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Attempting to measure latency on a photodiode.

This is the circuit I've put together: Source

enter image description here

  • TS7805 to give my 5V (from a 12V rail)
  • 1.2kHz square wave into first T1
  • D1 is a BPW41
  • scope on source and D1
  • D1 driving base on another transistor rather than driving load directly

Output is as follows (blue - siggen, yellow - D1):

enter image description here

As CH1 rises past 0.7V you can see a ripple on CH2 but at this point the LED will have hardly come on, so this is presumably T1 opening -> voltage drop being compensated for by the 7805 causing this ripple?

And from these measurements it presumably remains impossible to measure what sort of latency D1 has, as I haven't taken account of the time it takes T1 to open and the LED to start emitting, right?

edit - thanks to u/WhatRoughBeast

I'd indeed vastly overcomplicated it, decided to isolate both sides as follows (source):

isolated oscillator with led from receive side

This gives a nice rise time ~7us across 3V (first screenshot, which makes sense now given the posted ~100us response time across 100V of the BPW41 per data sheet).

Curious about the ~200us time it takes for the voltage to fall though (second image), I can't see where there would be any capacitance in the circuit (if that's what it is?)

7us rise across 3V

200us fall back to 0V

Ron
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    The internal delay of that photodiode is picoseconds, so what you're really measuring is the slow charging up and switching of the transistor. Is that what you want (title says photodiode)? – user1850479 Mar 19 '22 at 14:27
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    One other thing to consider, your photodiode is much faster than almost any LED, so if you switch to a faster read out circuit you'll actually be measuring the rise time of the LED. – user1850479 Mar 19 '22 at 15:45
  • What are the components in your schematics that you drew as a lightning bolt inside a circle? – The Photon Mar 19 '22 at 16:08
  • user1850479 - my mistake yes it’s in picos so yes it must be the LED. Photon - it’s the probes on the o-scope – Ron Mar 19 '22 at 16:11
  • @user1850479 the photodiode has internal capacitance and transit time, so even if photocarriers are generated within picoseconds of the light being applied, it can take nanoseconds or more before an appreciable current is produced at the terminals of the device. It takes care and trade-offs (for reverse voltage tolerance and active area size, mainly) to make a photodiode that can practically respond in picoseconds. – The Photon Mar 19 '22 at 16:11
  • @ThePhoton 20pF with reverse bias, so this is actually a pretty fast photodiode. Internal delay will be 100s of picoseconds, RC time constant of 1 ns into 50 ohms. – user1850479 Mar 19 '22 at 16:28
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    You've drawn forward bias on the photodiode. Is that really how you've connected it? For a fast detector, you want to reverse bias the diode. – John Doty Mar 19 '22 at 21:03
  • Sorry yes I drew it the wrong way on the chart – Ron Mar 20 '22 at 19:10

1 Answers1

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Don't measure the base drive (what you call, I think, siggen). Instead measure the voltage across the LED resistor, as this will show the actual current through the LED.

However, be warned - your setup cannot possibly measure the latency of the photodiode. What it can do is to measure the total latency of the LED and photodiode.

Also, don't worry about the response of the 7805. It is far, far too slow to respond within 20 or 30 nsec.

You are probably correct about the feasibility of your measurement. See here

If you really care about this sort of delay, transistor turn-on and turn-off will be dominant. Frankly, I'm not sure I trust the implied limit of 20 nsec on the turn-on of T1. If your signal generator is able to drive 50 ohms to 5 volts, that is producing 100 mA, which should be more than enough to drive most LEDs without using a transistor at all. With a pulse of this short a risetime, I'd recommend driving the LED with 50 ohm coax and a 50 ohm load, then providing the LED with a separate current limit resistor and limiting the LED current to something like 10 mA. Since this is 10% of the termination current, you'll get a decent coupling and a reasonably good risetime with very little ringing.

EDIT - Addressing the new circuit and traces.

You have asked about the delay in fall time. Your transistor is being driven into saturation, and it takes considerable time for a BJT to recover from that condition. If you look at the data sheet for your transistor, you should see turn-on and turn-off times. Since you haven't listed any component values or signal levels, it's impossible to tell just how bad the situation is.

I'd recommend starting a new question. Base this on your current circuit, but make two variations. The first version will have the photodiode simply drive a resistor to ground, preferably a fairly low value like 100 ohms. This will establish the response time of the PD itself. You need a fairly low resistor to prevent parasitic capacitances from slowing down the apparent response. When you do this, you'll need to put the PD in direct contact with the LED to collect as much light as possible and maximize your signal.

Once you know the actual response, then you can tack on your transistor and see what happens.

WhatRoughBeast
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  • Tried to mark your answer up but don't have enough points here - many thanks for replying. Have updated to simplify and drive the LED directly - indeed the output from the signal generator perfectly able to drive it as you suggested. – Ron Mar 19 '22 at 14:00
  • @Ron - See my edited version. – WhatRoughBeast Mar 19 '22 at 14:14
  • Thanks @WhatRoughBeast, legend. This has given me plenty food for thought. – Ron Mar 19 '22 at 14:22
  • @Ron - One more thing. It will be a good idea to protect your LED with a small-signal diode like a 1N914 or 1N4148 (NOT A 1N4001 to 1N4007!) connected in parallel with the LED, but of opposite polarity. See the negative on your LED waveform? That is caused by termination issues in your circuit, which you're not yet aware of. With enough on current in the LED, and a short enough fall time on your pulse generator, it's possible to produce enough voltage to damage the LED. Solder the diode directly to the LED leads. This is playing it safe, perhaps unnecessarily, but it's cheap insurance. – WhatRoughBeast Mar 22 '22 at 02:29