Amplifying a Photodiode using an OpAmp

[mazzo532]

Dear electrical engineers,

Here a question from a serious electronics n00b =D.

I tried to build a system as shown in the image. I have a photodiode which gives me a clock signal in the range of 200 to 500mV. Frequency will be in the range of 100 to 1000 Hz.
I would like to amplify this signal in order to have a voltage between C and D in the range of 2 to 5V, hence my goal gain should be around 10.

I have an LT1006 amplifier, which should be fine for this purpose (Documentation HERE). I'm trying to use it as a Non-Inverting OpAmp.

Useless to say that my system does not work. Somehow I get a constant output of about 2V regardless of the diode status.
There must be some gross mistake somewhere, but I have no idea. Could someone tell me what did I do wrong?

Thanks in advance,
Mazzo

 

[BiGyElLoWhAt]

Normally, with a constant voltage output, that means you're railing. Is your input voltage +/- 2V by any chance?
Try removing the gain and see what happens. Check for shorts, make sure your feedback loop is connected properly.

Op amps can never output a higher voltage than their source. If your gain function $g$ times the output voltage $V_{out}$ is higher than your source voltage (+/- 5V in the diagram), you will get your source voltage as your output.

 

[BiGyElLoWhAt]

Also, how are you reading it? I'm assuming an o-scope? Perhaps it's just ticking too fast? All diodes are not made equal, meaning that if you're trying to get it to break down, you need very close to what's called the "breakdown voltage", but you want it to breakdown only when photons hit it. So you want it to be below.

Another possibility, is you photodiode lighting up? You might have it in backwards, which just allows current to flow through it, as opposed to it being reverse biased.

 

[sophiecentaur]

I see you have no bias resistor from the diode to Ground. I would imagine that could define the Volts on the Diode better than relying on the OP Amp input resistor. Try a few tens of kOhm. Have you looked at the diode volts with a DVM? Could be interesting.
Also, why choose such low resistor values in your feedback loop? I'd be inclined to use at least 10 times those values.

 

[mazzo532]

Hello guys,
thanks both of you for your answers! Let's reply with order:

@BiGyElLoWhAt
- I'm using an adjustable power supply unit, set to 5V.
- I'm testing the photodiode just by intermittently illuminating it with a flashlight. By hand. The diode alone, plugged straight into an o-scope (or a multimeter) works fine. Ambient light gives about 200mV and direct flashlight gives 500mV. I would say it works.
- About the diode being backward. This is interesting . Diode receives light and gives me a voltage drop across its plugs: I plugged the hi-voltage side to the + side of my OpAmp. Negative side goes to D (that's my negative PSU plug). Is that wrong?
- Question: what do you mean by +-5V? Why would you say there's a negative voltage? Can't I just consider that my negative PSU connector is zero volts and the positive one is at +5V?

@sophiecentaur
- Grounding is in fact the muddiest point of the story for me. Gonna show you my infinite ignorance: why would I need to ground anything?
- I checked my diode health status with a multimeter. It works. Ambient light gives about 200mV and direct flashlight gives 500mV.
- About resistors, I have actually just set their ratio (that's 10) and I've randomly chosen their resistance magnitudes. How does that affects the system? By the way I'll test higher resistances as soon as I'm back in the lab.

Mazzo

 

[jim hardy]

here's why you can't get more than about 2 volts
from datasheet you posted

try 100X larger resistors.

old jim

EDIT i see Sophie caught it too

By the way I'll test higher resistances as soon as I'm back in the lab.

that should work

 

[mazzo532]

@jim hardy
Thanks for your help Jim! This is great, but still I have some doubt: my system was giving a 2V constant voltage. It was completely independent on the amount of light the diode was receiving. There must be some other mistake.

 

[Charles Link]

Suggest you use a simpler circuit where you connect the "+" terminal of the op-amp to ground. The one end of the photodiode is grounded and the other end connects to the "-" terminal of the op-amp. The photodiode polarity doesn't matter, you'll get the opposite output by reversing the polarity. The feedback resistor (R=100 kohm) connects from the "-" op-amp terminal to the op-amp output. A small capacitor, (10 pF) can be put in parallel with the feedback resistor to reduce noise. The output voltage is measured between the output terminal of the op-amp and ground. Oftentimes the op-amp requires +12 V and -12 V bias voltages. (This circuit amplifies the photocurrent with zero voltage across the diode. It is essentially a current amplifier.)

 

[jim hardy]

Thanks for your help Jim! This is great, but still I have some doubt: my system was giving a 2V constant voltage. It was completely independent on the amount of light the diode was receiving. There must be some other mistake.

keep us posted whether higher resistors helps ?

