How does feedback deal with the voltage induced on this photodiode?

Thread Starter

AronLloyd

Joined Jul 15, 2018
6
upload_2018-7-14_23-53-25.png
This circuit is a simple unity gain signal decoupler using an optical isolation chip. The left half of the circuit is a feedback system to account for fluctuations in the LED. The right half then translates the second photodiode's voltage back into the input voltage.
My misunderstanding is, for either the left or right side, how the op-amps can bring V- to ground (since V+ is grounded) by feedback when the photodiodes are directly connected to V- without resistance in-between. I don't know the way to approach this analysis.
To narrow my question, lets say that some input signal was producing oscillations on the second output diode (on the right). The op-amp does not rail. Does the op-amp just produce an equal an opposite voltage at that position in order to cancel by superposition? This intuitively seems wrong.

Some important things to note:
-The LED and photodiodes have a small capacitance on the order of 10's of pF.
-The circuit assumes the input is a positive only AC signal.
-I have tested the circuit. It works at unity gain as expected until some upper roll-off frequency.

The reason I still analyse the circuit is because it filters out higher frequencies, but I am finding trying to understand the theoretical frequency response so I can compare it to what I have measured.
Thanks in advance!!
 

Hymie

Joined Mar 30, 2018
819
When the output of IC1 is low (switching on transistor Q1), the second diode within U1 prevents the inverting input pin of IC1 from being taken negative. Note that this situation only occurs when the output of IC1 is low – at other times the input to IC1 can be a negative voltage.

One other observation on this circuit is that IC1 is configured as an integrator (and not an amplifier).
 
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Thread Starter

AronLloyd

Joined Jul 15, 2018
6
When the output of IC1 is low (switching on transistor Q1), the second diode within U1 prevents the inverting input pin of IC1 from being taken negative. Note that this situation only occurs when the output of IC1 is low – at other times the input to IC1 can be a negative voltage.

One other observation on this circuit is that IC1 is configured as an integrator (and not an amplifier).
So does that mean then,
since the second diode has a voltage difference, then so does the third.
It is a fact that IC2 outputs whatever Vin is. I can see how that is true. What I don't understand is how output voltage from IC2 can cancel the voltage applied to the inputs of IC2 by the third diode. Do voltages simply add like that in such a scenario? Do the electrons excited by photons on the diode just flow towards the positive potential created by IC2?
 

Alec_t

Joined Sep 17, 2013
11,274
My misunderstanding is, for either the left or right side, how the op-amps can bring V- to ground (since V+ is grounded) by feedback when the photodiodes are directly connected to V- without resistance in-between.
Assume the X1 input goes positive. IC1 output will tend to go low, turning on Q1 further and hence increasing the opto LED current. The opto photo-diodes are operating in the reverse-biased photoconductive mode, so their reverse current increases. The left-hand diode current flows through R1 and the resulting voltage drop across R1 pulls the inverting input of IC1 down towards ground. Hence the diode current is kept proportional to the X1 voltage. The right-hand diode current mirrors the left-hand diode current.
 
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Sensacell

Joined Jun 19, 2012
2,567
The voltage drop across the input photo diode should be zero, it's not working in photo-conductive mode, it's working in photo-voltaic mode.

The opamp IC1 servos the voltage at the inverting input to zero, through the optical feedback. The photo current is summed with the input current at the inverting input.
The transistor forms a voltage-controlled current source to linearize the response.

On the receiving side, the approximately equal photo current gets converted to a voltage by IC2.
C1 rolls-off the gain at higher frequencies to keep it stable.
 

Thread Starter

AronLloyd

Joined Jul 15, 2018
6
The voltage drop across the input photo diode should be zero, it's not working in photo-conductive mode, it's working in photo-voltaic mode.

The opamp IC1 servos the voltage at the inverting input to zero, through the optical feedback. The photo current is summed with the input current at the inverting input.
The transistor forms a voltage-controlled current source to linearize the response.

On the receiving side, the approximately equal photo current gets converted to a voltage by IC2.
C1 rolls-off the gain at higher frequencies to keep it stable.
You say the second photodiode produces a current. From my understanding, the electrons excited by photons in a photodiode cannot flow when zero-biased, therefore a (negative) voltage builds. Regardless, if you say a current flows, does that current initially flow into one of the op-amp inputs? Then the op amp outputs a voltage such that the current is redirected towards the output? How does one determine the output of the op-amp in terms of the current or voltage of the diode is my real question?
 

Thread Starter

AronLloyd

Joined Jul 15, 2018
6
The circuit frequency response is primarily determined by the integrator R1C1 time-constant.
Certainly not, the gain for R1C1 amplifier is well above 1 up to Megahertz range. We also don't necessarily care about the input signal after the first op-amp since we re-derive the signal from the diode's voltage on the right side. The roll-off observed is around 10-100kHz which is more in line with the low-pass ability of the op-amp to the right of the opto-coupler. The discrepancy I have discovered is about on the order of 2, so I am beginning to believe that C1 is causing this shift. I am trying to figure out quantitatively what this is but the three voltage sources all playing on the feedback makes this unclear.
 

ebp

Joined Feb 8, 2018
2,332
A photodiode, when operated "short circuited", as in these circuits, produces a current that is linearly proportional to illumination over a fairly wide range. What is fundamentally going on is that the inverting inputs of the op amps are current summing nodes - which is what they always are in inverting configurations.

