JPS62215834A - Photometric device - Google Patents

Abstract

(57) [Summary] This bulletin contains application data before electronic filing, so abstract data is not recorded.

Description

Detailed Description of the Invention (Industrial Field of Application) The present invention is a method of dividing a field into a plurality of regions, obtaining luminance information for each of the plurality of regions, and obtaining a photometric value by various calculations. Regarding photometric devices.

(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.

For example, Japanese Utility Model Publication No. 51-9271 proposes a photometry device that uses the arithmetic average value of the maximum value and the minimum value among the outputs from a plurality of charging elements to obtain an appropriate photometry value. However, with this metering device, if the background is particularly bright or dark, the subject may be underexposed or
Otherwise, there was a problem of overexposure.

In addition, in JP-A-55-114918, photometry is performed by dividing the field into a central area and a plurality of outer areas surrounding the central area, and the average value of the plurality of photometric values is output. A photometric device has been proposed that normalizes the plurality of photometric values, classifies the object scene based on the standardized output, and calculates an appropriate photometric output based on the classified output. However, since this photometry device cannot determine the size of the main subject, incorrect photometry values may be obtained.

(Objective of the Invention) The present invention is a suitable photometry device assuming that the main subject is in the center of the field, and in particular, it is possible to calculate appropriate photometric values by determining the size of the main subject. The purpose of the present invention is to provide a photometric device with the following features.

In order to achieve the above object, the present invention divides the field into at least three regions: a central region, a first outer region outside the central region, and a second outer region further outside the central region, and at least the outermost region is divided into three regions. The region is further divided into a plurality of small regions to obtain each region brightness information and small region brightness information, and for this outermost region, an arithmetic expression that includes at least one of the small region brightness information as a function is calculated for the small region. a light receiving means for selecting and obtaining region brightness difference information based on region brightness information; and a brightness difference detecting means for determining region brightness difference information of both adjacent regions from a plurality of region brightness information obtained by the light receiving means. , a calculation means in which a plurality of calculation formulas including at least one of the area brightness information as a function are set, and the calculation formula is calculated based on the at least one area brightness information and the at least two pieces of brightness difference information. The photometric device is characterized by: a selection means for selecting; and a photometric device.

(Example) FIG. 1 shows the results when the present invention is applied to a single-lens reflex camera.
It is a schematic diagram of an optical system. In the figure, l is a photographing lens, 2 is a 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+~2C4
It is divided into four parts. In this embodiment, as shown in FIG. 2, a plurality of light-receiving elements capable of receiving light in the field of view at positions corresponding to each region are arranged, and the field of view is divided into six regions 2A, 2A,
It is divided into 2B, 2C+ to 2C4, and 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, 2CI,
It is a silicon photodiode (SPD) corresponding to 2C2, 2C3° and 2C4, and the photocurrent iA corresponds to the brightness of each area.

iB, icl, ic2. Generate iC3 and iC4. 14 to 19 are VA and VB by logarithmically compressing these photocurrents.

This is a logarithmic compression circuit that outputs voltage values VCI, VC2, VCs, and VC4. VA, VB, V
CI, VC2゜V C3, VC4 are constants al,
a2. a3. a4. a5゜a6(>O)
, b (>O) and photocurrent iA, iB, i
C+.

It can be expressed as follows using i C2, i C3, and i C4.

■8 =al+b fniA VB =a2+b j7niB VC+ =a3+b I! nic+ VC2=a4+b fnic2 VC3=as+b ff1nic3 VC4=as+b ji! n1C4 However, al, a2. a3. a4. a
5. a6 is VA when the brightness of each area is equal.
= vn = vc I = VC2= VC3”V
It is assumed that the logarithmic compression circuits 14 to 19 have previously set the value C4. 20 is a logarithmic compression circuit 16~
19 output voltage VC1, VC2゜vc3. VC4 is connected to each input terminal In, 112. II3. 114, calculate the brightness value of the most peripheral part 2C of the object, and set it to 0.
This is a peripheral brightness value calculation circuit that outputs a voltage Vc from one output terminal. The configuration of this peripheral brightness value calculation circuit 20 is shown in FIG. In FIG. 4, 32. 33. 34°35,
36 is a magnitude comparison circuit which outputs the larger one from the OH output terminal and outputs the smaller one from the OH output terminal among the two voltage values inputted to II and 12. A specific example of the magnitude comparison circuits 32 to 36 is shown in FIG. Fifth
In the figure, 66 is a comparator with positive-phase and negative-phase input terminals and positive-phase and negative-phase output terminals. When the positive-phase input terminal voltage V+ and the negative-phase input terminal voltage V- are from the output end
H level voltage is output from the negative phase output terminal, L level voltage is output from the negative phase output terminal, and L level voltage is output from the positive phase output terminal and H level voltage from the reverse output terminal when V+<V-. It has become. Analog switches 67 to 70 are in a conductive state when the voltage applied to the control terminal is at H level, and are in an open state when the voltage is at L level. Assuming that the voltage input to the I1 input terminal is Vl+, and the voltage applied to the 12 input terminals is VI2, when VI+≧V12, the voltage at the positive phase input terminal of the comparator 66 is V+>the voltage at the negative phase input terminal V−. At this time, the positive phase output terminal voltage of the comparator 66 is at H level, the negative phase output terminal voltage is at L level, the control voltages of analog switches 67 and 70 are at H level, and the control voltages of analog switches 68 and 69 are at L level. Therefore, analog switches 67 and 70 are in a conductive state, and analog switch 68 . 69 is in an open state.

Therefore, the voltage Vll is outputted to the OH output terminal through the analog switch 67, and the voltage VI2 is outputted to the OH output terminal through the analog switch 70. Similarly, Vl+<v! 2, the positive phase input terminal voltage V+ of the comparator 66<the negative phase input terminal voltage V-, the positive phase output terminal voltage of the comparator 66 becomes L level, and the negative phase output voltage becomes H level. Therefore, the analog switch 67°70
Since the control voltage of analog switch 68.69 is at L level, analog switch 67.70 is in an open state, and the control voltage of analog switch 68.69 is at H level, so analog switch 68. 69 becomes conductive.

