic screen. ing. However, conventional photometering devices correct the average light metering value to give approximately the correct exposure for the entire subject, centering on a specific part of the subject. I couldn't consciously control highlights or shadows, which caused problems such as whitish subjects not appearing white or dark subjects not appearing black. (Object of the Invention) The present invention has been made to solve the problems of conventional photometering devices. Select a calculation method that can obtain exposure information that depicts highlights, and select a calculation method that can obtain exposure information that depicts shadows when the luminance information of the surrounding area is smaller than the second reference value on the low luminance side. Therefore, it is an object of the present invention to provide a photometric device that can photograph whitish objects as white and dark objects as black. (Example) FIG. 1 is a schematic diagram of an optical system when the present invention is applied to a single-lens reflex camera. In the same figure,
1 is the photographic lens, 2 is the quick return mirror,
3 is a focus plate, 4 is a pentaprism, 5 is an imaging lens, 6 is a light receiving section, and 7 is an image plane. In this embodiment, a subject image formed on a focus plate 3 by a photographing lens 1 is guided and imaged onto a light receiving section 6 by an imaging lens 5, and photometry is performed. FIG. 2 is an explanatory diagram of the light-receiving surface of the light-receiving section 6 shown in FIG. 1. In FIG. 2, 2A is an area at the approximate center of the field, 2B is an intermediate area surrounding area 2A, and 2C is an area at the periphery of the screen surrounding area 2B. , 2C ₁ to 2C ₄ . In this embodiment, as shown in FIG. 2, a plurality of light-receiving elements capable of receiving light in the field area at positions corresponding to each area are arranged, and the field is divided into six areas 2A, 2B, 2C ₁ to 2C. It is divided into 2C ₄ areas and the field brightness is photometered for each area. 3 to 6 are circuit diagrams for explaining the circuit configuration of the present invention. In Figure 3, 8, 9, 10, 11, 12,
13 is the above six areas 2A, 2B, 2C ₁ , 2
It is a silicon photodiode (SPD) corresponding to C ₂ , 2C ₃ , 2C ₄ and generates photocurrent i _A , i _B , i _C1 , i _C2 , i _C3 , i _C4 according to the brightness of each area. . 14 to 19 are logarithmic compression circuits that logarithmically compress these photocurrents and output voltage values V _A , V _B , V _C1 , V _C2 , V _C3 , and V _C4 . V _A , V _B , V _C1 ,
V _C2 , _C3 , V _C4 are constants a ₁ , a ₂ , a ₃ , a ₄ , a ₅ , a ₆ (
0), b (>0) and photocurrents i _A , i _B , i _C1 , i _C2 , i _C3 ,
It can be expressed as follows using i _C4 . V _A =a ₁ +b lni _A V _B =a ₂ +b lni _B V _C1 =a ₃ +b lni _C1 V _C2 =a ₄ +b lni _C2 V _C3 =a ₅ +b lni _C3 V _C4 =a ₆ +b lni _C4However , When a ₁ , a ₂ , a ₃ , a ₄ , a ₅ , and a ₆ have the same brightness in each area, V _A = V _B = V _C1 = V _C2 =
Logarithmic compression circuits 14 to 19 so that V _C3 = V _C4
It is assumed that the value is set in advance. 20 are output voltages V _C1 , V _C2 , and V C2 of the logarithmic compression circuits 16 to 19;
V _C3 and V _C4 are input terminals I11, I12, I13, respectively.
This is a peripheral brightness value calculation circuit which inputs the voltage to I14, calculates the brightness value of the peripheralmost part 2C of the object field, and outputs the voltage V _C from the output terminal O1. The configuration of this peripheral brightness value calculation circuit 20 is shown in FIG. In Fig. 4, 32, 33, 34, 35, and 36 are magnitude comparison circuits, which output the larger of the two voltage values input to I1 and I2 from the OH output terminal, and output the smaller one from the OL output terminal. This is a circuit that outputs from the end. A specific example of the magnitude comparison circuits 32 to 36 is shown in FIG.
In Fig. 5, 66 is a comparator with a positive phase/negative phase input terminal and a positive phase/negative phase output terminal, and when the positive phase input terminal voltage V+ and the negative phase input terminal voltage V- are V+V-, the positive phase output is An H-level voltage is output from the negative-phase output terminal, an L-level voltage is output from the negative-phase output terminal, and when V+<V-, an L-level voltage is output from the positive-phase output terminal, and an H-level voltage is output from the negative-phase output terminal. It's becoming like that. Analog switches 67 to 70 become conductive when the voltage applied to the control terminal is at H level, and open when the voltage is at L level. Assuming that the voltage input to the I1 input terminal is V _I1 and the voltage input to the I2 input terminal is V _I2 , when V _I1 V _I2 , the positive phase input terminal voltage of the comparator 66 is V +
The negative phase input terminal voltage becomes V-. At this time, the positive phase output terminal voltage of the comparator 66 becomes H level, the negative phase output terminal voltage becomes L level, and the analog switch 6
Since the control voltages of switches 7 and 70 are at H level and the control voltages of analog switches 68 and 69 are at L level, analog switches 67 and 70 are in a conductive state and analog switches 68 and 69 are in an open state. Therefore, the voltage V I1 is outputted to the OH output terminal through the analog switch 67, and the voltage V _I1 is outputted to the OL output terminal through the analog switch 70.
V _I2 is output. Similarly, when V _I1 < V _I2 ,
The voltage at the positive phase input terminal of the comparator 66 becomes V+<the voltage at the negative phase input terminal V-, and the voltage at the positive phase output terminal of the comparator 66 becomes L level and the voltage at the negative phase output terminal becomes H level. Therefore, since the control voltages of the analog switches 67 and 70 are at L level, the analog switches 67 and 70 are in the open state, and the control voltages of the analog switches 68 and 69 are at the H level.
level, so analog switches 68, 69
becomes conductive. At this time, the OH output terminal has
The voltage V _I2 is outputted through the analog switch 69, and the voltage V _I1 is outputted to the CL output terminal through the analog switch 68. In this way,
The larger of the voltages V _I1 and V _I2 is output to the OH output terminal, and the smaller one is output to the OL output terminal. Fourth
In the figure, using five magnitude comparison circuits, the voltages V _C1 ,
The largest of V _C2 , V _C3 , and V _C4 is V _H1
, the second largest one is V _H2 , the third largest one is V _H3 , and the smallest one is
The configuration is configured to output as V _H4 . The voltage V _C1 is input to the I1 input terminal of the magnitude comparison circuit 32, and the voltage V _C2 is input to the I2 input terminal, and the larger one is output from the OH output terminal, and the smaller one is output from the OL output terminal. are doing. Similarly, voltage V _C3 is input to the I1 input terminal of the magnitude comparison circuit 33, voltage V _C4 is input to the I2 input terminal, and the larger one is output from the OH output terminal, and the smaller one is input.
It is output from the OL output terminal. Size comparison circuit 34
Now, input the voltage output from the OH output terminals of the magnitude comparison circuits 32 and 33 to the I1 and I2 input terminals,
Comparing the sizes, the larger one is output from the OH output end, and the smaller one is output from the OL output end.
Here, the voltage output from the OH output terminal is V _C1 ,
This is the largest voltage among V _C2 , V _C3 , and V _C4 . Similarly, in the magnitude comparison circuit 35, the magnitude comparison circuit 3
The voltage output from the OL output terminal of 2 and 33 is I1,
The voltages are input to the I2 input terminal and compared, and the larger one is outputted from the OH output terminal and the smaller one is outputted from the OL output terminal. Here, the voltages output from the OL output terminal are V _C1 , V _C2 , This is the smallest voltage among V _C3 and V _C4 . The magnitude comparison circuit 36 is the magnitude comparison circuit 34
The OL output terminal voltage of the OL output terminal voltage and the OH output terminal voltage of the magnitude comparison circuit 35 are inputted to the I1 and I2 input terminals, and the magnitudes are compared. In the magnitude comparison circuit 36, V _C1 , V _C2 ,
It compares and outputs the two intermediate voltage values excluding the maximum and minimum of V _C3 and V _C4 , and the magnitude comparison circuit 36
2 of V _C1 , V _C2 , V _C3 , and V _C4 from the OH output terminal of
The second largest voltage is output, and from the OL output terminal,
The third largest voltage is output. In this way, V _C1 , V _C2 , V _C3 , and V _C4 are set to V _H1 in descending order.
