I'm working with a photodiode in photovoltaic mode. In order to amplify the signal I've done the following

Omitted power connections purposely.​

So this works great, it eliminated all the noise I'd get when I connected the photodiode directly to the non-inverting amplifier.
My question is what advantage, if any, does a Transimpedance amplifier circuit have over this circuit? Thanks!

 

One advantage a transimpedance amplifier has over your circuit is that it actually works: take a VERY close look at how you've connected U1 and the photodiode.

 

One advantage a transimpedance amplifier has over your circuit is that it actually works: take a VERY close look at how you've connected U1 and the photodiode.

Hmm I'm not seeing it, as far as I could tell last night this worked fine. Roughly 300mV across my PD and about 3.3V on the output of U2. The output followed as I changed the light exposure to the diode as well. What am I'm doing wrong here?

 

Ahh I think I've added an "extra" ground connection. Sorry, I drew this from memory. U1 inverting input should NOT be tied to ground, correct?

 

Still not right.

 

Still not right.

Could you tell me what is wrong?

 

I could, but it's something glaringly obvious-- and fundamental-- that you really should be able to see for yourself (and it has nothing to do with power supply connections).

 

I could, but it's something glaringly obvious-- and fundamental-- that you really should be able to see for yourself (and it has nothing to do with power supply connections).

ahhh my diode is backwards...I think. Still learning. Cathode to the non-inverting input?

 

Nope. Which way your diode is pointing affects only the polarity of your output, and nothing else.

Hint: what have you provided in your circuit to establish a reference potential for the inputs and output of U1? As you have it, they're undefined (i.e., floating).

 

Ok, I've taken the diode out of the feedback loop. In hindsight that doesn't make much sense. How about this? Thank you for the help by the way!

 

That's more like it!

Now take a look at the circuit and ask yourself, why is the first opamp even necessary? Why not eliminate it, and connect the anode of the photodiode directly to the (+) input of the second opamp?

 

That's more like it!

Now take a look at the circuit and ask yourself, why is the first opamp even necessary? Why not eliminate it, and connect the anode of the photodiode directly to the (+) input of the second opamp?

I tried this and ended up with an extremely noisy signal. I'll give it another go tonight! So back to my original question, is there any reason NOT to do it this way? Also, I shouldn't have an issue with current flowing through my diode right? I'm wondering if I should add a series resistor to ground.

 

Do you want an output that varies linearly with light level? Or do you want an output proportional to the log of the light level? As it is, your circuit will show a logarithmic response. To make it linear, you either have to shunt the photodiode with an appropriately-sized resistor, or switch to a transimpedance amplifier configuration.

Also, a voltage-mode circuit such as this will have very poor response speed, especially at low light levels; a properly-designed transimpedance amplifier will be much, MUCH faster.

 

Adding an amplifier adds that amplifier's noise and offset to the signal.
This will increase the amp noise by the √2 and can double the DC offset.

 

The second Op amp does need a low Z source to work.

https://www.thorlabs.com/tutorials.cfm?tabID=31760

Operating in Photoconductive mode lowers capacitance and improves bandwidth.

Look here too: https://www.google.com/url?sa=t&rct...part-1&usg=AFQjCNH2-G5-r6npn0w3h9s-I7a21jYs3A

 

I tried this and ended up with an extremely noisy signal. I'll give it another go tonight!

What kind of noise? Broadband thermal (i.e., "white") noise? RF interference? Pickup from power mains? You've got some very high impedances in this circuit, and it will be quite sensitive to any kind of interference; good shielding is a must.

 

What kind of noise? Broadband thermal (i.e., "white") noise? RF interference? Pickup from power mains? You've got some very high impedances in this circuit, and it will be quite sensitive to any kind of interference; good shielding is a must.

Not sure of the type of noise, but instead of a clean DC signal i'd get something that looked sinusoidal. The output followed the input by introducing an offset, not by changing the amplitude of the "AC" output signal.

 

The second Op amp does need a low Z source to work.

Ummm... why? The second opamp is operating in an ordinary non-inverting configuration and has a high-impedance input. As in all opamp circuits, the inputs must have a DC path to ground (or to some other voltage within the common-mode range of the opamp); but the input to a non-inverting amplifier such as this does NOT need to be driven from a low-impedance source.

 

Not sure of the type of noise, but instead of a clean DC signal i'd get something that looked sinusoidal. The output followed the input by introducing an offset, not by changing the amplitude of the "AC" output signal.

Sinusoidal, but at what frequency? If it was 60 Hz, that's power-line pickup. Use more shielding.

 

Hello,

Have a look at this PDF:
http://forum.allaboutcircuits.com/a...hotodiode-monitoring-with-op-amps-pdf.110000/

Bertus