At this time, the voltage v2 is outputted to the OH output terminal through the analog switch 69, and the voltage VI+ is outputted to the OH output terminal through the analog switch 68. In this way, the voltages Vll and VI2 are present at the OH output terminal.
The larger one is outputted to the OH output terminal, and the smaller one is outputted to the OH output terminal. In Figure 4, five size comparison circuits are used, and the voltage VC
The largest of I, VC2, VC3°VC4 is V
As H+, the second largest one is VH2, the third largest one is VH3, and the smallest one is VH.
The configuration is such that it is output as 4. The voltage VC+ is input to the l input terminal of the magnitude comparison circuit 32, the voltage VC2 is input to the I2 input terminal, and the larger one is
It is output from the H output end, and the smaller one is output from the OH output end. Similarly, the voltage VC3 is input to the Il input terminal of the magnitude comparison circuit 33, and the voltage VC4 is input to the I2 input terminal, and the larger one is outputted from the OH output terminal, and the smaller one is outputted from the OH output terminal. are doing. Size comparison circuit 34
Now, input the voltage output from the OH output terminal of the magnitude comparison circuit 32, 33 to the II, 12 input terminal, compare the magnitude, output the larger one from the OH output terminal, and output the smaller one from the
The voltage is output from the H output terminal, but the voltage output from the OH output terminal here is VCI, VC2, VC3, VC4 (
7) It is the largest voltage among them. Similarly, in the magnitude comparison circuit 35, the magnitude comparison circuit 32. Input the voltage output from the OH output terminal of 33 to the It and I2 input terminals, compare the magnitude, and output the larger one from the OH output terminal to the smaller one from the OH output terminal.
It is output from the output terminal, but here the voltage output from the OH output terminal is VCI, VC2, VC3, VC4 (
7) It has the smallest rhinovoltage among them. The magnitude comparison circuit 36 is
The OL terminal voltage of the magnitude comparator circuit 34 and the OH output terminal voltage of the magnitude comparator circuit 35 are inputted to the It12 input terminal and compared in magnitude. In the magnitude comparison circuit 36, VCI,
It compares and outputs the two intermediate voltage values excluding the maximum and minimum of VC2, VC3゜VC4, and the magnitude comparison circuit 3
From the 011 output terminal of 6, VCI, VC2, VC
3, VC4 (the second largest voltage among D is output, O
The third largest voltage is output from the H output terminal. In this way, VCI, VC2, VC3, and VC4 are rearranged in descending order of i: VH+>VH2>VH3≧VH4. 37 to 40 are resistors having the same resistance value.

This is for averaging VCI, VC2, VC3, VC, l. Vcm= (VC++ VC2+ VC3
+VC4)/4. 41 is a reference voltage generation circuit, which generates four reference voltages Vrl, Vr2 . Vr3.
It is generating Vz. Reference voltages Vz, Vr2. Vr3
．． The magnitude relationship of Vz is Vz>Vr2>Vr3>
It is Vz. 42 to 45 are comparators, and Vc
m, and the reference voltage VrIIVr2. Compare Vr31vr4 and find that Vcm>Vrl.

Vcm>Vr2. Vcm>Vr3. When Vcm>Vz, comparators 42, 43. 44. 45 each outputs an H level voltage, Vcm<VH, Vcm
<Vr3 Vcm<Vr3. When Vcm<Vr4, an L level voltage is output. 46. 47.
48°, 49 are inverters, 50. 51. 5
2 is an AND gate. 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, the input of the inverter 48 is connected to the output of the comparator 44, and the input of the impark 49 is connected to the output of the comparator 45. One input of the AND gate 50 is connected to the output of the inverter 46, the other to the output of the comparator 43, one input of the AND gate 51 is connected to the output of the inverter 47, and the other input is connected to the output of the comparator 43.
4, 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. The three outputs of the OR gate 53 are connected to the output of the comparator 42, the output of the AND gate 50, and the output of the AND gate 51, respectively. The four gates of the OR gate 54 are connected to the output of the comparator 42, the outputs of the AND gates 1 and 50, the output of the AND gate 51, and the output of the AND gate 52, respectively. The four gates of the OR gate 55 are connected to the output of the AND gate 50, the output of the AND gate 51, the output of the AND gate 52, and the output of the inverter 49, respectively. The three powers of the OR gate 56 are the output of the AND gate 51, the output of the AND gate 52, and the inverter 49.
connected to the output of 57-60 are analog switches, which are controlled by OR gates 53-56, respectively. 61 to 64 are resistors having the same resistance value. Resistors 61-64 are analog switches 57-
VH1, VH2, which is output by the conductive one among 60,
This is for averaging two, three, or four voltages of VH3 and VH4 (7) in descending order, or two or three voltages in descending order. 65 is an operational amplifier,
The output end and reverse input are connected and used as a voltage follower. The output terminal voltage of the operational amplifier 65 is equal to the positive phase input terminal voltage regardless of the circuit state after the output terminal. The output voltage of this operational amplifier 65 is Vc
and is output from the 01 output terminal. Next, V in Figure 4
r+, Vr2. Vr3. The characteristics of the field and the value of the voltage VC will be described based on the magnitude relationship between Vz and Vcm.

(1) Vr + Hill Vcm... Analog switch 57° 58 conduction At this time, a scene with a fairly bright background is assumed, so ignore the brightness of the relatively low brightness areas of 2CI, 2C2, 2C3, and 2C4. and Vcm(Vi(10VH2)/2).

(2) Vr2<Vcm<VH...Analog switch 57゜58, 59 conduction At this time, a scene with a slightly bright background, such as outdoors on a sunny day, is assumed. As such, inappropriate elements tend to appear as background luminance exhibiting low luminance values. Therefore, in this case, ignoring the brightness of the lowest brightness area, Vcm(VH++VH2+
VH3)/3.

(3) Vr3<Vcm<Vr2- Analog switch 57°5g, 59. 60 conduction At this time, a scene with a background of standard brightness is assumed, and in such a scene, both relatively bright and relatively dark parts of the subject cannot be ignored. Therefore, in such a case, Vc==(VH++V[ll+VH3+VH4
)/4.

(4) Vr, + < Vcm < Vr3 - Analog switch 58° 59, 60 conduction At this time, a scene with a somewhat dark background is assumed, so ignoring the highest brightness area, Vc = (V H2+
V H3 + V H4) / 3.

(5) Vcm < Vr+- analog switch 59.
60 conduction At this time, a scene with a fairly dark background, such as a night view, is assumed, and in such a scene, elements that are inappropriate for the brightness of the background that locally shows high brightness, such as electric lamp light, are likely to appear. Therefore, in this case, ignoring the relatively high brightness area, Vc=(VH3+VH4)/2 and t
Ru.

As described above, the peripheral brightness calculation circuit 20 calculates V H1 and V H2 according to the brightness state of the peripheral part of the object field.

Two or three of the largest of V H3 and V H4
The average value of two or four, or two or three of the smaller ones, is set as Vc and output from the O1 output terminal. 21 in FIG. 3 is a logarithmic compression circuit 14. 15 output voltage V
A, VB and the output voltage Vc4 of the peripheral brightness calculation circuit 20, respectively, at the input terminal 121. 122. This is a selection circuit that is input to I23 and determines which arithmetic expression to select from among a plurality of arithmetic expressions to be described later. In Figure 3, 22, 2
3. 24 is a resistor having the same resistance value, and 25 is an analog switch. Resistance 22. 23. 24 and the analog switch 25 constitute an average value circuit, and when an 11-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 25 becomes conductive. Become,
The output voltage Vl of the average value circuit is (VA + VB + V
C) /3 t: becomes. On the other hand, the A/P of the selection circuit 21
When an L level voltage is output from the output terminal and input to the control terminal of the analog switch 25, the analog switch 25 becomes open, and the output voltage V of the average value circuit
l becomes (VA+VB)/2.