It is rearranged so that it becomes V _H2 V _H3 V _H4 . 37
~40 are resistors with the same resistance value,
This is for averaging V _C1 , V _C2 , V _C3 , and V _C4 . Vcm=(V _C1 +V _C2 +V _C3 +V _C4 )/4.
41 is a reference voltage generation circuit, which generates four reference voltages.
Vr ₁ , Vr ₂ , Vr ₃ , and Vr ₄ are generated. Reference voltage
The magnitude relationship of Vr ₁ , Vr ₂ , Vr ₃ , and Vr ₄ is Vr ₁ > Vr ₂ >
Vr ₃ > Vr ₄ . 42 to 45 are comparators, which compare Vcm with reference voltages Vr ₁ , Vr ₂ , Vr ₃ , Vr ₄ , and compare Vcm with reference voltages Vr ₁ , Vr 2 , Vr ₃ , Vr ₄ ,
When VcmVr _{is 4} , comparators 42, 4
3, 44, and 45 output H level voltages, Vcm<Vr ₁ , Vcm<Vr ₂ , Vcm<Vr ₃ , Vcm
When <Vr ₄ , an L level voltage is output. 46, 47, 48, and 49 are inverters, and 50, 51, and 52 are AND gates. The input of the inverter 46 is connected to the output of the comparator 42, the input of the inverter 47 is connected to the output of the comparator 43, and the input of the inverter 48 is connected to the output of the comparator 42.
The input of the inverter 49 is connected to the output of the comparator 45. and gate 5
One input of the AND gate 51 is connected to the output of the inverter 46, the other to the output of the comparator 43, and one input of the AND gate 51 is connected to the output of the inverter 47.
The other is connected to the output of the comparator 44, one input of the AND gate 52 is connected to the output of the inverter 48, and the other is connected to the output of the comparator 45. 53 to 56 are or gates. Each of the three inputs of the OR gate 53 is connected to the comparator 4.
2, the output of AND gate 50, and the output of AND gate 51. or gate 54
The four inputs are the output of the comparator 42, the output of the AND gate 50, and the AND gate 51, respectively.
The output of the AND gate 52 is connected to the output of the AND gate 52. The four inputs of the OR gate 55 are the output of the AND gate 50, the output of the AND gate 51, and
It is connected to the output of the AND gate 52 and the output of the inverter 49. Three inputs of the OR gate 56 are connected to the output of the AND gate 51, the output of the AND gate 52, and the output of the inverter 49. 5
Analog switches 7 to 60 are controlled by OR gates 53 to 56, respectively. 61 to 64 are resistors having the same resistance value. Resistors 61-64 are analog switches 57-6
V _H1 , V H2 , V _H2 , which is output by the conductive one among 0
This is for averaging the voltages of two, three, or four of V _H3 and V _H4 in descending order, or two or three voltages in ascending order of magnitude. 65 is an operational amplifier, the output end and the negative phase input end are connected, and it is used as a voltage follower. operational amplifier 6
The output terminal voltage of No. 5 is equal to the positive phase input terminal voltage regardless of the circuit state after the output terminal. Let the output voltage of this operational amplifier 65 be V _C ,
It is output from the O1 output terminal. Next, in Figure 4
From the magnitude relationship of Vr ₁ , Vr ₂ , Vr ₃ , and Vr ₄ with Vcm,
The characteristics of the field and the value of voltage V _C will be described. (1) Vr ₁ Vcm
...Analog switches 57 and 58 conduction At this time, since a scene with a fairly bright background is assumed, the brightness of the relatively low brightness areas of 2C ₁ , 2C ₂ , 2C ₃ , and 2C ₄ is ignored, V _C =
(V _H1 +V _H2 )/2. (2) Vr ₂ Vcm＜Vr ₁
...Analog switches 57, 58, 59 conduction At this time, a scene with a slightly bright background, such as outdoors on a sunny day, is assumed. Inappropriate elements tend to appear as brightness. Therefore, in this case, the brightness of the lowest brightness area is ignored and V _C =(V _H1 +V _H2 +V _H3 )/3. (3) Vr ₃ Vcm＜Vr ₂
...Analog switches 57, 58, 59, and 60 conduction At this time, a scene with a background of standard brightness is assumed, and in such a scene, relatively bright and relatively dark parts of the subject cannot be ignored. do not have. Therefore, in such a case, V _C = (V _H1 + V _H2 +
V _H3 +V _H4 )/4. (4) Vr ₄ Vcm＜Vr ₃
...Analog switches 58, 59, 60 conduction At this time, since a scene with a somewhat dark background is assumed, ignoring the highest brightness area, V _C = (V _H2
+V _H3 +V _H4 )/3. (5) Vcm＜Vr ₄
...Analog switches 59 and 60 conduction At this time, a scene with a fairly dark background, such as a night view, is assumed, and in such a scene, the brightness of a locally high-luminance background, such as electric lamp light, is inappropriate. elements are likely to appear. Therefore, in this case, ignoring relatively high brightness regions, V _C
=(V _H3 +V _H4 )/2. As described above, the peripheral brightness calculation circuit 20 calculates V _H1 , V _H2 , and
2 or 3 or 4 from the larger of V _H3 or V _H4 , or 2 or 3 from the smaller
The average value of the two is set as V _C and output from the O1 output terminal. 21 in FIG. 3 is a logarithmic compression circuit 14, 15
_The output voltages V _A and V _B of
3 is a selection circuit that determines which arithmetic expression to select from a plurality of arithmetic expressions to be described later. In FIG. 3, 22, 23, and 24 are resistors having the same resistance value, and 25 is an analog switch. An average value circuit is configured by the resistors 22, 23, 24 and the analog switch 25, and when an H level voltage is output from the A/P output terminal of the selection circuit 21 and input to the control terminal of the analog switch 25, the analog switch is activated. 25 becomes conductive, and the output voltage V ₁ of the average value circuit is
(V _A + V _B + V _C )/3. On the other hand, selection circuit 2
When an L level voltage is output from the A/P output terminal of 1 and input to the control terminal of the analog switch 25, the analog switch 25 becomes open and the output voltage V ₁ of the average value circuit becomes (V _A +
V _B )/2. Reference numeral 26 designates an operational amplifier, whose output end and negative phase input end are connected to be used as a voltage follower, and the output voltage V ₁ of the average value circuit is input to the positive phase input end. The output terminal voltage of the operational amplifier 26 is V ₁ regardless of the circuit state after the output terminal. 27, 28, 29, 3
0 is a resistor having the same resistance value, and 31 is an operational amplifier. The resistors 27, 28, 29, 30 and the operational amplifier 31 constitute a subtraction circuit.
If the O2 output end voltage of the selection circuit 21 is _V2 , the output voltage of this subtraction circuit is _V1 - _V2 . In this embodiment, this voltage V ₁ -V ₂ represents a photometric value determined by a plurality of calculation formulas described later. FIG. 6 is a circuit diagram of the selection circuit 21 in FIG. 3. Resistors 71, 72, 73, and 74 have the same resistance value, and 75 is an operational amplifier, which constitutes a first subtraction circuit. Similarly, 76, 77, 7
8 and 79 are resistors having the same resistance value, and 80 is an operational amplifier, which constitutes a second subtraction circuit.
The voltages V _A and V _B are input to the first subtraction circuit,
Its output voltage is V _B −V _A. The voltages V _B and V _C are input to the second subtraction circuit, and its output voltage is
V _C −V _B. 81 is a reference voltage generation circuit which generates reference voltages Vr ₅ , Vr ₆ , Vr ₇ (Vr ₅ >Vr ₆ >Vr ₇ ). 82, 83, 84 are comparators. A voltage V _C is input to each positive phase input terminal, and reference voltages Vr ₅ , Vr ₆ , Vr ₇ are input to each negative phase input terminal, and when the comparator 82 is V _C Vr ₅ ,
The comparator 83 outputs an H level voltage, and outputs an L level voltage when V _C < Vr ₅ , and the comparator 83 outputs an L level voltage when V _C < Vr _5.