As Sophie pointed out
your circuit allows no current through the photodiode
so the first photon to strike it drives it to full voltage
and its capacitance holds it there
That's an exagggeration of course but only a mild one
a resistor across diode gives the charges someplace to go
try a megohm ?
The opamp has a few nanoamps of input current
see parameter "Input Bias Currrent", 9 to 15 na,
and your circuit must provide a path for that trickle . You want it not to flow through your photodiode.

 

[mazzo532]

keep us posted whether higher resistors helps ?

As Sophie pointed out
your circuit allows no current through the photodiode
so the first photon to strike it drives it to full voltage
and its capacitance holds it there

a resistor across it gives the charges someplace to go
try a megohm ?

Interesting. You mean I should have a path to the Ground placing a MOhm resistor? Should i put between points A and D?

 

[Charles Link]

Interesting. You mean I should have a path to the Ground placing a MOhm resistor? Should i put between points A and D?

Please read my suggestion in post #8. I have used this op-amp circuit with photodiodes countless times and it is both simple and reliable.

 

[jim hardy]

Interesting. You mean I should have a path to the Ground placing a MOhm resistor? Should i put between points A and D?

I think so. Sophie called it a "bias" resistor.

megohm was a USWAG, acronym for Unscientific_wild-a**guess
it'll depend on what is the current produced by your photodiode

can you measure that ?

EDIT late addition
@mazzo532
see polarity issue in post # 14 and 16

 

[f95toli]

Interesting. You mean I should have a path to the Ground placing a MOhm resistor? Should i put between points A and D?

What type of photodiode is it? Is it a packaged device that actually gives a voltage out? Or is is just a "plain" diode?
In the latter case you certainly need a path to ground (the light generates a current and that current needs to go somewhere).
I would have thought a few kOhms would be a good starting point (Mohm sounds way too high), but the best thing to do is to check the datasheet of the photodiode, there you will probably find some example circuits and you can just use whatever value they are using.

 

[jim hardy]

CharlesLink's suggestion is a good one

AHA because it looks like you are driving your single supply opamp's input negative in non-inverting mode, asking it to do the impossible - produce a negative output...

Charles suggests instead inverting mode like this
from LM324 datasheeet at http://www.ti.com/lit/ds/symlink/lm124-n.pdf

that might work with a single supply opamp in the polarity shown

easy enough to try.

 

[Charles Link]

CharlesLink's suggestion is a good one

AHA because it looks like you are driving your single supply opamp's input negative in non-inverting mode, asking it to do the impossible - produce a negative output...

Charles suggests instead inverting mode like this
from LM324 datasheeet at http://www.ti.com/lit/ds/symlink/lm124-n.pdf
View attachment 102981
that might work with a single supply opamp in the polarity shown

easy enough to try.

Thank you Jim! I would have included a circuit diagram with my description but I haven't yet learned how to upload files onto PF. That's precisely the circuit I was referring to. The circuit gives an output voltage equal to $V_{out}=I_p R_f$ where $I_p$ is the photocurrent at zero voltage across the photodiode.

 

[jim hardy]

but I haven't yet learned how to upload files onto PF

i use the "snip" , click " copy" , and paste into Paint, add notes & arrows then save to disk (made a folder FOR_PF for such things)
then use UPLOAD button lower right
i save as jpg not png because the files are smaller that way
.......................

.regarding the circuit ----

i missed his reverse polarity/single-supply conflict until i read your post carefully.
do you think he'd get away his non-inverting configuration if he just flipped his diode over?

Ahhh it's the small things of the Earth that confound the mighty, eh ?

 

[BiGyElLoWhAt]

USWAG, acronym for Unscientific_wild-a**guess

When will NASA stop with all the acronyms?

 

[Charles Link]

i use the "snip" , click " copy" , and paste into Paint, add notes & arrows then save to disk (made a folder FOR_PF for such things)
then use UPLOAD button lower right
i save as jpg not png because the files are smaller that way
.......................

.regarding the circuit ----

i missed his reverse polarity/single-supply conflict until i read your post carefully.
do you think he'd get away his non-inverting configuration if he just flipped his diode over?

Ahhh it's the small things of the Earth that confound the mighty, eh ?

Flipping the photodiode over might make it work, but I am not a EE and the original circuit is overly complex. The op-amp circuit with the circuit diagram @jim hardy provided is a standard one that is used in photoradiometric measurements throughout the industry. The "-" terminal of the op-amp as a virtual ground along with the near infinite input impedance of the op-amp makes it a very simple circuit to analyze. .. There is one puzzle that did surface (the technicians at the workplace asked me this question) with the analysis of this circuit, particularly when a load resistor is placed at the output. How come the (current) goes into is not equal to the comes out of? And the answer is evidently the current in the two bias lines is not necessarily equal but will be such that the current into the device is the current out of the device.

 

[jim hardy]

There is one puzzle that did surface (the technicians at the workplace asked me this question) with the analysis of this circuit, particularly when a load resistor is placed at the output. How come the (current) goes into is not equal to the comes out of? And the answer is evidently the current in the two bias lines is not necessarily equal but will be such that the current into the device is the current out of the device.

current into and out of where ?

The two signal input lines , + and - , ideally would both have zero current , infinite impedance , sensing voltage perfectly ..

In reality they have a few nanoamps (or pico or femto amps in better opamps) which we generally dismiss
but we do know it exists and it is stated in the datasheet as ""Input Bias Current". It can be either polarity.
The difference in those bias currents is spec'd also. it's "Input Offset Current"

Current coming out of the amplifier via the output pin got into the device through the power supply pins
so Kirchoff is neither disproven nor disobeyed , just we have to keep our thinkin' sraight.
Suggest your tech draw his circle around it enclosing ALL the pins.

any help ?

old jim

 

[Charles Link]

current into and out of where ?

The two signal input lines , + and - , ideally would both have zero current , infinite impedance , sensing voltage perfectly ..

In reality they have a few nanoamps (or pico or femto amps in better opamps) which we generally dismiss
but we do know it exists and it is stated in the datasheet as ""Input Bias Current". It can be either polarity.
The difference in those bias currents is spec'd also. it's "Input Offset Current"

Current coming out of the amplifier via the output pin got into the device through the power supply pins
so Kirchoff is neither disproven nor disobeyed , just we have to keep our thinkin' sraight.
Suggest your tech draw his circle around it enclosing ALL the pins.

any help ?

old jim

Yes Jim, that is the solution we figured out. I think the technician raised a good question.(this dates back a few years). They originally saw the op-amp much like I did as a device basically with an input and output port. A good EE like yourself might find it trivial, but it puzzled us at first.

 

[Charles Link]

I read post #1 more carefully. It mentions a clock signal. Is the idea to use an optical chopper on the light source to create the clock signal? If that is the case, the op-amp circuit I have proposed should work for this application.

 

[jim hardy]

If that is the case, the op-amp circuit I have proposed should work for this application.

agreed

 

[mazzo532]

Hello Gentlemen,

the schematic you proposed works (and it's simpler, as you pointed out). Thank you very much for the suggestion!
I take the liberty to ask for some help more:
Now the system works, but only if I connect the V- power line with Ground.

I am using a typical "regulated DC power supply", which has V+, V- and Ground. Therefore I need to use both V+ and V- to power the amplifier. I am measuring the voltage between OpAmpOutput and Ground.I'm trying to understand why do I have to short-circuit V- and Ground, and to the consequences in my whole trigger acquisition system.
Is this related with the concept of "Single power supply"?

Thanks again for your generous help!Mazzo

 

[Charles Link]

Hello Gentlemen,

the schematic you proposed works (and it's simpler, as you pointed out). Thank you very much for the suggestion!
I take the liberty to ask for some help more:
Now the system works, but only if I connect the V- power line with Ground.

I am using a typical "regulated DC power supply", which has V+, V- and Ground. Therefore I need to use both V+ and V- to power the amplifier. I am measuring the voltage between OpAmpOutput and Ground.I'm trying to understand why do I have to short-circuit V- and Ground, and to the consequences in my whole trigger acquisition system.
Is this related with the concept of "Single power supply"?

Thanks again for your generous help!Mazzo

I have seen power supplies that have a +V and -V but are not dual. You need a dual power supply and on one side you ground the "-" to get +12V on the +, and on the other side, you ground the "+" to get -12V on the "-". You may find one that gives you "+" and "-" with the ground in the middle, but this is usually not the case... And glad to hear you liked the simpler circuit. In a pinch, you could use two single +12V supplies (simple AC to DC (conversion) power packs). Usually the DC voltages on them float, so to get a -12V, you just ground the "+", etc. and for +12 V, be sure and ground the "-". Don't know if these typically are high quality, e.g. containing low ac ripple, etc., but they should work. You can cut the pins (the output jack) off and work with the two wires in the output cord if you don't have the necessary adaptors. Op-amps I don't think typically require much current unless you have a very small load resistor at the output.

 

[jim hardy]

I'm trying to understand why do I have to short-circuit V- and Ground, and to the consequences in my whole trigger acquisition system.

"Ground" ? To what is this little opamp circuit connected ? If whatever it is connected to needs for you circuit to be "grounded", then you'll have to "ground" it.

But that is a requirement imposed on your circuit from outside.
Your circuit would work by itself on the moon and powered by a 9v battery., 240 thousand miles from "ground".

Can you give a clue about this "whole trigger acquisition system" ?

 

[Charles Link]

"Ground" ? To what is this little opamp circuit connected ? If whatever it is connected to needs for you circuit to be "grounded", then you'll have to "ground" it.