Reference numeral 26 denotes an operational amplifier, which is used as a voltage follower by connecting its output terminal and negative phase input terminal, and the output voltage v1 of the average value circuit is input to the positive phase input terminal.

The output terminal voltage of the operational amplifier 26 is Vl regardless of the circuit state after its output. 27.degree.28.29.30 are resistors having the same resistance value, and 31 is an operational amplifier. Resistance 27. 28. 29. 30 and operational amplifier 3
1 constitutes a subtraction circuit.

If the 02 output terminal voltage of the selection circuit 21 is v2, the output voltage of this subtraction circuit is Vl-V2. This voltage V+-
In this embodiment, V2 represents a photometric value determined by a plurality of arithmetic expressions to be described later.

FIG. 6 is a circuit diagram of the selection circuit 21 in FIG. 3.

71. 72. 73. 74 is a resistor having the same resistance value, and 75 is an operational amplifier, which constitutes a first subtraction circuit. Similarly, 76, 77, 78, and 79 are resistors having the same resistance value, and 80 is an operational amplifier, which constitutes a second subtraction circuit. The first subtraction circuit has voltages VA and V
B is input, and its output voltage is VB-VA. The voltages VB and VC are input to the second subtraction circuit, and its output voltage is VC-VB. A reference voltage generation circuit 81 generates a reference voltage Vrs. 82 is a comparator. The voltage Vc is input to the positive phase input terminal, the reference voltage Vrs is input to the negative phase input terminal, and the comparator 82
5, the I-level voltage is output, and Vc<Vrs
When this happens, an L level voltage is output. A reference voltage generation circuit 85 generates reference voltages VPa, vpb, VQa, and VQb. The reference voltage generation circuit 85 has a control terminal B.
/D, and the output voltage of the comparator 82 is input as a control voltage. The reference voltage when an H level control voltage is input to B/D is VPa=V
Pt, Vpb=vp2. VQa=VQ+.

VQb=VQ2, while the terminal B/D
f/I/ c7) Set the reference voltage when the control voltage is input as VPa=VP3. VPb=VP4. VQa
=VP3. Let VQb=VQ4. Reference voltage Vr'+
-VF6. VQI ~ VQ4 (7) The code is VP
I, VF6. VQI, VO3 are positive, VF6. VF
6. VO2, VO2 is 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 VP+ or V P 3 is input to the negative phase input terminal of the comparator 86 depending on the I (level or L level) of the control voltage input to the control terminal B/D of the reference voltage generation circuit 85. Comparator 87 Similarly, a reference voltage VP2 or VF6 is inputted to the negative phase input terminal of 90.91 is a comparator, and 92.93 is an inverter.

The output voltage VC-VB of the operational amplifier 80 of the second subtraction circuit is input to the positive phase input terminals of the comparators 90 and 91. The negative phase input terminal of the comparator is connected to the reference voltage VQI depending on the I (level, L level) of the control voltage input to the control terminal B/D of the reference voltage generation circuit 85.
Alternatively, VQ3 is input. Similarly, the reference voltage VQ2 or VQ4 is input to the negative phase input terminal of the comparator 90. Comparator 86. 87. 90.91
The output terminal of the positive phase input terminal voltage V+ and the negative phase input terminal voltage V
- Depending on the magnitude of V-, H level voltage, V
When +<V-, an L level voltage is output. 94-1
02 is an AND gate, and comparator 86. 87
．． Depending on the combination of H level and L level at the output terminals of 90°91, one of the AND gates outputs an I] level voltage, and the other AND gates output an L level voltage. VB-VA and VPa and vpb.

The following describes which AND gate among the AND gates 46 to 54 outputs the H level based on the magnitude relationship between VC-VB and VQa and VQb.

1) VB-VA>Vpa i) vc-VB>VQa ・=7nd gate 94ii) VQa>VC-VB>VQb -
= 〃95iii) VQb > VC-VI3
- ” 962) VPa>VB-VA
>VPb i) vc-VB>VQa -7nd gate 97ii) VQa>VC-VB>VQb -”
'98iii) vQb>vc-VB
・= l/993) vPb>vB-V, A i) VC-VB>VQa ---AND gate 100ii) VQa>VC-VI13>VQ
b---/l 101iii) VQb>V
C-VB-" 102103 is an inverter that inverts the output state of the comparator 82.

108 and 109 are OR gates, tto and ttt are AND gates, and 112 is an OR gate. orgate lO8
The four inputs of the AND gate 98.99. 101.
It is connected to the output of 102, and the AND gate 98.9
9゜One AND gate of 101 and 102 is H
When the level voltage is output, the output terminal voltage of the 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 to 102, and when any one of the AND gates 96, 98 to 102 outputs an H level voltage, the output of the OR gate 109 The output terminal voltage becomes I (level. One of the two inputs of the AND gate 110 is the OR gate 10
The other end is connected to the output end of the comparator 82.

When the comparator 82 is outputting H level,
The output of AND gate 110 is equal to the output state of OR gate 108. At this time, since one of the two inputs of the AND gate 111 is connected to the output of the inverter 103, one of the inputs of the AND gate 111 h<r,
Since there is a level ``?'', its output is the level.

On the other hand, one of the two inputs of the AND gate 111 is connected to the output of the OR gate 109, and the other is connected to the output of the OR gate 109.
It is connected to the output of inverter 103. When the comparator 82 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, the output voltage of the inverter 103 is at H level, and since the H level voltage is input to one of the two inputs of the AND gate 111, this output is as follows.
It is equal to the output state of OR gate 109. and gate 1
The outputs of 10°111 are respectively input to OR gates 112, and AND gates 110. When at least one of the gates 111 is outputting an H level voltage, the OR gate 112 outputs an H level voltage. The output terminal of the OR gate 112 is connected to the A/I' output terminal.

113-117, 119-123 are Antogame-1,■
18°124 is the OR gate, and 125 is the NOAH gate. 126 is a reference voltage generation circuit, which generates reference voltages Vrg,
Vrc+.

Vr+o, Vro, -Vr+2. -Vr+3. -V
r+4. -Vr+s is generated. 127-135 are analog switches. 2-person AND gate 113-1
One input of the AND gates 17 is connected to the output of the comparator 82, and one input of the two AND gates 119 to 123 is connected to the output of the inverter 103. The other input of the AND gate 113 is connected to the output of the AND gate 94, the other input of the AND gate 114 is connected to the output of the AND gate 95, and the other input of the AND gate 115 is connected to the AND gate 96. The other input is connected to the output of AND gate 99, and the other input of AND gate 117 is connected to the output of AND gate 102. 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, and the other input of AND gate 121 is the output of AND gate lOO, and The other input of the gate 122 is connected to the output of the AND gate 101, and the other input of the AND gate 123 is connected to the output of the AND gate 102.

Therefore, when an H level voltage is output to the output terminal of the comparator 82, an H level voltage is input to one input of AND gates 113 to 117, and an L level voltage is input to one input of AND gates 119 to 123. The output of 113 is equal to the output state of AND gate 94;
The output of AND gate 114 is equal to the output state of 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 the output of AND gate 117 is equal to the output state of AND gate 102. equal to the output state of AND gate 119
The output voltage of 123 is at L level. Conversely, when an H level voltage is output to the output terminal of the inverter 103, an L level voltage is input to one input of AND gates 113 to 117, and an H level voltage is input to one input of AND gates 119 to 123. The output voltage of AND gates 113 to 117 is L
The output of AND gate 1!9 is equal to the output state of AND gate 94, the output of AND gate 120 is equal to the output state of AND gate 97, and 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. orgate 118-2
One of the two inputs is connected to the output of AND gate 114 and the other to the output of AND gate 115.

The output of the OR gate 118 is the output of the AND gates 115, 11
It becomes H level when at least one of the output voltages of No. 6 is at H level. Of the two inputs of the OR gate 124, one is connected to the output of the AND gate 121 and the other is connected to the output of the AND gate 122. The output of the OR gate 124 is the H level when at least one of the output voltages of the AND gates 121 and 122 is at the H level.The eight inputs of the NOR gate 125 are the AND gates. 118 output, AND gate 116 output, AND gate 117 output,
It is connected to the output of AND gate l19, the output of AND gate 120, the output of OR gate 124, and the output of AND gate 123, respectively. ANDGATE 113-11
7. When the outputs of 119 to 123 are all at L level, the output voltage of NOR gate 125 is at H level, and at other times, the output voltage of NOR gate 125 is at L level.

A reference voltage is applied from a reference voltage generation circuit 126 to the input terminals of analog switches 127 to 130 and 132 to 135. The input terminal of the analog switch 127 has a reference voltage VrB, the input terminal of the analog switch li8 has a reference voltage Vrg, the input terminal of the analog switch +29 has a reference voltage V r 10 s, and the input terminal of the analog switch 130 has a reference voltage V r 11, the input terminal of the analog switch 132 has a reference voltage V r 12, and the analog switch +
A reference voltage V r 13 is applied to the input end of the analog switch 133 , a reference voltage V r 14 is applied to the input end of the analog switch 134 , and a reference voltage V r Is is applied to the input end of the analog switch 135 .

The input terminal of analog switch 131 is Ov. The output terminals of the analog switches 127 to 137 are connected to each other and become the 02 output terminal. The control terminal of the analog switch 127 is connected to the output of the AND gate 113, the control terminal of the analog switch 128 is connected to the output of the OR gate 118, the control terminal of the analog switch 129 is connected to the output of the AND gate 116, and the analog switch 130
The control terminal of the analog switch 131 is the output of the AND gate l17, the control terminal of the analog switch 131 is the output of the NOR gate 125, and the analog switch ! The control terminal of 32 is connected to the output of AND gate 119, the control terminal of analog switch 133 is connected to the output of AND gate 120, the control terminal of analog switch 134 is connected to the output of OR gate 124, and the control terminal of analog switch 135 is connected to the output of AND gate 123. connected to the output. Three inputs VA, VB of the selection circuit in Fig. 4
, regardless of the voltage level of VC, the AND gate 113, the OR gate 118, the AND gate 1-116,
Antgame 1-117, ANDGATE 125, ANDGATE 119, ANDGATE 120, ORGATE 124,
Since the output terminal of one of the AND gates 123 becomes ■ level, and the output terminals of the other gates become L level,
Among the analog switches 127 to 135, only one analog switch to which the H level control voltage is applied becomes conductive, and the other analog switches become open.

When the output of AND gate l13 is H level voltage,
The analog switch 127 becomes conductive, and the voltage Vra
is output from the 02 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 Vr9 is output from the 02 output terminal,
When the output of the AND gate 116 is at the [■ level voltage, the analog switch 129 becomes conductive and the voltage V
r + o is output from the 02 output terminal, and the AND gate 11
When the 7 output is the I- level voltage, the analog switch 130 becomes conductive and the voltage Vro is output from the 02 output terminal, and when the NOR gate 125 is the I] level voltage, the analog switch 131 becomes conductive and Ov becomes 0.
2 output terminal, and the output of AND gate 119 is H.
When the voltage is at the H level, the analog switch 132 becomes conductive and the voltage -V r 12 is output from the 02 output terminal, and when the output of the AND gate 120 is at the H level voltage, the analog switch 133 becomes conductive and the voltage -V r 12 is output from the 02 output terminal.
V r 13 is output from the o2 output terminal, and the OR gate 1
When the output of AND gate 123 is I (level voltage), the analog switch 134 becomes conductive and the voltage -V r 14 is output from the 02 output terminal, and when the output of AND gate 123 is H level voltage, the analog switch 135 becomes conductive. A voltage -V r +s is output from the 02 output terminal.

however. VrB, VF9. VrlO, Vrll,
> O, -VF12° -Vr+3. -Vru, -Vr
+s, < O.

Next, the circuit operations in FIGS. 3 and 6 will be explained with reference to FIGS. 7 to 9.

(+) The outermost area 2c of the field area 6 shown in FIG.
When the luminance signal Vc obtained from (2C+ to 2C4) is larger than the reference voltage Vr5, that is, Vc > Vr
When it is determined that the subject is outside as s (when there is a bright subject such as the sky in the background and the peripheral area of the screen is determined to be bright), in this case, the luminance signal difference VB-VA (hereinafter referred to as ΔBA) is further calculated. ) and luminance signal difference VC-VB (hereinafter referred to as ΔC
Depending on the value of (abbreviated as B), the reference voltage VPI can also be set as a constant.
, VF6. VQl, VQ2 (Reference voltage relationship is VF6 (0<VPI, VQ2<0<VQI)
Using this, the photometric value Vl-V2 is determined by the following operation.

Specifically, as shown in FIG. 8(a), the luminance signal difference ΔBA between the central area 2A and the first outer area 2B of the object field area becomes small; The luminance signal difference ΔCB between the outermost region 2C and
It is thicker than + (in this case, it can be determined that the main subject is often present in both area 2A and area 2B).

Therefore, in order to give appropriate exposure to the main subject, the photometric value Vl-V2 is determined by the luminance signals VA and VB of the two areas and area 2B.
, and the correction coefficient is O, the following calculation formula (1)-
■Seek more.

Vl-V2= (VA+VB)/2-(t)-■ Then, regarding the circuit operations in FIGS. 3 and 6, first we will explain the selection circuit 21 in FIG. 6.
+-2C4) photometric signal (voltage) Vc is the reference voltage Vr5
Since it is larger, the comparator 82 outputs an 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 voltage of the circuit 85 is VPa=VP+.