The comparator 84 outputs an H level when V _C <Vr ₆ , and an L level when V _C <Vr ₇ , and outputs an L level when V _C <Vr ₇ . 85 is a reference voltage generation circuit that generates reference voltages V _Pa , V _Pb , V _Qa ,
Generates V _Qb . The reference voltage generation circuit 85 has a control terminal B/D, and the output voltage of the comparator 83 is inputted as a control voltage. The reference voltage when an H level control voltage is input to B/D is V _Pa = V _P1 , V _Pb = V _P2 ,
V _Qa = V _Q1 , V _Qb = V _Q2 , while L is applied to the terminal B/D.
The reference voltage when the level control voltage is input is V _Pa = V _P3 , V _Pb = V _P4 , V _Qa = V _Q3 , V _Qb
Let = V _Q4 . The signs of the reference voltages V _P1 to V _P4 , V _Q1 to V _Q4 are positive for V _P1 , V _P3 , V _Q1 , and V _Q3 , and positive for V _P2 , V _P4 ,
V _Q2 and V _Q4 are negative. 86 and 87 are comparators, and 88 and 89 are inverters. The output voltage V _B −V _A of the operational amplifier 75 of the first subtraction circuit is input to the positive phase input terminals of the comparators 86 and 87 . The reference voltage V P1 or _{V P3 is} input to the negative phase input terminal of the comparator 86 depending on the H level or L level of the control voltage input to the control terminal B/D of the reference voltage generating circuit 85.
Similarly, the reference voltage V _P2 or V _P4 is input to the negative phase input terminal of the comparator 87. 90,91
is a comparator, and 92 and 93 are inverters. At the positive phase input terminals of the comparators 90 and 91,
The output voltage V _C of the operational amplifier 80 of the second subtraction circuit
−V _B is input. A reference voltage V _{Q1 or V Q3 is} applied to the negative phase input terminal of the comparator depending on the H level or L level of the control voltage input to the control terminal B/D of the reference voltage generation circuit 85.
is input. Similarly, the reference voltage V _Q2 or V _Q4 is input to the negative phase input terminal of the comparator 90. The output terminals of the comparators 86, 87, 90, and 91 have an H level voltage when V+V-, and an L level voltage when V+<V-, depending on the magnitude of the positive phase input terminal voltage V+ and the negative phase input terminal voltage V-. Voltage is output. 94 to 102 are AND gates, and depending on the combination of H level and L level at the output terminals of the comparators 86, 87, 90, and 91, one of the AND gates outputs an H level voltage, and the other AND gates output an H level voltage. The gate outputs an L level voltage. V _B −V _A and V _Pa and V _Pb , V _C −V _B and V _Qa and
Which AND gate among the AND gates 46 to 54 outputs an H level in relation to V _Qb will be described below. 1 V _B −V _A V _Pa V _C −V _B V _Qa …And gate 94 V _Qa ＞V _C −V _B V _Qb … 〃 95 V _Qb ＞V _C −V _B … 〃 96 2 V _Pa ＞V _B − V _A V _Pb V _C −V _B V _Qa …And gate 97 V _Qa ＞V _C −V _B V _Qb … 〃 98 V _Qb ＞V _C −V _B … 〃 99 3 V _Pb ＞V _B −V _A V _C −V _B V _Qa …AND gate 100 V _Qa >V _C −V _B V _Qb … 〃 101 V _Qb >V _C −V _B … 〃 102 103, 104, 105 are inverters,
The output states of comparators 82, 83, and 84 are each inverted. 106 and 107 are AND gates, and the two inputs of the AND gate 106 are connected to the output of the inverter 103 and the output of the comparator 83, and the two inputs of the AND gate 107 are connected to the output of the inverter 104 and the output of the comparator 84. . In this way, comparator 8
2, 83, 84, and inverters 103, 104,
The voltage V _C is compared with reference voltages Vr ₅ , Vr ₆ , Vr ₇ by AND gates 106 and 107, and depending on the output states of the comparator 82, AND gate 106, AND gate 107, and inverter 105, the following is determined. The comparison results are classified as follows.

[Table] 108, 109 is or gate, 110, 111
is an AND gate, and 112 is an OR gate. The four inputs of OR gate 108 are AND gate 9
It is connected to the outputs of OR gates 8, 99, 101, and 102, and when any one of AND gates 98, 99, 101, and 102 outputs an H level voltage, the output terminal voltage of OR gate 108 becomes H level. . Similarly, the six inputs of the OR gate 109 are connected to the outputs of the AND gates 96, 98-102.
When any one AND gate outputs an H level voltage, the output terminal voltage of the OR gate 109 becomes H level. One of the two inputs of the AND gate 110 is connected to the output of the OR gate 108;
The other end is connected to the output terminal of AND gate 106. When AND gate 106 is outputting H level, the output of AND gate 110 is equal to the output terminal state of OR gate 108. At this time, since one of the two inputs of AND gate 111 is connected to the output of AND gate 107, one of the inputs of AND gate 111 is at L level, so its output is at L level. On the other hand, one of the two inputs of AND gate 111 is connected to the output of OR gate 109, and the other is connected to the output of AND gate 107. When the AND gate 106 is outputting an L level voltage, the L level voltage is input to one input of the AND gate 110, so the output of the AND gate 110 is an L level voltage.
At this time, if the output voltage of the AND gate 107 is H level, the H level voltage is input to one of the two inputs of the AND gate 111, so this output is equal to the output state of the OR gate 109. . Outputs of AND gates 110 and 111, output of comparator 82, and inverter 105
The outputs of are respectively input to the OR gate 112,
AND gates 110, 111, comparator 8
2. When at least one of the inverters 105 is outputting an H level voltage, the OR gate 112 outputs an H level voltage. The output terminal of the OR gate 112 serves as an A/P output terminal. 113-117, 119-123 are AND gates, 118, 124 are OR gates, 1
25 is Noah Gate. 126 is a reference voltage generation circuit, which generates reference voltages Vr ₈ , Vr ₉ , Vr ₁₀ , Vr ₁₁ , -
Vr ₁₂ , −Vr ₁₃ , −Vr ₁₄ , −Vr ₁₅ , Vr ₁₆ , −Vr ₁₇ are generated. 127-137 are analog switches. One input of the two-input AND gates 113 to 117 is connected to the output of the AND gate 106, and one input of the two-input AND gates 119 to 123 is connected to the output of the AND gate 107. The other input of the AND gate 113 is the output of the AND gate 94 and the AND gate 1.
The other input of AND gate 14 is the output of AND gate 95, the other input of AND gate 115 is the AND gate 96, the other input of AND gate 116 is the output of AND gate 99, and the other input of AND gate 117. The input is AND gate 102
is connected to the output of The other input of AND gate 119 is the output of AND gate 94, the other input of AND gate 120 is the output of AND gate 97, the other input of AND gate 121 is the output of AND gate 100, and the other input of AND gate 121 is the output of AND gate 100. The other input of gate 122 is AND gate 1
The output of AND gate 123 and the other input of AND gate 123 are connected to the output of AND gate 102. Therefore, when an H level voltage is output to the output terminal of the AND gate 106, the AND gates 113 to
H level voltage to one input of 117, 119 to 1
23, the output of the AND gate 113 is equal to the output state of the AND gate 94, and the output of the AND gate 114 is equal to the output state of the AND gate 95.
The output of AND gate 115 is equal to AND gate 96, the output of AND gate 116 is equal to the output state of AND gate 99, and AND gate 117 is equal to the output state of AND gate 99.