You refer to a third photodiode - there are only two. One is an infrared emitter, the other two are photodiodes. The two photodiodes are not particularly well matched, but the proportionality of mismatched is quite constant for any particular unit. Variations with aging, temperature, et cetra, track well between the input side diode and the output side diode. The input side circuit, by closing the feedback loop around the input side diode largely removes the non-linearity of the diodes, the effects of temperature and aging AND the uncertainties of the performance of the IRED.
 
IC2 is the classic current to voltage converter. The cap does provide some rolloff and adds stability.
The circuit relies on the fact that the photo-transistors are on the same die, thus they are closely matched.

The front end almost looks like a current to voltage converter too.
 

Thread Starter

AronLloyd

Joined Jul 15, 2018
6
A photodiode, when operated "short circuited", as in these circuits, produces a current that is linearly proportional to illumination over a fairly wide range. What is fundamentally going on is that the inverting inputs of the op amps are current summing nodes - which is what they always are in inverting configurations.

You refer to a third photodiode - there are only two. One is an infrared emitter, the other two are photodiodes. The two photodiodes are not particularly well matched, but the proportionality of mismatched is quite constant for any particular unit. Variations with aging, temperature, et cetra, track well between the input side diode and the output side diode. The input side circuit, by closing the feedback loop around the input side diode largely removes the non-linearity of the diodes, the effects of temperature and aging AND the uncertainties of the performance of the IRED.
So your saying that these photodiodes are not operating in zero-biased photovoltaic mode? They are zero-biased. Is it that some current does flow in photovoltaic mode? From what I have read, the current is restricted, which must not necessarily mean prevented, and a voltage builds up due to this restriction.
Or is it because of the short-circuit that this definition doesn't necessarily apply?
 

ebp

Joined Feb 8, 2018
2,332
They are operating at zero bias or "short-circuit." This is called photovoltaic mode, but it is something of a misnomer because the voltage across the diode is held at zero. When the diode is operated in this way, as I said before, the photocurrent is highly linear over at least a few decades. I have no idea what you are referring to by saying current is "restricted". If the diode were instead operated open-circuit (that is, zero current flow), the response would be accurately logarithmic over at least a few decades. This is all very well described in lots of documents available on the web. In reality some very small voltage will be allowed across the diode due to the input offset voltage of the op amp and due to the fact that the gain of the op amp is not infinite. These voltages contribute to error, so they must be minimized by selection of a suitable op amp.

The input circuit closes the feedback loop around the op amp via the input side photodiode. If the op amp were perfect (zero input offset voltage, zero input bias current and infinite gain) and the small feedback capacitor were ignored (it is there for loop stability reasons because the transistor and the other parts add gain and introduce some phase shift), then the current that flowed through R1 would be exactly equal to the current in the photodiode . The current in the photodiode is due to the photons from the IR emitter, but the circuit drives the IR emitter at the current required for the necessary photodiode photocurrent to bring the sum of the currents at the inverting input of the amplifier to zero. For example, if you apply 1.0 V at the input, the current through R1 will be 10 µA because the feedback around the amp holds the inverting input at the same potential as the non-inverting input - zero volts. The op amp and transistor will drive sufficient current into the IR emitter to make the current in the input side photodiode 10 µA.

As mentioned before, the input side and output side photodiodes may not have exactly the same gain in terms of illumination to current, but the ratio is very stable and tracking between the two is very good. It is this stability of ratio that allows the circuit to achieve the performance that it does.

The output amplifier is a transimpedance amplifier with gain expressed in terms of volts per ampere - a current to voltage converter. The gain is expressed in ohms and is equal to the -R5 for the circuit in the scehmatic (negative because it is inverting). Again, the voltage across the photodiode is held at zero so it behaves the same way the input side diode does.

The overall voltage gain of the circuit, if the photodiodes were perfectly matched, would be the ratio of R5 to R1. The gain must be made adjustable using a trimmer potentiometer if it needs to be precise because the photodiodes are not perfectly matched.

I don't know what you have been reading, but you need to read the applications note for the coupler:
http://www.vishay.com/docs/83708/appnote50.pdf
 
I worked in the industry and built a QE (Quantum Efficiency) system including an I_V converter for solar cells.

Photovoltaic means zero bias. Photoconductive means reverse biased. See: https://www.thorlabs.com/tutorials.cfm?tabID=787382ff-26eb-4a7e-b021-bf65c5bf164b

The solar cell does have a series and shunt resistance in the model as well as a current generator that has a temperature term and a diode factor.

The second OP amp has a stray wire.

If you want, you can apply a bias to the (+) instead of connecting to ground. The transfer equation then goes from V=-I*Rf to V = V(bias) - I * Rf My latest generation of an I-V converter was autorangeable with LabView. The ranges were bipolar 100 mA, 10 mA, 1 mA, 0.1 mA full scale + -10V out. Bias and current output were read by two 5 digit meters. The design allowed +- 10V bias. If you suppressed up to +-50 mA, you were limited to +-5V
It was a 2 terminal/4 terminal device meaning Kelvin probing was possible. It also had a bipolar peak detector that lit a bipolar LED for 1 second.



Solar cells with either no IR filter or with filter that spectral shaped the response was our standard method to calibrate intensity. A few cells were calibrated at NREL and we calibrated about 10 others. Intensity calibration was done using three cells. One you used the day before and two others. They were I-V swept and Jsc/Isc was used to verify intensity AND spectral shifts. As our arc lamp aged, it put out less blue light. We actually helped the light source manufacturer shape the spectrum for their sources.

There is a slope in reverse bias which should be a high resistance.
 
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