VPb=VP2. VQa=VQl and VQb=VQ2.

On the other hand, since the output signal (voltage) VB-VA of the operational amplifier 75 satisfies the condition vp2<ΔBA<VP+, the output of the comparator 86 becomes L level, the output of the comparator 87 becomes H level, and the output signal of the operational amplifier 80 ( Voltage) VC-VB is under the condition of VQ+ <ΔCH, so the output of comparator 90 is H level,
The output of 1 becomes I] level. Therefore, and gate 9
Only 7 becomes the output of 14, and the H of this AND gate 97
Level output, AND gates 94-96, 98-102
The L level output of the AND gate 106 and the H level output of the AND gate 106 cause the OR gate 112 to be set to the L level and the output terminal A/
P is set to L level, only the NOR gate 125 is set to H level, the other AND gates 113 to 123 are set to L level, and the voltage at the 02 output terminal is set to 0. Therefore, the voltage (VA + VB)/2 is output from the operational amplifier 26 in FIG. 3, and Ov is supplied to the negative phase input terminal of the operational amplifier 31. (VA+VB)/2.

Specifically, as shown in FIG. 8(b), the luminance signal difference ΔBA between the area 2A and the area 2B is larger than the predetermined value VPl on the + side, and the luminance signal difference ΔCB between the area 2B and the area 2C is also larger. Since it is larger than the predetermined value VQ+ on the + side, it can be determined that in this case, the main subject is often present in the entire area 2A and a part of the area 2B. In this case, it is possible to calculate the photometric value V r -V 2 using only the luminance signal VA of area 2A, but empirically it has been found that a better exposure value can be obtained by taking the background luminance signal into account to some extent. Become. Therefore, the photometric value Vl-V2 is obtained from the following arithmetic expression (1)-0 using the correction coefficient VrB for the main subjects VA and VB.

Vl-V2= (VA+VB)/2-Vrs ・(
1)-■Then, the circuit operation is as follows: Coverator 88.89.
92.degree. 93 and comparator 82 all output I] level, only AND gate 94 outputs H level, and the other AND gates 95 to 102 output L level.

Therefore, the output of the AND gate 112 is set to L level, the output of AND gate 113 is set to H level, and the operational amplifier 2
The output of the operational amplifier 31 becomes a voltage (VA+VB)/2, and the output v, -v2 of the operational amplifier 31 becomes (VA+VB)/2-VrB.

Specifically, as shown in FIG. 8(C), the luminance signal difference ΔBA between the area 2A and the area 2B is a predetermined value V on the + side.
On the other hand, since the luminance signal difference ΔCB between area 2B and area 2C is small, in this case it can be determined that the main subject exists in the entire area 2A, or that the main subject is particularly small. In this case, the photometric values v, -V2 can be calculated for the main subject in area 2A, but it is necessary to take into account the luminance signal vn of area 2B, which corresponds to the background, and set an appropriate exposure for the main subject. In order to give the photometric value Vl-V2, the areas 2A and 2B are targeted, and the correction coefficient Vr
Using g (Vrg>VrB), the following calculation formula (1) -
■Seek more.

Vl −V2 = (VA + Vs)/2− Vr9
・(1)-■ And the circuit operation is performed by the comparator 86.
87 and 90 all output H level, only AND gate 95 outputs H level, and other AND gates 94
゜96 to 102 output L level and comparator 82
outputs [I level. Therefore, the output of the OR gate 112 is set to the L level, the output of the OR gate 118 is set to the H level, and the output of the operational amplifier 26 is set to the voltage (VA+VB).
)/2, and the output Vl-V2 of the operational amplifier 31 is (V
A+VI3)/2-Vr9.

Specifically, as shown in FIG. 8(d), the luminance signal difference ΔBA between the two regions and the region 2B becomes larger than the predetermined value VP+ on the + side, while the difference between the two regions and the region C Since the luminance signal difference ΔCB between the two and It can be determined that there is a subject (for example, the sun, reflection on the sea surface, etc.), or there is a subject with considerably high brightness in the area 2B in a landscape photograph or the like. In this case, data-wise, all luminances in areas 2A, 2B, and 2C are targeted, and the photometric value V+-Vz is determined by setting the correction coefficient to 0 (V+-V2= (VA+VA
+VC)/3) was also recognized by the inventor as a method for obtaining good results, but in this example, another method for obtaining good results, that is, as in the case of (1-3) above, was used. , is calculated from the following arithmetic expression (1)-0 using the correction coefficient vr9, targeting the brightness of the area 2A and the area 2B.

V+ -V2 = (VA + VB)/2- Vr9
- (1) -■ And the circuit operation is performed by the comparator 86
, 86 and 87 output H level, while comparators 90 and 91 output L level, and AND gate 96
is I (output level, other anime games) 94.
95°97 to 102 output L level, and comparator 82 outputs H level. Therefore, or gate 11
2 is set to L level, and the output of OR gate 118 is set to H level.
As a level, the output of the operational amplifier 26 is a voltage (VA0v
n)/2, and the output of the operational amplifier 31 is Vl-V2(
V A + V B ) / 2- V r 9.

Specifically, as shown in FIG. 8(e), the luminance signal difference ΔBA between the two areas and area 2B is small, and
Since the luminance signal difference ΔCB between B and area 2C is a negative value and has a larger absolute value than VQ2, this is a large case in which the main subject exists in both areas 2 and 2B, but it can be determined that it is whitish. In this case, the photometric value may be calculated based on the brightness of area 2A and area 2B, but if the peripheral part of the screen has lower brightness than the central part, then
If you calculate the photometric value by taking the brightness of the subject into account to some extent, you can reliably capture the whitish main subject as white (highlight control), and it is known from the data that a good exposure value can be obtained. Therefore, in this case, in order to provide an exposure that highlights the main subject, all of the luminance signals VA, VB, and VC are targeted, and the photometric value V+-
V2 is calculated using the following calculation formula (1
)−〇.

V+-V2=(VA+VB+VC)/3-Vrto
-(1)-〇Then, the circuit operation is performed by the comparator 87
outputs H level, comparators 86, 90 and 91 output L level, AND gate 99 outputs H level, and other AND gates 94, 98, 100 output H level.
~102 output L level, and comparator 82 outputs H level.

Therefore, the output of the OR gate 112 is set to H level, the output of the operational amplifier 26 becomes the voltage (VA+VB+VC)/3, and the output Vl-V2 of the operational amplifier 31 becomes (VA + V13).
+VC) /3-Vrto.

Specifically, as shown in FIG. 8(f), area 2
The luminance signal difference ΔBA between A and area 2B is a negative value and a predetermined value V
The absolute value is thicker than P2 (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 VQ2, so in this case, the main subject is the entire area 2A and area 2).
It can be determined that this is a medium-sized case that exists in a part of B and that the subject is whitish. In this case, areas 2A, 2B, and 2
Targeting the brightness of C, the photometric value V+-V21;! Correction coefficient vrH (Vr.