The output of AND gate 102 is equal to the output state, and the output voltage of AND gates 119 to 123 is L.
level. Conversely, when an H level voltage is output to the output terminal of AND gate 107, one input of AND gates 113 to 117 becomes an L level voltage, 1
Since the H level voltage is input to one input of AND gates 19 to 123, the output voltage of AND gates 113 to 117 is L level, the output of AND gate 119 is equal to the output state of AND gate 94, and the output state of AND gate 120 is The output of AND gate 121 is equal to the output state of AND gate 100, the output of AND gate 122 is equal to the output state of AND gate 101, and the output of AND gate 123 is equal to the output state of AND gate 102. is equal to the output state of or gate 11
One of the two inputs of 8 is AND gate 114
and the other is connected to the output of the AND gate 115. The output of the OR gate 118 is
It becomes H level when at least one of the output voltages of AND gates 115 and 116 is at H level. Of the two inputs of the OR gate 124,
One side is connected to the output of AND gate 121, and the other side is connected to the output of AND gate 122. The output of AND gate 124 is
It becomes H level when at least one of the output voltages of 1 and 122 is at H level. The 10 inputs of the NOR gate 125 are the output of the AND gate 113, the output of the OR gate 118, and the AND gate 1.
16 output, AND gate 117 output, AND gate 119 output, AND gate 120 output, OR gate 124 output, AND gate 12
3, the output of the comparator 82, and the output of the inverter 105, respectively. When the outputs of AND gates 113 to 117, 119 to 123, comparator 82, and inverter 105 are all at L level, the output voltage of NOR gate 125 is at H level; otherwise, the output voltage of NOR gate 125 is at L level. become. A reference voltage is applied from a reference voltage generation circuit 126 to the input terminals of analog switches 127-130 and 132-137. A reference voltage Vr ₈ is applied to the input terminal of the analog switch 127, a reference voltage Vr ₉ is applied to the input terminal of the analog switch 128, a reference voltage Vr ₁₀ is applied to the input terminal of the analog switch 129, and the analog switch 1
A reference voltage Vr ₁₁ is applied to the input terminal of the analog switch 130, a reference voltage Vr ₁₂ is applied to the input terminal of the analog switch 132, and a reference voltage Vr ₁₃ is applied to the input terminal of the analog switch 133.
The input terminal of the analog switch 134 has a reference voltage.
Vr ₁₄ , the input terminal of analog switch 135,
Reference voltage Vr ₁₅ and analog switch 136 have reference voltage Vr ₁₆ and analog switch 137 at the input terminal.
A reference voltage Vr ₁₇ is applied to the input terminal of.
The input terminal of analog switch 131 is OV.
The output terminals of analog switches 127-137 are connected to each other and become O2 output terminals. The control terminal of analog switch 127 is connected to the output of AND gate 113, the control terminal of analog switch 128 is connected to the output of OR gate 118, the control terminal of analog switch 129 is connected to the output of AND gate 116, and analog switch 130
The control terminal of the analog switch 131 is the output of the AND gate 117, the control terminal of the analog switch 131 is the output of the NOR gate 125, and the control terminal of the analog switch 132 is the AND gate 119.
The control terminal of the analog switch 133 is the output of the AND gate 120, the control terminal of the analog switch 134 is the output of the OR gate 124, the control terminal of the analog switch 135 is the output of the AND gate 123, and the control terminal of the analog switch 136. The terminals are the output of the comparator 82 and the analog switch 137.
The control terminal of is connected to the output of the inverter 105. Three inputs of the selection circuit in Figure 4
Regardless of the voltage levels of V _A , V _B , and V _C , regardless of the magnitude relationship,
AND gate 113, OR gate 118, AND gate 116, AND gate 117, NOR gate 125, AND gate 119, AND gate 12
0, the output terminal of one of the OR gate 124, the AND gate 123, the comparator 82, and the inverter 105 becomes H level, and the output terminals of the other gates become L level, so the analog switch 1
Among the analog switches 27 to 137, only one analog switch to which an H level control voltage is applied becomes conductive, and the other analog switches become open. When the output of the AND gate 113 is an H level voltage, the analog switch 127 becomes conductive and the voltage Vr ₈ is output from the O2 output terminal, and when the output of the OR gate 118 is an H level voltage,
The analog switch 128 becomes conductive and the voltage
Vr ₉ is output from O2 output terminal, AND gate 11
When the output of Vr 6 is an H level voltage, the analog switch 129 becomes conductive, and the voltage Vr ₁₀ becomes O2.
is output from the output terminal, and when the AND gate 117 output is H level voltage, the analog switch 13
0 becomes conductive, voltage Vr ₁₁ is output from the O2 output terminal, and when the NOR gate 125 is at H level voltage, the analog switch 131 becomes conductive, OV is output from the O2 output terminal, and AND gate 1
When the output of Vr 19 is an H level voltage, the analog switch 132 becomes conductive, and the voltage Vr ₁₂ becomes O2.
is output from the output terminal, and when the output of the AND gate 120 is an H level voltage, the analog switch 1
33 becomes conductive and the voltage Vr ₁₃ is output from the O2 output terminal, and when the output of the OR gate 124 is an H level voltage, the analog switch 134 becomes conductive and the voltage Vr ₁₄ is output from the O2 output terminal. When the output is H level voltage, the analog switch 135 becomes conductive and the voltage
Vr ₁₅ is output from the O2 output terminal and comparator 8
When the output of the inverter 105 is an H level voltage, the analog switch 136 becomes conductive and the voltage Vr ₁₆ is output from the O2 output terminal, and when the output of the inverter 105 is an H level voltage, the analog switch 137 becomes conductive and the voltage Vr 16 becomes conductive. ₁₇ is output from the O2 output terminal. However, Vr ₈ , Vr ₉ , Vr ₁₀ , Vr ₁₁ , Vr ₁₆
>0, −Vr ₁₂ , −Vr ₁₃ , −Vr ₁₄ , −Vr ₁₅ , −Vr ₁₈ <0
It is. Next, the circuit operations in FIGS. 3 and 6 will be explained with reference to FIGS. 7 to 9. (1) V _c > Vr _5Outermost area 2C of the field of view shown in Figure 2
When the brightness signal V _C obtained from the camera is greater than the predetermined value Vr ₅ , that is, when V _C > Vr ₅ , there may be a bright sky on a clear day, or a subject with sufficiently high brightness at the periphery of the screen. be judged. In such cases, if the brightness output of the subject is used as the exposure value, it will often be considerably underexposed, so the entire screen will be slightly highlighted (also known as highlight control), if the subject is whitish. Exposure compensation is performed on the brightness output of the subject to the side of overexposure so as to give an exposure that will ensure that the subject appears white. Therefore,
The photometric value V ₁ −V ₂ is for all areas 2A, 2B, 2C
The luminance signals V _A , V _B , V _C of
Using Vr ₁₆ , it is calculated using the following arithmetic expression (1). V ₁ −V ₂ =(V _A +V _B +V _C )/3−Vr ₁₆ …(1)− Then, regarding the circuit operations in FIGS. 3 and 6, first we will explain the selection circuit 21 in FIG. 6. Since the luminance signal (photometric signal) V _C obtained from the outermost region 2C (2C ₁ to 2C ₄ ) is larger than the reference voltage Vr ₅ , the comparator 82 outputs an H level,
Therefore, the OR gate 112 outputs an H level to set the output terminal A/P to an H level, and only the analog switch 136 is turned on, making the voltage at the O2 output terminal the reference voltage Vr ₁₆ from the reference voltage generating circuit 126. Therefore, the analog switch 25 in FIG. 3 becomes conductive, and the voltage (V _A +V _B +V _C )/3 is output from the operational amplifier 26.
1, the voltage Vr ₁₆ is supplied to the negative phase input terminal, so
The output V ₁ −V ₂ of the operational amplifier 31 is the voltage (V _A +
V _B +V _C )/3-Vr ₁₆ . (2) When the luminance signal V _C is greater than or equal to the predetermined value Vr ₆ but smaller than the predetermined value Vr ₅ , that is, Vr ₆ V _C <
Vr ₅ when the subject is determined to be a normal sunny day (does not include sunny days with very bright skies such as midsummer) or outdoors on cloudy days (when the sky is in the background, etc.) (when a bright subject enters the picture and it is determined that the peripheral area of the screen is bright), in this case, the luminance signal difference V _B - V _A (hereinafter abbreviated as ΔBA) and the luminance signal difference V _C - _{V B} (hereinafter abbreviated as ΔCB) The reference voltages as constants V _P1 , V _P2 , V _Q1 ,
V _Q2 (Relationship between reference voltages is V _P2 < 0 < V _P1 , V _Q2 <
0<V _Q1 ), the photometric value is calculated using the following calculation formula.
Find V ₁ − V ₂ . (2-1) When V _P2 <ΔBA<V _P1 V _Q1 <ΔCB, specifically, the luminance signal difference between the central area 2A and the first outer area 2B of the field area as shown in Figure 8a ΔBA becomes smaller, and on the other hand, the luminance signal difference ΔCB between the first outer region 2B and the outermost region 2C
becomes larger than the predetermined value V _Q1 on the + side, and in this case, it can be determined that the main subject is often present in both area 2A and area 2B. Therefore, the photometric value
V ₁ −V ₂ is set to give the main subject appropriate exposure.
Targeting the luminance signals V _A and V _B of the region 2A and the region 2B, the correction coefficient is set to 0, and is calculated using the following arithmetic expression (2). V ₁ -V ₂ =(V _A +V _B )/2...(2)- Then, regarding the circuit operations in FIGS. 3 and 6, first we will explain the selection circuit 21 in FIG. 6. Since the photometric signal (voltage) V _C of 2C (2C ₁ to 2C ₄ ) is larger than the reference voltage Vr ₆ and smaller than Vr ₅ , the comparators 83 and 84 output H level, and the comparator 82 outputs L level, The output of AND gate 106 is set to H level. Therefore, the control terminal B/ of the reference voltage generation circuit 85
An H level signal is supplied to D, and the reference voltages of the circuit 85 are V _Pa = V _P1 , V _Pb = V _P2 , V _Qa = V _Q1 , V _Qb
= V _Q2 . On the other hand, the output signal (voltage) V _B −V _A of the operational amplifier 75 satisfies the condition V _P2 <ΔBA < V _P1 , so the output of the comparator 86 is L level, the output of the comparator 87 is H level, and the operational amplifier 80 output signal (voltage) V _C −V _B is V _Q1
Since the condition is <ΔCB, the output of comparator 90 becomes H level, and the output of comparator 91 becomes H level. Therefore, only AND gate 97 is H.
The H level output of AND gate 97, the L level outputs of AND gates 94 to 96, 98 to 102, and the H level output of AND gate 106 cause OR gate 112 to go to L level and output terminal A/ to L level. At the same time, only the NOR gate 125 is set to H level, and the other AND gates 1
13 to 123 are set to L level, and the voltage at the O2 output terminal is set to OV. Therefore, operational amplifier 2 in Figure 3
6 outputs voltage (V _A + V _B )/2, and OV is supplied to the negative phase input terminal of operational amplifier 31, so the output V ₁ - V ₂ of operational amplifier 31 is the voltage (V _A + V _B )/2. (2-2) When V _P1 <ΔBA V _Q1 <ΔCB, specifically, as shown in Figure 8b, the luminance signal difference ΔBA between area 2A and area 2B is a predetermined value on the + side _V
Furthermore, since the luminance signal difference ΔCB between area 2B and area 2C is also larger than the predetermined value V _Q1 on the + side, it is determined that the main subject is often present in all of area 2A and part of area 2B. can. In this case, it is possible to calculate the photometric value V ₁ - V ₂ using only the luminance signal V _A of area 2A, but empirically, it is better to take the background luminance signal into account to some extent to obtain a better exposure value. become. Therefore, the photometric value V ₁ −V ₂
targets the main subjects V _A and V _B , and the correction coefficient Vr ₈
is calculated using the following arithmetic expression (2). V ₁ −V ₂ =(V _A +V _B )/2−Vr ₈ …(2)− Then, the circuit operation is performed by the comparators 88, 8
9, 92, 93 and AND gate 106 are all H
level, only the AND gate 94 outputs the H level, and the other AND gates 95 to 102
outputs L level. So, or gate 11
The output of the operational amplifier 26 becomes the voltage (V _A +V _B )/2, and the output of the operational amplifier ₃₁ becomes (V _A +V _{B )} . /2-Vr ₈ . (2-3) When V _P1 <ΔBA V _Q2 <ΔCB<V _Q1 , specifically, as shown in FIG. On _the other hand, since the luminance signal difference ΔCB between the area 2B and the area 2C is small, it can be determined that the main subject is present in the entire area 2A, or that the main subject is particularly small. In this case, the photometric value V ₁ −V ₂ may be calculated for the main subject in area 2A, but considering the luminance signal V _B of area 2B, which corresponds to the background,
In order to give appropriate exposure to the main subject, the photometric value V ₁ - V ₂ targets areas 2A and 2B, and using the correction coefficient Vr ₉ (Vr ₉ > Vr ₈ ), the following calculation formula (2) is calculated. −
Seek more. V ₁ −V ₂ =(V _A +V _B )/2−Vr ₉ …(2)− Then, the circuit operation is performed by the comparators 86, 87
and 90 all output H level, only AND gate 95 outputs H level, the other AND gates 94, 96 to 102 output L level, and AND gate 106 outputs H level. Therefore, the output of OR gate 112 is set to L level, the output of OR gate 118 is set to H level, and operational amplifier 2
The output of the operational amplifier 31 is the voltage (V _A + V _B )/2, and the output V ₁ - V ₂ of the operational amplifier 31 is (V _A + V _B )/2 - Vr ₉
becomes. (2-4) When V _P1 < ΔBA ΔCB < V _Q2 , specifically, as shown in _FIG . Larger, on the other hand, area 2B and area 2
Since the luminance signal difference ΔCB with C is smaller than the predetermined value V _Q2 on one side, in this case, the main subject is the above-mentioned (2-
When there is a high-brightness subject (e.g. the sun, reflection on the sea surface, etc.) that is similar in size to the case described in 3) and corresponds to area 2B, or when photographing a landscape, area 2B is used.
It can be determined that there is a subject with high brightness corresponding to . In this case, the photometric value V 1 −V ₂ is calculated by targeting all the luminances of areas 2A, 2B, and 2C in terms of data and setting the correction coefficient to 0 [ _{V 1 −V 2} =(V _A +V _A
+ _VC , area 2A and area 2B
It is calculated from the following equation (2) using the correction coefficient Vr ₉ . V ₁ −V ₂ =(V _A +V _B )/2−Vr ₉ …(2)− Then, the circuit operation is performed by the comparators 86, 86
and 87 output H level, on the other hand, comparators 90 and 91 output L level, AND gate 96 outputs H level, and other AND gates 94, 95, 97 to 102 output L level, AND gate 106 outputs H level.
Therefore, the output of OR gate 112 is set to L level, and the output of OR gate 118 is set to H level,
The output of the operational amplifier 26 is a voltage (V _A +V _B )/2, and the output V ₁ −V ₂ of the operational amplifier 31 is (V _A +
V _B )/2-Vr ₉ . (2-5) When V _P1 <ΔBA<V _P1 ΔCB<V _Q2 , specifically, as shown in FIG. 8e, the luminance signal difference ΔBA between area 2A and area 2B is small;
The luminance signal difference ΔCB between area 2B and area 2C is a negative value and the absolute value is larger than V _Q2 , so in this case it is determined that the main subject is large and exists in both areas 2A and 2B, and that it is whitish. can. In this case, the photometric value may be calculated based on the brightness of area 2A and area 2B, but if the peripheral area of the screen has lower brightness than the central area, the brightness of area 2C at the periphery of the screen may be considered to some extent. It is known that by calculating the photometric value, you can reliably capture a whitish main subject in white (highlight control), and that this will give you a good exposure value based on the data. Therefore,
At this time, all luminance signals V _A , V _B , and V _C are targeted in order to provide an exposure that highlights the main subject, and the photometric value V ₁ − V ₂ is calculated using a correction coefficient Vr ₁₀
is calculated using the following arithmetic expression (2). V ₁ −V ₂ =(V _A +V _B +V _C )/3−Vr ₁₀ …(2)− Then, the circuit operation is such that the comparator 87 outputs the H level, while the comparators 86, 90, and 91 output the L level. , the AND gate 99 outputs H level, and the other AND gates 94,
98, 100 to 102 output L level, and AND gate 106 outputs H level. Therefore, the output of the OR gate 112 is set to H level, and the output of the operational amplifier 26 is a voltage (V _A +V _B +
V _C )/3, and the output of the operational amplifier 31 is V ₁ −V ₂
is (V _A +V _B +V _C )/3-Vr ₁₀ . (2-6) When ΔBA<V _P2 ΔCB<V _Q2 , specifically, as shown in _FIG . The absolute value is large, and the luminance signal difference ΔCB between area 2B and area 2C is also a negative value, and the absolute value is larger than the predetermined value V _Q2 , so in this case, the main subject exists in all of area 2A and part of area 2B. It can be determined that this is a medium-sized case and a whitish subject. In this case, the above (2-5)
In order to provide an exposure that highlights the main subject, areas 2A, 2B, and 2
The brightness of C is the target, and the photometric value V ₁ − V ₂ is the correction coefficient
Using Vr ₁₁ (Vr ₁₁ < Vr ₁₀ ), the following calculation formula (2) −
Seek more. V ₁ −V ₂ =(V _A +V _B +V _C )/3−Vr ₁₁ …(2)− Then, the circuit operation is performed by the comparators 86 to 91
outputs L level, AND gate 102 outputs H level, and other AND gates 94 to 10
1 outputs the L level, and the AND gate 106 outputs the H level.
