# Photodiode in a log amp

J

#### John Popelish

Jan 1, 1970
0
Jack said:
Hi!

I'm trying to construct a battery operated light meter using

IPL 10020 photodiode
LM301AN op amp
plus various bits and pieces...

I found the following schematic for a logarithmic exposure meter that
should work with that photodiode.

.----------------------------.
| |
.-------------. |
| | | |
.-|\ | |
| \-->|--M--+--/\/\--+--/\/\--+
| / | |
.-|/| | ___
| | | -
| | \\ | |
.------>|--------------+--/\/\--+
| |
.---------------------------.

The datasheet says that feeding the photodiode into a high impedance
gives a logarithmic voltage/illumination responce. The upper diode is
for very low light levels where the amplifier bias current may cause the
output to go negative.

I have three questions.

1. How does that circuit operate?

It looks a lot more complicated than it is. Think of the right end of
the photo diode as zero volts. This voltage has been biased within
the supply range by the two resistors on the right.

The opamp acts as a follower (gain of 1 amplifier) to copy the photo
diode voltage at the point that its feedback is taken (the left end of
the resistor above the photo diode. This forces a current through
that resistor than is proportional to the photo diode voltage (which
is proportional to the log of illumination). That current passes
through the meter.
I know that there is an current to voltage converter.

Actually it is a photo diode voltage to (meter) current converter.
I know that current through diodes increases exponentially when voltage
increases linearly.

That is the photo current which is proportional to illumination. So
the diode voltage is proportional to the log of that photo current.
How is the current controlled so that the voltage is increased
logarithmically?

The properties of a PN junction do that, automatically, as long as the
temperature remains constant.
2. I'm planning to connect it to an ADC. Over what should the
measurement be made? (M stands for an ampere meter in the original
schematic)

Then measure the voltage drop across the resistor above the photo
diode.
You can then leave out the diode and meter.
3. How accurate is this approach? Should I consider a linear

The biggest error is probably the temperature effect on the PN
junction. But you can measure temperature (with another diode biased
by a constant current) and calculate the correction.

J

#### Jack Middleton

Jan 1, 1970
0
Hi!

I'm trying to construct a battery operated light meter using

IPL 10020 photodiode
LM301AN op amp
plus various bits and pieces...

I found the following schematic for a logarithmic exposure meter that
should work with that photodiode.

.----------------------------.
| |
.-------------. |
| | | |
.-|\ | |
| \-->|--M--+--/\/\--+--/\/\--+
| / | |
.-|/| | ___
| | | -
| | \\ | |
.------>|--------------+--/\/\--+
| |
.---------------------------.

The datasheet says that feeding the photodiode into a high impedance
gives a logarithmic voltage/illumination responce. The upper diode is
for very low light levels where the amplifier bias current may cause the
output to go negative.

I have three questions.

1. How does that circuit operate?
I know that there is an current to voltage converter.
I know that current through diodes increases exponentially when voltage
increases linearly.
How is the current controlled so that the voltage is increased
logarithmically?

2. I'm planning to connect it to an ADC. Over what should the
measurement be made? (M stands for an ampere meter in the original
schematic)

3. How accurate is this approach? Should I consider a linear

Jack

J

#### Jack Middleton

Jan 1, 1970
0
John said:
It looks a lot more complicated than it is. Think of the right end of
the photo diode as zero volts. This voltage has been biased within
the supply range by the two resistors on the right.

The opamp acts as a follower (gain of 1 amplifier) to copy the photo
diode voltage at the point that its feedback is taken (the left end of
the resistor above the photo diode. This forces a current through
that resistor than is proportional to the photo diode voltage (which
is proportional to the log of illumination). That current passes
through the meter.

Thanks, I think I got it now. I had been staring that diagram for
ages without making a head or tail of it.
The biggest error is probably the temperature effect on the PN
junction. But you can measure temperature (with another diode biased
by a constant current) and calculate the correction.

Yes, that might be a problem, winters are cold over here. I think I've
seen a compensating circuit somewhere, need to look at it before the fall.

Many thanks for the help,

Jack

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