Hi everyone. I have this charge amplifier circuit which is configured in a way that I don't understand. Instead of the noninverting terminal of the opamp connected to ground, the schematic is pulling it up to a voltage Vref. Can anyone explain this reasoning? Thank you!

 

I think the op amp is used as a comparator to monitor the battery's voltage.
Post a schematic to clear things up.

 

Sorry, here is a rough sketch of the circuit.

 

It's a transimpedance amplifier with current from a photodiode as an input, and Vref1 is just to level-shift the output (by Vref1), probably because of a unipolar supply. Vref2 would be a little above Vref1, to take into account ambient light and noise levels.

The feedback cap is generally necessary in PD TI amps to eliminate gain peaking from the PD junction capacitance across the inverting input of the op-amp, which forms a pole with the feedback resistor.

 

Thanks for the info. First I am just trying to build a charge amplifier with the op amp and a photodetector. I attached the circuit that I am working with. I selected 100k as an arbitrary value for the feedback resistor, and I calculated the feedback cap with the equation shown in the schematic, derived from the transfer function of a transimpedance amplifier. The photodiode I am using is a UDT PIN 10d, and the output capacitance is 300pf. So in my circuit, Cin is the 300pf + 1pf because I read that most opamps have an input capacitance of 1pf. Are there calculations correct? Because I powered up my circuit with these paramaters and I did not get anything on the scope. Is the max473 a suitable opamp for this application?

 

You need a negative voltage source, either to make the op-amp supplies split-rail, or you could keep a single supply rail on the op-amp and use a negative voltage to reverse bias the photodiode (turned around from as depicted in the circuit). Then, as long as the input offset currents of the op-amp aren't excessive, it should work. As it is, the output wants to swing below 0V, and it can't. I've been caught out by photodiode current direction before, and I'm sure I'll be caught out again.

100k is a pretty good starting value, and your compensation cap methodology is sound.

Here's a handy link from when Nat Semi was great. Check out also the writings of Bonnie Baker (EDN) and Bob Pease (Electronic Design) on the subject.

 

Yeah I was using a 741 with +/- 15 V rails but the output was just being pulled low to -15 V so I decided to switch opamps and see how this one works.

Could this be because my bias current is greater than the photodiode current? What would be an acceptable resistor value to compensate for this?

I'll give it another try and let you know what I am getting.

 

You're quite right, there will be an voltage offset on the op-amp output that is a function of input bias current X Rf, but even on a 741 this should only account for a few mV of output offset.

Try blocking all light to the PD (a blob of Blu-Tack is ideal) and see what the output looks like. If it still saturates, then it might do to check the circuit board for flux contamination - that's conductive enough to leak enough current to saturate a sensitive circuit such as this. You'll still need either a split op-amp supply, or alternative PD biasing, as mentioned before.

 

I removed my PD from the circuit and reversed biased the cathode 5V then connected the anode to the scope to see what I was getting, but it was just getting pulled high. Using pulsed green LED.

I changed detectors and it works better now, I get about 50mV pulses on the scope when it is not in the circuit. I am now using a Thorlabs DET36A (without built in amp).

My circuit is the same as what I posted except now I am using a 741 with positiveand negative supply rails (15V). When I hookup my detector to the inverting terminal, the output remains at 0V. What could be wrong this time? I am very confused.

 

Troubleshooting is much quicker with all the physical presences in the same room...

If you're biasing a lone photodiode and feeding that straight into a 'scope, then it's normally best to terminate the 'scope input with 50 Ω, else the dark current (plus any current from ambient light) will set up a large offset voltage across the default 1 MΩ input impedance. Try that with the Thorlabs PD module and the original PD (with bias). The circuit on the Thorlabs datasheet is a good one.

It's time to start narrowing down the source of your problems with the op-amp circuit. A good start would be to isolate the PD from the op-amp, so at least one or the other can be blamed. You can make sure your op-amp circuit is working by removing the PD and injecting a signal through a 100 kΩ resistor into the inverting input - if the op-amp circuit is OK, then the output waveform is simply the inverse of the input, within bounds of frequency response, saturation etc. If you don't have a signal generator handy, then inject a bit of mains hum with a judicious finger.

If the op-amp circuit works OK as an ordianry inverting amp, then the problem is more than likely with the PD/bias section. Could you give a part number for the PD? It may be something like a low sensitivity to green light. If in doubt, try exposure to an incandescent light bulb (they're pretty broad-spectrum), that should make something move on the output.

 

I just checked the op amp circuit and Thorlabs PD in isolation, and they both seem to be working ok. My signal generator supplies the inverting terminal of the opamp with 5V square pulses and I see inverse 5V pulses on the scope. For the PD, I supplied it pulses of light from a red LED (red seems to have better spectral response then green for this PD says the datasheet) and scoped the output via 100 ohm resistor and I am seeing pulses of about 80mV peaks. Add them two together and nothing is on the scope.

It is a Thor labs DET 36A. Is the PD not supplying enough current?

 

robby991, when you say no output with the PD/op-amp combo, do you mean no signal output and the DC level at 0V, or the output stuck at a rail? If your PD is giving out 80 mV into 100 Ω, then the same signal will try and stick out 80 V with 100 kΩ of gain on your amp. There'll be enough dark current and ambient light to produce saturation of the amp output and you won't see any signal in this.

If this is the case, crank down your gain to 10 kΩ (adjusting Cf accordingly), and if you see a signal on the output, fettle to taste.

 

Darren, I haven't had a chance to get back to this circuit yet. I will try this weekend and let you know what I am getting.