Output the level. Therefore, when the output of the OR gate 112 is set to H level and the output of the AND gate 117 is set to H level, the output of the operational amplifier 26 becomes the voltage (V _A + V _B + V _C )/3, and the output of the operational amplifier 31 becomes V ₁ - V ₂ (V _A +V _B +V _C )/3-Vr ₁₁ . (2-7) When ΔBA<V _P2 V _Q2 <ΔCB<V _Q1 , specifically, as shown in Figure 8g, the luminance signal difference ΔBA between area 2A and area 2B is a negative value and a predetermined value Since the absolute value is larger than V _P2 and the luminance signal difference ΔCB between area 2B and area 2C is small, in this case the main subject exists in the entire area 2A, or the main subject is particularly small, and the main subject is It can be determined that the subject is whitish. In this case as well, in order to obtain an exposure that highlights the main subject part as described above, target areas 2A, 2B, and 2C, set the correction coefficient to 0, and calculate from the following calculation formula (2). . V ₁ -V ₂ = (V _A + V _B + V _C )/3...(2)- Then, the circuit operation is such that the comparator 90 outputs an H level, while the comparators 86, 87, and 91 output an L level. , andgate 101
outputs H level, and the other AND gates 94
~100, 102 output L level, and AND gate 106 outputs H level. Therefore, the output of the OR gate 112 is set to H level, and the output of the NOR gate 125 is set to H level,
The output of the operational amplifier 26 is the voltage (V _A +V _B +V _C )/
3, and the output V ₁ −V ₂ of the operational amplifier 31 is (V _A
+V _B +V _C )/3. (2-8) When ΔBA<V _P2 V _Q1 <ΔCB, specifically, as shown in _FIG . Since the absolute value is large and the luminance signal difference ΔCB between area 2B and area 2C is larger than the predetermined value V _Q1 , in this case, the main subject is as described in (2-1) above.
In cases where the size is similar to the case described above and there is a difference in brightness in the main subject part, and area 2A is slightly high brightness, or in a landscape photograph, a subject with considerably low brightness occupies area 2B. It can be determined that this is the case. In this case, data-wise areas 2A, 2
Obtain the photometric value V ₁ −V 2 using all luminances of B and 2C as targets and _setting the correction coefficient to 0 [V ₁ −V ₂ = (V _A
+V _B +V _C )/3] was also recognized by the inventor as a method for obtaining good results, but in this example, another method for obtaining good results, namely, the above case (2-1) was used. Similarly, area 2A and area 2
Targeting the brightness of B, the correction coefficient is set to 0, and is calculated using the following arithmetic expression (2). V ₁ -V ₂ = (V _A + V _B )/2...(2)- Then, the circuit operation is performed by the comparators 90 and 91.
outputs H level, while comparators 86, 8
7 outputs L level and AND gate 100 outputs H level.
Output the level and other AND gates 94-9
9, 101, and 102 output L level, and AND gate 106 outputs H level. Therefore, the output of the OR gate 112 is set to the L level, the output of the NOR gate 125 is set to the H level, and the operational amplifier 2
The output of the operational amplifier 31 becomes a voltage (V _A +V _B )/2, and the output V ₁ -V ₂ of the operational amplifier 31 becomes (V _A +V _B )/2. (2-9) When V _P2 <ΔBA<V _P1 V _Q2 <ΔCB<V _Q1 , specifically, as shown in FIG. 8i, the luminance signal difference ΔBA between area 2A and area 2B is small;
Also, since the luminance signal difference ΔCB between area 2B and area 2C is small, it is determined that this is the case when the main subject occupies the entire scene, or when there is no intention to set the main subject, such as in landscapes. can. In this case, the following calculation formula (2)-
Seek more. V ₁ -V ₂ =(V _A +V _B +V _C )/3...(2)- Then, the circuit operation is performed by the comparators 87 and 9.
0 outputs H level, while comparator 86
and 91 output L level, AND gate 98
outputs H level, and the other AND gates 94
-97, 99-102 output L level, and AND gate 106 outputs H level. Then,
When the output of the OR gate 112 is set to H level and the output of the NOR gate 125 is set to H level, the output of the operational amplifier 26 becomes the voltage (V _A + V _B + V _C )/3,
The output V ₁ −V ₂ of the operational amplifier 31 is (V _A +V _B +
V _C )/3. (3) When the luminance signal V _C is greater than or equal to the predetermined value Vr ₇ but smaller than the predetermined value Vr ₆ , that is, Vr ₇ V _C <
As Vr ₆ , when it is determined that the subject is in a general case such as outdoors or indoors in the shade (when there is a subject such as an indoor wall in the background and it is determined that the peripheral area of the screen is dark), this When the luminance signal difference V _B −V _A
(ΔBA) and the luminance signal difference V _C −V _B (ΔCB), V _P3 , V _P4 , V _Q3 , V _Q4 as constants (reference voltage V _P4 <0<V _P3 , V _Q4 <0 <V _Q3 ), the photometric value V ₁ −V ₂ is determined by the following calculation formula. (3-1) When V _P4 <ΔBA<V _P3 V _P3 <ΔCB, specifically, as shown in Figure 9a, area 2A
The brightness signal difference ΔBA between area 2B and area 2B becomes small, and the brightness signal difference ΔCB between area 2B and area 2C reaches the predetermined value V _Q3.
In this case, the main subject is area 2A.
It can be determined that this is a large case that exists in both 2B and 2B, and that it is a case of a darkish subject. In this case, the photometric value may be calculated by simply targeting the brightness of areas 2A and 2B and setting the correction coefficient to 0, but it is necessary to ensure that the dark main subject appears black (shadow control). More preferably, in this embodiment, in order to provide exposure such that the main subject portion is depicted in shadow, areas 2A and 2 are
Targeting the luminance of B, the photometric value V ₁ −V ₂ is obtained from the following arithmetic expression (3) using the correction coefficient −Vr ₁₃ on one side. V ₁ −V ₂ = (V _A + V _B )/2 + Vr ₁₃ …(3)− Then, the circuit operation is performed in the outermost region 2C (2
C ₁ to 2C ₄ ) photometric signal (voltage) V _C is the predetermined voltage Vr ₆
Since it is smaller than Vr ₇ but larger than Vr 7, the comparator 83 outputs an L level signal, and therefore an L level signal is supplied to the control terminal B/D of the reference voltage generation circuit 85, and the reference voltage of the circuit 85 is V _Pa =
V _P3 , V _Pb = V _P4 , V _Qa = V _Q3 , V _Qb = V _Q4 .