<Vrto) using the following arithmetic expression (1)-〇.

Vl −V2 = (VA+VB+VC)/3−Vrll
・(1)-〇Then, the circuit operation is as follows: comparator 8
6 to 91 output the L level, and the AND gate 102 outputs the H level.
The other AND gates 94 to 101 output L level, and the comparator 82 outputs H level. Therefore, when the output of the OR gate l12 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 (■Δ+Vl'3+VC)/3, and the output v, -v2 of the operational amplifier 31 becomes (VA+VB
1VC)/3-Vrll.

Specifically, as shown in FIG. 8(g), the area 2
The luminance signal difference ΔBA between the person and the area 2B is a negative value and a predetermined value V.
Since the absolute value is larger than 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, 213, and 2C, and set the correction coefficient to 0.
is obtained from the following arithmetic expression (1)-■.

Vl −V2 = (VA+VB+VC)/3
- (1) -〇 And the circuit operation is performed by the comparator 90
outputs H level, while comparator 86. 87
and 91 output the L level, and the AND gate lot outputs the H level.
output the level and other AND gates 94-100.
102 outputs L level, and comparator 82 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 a voltage (VA + VB + VC) /
3, and the output V l - V 2 of the operational amplifier 31 is (V
A+Vn+VC)/3.

(t-S), specifically, as shown in FIG. 8(h), region 2
The luminance signal difference ΔI3A between area A and area 2B is a negative value and has a larger absolute value than the predetermined value VP2, and the luminance signal difference ΔCB between area 2B and area 2C is greater than the predetermined value V(H, so in this case, the main If the subject is about the same size as described in (1-1) above and there is a difference in brightness in the main subject part, and area 2A is slightly high brightness, or if the area 2B is a landscape photograph, etc. It can be determined that this is a case where a subject with a considerably low luminance occupies the area.

In this case, data-wise, all luminances of areas 2A, 2B, and 2C are targeted, and the correction coefficient is set to 0, and the photometric value Vl −
Find V2 (Vl-V2= (VA+VB+VC)/
3) was also recognized by the inventor as a method for obtaining good results, but in this example, another method for obtaining good results, that is, as in the case of (1-1) above, is applied to area 2A. and the brightness of area 2B, and the correction coefficient is set to 0.

Vl-V2= (VA+VB)/2-(1)
-@Then, the circuit operation is such that comparators 90 and 91 are high.
On the other hand, the coparators 86 and 87 output the L level, the AND gate 100 outputs the H level, and the other gates 1-94 to 99, 101, and 102 output the L level.
The comparator 82 outputs the H level. Therefore, the output of the OR gate 112 is set to L level,
When the output of the NOR gate 125 is set to H level, the output of the operational amplifier 26 becomes a voltage (VΔ+VB)/2, and the outputs v, -v2 of the operational amplifier 31 become (VA+Vn)/2.

Specifically, as shown in FIG. 8(i), the area 2
The luminance signal difference ΔBA between A and area 2B is small, and the luminance signal difference CB between area 2B and area 2C is also small. It can be determined that this is a case where there is no intention to set the main subject. In this case, areas 2A, 2B and 2
Targeting the brightness of C, and setting the correction coefficient to O, it is calculated using the following arithmetic expression (1)-0.

V+-V2= (VA+VB VC)/3-(
1)-〇Then, the circuit operation is performed by comparators 87 and 9.
o outputs I] level, on the other hand, comparators 86 and 91 output L level, AND gate 98 outputs H level, and other AND gates 94 to 97.99 to 1
02 outputs L level, and comparator 82 outputs H level. Therefore, 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 (VA + VB + VC)/3, and the output of the operational amplifier 31 becomes V+ - V2 + (VA +
VB 1VC)/3.

(2) The outermost area 2C of the field of view shown in Figure 2 (
If the luminance signal VC obtained from 2C+ to 2C4) is smaller than the reference voltage Vr5, that is, if VC<Vrs and it is determined that the room has walls or the like in the background, then the above-mentioned (1 ), the luminance signal difference Δ
A constant VP3.B is determined by the values of BΔ and luminance signal difference ΔCB.
VF6. VO2, VO4 (VF6 < O < VF6
Using ゜VO4<O<VO2), the photometric values v and -V2 are determined by the following operation.

Specifically, as shown in FIG. 9(a), the luminance signal difference ΔBA between area 2A and area 2B becomes small, and
Since the luminance signal difference ΔCB between the area 2C and the area 2C is larger than the predetermined value VQ3, it can be determined that the main subject is present in both areas 2A and 2B, is large, and is a darkish subject.

In this case, you can simply calculate the photometric value using the brightness of areas 2A and 2B as targets and setting the correction coefficient to 0, but you can make sure that the dark main subject is captured in black (shadow control).
In this embodiment, in order to provide exposure such that the main subject is depicted in shadow,
Targeting the brightness of areas 2A and 2B, the photometric value Vl-V2 is calculated using the following calculation formula (
2) Determine from −■.

Vl-V2=(VA+VI3)/2+Vr13-(
2)-■Then, the circuit operation is performed in the outermost region 2C (2)
C+~2C4) photometric signal (voltage) VC is a predetermined voltage Vr
5, the comparator 82 outputs L level, and therefore the control terminal B of the reference voltage generation circuit 85
/D is supplied with an L level signal, and the reference voltage of the circuit 85 is VPa = VF6. vpb=VF6. V
Qa = VO2, VQb = VQ4. On the other hand, the comparator 86 outputs the L level, the comparators 87, 90 and 91 output the ■ level, the AND gate 97 outputs the 11 level, and the other AND gates 94
~96°98~102 outputs L level, and inverter 103 outputs H level. Therefore, or gate 11
The output of the operational amplifier 26 is set to the L level, the output of the AND gate 120 is set to the H level, and the output of the operational amplifier 26 is set to the voltage (V A
+VB)/2, and the operational amplifier 31 (7) output Vl-
V2 becomes (VA+VB)/2+Vr+3.

Specifically, as shown in FIG. 9(b), the luminance signal difference ΔBA between the area 2A and the area 2B is larger than the predetermined value vP3.
The luminance signal difference ΔCB between area 2B and area 2C is a predetermined value vQ3
Since the main subject is larger, it can be determined that the main subject is present in all of the area 2A and part of the area 2B, and that it is a blackish subject. In that case, in order to give the main subject an exposure that will render a shadow in the same way as in (2-1) above, the brightness of areas 2A and 2B will be targeted, and the photometric value V + - V2 will be set on one side. The correction coefficient -Vr+2 of ・(
IVr121x1vr131), the following calculation formula (
2) Determine from −■.

Vl-V2=(VA+VB)/2+Vr+2-(2)-
■Then, the circuit operation is such that all comparators 86 to 91 output I (level), AND gate 94 outputs level ■], other AND gates 95 to 102 output L level, and inverter 103 outputs 1 ( Output the level.

