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In this paper there is a section as follows:

enter image description here

It says "I0 is the amplitude of the photocurrent pulse generated by the photodiode..."

But what about the formula for Vout(t)? Isnt it wrong? I thought current pulse I0 first flows into C and then the current across R would flow by discharging of the C.

I also tied to simulate this:

enter image description here

And here is simulation plots for:

enter image description here

If the paper's formula were true the Vout should have started from 4V. But it starts from 50mV. (?)

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    \$\begingroup\$ Given the equation for exponential decay starting at T=0, the authors seem to be assuming that the diode is hit with a light pulse that is so fast and strong that it causes the diode to nearly instantly charge the capacitor to some significant initial voltage. The equation then models what happens after that point. \$\endgroup\$
    – user4574
    Apr 13 at 0:54
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    \$\begingroup\$ One or more members who seem to have a very poor understanding of the concept "this answer is useful" has downvoted both the answers given so far. They may not end up answering the question but they are genuine and useful attempts to do so. \$\endgroup\$
    – Russell McMahon
    Apr 13 at 4:14

3 Answers 3

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If the paper's formula were true the Vout should have started from 4V. But it starts from 50mV. (?)

Am ignoring the fact that OP's diode polarity is incorrect, and am assuming photocurrents produce +ve voltages across Rload.

Vout starts from zero volts before optical pulse arrives. The SPICE pulse construction includes a ramping rise time of 0.4 nanoseconds, up toward 0.78A, a 0.2ns flat at full amplitude, and a similar 0.4ns fall to zero. For such a short pulse, the dominant component is that huge capacitance - the 50 ohm load resistor only really matters when discharging the capacitor.

To find capacitor voltage at the completion of the optical pulse, you might instead regard the current source and capacitor as an integrator, where the equation \$I = C{{\Delta V}\over{\Delta t}}\$. You can do a piece-wise integration:

  • during the 0.4ns rise, \${\Delta V} = {{1} \over {2}} \times {{I}\over{C}} \times {\Delta t}\$ =0.0156 V
  • during the 0.2 flat, \${\Delta V} = {{I}\over{C}} \times {\Delta t}\$ = .0156 V
  • during the 0.4ns fall, another 0.0156 V

So at the end of the pulse (after 1ns), capacitor voltage has risen 3 x 0.0156 V.
This is close to the SPICE result of 0.046753 V...the effect of the 50 ohm resistor is not greatly significant.

From this point on, the current source is idle, and the capacitor discharges via Rload, with a time constant of 50 ohms x 10nf.


You can also use the original exponential equation quoted by OP. The trick here is to find an equivalent square pulse whose I x t identical to the ramp-up, flat, ramp-down. This too is an approximation that doesn't give an exact result, but the error is small.

By inspection, ramp-up and ramp-down add to a total of 0.78A x 0.4ns. Add to this the flat section 0.78A x 0.2ns. The sum of these is equivalent to a square pulse of 0.78A with duration of 0.6ns:
\$IR(1-e^{{-0.6ns}\over{RC}})\$

\$ 0.78 \times 50 \times(1-e^{-0.0012})\$

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Check the PULSE command in your simulation. The duration (width) of the current pulse is probably too short, only 0.2ns (ignoring rise and fall times). Did you mean 0.2us (microseconds rather than nano-seconds)?

The paper also states that:

the maximum level of the output signal, Vout = RI0, is achieved for relatively long laser pulses, with durations topt ≥ 3RC.

So, for Vout to reach the theoretical maximum in your simulation, the current pulse duration must be at least 3 times the RC time constant (relevant part of the paper is shown in the screenshot below). The RC time constant for your circuit is 500ns (50ohms x 10nF), so the current pulse duration should be at least 1500ns, ie: 1.5us.

Also, the direction of the current pulse in your simulation is incorrect. The current pulse should be in the direction that will forward bias the diode if the voltage gets high enough to do so. To get 0.6V across a 50ohm resistor requires a current of 12mA. Since your current pulse has an amplitude of 780mA, the diode should conduct the majority of this current once the output has reached steady- state.

enter image description here


Responding to this part of the OP:

If the paper's formula were true the Vout should have started from 4V. But it starts from 50mV. (?)

Assuming that the current pulse starts at t=0, then the formula gives Vout = 0V at t=0, and not 50mV or some other voltage. This is indeed correct for situations where the diode is not biased to any voltage prior to the light pulse hitting the diode; this mode is called "photovoltaic mode". As described in the paper in the snippet of the paper copied to the OP, the paper states that this formula is for the zero-bias case:

The photodiode is used in zero DC bias (photovoltaic) mode

The 50mV you mentioned may be due to the "dark current" of the photo detector; this is the current flowing in the detector when it is in the dark, ie: no light is hitting it. For Vout to be at 50mV means there is 1mA of current flowing in the 50Ω load resistor, so it is dissipating 50uW of power. This power must come from somewhere; since the photodetector is not providing this power (since there is no light falling on it), then where is this power coming from? Perhaps your real-world circuit has an external power supply that is not shown in either Fig 1 of the paper you referenced, nor the circuit you simulated. If that is the case, then the formula from the paper, and your simulation, are not applicable to the circuit you have built.

Unless there is an external power supply connected to your real-world circuit, I am at a loss to explain how Vout can be at 50mV when there is no light falling on the detector.

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  • \$\begingroup\$ 1) I didn't mean microseconds, But i got the point about his assumption. 2) The diode is ideal diode. It has no effect on my simulation you can just remove it. \$\endgroup\$
    – ty_1917
    Apr 13 at 0:56
  • \$\begingroup\$ 2) The ideal diode seems to have no effect on your simulation only because the duration of the current pulse in your sim is too short; Vout does not get high enough to turn it on. \$\endgroup\$ Apr 13 at 1:16
  • \$\begingroup\$ I never saw a diode is used in photodiode simulations. See examples:electronics.stackexchange.com/questions/423982/… and ti.com/diagrams/circuit060018_circuit060018-sboa220.jpg Where are diodes? Equivalent model is SPICE is made up of Current source Cj and Rsh only. \$\endgroup\$
    – ty_1917
    Apr 13 at 1:35
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    \$\begingroup\$ @ty_1917 I'm a moderator. I've deleted a few of your comments that were both rude and unproductive. Comments may not suit your needs, but people are without exception trying to help and are spending time and effort attempting to help you. You have been a member here long enough to know that rudeness is unacceptable, and hopefully know that biting the hand that feeds you is liable to reduce help you receive in future. All the comments appear pertinent - perhaps because the requirement is not fully clear. \$\endgroup\$
    – Russell McMahon
    Apr 13 at 4:10
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    \$\begingroup\$ @ty_1917 Well, if you are now saying that your "real" circuit has 3V reverse bias, then this does not align with your original post. The part of the paper that you copied and pasted to your OP was specifically for a photodiode used in photovoltaic mode, ie: zero bias. So then we have to ask: what exactly is your question? \$\endgroup\$ Apr 13 at 13:55
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In this situation, the authors are assuming small values of current generated from the photodiode, so the voltage never gets big enough to turn on the diode. This would also be the case if the diode were input to a trans conductance connected op amp, perhaps with a capacitor across the feedback resistor, resulting in the one-pole response of the formula.

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