On the other hand, comparator 86 outputs L level, comparators 87, 90 and 91 output H level, AND gate 97 outputs H level, and other AND gates 94-96, 98-1
02 outputs L level, and AND gate 107 outputs H level. Therefore, or gate 112
When the output of the AND gate 120 is set to L level and the output of the AND gate 120 is set to H level, the output of the operational amplifier 26 becomes the voltage (V _A +V _B )/2, and the output V ₁ -V ₂ of the operational amplifier 31 becomes (V _A +V _B )/2. 2-Vr becomes ₁₃ . (3-2) When V _P3 <ΔBA V _Q3 <ΔCB, specifically, as shown in FIG. 9b, the luminance signal difference ΔBA between area 2A and area 2B is larger than the predetermined value V _P3 , and Since the luminance signal difference ΔCB of 2C is larger than the predetermined value V _Q3 , it can be determined that the main subject is present in all of the area 2A and part of the area 2B, and is a blackish subject.
In that case, the above (3-1) should be applied to the main subject.
In order to give an exposure that depicts shadows in the same way as above, the brightness of areas 2A and 2B is targeted, and the photometric value V ₁ −V ₂ is set by the correction coefficient −Vr ₁₂ on one side (|Vr ₁₂
|<|Vr ₁₃ |), it is calculated from the following arithmetic expression (3)-. V ₁ −V ₂ = (V _A + V _B )/2 + Vr ₁₂ …(3)− Then, the circuit operation is performed by the comparators 86 to 91
all output H level, and the AND gate 94 outputs H level.
Output the level and other AND gates 95-1
02 outputs L level, and AND gate 107 outputs H level. Therefore, or gate 112
When the output of the AND gate 120 is set to L level and the output of the AND gate 120 is set to H level, the output of the operational amplifier 26 becomes a voltage (V _A +V _B )/2, and the output V ₁ -V ₂ of the operational amplifier 31 becomes (V _A +V _B )/2. 2+Vr ₁₂ . (3-3) When V _P3 <ΔBA V _Q4 <ΔCB<V _Q3 , specifically, as shown in Figure 9c, the area 2
The luminance signal difference ΔBA between area 2A and area 2B is larger than the predetermined value V _P3 , while the luminance signal difference ΔCB between area 2B and area 2C is small. It can be determined that this is a case where the subject is small and has a dark appearance. In this case, in order to provide exposure such that the main subject part is depicted in shadow, the luminance of areas 2A and 2B is targeted, and the photometric value V ₁ - V ₂
is calculated from the following equation (3) with the correction coefficient set to 0. V ₁ -V ₂ =(V _A +V _B )/2...(3)- Then, the circuit operation is as follows: comparator 91 outputs L level, comparators 86 to 90 output H level, and AND gate 95 outputs H level. outputs H level, and the other AND gates 94, 96 to 1
02 outputs L level, and AND gate 107 outputs H level. Therefore, or gate 112
is set to L level, NOR gate 125 is set to H level, and the output of operational amplifier 26 is set to voltage (V _A +
V _B )/2, and the output of the operational amplifier 31 is V ₁ −V ₂
Also becomes (V _A + V _B )/2. (3-4) When V _P3 < ΔBA ΔCB < V _Q4 , specifically, as shown in Figure 9d, the area 2
The luminance signal difference ΔBA between area A and area 2B is larger than the predetermined value V _P3 , and the luminance signal difference ΔCB between area 2B and area 2C
is a negative value and its absolute value is larger than the predetermined value V _Q4 , so
It can be determined that the main subject is a normal subject of approximately the same size as in the case (3-3) above, and that there is a high-luminance subject (for example, an electric light) in area 2B. As in this case, when there is a high-brightness subject indoors in area 2B, the outdoor sun etc.
It can be said from the data that the influence of this high brightness is less compared to when the photometer value is V ₁
-V ₂ is calculated from the following arithmetic expression (3), with the brightness of regions 2A, 2B, and 2C as targets, and the correction coefficient set to 0. V ₁ −V ₂ =(V _A +V _B +V _C )/3...(3)− Then, the circuit operation is performed by the comparators 90 and 9
1 outputs L level, while comparator 8
6 and 87 output H level, AND gate 96
outputs H level, and the other AND gates 9
4, 95, 97 to 102 output L level, and AND gate 107 outputs H level. Therefore, the OR gate 112 becomes H level, the NOR gate 125 is set to H level, the output of the operational amplifier 26 becomes the voltage (V _A +V _B , V _C )/3, and the output V ₁ -V ₂ of the operational amplifier 31 also becomes (V _A +V _B + V _C )/3
becomes. (3-5) When V _P4 <ΔBA<V _P3 ΔCB<V _Q4 , specifically, as shown in FIG. 9e, the luminance signal difference ΔBA between area 2A and area 2B is small;
The luminance signal difference ΔCB between area 2B and area 2C is a negative value and the absolute value is larger than the predetermined value _VQ4 , so this is a case where the main subject exists in both areas 2A and 2B, and only areas 2A and 2B are present. It can be determined that this is a case where the area is illuminated by a light or the like. In that case, there is an idea of calculating the photometric value based on the brightness of only the areas 2A and 2B, but in this embodiment, the dark peripheral area of the object, that is, the area 2C, is also taken into account to some extent, and the photometric value V ₁ - V ₂ is calculated using the following arithmetic expression (3), with the brightness of areas 2A, 2B, and 2C as targets, and the correction coefficient set to 0. V ₁ -V ₂ =(V _A +V _B +V _C )/3...(3)- Then, the circuit operates as follows: comparators 86, 90, and 91 output L level, while comparator 87 outputs H level, And gate 99 is H
Output the level and other AND gates 94-9
8, 100 to 102 output L level, and AND gate 107 outputs H level. Therefore, ORGATE 112 becomes H level,
When the NOR gate 125 is set to H level, the output of the operational amplifier 26 becomes the voltage (V _A + V _B + V _C )/3,
The output V ₁ −V ₂ of the operational amplifier 31 is also (V _A +V _B +
V _C )/3. (3-6) When ΔBA<V _P4 ΔCB<V _Q4 , specifically, as shown in Figure 9f, the area 2
The luminance signal difference ΔBA between A and area 2B is a negative value, and the absolute value is larger than the predetermined value V _P4 .
The luminance signal difference ΔCB with C is also a negative value and the absolute value is larger than the predetermined value V _Q4 . It can be determined that this is a case where part of the region 2B is illuminated by a light or the like. In this case, unlike the case (3-5) above, it is necessary to correct using the correction coefficient in order to give appropriate exposure to the main subject, and areas 2A, 2B and 2
Targeting the brightness of C, the photometric value V ₁ -V ₂ is calculated using the following arithmetic expression (3) using a correction coefficient -Vr ₁₅ on one side. V ₁ −V ₂ = (V _A + V _B + V _C )/3 + Vr ₁₅ ...(3)- Then, the circuit operation is as follows: comparators 86 to 91 all output L level, and AND gate 102 outputs H level.
Output the level and other AND gates 94-1
01 outputs L level, and AND gate 107 outputs H level. Therefore, or gate 112
becomes H level, and the AND gate 123 is set to H level, and the output of the operational amplifier 26 becomes the voltage (V _A +
V _B +V _C )/3, and the output of the operational amplifier 31 is V ₁
−V ₂ becomes (V _A +V _B +V _C )/3+Vr ₁₅ . (3-7) When ΔBA<V _P4 V _Q4 <ΔCB<V _Q3 , specifically, as shown in Figure 9g, the luminance signal difference ΔBA between area 2A and area 2B is a negative value and a predetermined value. Since the absolute value is larger than V _P4 and the luminance signal difference ΔCB between area 2B and area 2C is small, in this case, the main subject exists in the entire area 2A, or the main subject is small and exists in a part of area 2A. It can be determined that this is the case. In this case, in order to give appropriate exposure to the main subject, areas 2A, 2B and 2
Targeting the luminance of C, the photometric value V ₁ −V ₂ is obtained by the following arithmetic expression (3) using one side of the correction coefficient −Vr ₁₄ (|Vr ₁₄ |>|Vr ₁₅ |). V ₁ −V ₂ =(V _A +V _B +V _C )/3+Vr ₁₄ …(3)− Then, the circuit operation is performed by the comparators 86, 8
7 and 91 output L level, while comparator 90 outputs H level, and AND gate 1
01 outputs H level, the other AND gates 94 to 100 and 102 output L level, and AND gate 107 outputs H level. Then,
The OR gate 112 outputs an H level, the OR gate 124 outputs an H level, and the operational amplifier 26
The output of the operational amplifier 31 is the voltage (V _A + V _B + V _C )/3, and the output V ₁ −V ₂ of the operational amplifier 31 is (V _A + V _B + V _C )/3.
