Opamp photodiode analysis...

Hello all,

I have confusion when analyzing this simple basic circuit:

I have a circuit that amplify the signal from a photodiode (D1) with a opamp U1 and R1. Here is my calculation for Vo: (suppose ideal opamp)

When there is light, D1 creates current Id, whose direction goes from D1 to Vn (cant go to opamp) and goes to Vo. So I have:

Vn-Vo=R1*Id;
Vn=Vp=0 (virtual ground)

=>Vo= -R1*Id

That's my result!!! So why in the book, it says Vo=R1*Id (not inverting) ? If Vo=R1*Id, there is a current will flow from Vo to Vn to cancel Id???

In your circuit the photo-diode does not have a reverse bias voltage so it behaves as a tiny solar-cell. I think light will cause the output of the opamp to go negative.
The current in the photo-diode is the same as the current in R1.

Actually, the circuit has a positive output voltage. For a vivid explanation, I suggest two different ways:
- assume zero current flow at the beginning. Then the photo current will build up a photoelement voltage, negative at the
cathode. The OP gain causes a positive voltage at the output.
- consider, that the photocurrent is a reverse current. Only a positive output voltage can cause a reverse current through
R1 and D1.

To anticipate a possible question. Yes, the photocrrent has the same direction in all photo diode operation modes.
But in photoelement mode, the photo current is cancelled by a forward current, if no external load exists.

I notice that this basic circuit often causes confusion. It has been discussed repeatedly at edaboard, see e.g.:

So the book was wrong!!! The confusion here is the direction of current. I was confused.

Ok, so I have 1 more analysis for opmap. V1, V2 are inputs. Assume ideal opamp, and here is my calculation:

(Vo-Vn)/R1 = (Vn-V1)/R2 <-----Here is the confusion part, what should it be Vn-V1 or V1-Vn?

So what is the direction of current flows from V1? Will it be from (V1->Vn->Vo) or (Vo->Vn->V1) or (V1->Vn, Vo->Vn).

Next question is:
if direction is V1->Vn->Vo, how do I know that V1>Vn so current will flow in that direction. Same question for other direction...

So what is the direction of current flows from V1? Will it be from (V1->Vn->Vo) or (Vo->Vn->V1) or (V1->Vn, Vo->Vn).

Click to expand...

Clearly no current flows into Vn for an ideal OP. So (V1->Vn->Vo) or (Vo->Vn->V1) are both correct, depending on the input voltage polarity.

So the book was wrong!

Click to expand...

I don't know. Id polarity isn't shown in the schematic. But the the absolute output
voltage is positive, when the photo diode is illuminated.

If Id designates the (reverse) photo current, Vo=R1*Id is correct.

I think I have an answer for the PD problem. As circuit shows: Vn=Vp=0, which creates zero voltage across the D1 (and always be, because of the opamp). Which means PD is in reversed bias, or in photo conducting mode. Therefore, when a light strike the PD, it will create photocurrent (Id) flows from cathode(N-type) to anode(P-type). The formula will be:

Vo-Vn=R1*Id
Vn=Vp=0
=> Vo=R1*Id !!! (the book is right!!! and I was wrong )

If my though is wrong, please correct! Thanks

OK, so now I have phototransistor (PT) instead of photodiode (PD)...Can anyone tell me if this circuit is wrong?

Here is my analysis, when a light strikes PT, it will create a Photo current from collector to emitter. Direction of current should be Vo->Vn->V1

(Vo-Vref)/R1 = (Vref-V1)/R2
<=>Vo = R1/R2*Vref - R1/R2*V1 + Vref
<=>Vo = 5.7*Vref - 4.7*V1

Which mean the output will be 5 times the input and is inverted. Also it oscillates at 5.7*Vref (V). Am I right? Can PT be used in this way, like PD?

The circuit is (partly) working, but the description isn't correct. Basically, a phototransistor is a current rather than
a voltage source. R2 could be omitted without changing the basic operation, a small series resistor may be possibly
required to achieve stability, a capacitor in parallel to R1 can do the same. The output has it's bias level at Vref rather
than 5.7 * Vref.

The circuit operation can be described best by this equation:

Vo = Ic*R1 + Vref

What do you mean by "partly working"? My purpose is to detect the intensity of a light.

So I should take out R2 and replace R1 to 5k to have the gain of 5 of the photo current? And if I put Vref=0, it should be exactly as the circuit above (Photodiode circuit)?

1 more question. I have seen this kind of circuit but with the power supply connect to R then connect to PT (Vs->R->PT). So what is a difference ?

Just quick question...


What is the output of this circuit? (suppose Vdac=0=GND)
Ip=Current Photo

will it be Vo= -Ip * R
or Ip*R?

And why? Because someone told me it should be -Ip*R not Ip*R. And most of transimpedance circuit use in different configuration.

I am getting confused here. Please help!

I don't see a reasonable purpose of the circuit. A phototransistor doesn't generate a photocurrent Ip at Vce = 0. So you get effectively
not output with VDAC = 0.

The equation given with above circuit is correct as well, but with the reservation, that a minimum Vref respectively VDAC of
0.1...0.3 V is required to generate a photocurrent.

Vo = Ic*R1 + Vref

See a the Vce/Ice charateristic from a phototransistor datasheet: