So we are trying to detect very low levels of UV light using the circuit below. One of the problems I am having is that the response time for the output at "signal" is as long as 1-2 seconds. If we pulse our light at 10 Hz the output is 1.8 V but if we have the LED on constantly we are getting 5V. I assume this is becuase when we pulse the signal the output hasn't had time to rise yet.

From the research I have done so far it looks like a smaller capacitor is needed but with so much gain do I also need an op amp with a higher GBP? The AD820 has a GBP of 1.8MHz. Is there a simple calculation for what C45 should be or is the only one F = 1/2.pi.C.R?

Any advice/suggested reading material would be much appreicated.

Photodiode Amplification Circuit

  • \$\begingroup\$ Get rid of C45, or cut it down to 1pF. What's the capacitance of the photodiode? (You could try and bias the PD.) \$\endgroup\$ – George Herold Mar 19 '18 at 15:01
  • \$\begingroup\$ @GeorgeHerold the capacitance of the photodiode is 1000pF i beleive. if you can just get rid of that capacitor what is it's function \$\endgroup\$ – Revilo Engineering Mar 19 '18 at 15:18
  • \$\begingroup\$ Analog devices photodiode wizrd \$\endgroup\$ – Andy aka Mar 19 '18 at 17:13
  • \$\begingroup\$ @ReviloEngineering, there is a bunch of stuff about compensating TIA PD amps. Here's one, ti.com/lit/an/snoa515a/snoa515a.pdf Equation 3 is useful in determining the compensating cap value. \$\endgroup\$ – George Herold Mar 20 '18 at 15:12

Using a 264M feedback resistor is asking for trouble, since you will need to take great care in cleaning solder flux from your board. You would do better with


simulate this circuit – Schematic created using CircuitLab

This will give you a time constant of 1 msec, which is much more appropriate for your modulation.

As for the why of C45, you should Google "photodiode transimpedance amplifier stability". Briefly, the capacitance of the photodiode at the input to the op amp will cause the amp to start oscillating. Feedback capacitance compensates for the input capacitance. The exact relationship is beyond the scope of an answer here, but do some research.

EDIT - Since you don't know how transimpedance amps work, let me try a few numbers. Assume 10 nA out of the PD. Pushing this through the 264 M resistor will, by Ohm's Law, produce 2.64 volts. Reducing the feedback resistor to 10 M will produce 0.1 volts. Multiplying this by 26 (the gain of the second op amp) will produce 2.6 volts, which I assume will be close enough.

If you were trying for a combination of maximum sensitivity and bandwidth, the revised circuit would suffer from slightly increased noise, but this should not be a problem in your case.

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  • \$\begingroup\$ the size of the resistor was not by choice, the light we are measuring is very low and this is the only way we can see a signal. Would the circuit you proposed have similiar gain? \$\endgroup\$ – Revilo Engineering Mar 19 '18 at 16:17
  • \$\begingroup\$ The circuit above shows a Transimpedance gain of 10M followed by a non-inverting amplifier of A=26, giving a full gain of 260M. This is a fine approach assuming you don't need very High SNR at DC. \$\endgroup\$ – Luke Gary Mar 19 '18 at 16:23
  • \$\begingroup\$ @LukeGary wouldn't that make the gain of my original circuit higher than 264M \$\endgroup\$ – Revilo Engineering Mar 19 '18 at 16:24
  • \$\begingroup\$ sure, but with a much higher bandwidth. The Large Rf in the TIA front end creates a nasty pole in the feedback loop with the parasitic capacitance AND feedback capacitance that will limit the bandwidth of the amplifier severely. You can get the bandwidth you need with the original implementation but your layout will need some serious tuning. A difference of 4M in transimpedance gain may seem like a lot, but you will see that with the original circuit just from component tolerance. \$\endgroup\$ – Luke Gary Mar 19 '18 at 16:27
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    \$\begingroup\$ @ReviloEngineering - No. The second op amp, with no R2, is a follower, not an amplifier, and will have a gain of 1 regardless of the resistor value. Adding the second resistor produces a non-inverting amplifier with gain (1+(R3/R2)). \$\endgroup\$ – WhatRoughBeast Mar 19 '18 at 16:48

So the large problem I see is the 10nF feedback capacitor. Your Bandwidth is approximated with 1/2piRfCf in this case, so you are looking at BW = 1/2pi*264M*10nF = 0.06Hz!!. Try reducing that Cf first, but a more robust solution would be to reevaluate how much front end gain you really need in comparison to you bandwidth requirement. In general the gain bandwidth product requirement should be calculated for a TIA with Fgbw > (Cin + Cf)/(2piRfCf*Cf) where Cin is the input capacitance of the amplifier plus the photodiode capacitance and an approximation of the parasitic capacitance. Similarly, Cf would be the Feedback capacitance plus the approximate parasitic element on the pcb.

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