Understanding transimpedance amplifiers

[new299]

I've been trying to better understand trans-impedance amplifiers specifically this configuration:

http://upload.wikimedia.org/wikipedia/en/thumb/f/f1/TIA_simple.svg/375px-TIA_simple.svg.png

I assume the photodiode is effectively a current source. So a current will flow out of the photodiode, into the feedback resistor to Vout.

But what voltage does the photodiode produce at the negative input? If anyone has any tips that could help push me in the right direction I'd be most grateful.

[w2aew]

I've been trying to better understand trans-impedance amplifiers specifically this configuration:

http://upload.wikimedia.org/wikipedia/en/thumb/f/f1/TIA_simple.svg/375px-TIA_simple.svg.png

I assume the photodiode is effectively a current source. So a current will flow out of the photodiode, into the feedback resistor to Vout.

But what voltage does the photodiode produce at the negative input? If anyone has any tips that could help push me in the right direction I'd be most grateful.

In an ideal sense = zero volts, the negative (inverting) input will be equal to the positive (non-inverting) input.

In simplest terms - a very tiny voltage.  Just enough to cause the op amp to swing the output far enough to soak up the photocurrent (+/- input leakage) in the feedback resistor (not taking into account the input offset voltage). 

[new299]

In an ideal sense = zero volts, the negative (inverting) input will be equal to the positive (non-inverting) input.

In simplest terms - a very tiny voltage.  Just enough to cause the op amp to swing the output far enough to soak up the photocurrent (+/- input leakage) in the feedback resistor (not taking into account the input offset voltage).

Thanks, yes I understand that the inputs of the opamp will be the same and assume the negative terminal is virtual ground. However, I assume there is a contribution to this voltage from the photodiode. What is this contribution? (I guess I don't fully understand how a current becomes a voltage here).

[w2aew]

In an ideal sense = zero volts, the negative (inverting) input will be equal to the positive (non-inverting) input.

In simplest terms - a very tiny voltage.  Just enough to cause the op amp to swing the output far enough to soak up the photocurrent (+/- input leakage) in the feedback resistor (not taking into account the input offset voltage).

Thanks, yes I understand that the inputs of the opamp will be the same and assume the negative terminal is virtual ground. However, I assume there is a contribution to this voltage from the photodiode. What is this contribution? (I guess I don't fully understand how a current becomes a voltage here).

Think of it this way... The photocurrent pulls the inverting input very slightly below ground.  The resulting input voltage difference causes the op amp output to rise.  As it rises, the photocurrent is drawn through the feedback resistor because of the increased voltage across it.  Since the voltage gain of the op amp is extremely high - a miniscule input voltage difference causes a large output voltage swing.  Also, ideally the op amp inputs draw no current - thus all of the current must flow in Rf.

[new299]

Think of it this way... The photocurrent pulls the inverting input very slightly below ground.  The resulting input voltage difference causes the op amp output to rise.  As it rises, the photocurrent is drawn through the feedback resistor because of the increased voltage across it.  Since the voltage gain of the op amp is extremely high - a miniscule input voltage difference causes a large output voltage swing.  Also, ideally the op amp inputs draw no current - thus all of the current must flow in Rf.

Thanks that helped a lot. I guess I don't really need to think in terms of voltage from the photodiode. Rather just that there is some voltage which will then cause the opamp to generate a current through Rf to balance the current of the photodiode thereby resulting in 0volts/amps at the negative terminal.

I think I've understood things a little better now. Thanks again!

[KeepItSimpleStupid]

It's like the sum of the current at the inverting side is zero,  Draw the arrows in the same direction (toward or away from the dot) for positive.  No current flows into the inverting input.

So, some I generated by the photodiode is counter balanced by the same current in the opposite direction =I*Rf to make that node 0V.

The OP amp holds it at the value of the positive input.  Since OP amps aren't ideal, it holds it at Vos.  Vos is a strong function of temperature.

the transfer function is more like Vout = V(non-inverting) - I*Rf; but V(non-inverting) is zero.