Therefore, the output of the OR gate 112 is set to L level, the output of the AND gate 120 is set to H level, and the operational amplifier 2
The output of operational amplifier 31 (7) becomes the voltage (VA+VB)/2, and the output of operational amplifier 31 (7) V+-V2 becomes (VA+VB)/2+Vz
It becomes z.

Specifically, as shown in FIG. 9(C), the area 2A
The luminance signal difference ΔBA between the regions 2B and 2B is larger than the predetermined value VP3, and the luminance signal difference ΔCB between the regions 2B and 2C is small. It can be determined that this is a case where the distance is small and the subject is dark. In this case, in order to give an exposure that renders the main subject part in shadow, the luminance of areas 2A and 2B is targeted, and the photometric value V +
-V2 is obtained from the following arithmetic expression (2)-■, with the correction coefficient being O.

Vl −V2 = (VA + Va)/2
-= (2) -■ Then, the circuit operation is such that the comparator 91 outputs the L level, while the comparator 86
~90 outputs I(level, and AND gate 95 outputs l(
level and other AND gates 94. 96~
102 outputs L level, and inverter 103 outputs 1-level. Therefore, the OR gate 112 is set to L level, the NOR gate 1'25 is set to H level, the output of the operational amplifier 26 becomes the voltage (VA+VB)/2, and the output 'rJ-V2 of the operational amplifier 31 also becomes (VA+VB)/2.

Specifically, as shown in FIG. 9(d), the luminance signal difference ΔBA between the area 2A and the area 2B is larger than the predetermined value VPa, and the luminance signal difference ΔCB between the area 2B and the area 2C is a negative value. Since the absolute value is larger than the predetermined value VQ4, the main subject is as described above (
It can be determined that this is a case where the object is a normal object of approximately the same size as in case 2-3) and there is a high-luminance object (for example, an electric light) in the region 2B. Area 2 indoors as in this case
When there is a high brightness subject in B, it can be said from the data that the influence of this high brightness is less than when the outdoor sun etc. is located in area 2B, so the photometric values v and -v2 are Area 2A.

Targeting the brightness of 2B and 2C, and setting the correction coefficient to 0,
It is calculated using the following arithmetic expression (2)-■.

Vl −V2 = (VA + VB + Vc)/3
-(2)-■Then, the circuit operation is such that comparators 90 and 91 output L level, comparators 86 and 87 output H level, and AND gate 96 outputs H level.
level and other AND gates 94. 95.
97 to 102 output L level, and inverter 103 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 77 is the voltage (VA+VB+VC)/3, and the output V+-V2 of the operational amplifier 31 is also (VA+VB
+vC)/3.

Specifically, as shown in FIG. 9(e), the luminance signal difference ΔBA between area 2A and area 2B is small, and area 2
Since the luminance signal difference ΔCBC between B and area 2C is a negative value and the absolute value is larger than the predetermined value VQ4, this is a case where the main subject exists in both areas 2A and 2B, and in area 2A.

It can be determined that this is the case where only 2B 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 + - V 2 is area 2A, 2
Targeting the brightness of B and 2C, and setting the correction coefficient to O, it is calculated from the following arithmetic expression (2)-〇.

V+-V2=(VA+VB+VC)/3-(2)-
〇 And, the circuit operation is performed by comparators 86, 90 and 91.
outputs L level, on the other hand, comparator 87 outputs H level, AND gate 99 outputs H level, other AND gates 94 to 98, 100 to 102 output L level, and inverter 103 outputs H level. Level output.

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 (VA + VB + Vc)/3, and the output VB - V2 of the operational amplifier 31 also becomes (VA + VB + VC)/3. .

Specifically, as shown in FIG. 9 [f], area 2
The luminance signal difference ΔBA between A and area 2B is a negative value and a predetermined value V
The absolute value is larger than P4, and the luminance signal difference ΔCB between areas 2B and 20 is also a negative value, and the absolute value is larger than the predetermined value VQ4. It can be determined that this is the case where all of the area 2A and a part of the area 2B are illuminated by a light or the like. In this case, unlike the case (2-5) above, it is necessary to correct using a correction coefficient in order to give appropriate exposure to the main subject, and the brightness of areas 2A, 2B and 2C is targeted. , the photometric value V+-V2 is the correction coefficient of one side -V
It is obtained from the following arithmetic expression (2)-〇 using r15.

V+ -V2 = (VA+Vn+VC)/3-Vrls
-(2)-〇And the circuit operation is the comparator 86
~91 all output L level, AND gate 102 outputs H level, and other AND gates 94~lot
outputs L level, and inverter 103 outputs II level. Therefore, OR gate 112 becomes H level, AND gate 123 is set to H level, and operational amplifier 2
The output of the operational amplifier 31 is the voltage (VA+VB+VC)/3, and the output Vl-V2 of the operational amplifier 31 is (VA+VB+VC).
/3 +V r +s.

Specifically, as shown in FIG. 9(g), area 2
The luminance signal difference ΔBA between A and area 2B is a negative value and a predetermined value V
The absolute value is larger than P4, and the luminance signal difference ΔCB between areas 2B and 20 is small, so in this case, the main subject is in area 2A.
It can be determined that the main subject exists in the entire area 2A, or the main subject is small and exists in a part of the area 2A.

In this case, in order to give the main subject an appropriate exposure,
For the brightness of areas 2A, 2B and 2C, the photometric value V
+ -V2 is the correction coefficient on one side -Vr14(l Vr+4
1>) Vr+51) is calculated from the following arithmetic expression (2)-〇.

Vl −V2 = (to V+VB+VC)/3−Vr+4
- (2) -■ And the circuit operation is performed by the comparator 86. 87° and 91 output L level, while comparator 90 outputs ■] level,
101 outputs H level, and the other AND gates 94
~100°102 outputs L level and inverter 10
3 outputs H level. Therefore, the OR gate 112 outputs an H level, the OR gate 124 outputs an H level, and the output of the operational amplifier 26 becomes the voltage (VA+VB+VC
)/3, and the output Vl-V2 of the operational amplifier 31 (7) becomes (VA+VB+VC)/3+Vr14.

Specifically, as shown in FIG. 9(h), the area 2
The luminance signal difference ΔBA between A and area 2B is a negative value and a predetermined value V
The absolute value is larger than P4, and the luminance signal difference ΔCB between area 2B and area 2C is larger than the predetermined value VQ3, so in this case, the main subject is about the same size as in the case (2-1) above, and the main subject is This is determined to be the case when there is a difference in brightness and the area 2A has a slightly high brightness, but the subject as a whole is blackish, or when the subject is a landscape photo and the area 2B is occupied by a subject with considerably low brightness. can. In this case, the photometric value V1-V2 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
I-V2= (VA+VB+VC)/3) Although it is known that good results can be obtained depending on the conditions of the scene, in the example, the method that gave good results overall,
That is, all luminances of areas 2A, 2B, and 2C are targeted, and the photometric value Vl -V2 is the correction coefficient -Vr+4 on one side.
is calculated using the following arithmetic expression (2)-■.