3 + Vr ₁₄ . (3-8) When ΔBA<V _P4 V _Q3 <ΔCB, specifically, as shown in FIG _. Since the absolute value is large and the luminance signal difference ΔCB between area 2B and area 2C is larger than the predetermined value V _Q3 , in this case, the main subject is about the same size as in the case (3-1) above, and the main subject part If there is a difference in brightness and darkness, and area 2A has a slightly high brightness, but the subject as a whole is dark, or area 2B
It can be determined that this is a case where a subject with considerably low brightness occupies the area. In this case, data-wise, area 2
Obtain the photometric value V ₁ −V 2 using all luminances of A, 2B _, and 2C as targets and setting the correction coefficient to 0 [V ₁ −V ₂
= (V _A + V _B + V _C )/3] It is also known that good results can be obtained depending on the conditions of the subject, but in the example, the method that obtained overall good results was That is, the photometric values V ₁ -V ₂ are calculated using the following arithmetic expression (3) using the correction coefficient -Vr ₁₄ on one side, targeting all the luminances of the areas 2A, 2B, and 2C. V ₁ -V ₂ = (V _A + V _B + V _C )/3 + Vr14...(3)- Then, the circuit operation is performed by the comparators 86 and 8.
7 outputs L level, while comparator 90
and 91 output H level, AND gate 10
0 outputs H level, and the other AND gates 9
4 to 99, 101, and 102 output L level,
AND gate 107 outputs H level. Therefore, the OR gate 112 outputs H level and the OR gate 124 outputs H level, so the output of the operational amplifier 26 becomes the voltage (V _A + V _B + V _C )/3, and the output V ₁ - V ₂ of the operational amplifier 31 becomes (V _A +V _B +
V _C )/3 + Vr ₁₄ . (3-9) When V _P4 <ΔBA<V _P3 V _Q4 <ΔCB<V _Q3 , specifically, as shown in FIG. 9i, the luminance signal difference ΔBA between area 2A and area 2B is small;
Also, since the luminance signal difference ΔCB between area 2B and area 2C is small, this is the case when the main subject occupies the entire field, or when the main subject is not particularly set, such as in a landscape photograph. One thing can be determined. In this case, in order to give appropriate exposure to the entire regions 2A, 2B and 2C,
Targeting all luminances from A to 2C, the photometric value V ₁ −
V ₂ is calculated using the following equation (3) with the correction coefficient set to 0. V ₁ -V ₂ =(V _A +V _B +V _C )/3...(3)- Then, the circuit operation is performed by the comparators 86 and 9.
1 outputs L level, comparators 87 and 90 output H level, AND gate 98 outputs H level, and other AND gates 94 to
97, 99 to 102 output L level, and AND gate 107 outputs H level. Therefore, the OR gate 112 outputs an H level, and the NOR gate 125 outputs an H level, so the operational amplifier 26
The output of the operational amplifier 31 is a voltage (V _A + V _B + V _C )/3, and the output V ₁ −V ₂ of the operational amplifier 31 is also (V _A + V _B + V _C )/3.
It becomes 3. (4) When the luminance signal V _C is smaller than the predetermined value Vr ₇ , that is, when V _C < Vr ₇ , and the subject scene is determined to have considerably low luminance, such as a night scene, in this case, the subject If the brightness output of the photographic field is used as the exposure value, it will often result in considerable overexposure. Therefore, in this case, the exposure should be made so that the entire screen is slightly shadowed (also known as shadow control, to ensure that a dark subject appears black), based on the brightness output of the subject. Perform underexposure compensation. Therefore, the photometric value V ₁ -V ₂ targets the luminance signals V _A , V _B , V _C of all areas 2A, 2B, 2C, and using the correction coefficient -Vr ₁₇ on one side, the following calculation is performed. Obtained from equation (4). V ₁ −V ₂ = (V _A + V _B + V _C )/3 + Vr ₁₇ ...(4) Area 2
Assuming A), the size of the main subject can be determined, and correction can be performed in accordance with the size of the main subject. Furthermore, when the photometry device of this embodiment detects that the subject is whitish or dark, it consciously draws highlights so that the whitish subject appears white and the dark subject appears black. This is done by correcting (highlight control) or shadow depiction (shadow control), and since this correction changes the amount of correction depending on the size of the main subject, extremely effective photometric values can be obtained. It is possible to control the In addition, in the explanatory diagrams of FIG. 8 and FIG. 9 used in the explanation of the above-mentioned embodiment, the brightness level values of each area 2A to 2C are at the same level when the difference in brightness between adjacent areas is small. Of course, this shows a slight difference in actual photometry (meaning that the brightness difference is smaller than the predetermined value compared, for example, V _P1 ), and Figures 8 and 9 are just It serves only as an explanatory diagram to facilitate understanding of the present invention. In the example, the calculation method for the photometric value V ₁ −V ₂ was divided into four cases (V _C1 to V _C4 ) depending on the brightness level at the periphery of the screen. Alternatively, the peripheral area (V _C ) of the screen may be used as one area for photometry. In addition, the calculation method for the photometric value V ₁ - V ₂ is based on two cases: when the target photometry area is only V _A and V _B in the center and middle area, and when V _A , V _B , V _C of the entire screen.
Although the two cases have been described, the photometric value may be determined in the same manner as described above by always targeting part or all of the multi-divided peripheral area. Alternatively, the field of view may be divided into three or more annular regions, and differences in brightness signals between adjacent regions may be used. It should be noted that the present invention can be well applied not only to single-lens reflex cameras but also to lens shutter cameras and the like. In this embodiment, the selection circuit is constructed of a logic circuit, but it goes without saying that the present invention may also be implemented by processing the selection circuit using software using a microcomputer. (Effects of the Invention) As described above, the present invention provides a calculation method that can obtain exposure information that depicts a highlight when the brightness information of the peripheral area of the object is larger than the first reference value on the high brightness side. By selecting a calculation method that can obtain exposure information that depicts shadows when the luminance information of the surrounding area is smaller than the second reference value on the low luminance side, whitish subjects can be photographed without fail. It is possible to provide a photometry device that can obtain photometry values that can reliably make a dark subject appear white and blackish.

[Brief explanation of the drawing]

FIG. 1 is a schematic diagram of an optical system as an embodiment when the present invention is applied to a single-lens reflex camera. 2 is an explanatory diagram showing a plurality of photometric areas on the light receiving surface of the light receiving means shown in FIG. 1; FIG. Figure 3 is a circuit diagram of the photometric device. FIG. 4 is a detailed circuit diagram of the peripheral brightness calculation circuit shown in FIG. 3. FIG. 5 is a detailed circuit diagram of the size discrimination circuit shown in FIG. 4.
FIG. 6 is a detailed circuit diagram of the selection circuit of FIG. 3. 7th
The figure is an explanatory diagram illustrating the photometric value calculation formula selected in FIG. 3. 8 and 9 are explanatory diagrams showing the levels of photometric values obtained using the photometric value calculation formula selected in FIG. 3. FIG. 2A, 2B, 2C ₁ , 2C ₂ , 2C ₃ , 2C ₄ ... Each area for receiving light, 6 ... Light receiving section, 20 ... Surrounding brightness calculation circuit, 21 ... Selection circuit.

Claims (1)

[Scope of Claims] 1. A photometric device that divides a field into at least a central area and a peripheral area, and obtains brightness information of the plurality of areas; a comparison means for comparing a first reference value set on the luminance side and a second reference value set on the low luminance side, based on at least the luminance information level of the peripheral area determined by the comparison means, Exposure information is calculated by selecting a specific calculation method.
When the brightness information level is higher than the first reference value, a calculation method that can obtain exposure information for highlighting is selected, and when the brightness information level is lower than the second reference value, a calculation method is selected to obtain exposure information for highlighting. A photometric device comprising: exposure information calculation means for selecting a calculation method that provides exposure information to be depicted.