But that is a requirement imposed on your circuit from outside.
Your circuit would work by itself on the moon and powered by a 9v battery., 240 thousand miles from "ground".

Can you give a clue about this "whole trigger acquisition system" ?

@jim hardy I think I understand your input, but I think I also follow the OP's input. The circuit does not require a ground to any external ground, but the non-inverting "+" terminal of the op-amp(I think it is pin 3) is reference point for which a +12 V bias voltage and a -12 V bias voltage (The 12 volts isn't fussy=any voltage between 9 Volts and 12 volts should work) needs to be connected to the bias lines (I think these are pins 4 and 7 on an op-amp that I was familiar with, but the OP can look at the pin diagram for the op-amp that he is using). The line this non-inverting terminal (pin 3) is on needs to be connected to the minus end of a +12 V DC bias voltage that goes to the + voltage bias pin of the op-amp and similarly, the "+" end of a -12 V DC bias needs to be connected to the same non-inverting terminal (pin 3) of the op-amp. Yes, it would work anywhere (e.g. on the moon) and doesn't require a permanent ground, but the reference point (the non-inverting terminal) does need to be connected to the bias voltage supplies. The output voltage (normally out of pin 6) is also measured between pin 6 and pin 3. The circuit will normally not work properly if you fail to properly connect pin 3 to the bias voltage lines as described above. If you have multiple op-amps on the same circuit card, normally this pin 3 is a reference line=ground for the entire circuit card, even if you let it float.

 

[jim hardy]

@jim hardy I think I understand your input, but I think I also follow the OP's input.

I take it you and op are both speaking if this circuit ?

which is from the datasheet fpr a single supply opamp .
It doesn't show power but it'll work with a single supply connected between the amp's two supply pins, +to V+ and - to V-.
The V- pin becomes the "gnd" , or more properly it becomes 0V reference or circuit common.

That's what's special about single supply opamps like OP chose or like LM324 that I'm so fond of - its 0V reference doesn't need to be halfway between V+ and V-, it can be all the way down at V- .

You correctly described a dual supply opamp where 0V reference must lie some distance above V-and some distance below V+. That distance i call "Headroom"..

His LT1006 needs no headroom to V-

though it need about 1.5V of headroom to V+ , as does LM324.

The circuit will normally not work properly if you fail to properly connect pin 3 to the bias voltage lines as described above.

in this circuit, amp's inverting input pin gets tied to its negative power pin and both could be connected to - pin of a plain 9V battery. That point becomes 0V reference.
OP says he has to Ground his 0V reference, i don't know to what he has to "ground" it, but presumably to some other piece of gear.

Make sense ?

 

[Charles Link]

I take it you and op are both speaking if this circuit ?

which is from the datasheet fpr a single supply opamp .
It doesn't show power but it'll work with a single supply connected between the amp's two supply pins, +to V+ and - to V-.
The V- pin becomes the "gnd" , or more properly it becomes 0V reference or circuit common.

That's what's special about single supply opamps like OP chose or like LM324 that I'm so fond of - its 0V reference doesn't need to be halfway between V+ and V-, it can be all the way down at V- .

You correctly described a dual supply opamp where 0V reference must lie some distance above V-and some distance below V+. That distance i call "Headroom"..

His LT1006 needs no headroom to V-
View attachment 103140
though it need about 1.5V of headroom to V+ , as does LM324.in this circuit, amp's inverting input pin gets tied to its negative power pin and both could be connected to - pin of a plain 9V battery. That point becomes 0V reference.
OP says he has to Ground his 0V reference, i don't know to what he has to "ground" it, but presumably to some other piece of gear.

Make sense ?

Question for @jim hardy . With your op-amp, are you able to get negative output voltages? Normally in the op-amp's that I have seen, the voltage swing is determined by the bias voltages. If you use +/-12 Volts, your voltage swing is basically +/- 12 Volts. With a photodiode, you won't need the negative voltage unless you hook it up backwards, so yes, you have a simpler solution that requires only one power supply. Note: I'm basically familiar with two op-amps: the 741 and the LF356.

 

[jim hardy]

Question for @jim hardy . With your op-amp, are you able to get negative output voltages?

Nope . With single + supply everything must be positive. Input can sense all the way down to zero, maybe a very few millivolts less, see applications info on LT1006 linked by OP in first post.
I think his first circuit would have worked had he flipped his photodiode turned upside down
but i was slow to catch on to his "noninverting follower with gain" circuit with negative input. It couldn't drive output negative, of course, there's not a negative supply.Check out
https://courses.cit.cornell.edu/bionb440/datasheets/SingleSupply.pdf
and
http://www.ti.com/lit/an/sloa030a/sloa030a.pdfold jim