Vl-V2=(VA+VB+VC)/3-+Vr14
-(2)-〇And the circuit operation is as follows: Comparators 86 and 87 output L level, comparators 90 and 91 output H level, AND gate lOO outputs H level, and other AND gates output L level. 94-99.10
1゜102 outputs L level, and inverter 103 outputs H level.
Output the level. Therefore, the OR gate 112 outputs an H level, and the OR gate 124 outputs an H level, so the output of the operational amplifier 26 is the voltage (VA+VI3+VC
) /3, operational amplifier 31f7) output V+-V2
becomes (VA+VI3+VC)/3+Vr14.

Specifically, as shown in FIG. 9(i), area 2
Since the luminance signal difference ΔBA between area A and area 2B is small, and the luminance signal difference ΔCB between area 2B and area 2C is also small, this is the case when the main subject occupies the entire field, or when shooting a landscape photograph, etc. It can be determined that this is a case where the main subject is not particularly set. In this case, area 2
In order to give appropriate exposure to the whole of A, 2B and 2C,
Targeting all the brightness of areas 2A to 2C, the photometric value V+-
V2 is determined by the following equation (2)-■ with the correction coefficient set to 0.

Vl-V2= (VA+VB+VC)/3-(2
)-〇Then, the circuit operation is performed by comparators 86 and 91.
outputs L level, comparators 87 and 90 output I-level, AND gate 98 outputs H level, and other AND gates 94-97, 99-102
outputs L level, and inverter 103 outputs H level. Therefore, the OR gate 112 outputs an H level, and the NOR gate 125 outputs a T level, so the output of the operational amplifier 26 becomes a voltage (VΔ + VB + Vc) /3, and the output Vt - V2 of the operational amplifier 31 also becomes (VA + VB + VC) /3. 3 and Naru.

The features of the photometric device according to the embodiments described above are as follows:
Assuming that the main subject is located in the center of the field area (area 2A), the size of the main subject can be determined, and correction can be performed in accordance with the size of the main subject.

In addition, in this embodiment, the outermost area 2C of the field area
When the average value of the luminance signals obtained from each of the small regions 2C+ to 2C4 is on the high luminance side, the small regions 2C+ to 2C4
By ignoring the output of a small area that has a low luminance signal within the background, elements that are inappropriate as background luminance can be removed, such as a shaded ground outdoors on a sunny day. or,
If the average value of the luminance signals obtained from each of the small regions 2C+ to 2C4 is on the low luminance side, the small regions 2C+ to 2C
By ignoring the output of a small area that has a high luminance signal within 4, elements that are inappropriate as background luminance, such as electric lamp light in a night scene, can be removed.

In addition, when the photometry device of this embodiment detects that the subject is whitish or blackish, it intentionally depicts highlights (highlight control) or shadows so that whitish subjects appear white and dark subjects appear black. This is done by correcting the depiction (shadow control), and this correction is made by changing the amount of correction depending on the size of the main subject, making it possible to control the photometric value extremely effectively.

In addition, in the explanatory diagrams of FIG. 8 and FIG. 9 used in the explanation of the above-mentioned embodiment, the values of the brightness level of each area 2A to 2C are as follows.
When the brightness difference between adjacent areas is small, the same level is shown, but this is a thermal theory, and in actual photometry there will be some difference (the brightness difference is smaller than the predetermined value compared, for example VP+). 8 and 9 serve only as explanatory diagrams to facilitate understanding of the present invention.

In addition, in the embodiment, there are four calculation methods for the photometric value Vl-V2 depending on the brightness level at the periphery of the screen (V
Although the area is divided into C+ to VC4), the photometric value may be obtained by dividing the area into smaller areas, or conversely, the peripheral area (Vc) of the screen may be used as one area for photometry. In addition, the calculation method for the photometric value Vl-V2 is determined in two cases: when the target photometric area is only VA and VB in the center and intermediate areas, and when the photometric area is
Although we have divided into two cases, A, VB, and VC, we always target part or all of the multi-divided peripheral area and obtain the photometric value in the same manner as above. Also good. Alternatively, the field of view may be divided into three or more annular regions, and the luminance signal difference between adjacent regions may be used.

Note that the present invention is not limited to single-lens reflex cameras, but can also be applied favorably to lens-based cameras and the like. Note that in this embodiment, the selection circuit is configured with a logic circuit, but it may also be processed using software using a microcomputer.
Naturally, it is a matter of enthusiasm that the present invention will be put into practice.

(Effects of the Invention) As explained above, the present invention is a suitable photometry device that assumes that the main subject is in the center of the field, and is particularly capable of determining the size of the main subject. Furthermore, from among the multiple set photometric value calculation formulas, a suitable calculation formula is selected according to the size of the main subject, and brightness information of the outermost area of the subject, where inappropriate elements are particularly likely to be added, is selected. Since the brightness information is obtained by removing the brightness of inappropriate elements, it is possible to provide a photometric device that can calculate a photometric value appropriate for the main subject.

[Brief explanation of drawings]

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. FIG. 7 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+・2C2・2C3・2C4・・・・
. . . Each area that receives light, 6 . . . Light receiving section, 20 . . . Peripheral luminance calculation circuit, 21
......Selection circuit. Agent: Gi Marushima -:□゛: Shiharu Figure 2 T/70 ■ VC and Vr5 ■Vζ<VK (Q) Cb) (C)
Cd)Cq)
(h) ζko) (0) (b) (C) Cd) (e) (5) C(j)
(hntL)

Claims (1)

[Claims] (1) The field of view is divided into at least three regions: a central region at the center, a first outer region outside the central region, and a second outer region outside the first outer region, and the plurality of regions and at least the outermost region is composed of a plurality of light receiving parts that further divides the region into a plurality of small regions and obtains the luminance information of each region and the luminance information of the small region, and the plurality of small regions are divided into each small region. At least one of the brightness information of each surrounding small area
Light-receiving means for determining surrounding area brightness information by selecting a peripheral brightness calculation circuit corresponding to one calculation formula based on the small area brightness information from peripheral brightness calculation circuits corresponding to a plurality of calculation formulas including one brightness information as a function. and brightness difference detection means for determining area brightness difference information between adjacent areas of the area from the plurality of area brightness information obtained from the light receiving means; a calculation means preset with a photometric value calculation circuit corresponding to a plurality of calculation formulas including at least one of the area brightness information as a function; at least one of the plurality of area brightness information; and the brightness difference detection. 1 from the calculation means based on at least two pieces of brightness difference information obtained by the calculation means.
A photometric device comprising: a selection means for selecting a photometric value calculation circuit corresponding to two calculation formulas.