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I am a student in physics, and I have a project where I need to pulse LEDs with short pulse widths. I have a SRS DG645 pulse generator and using these LEDs.

I have an RG58 coax cable with a BNC fitting on one connected to the pulser output. The other end I've cut and joined the central conductor to the positive lead of the LED and the grounding braid to the negative lead of the LED. This works, but my advisor suggested adding a series resistance to protect the LED. For testing purposes, I can do this with a breadboard, but what options are there for a more "production-ready" product? And what about reflections? How would I handle the 50 ohm termination in this case?

My understanding of electronics and circuits is pretty limited, and my advisor does not answer questions he thinks we should already know and instead tells us to go find the answer for ourselves.

Thank you for any help you can provide.

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  • \$\begingroup\$ What's the rise/fall time on your pulse and length of cable? Slow enough and you don't need to worry about impedance matching. \$\endgroup\$
    – user1850479
    Commented Sep 18, 2020 at 22:50
  • \$\begingroup\$ @user1850479 interesting, but why worry at all? If anything, maybe power transfer efficiency from coax into the load (which unlikely an issue here), but why worry about reflections back? \$\endgroup\$
    – P2000
    Commented Sep 18, 2020 at 23:13
  • \$\begingroup\$ @user1850479 We are looking at ~1 ns rise and fall times, with 10 ns widths. The cable length needed in our final configuration will be 78 feet, but I'm currently doing tests with shorter cables of about 3 feet. \$\endgroup\$
    – Will
    Commented Sep 20, 2020 at 18:10

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Assuming you're pulsing it fairly hard, if you're only ever going to use a source with a \$50\Omega\$ source impedance and a \$50\Omega\$ cable, then the LED is as protected as it's going to get unless you do something fancy with an RC network at the LED -- and if you do that, you'll just be protecting it against over-long pulses, and you'll be really complicating your life as far as acceptable duty cycles and the interaction between duty cycle, frequency, and light output.

If you're pulsing it hard (i.e., if you're operating it at above it's designed maximum current rating) then anything that you do to "protect" the LED will fail to protect it if you accidentally give it too high of a duty cycle.

C'mon -- it's a physics lab. You're expected to blow up electronics components.

OTOH, if you're not pushing much current through it, then some series resistance will help stabilize the LED current, because diode voltages depend heavily on everything (temperature, part to part variation, possibly the phase of the moon, aging, etc.). Having a series resistor puts more drop across the resistor, which has the effect of moderating the influence of he diode voltage on the current level.

Reflections shouldn't be a problem. The voltage wave will travel down the cable, hit the LED and be reflected, travel back up to the source and -- because the source is matched to the cable -- be absorbed. Your LED will just see a constant voltage going through a constant \$50\Omega\$ resistance.

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  • \$\begingroup\$ We are sending 2-5 V amplitude pulses. I think the reflection is doubling the voltage at the LED, so that when the pulser output is set to 2.5 V, the LED is seeing 5 V. We are actually aiming for very short, small pulses with minimal light - 10 ns pulse width with 1 kHz pulse frequency. I guess my advisor was thinking we should add a series resistor such that the maximum output of the pulser (5 V) is incapable of generating enough current to burn out the LED. \$\endgroup\$
    – Will
    Commented Sep 20, 2020 at 18:13
  • \$\begingroup\$ That doesn't seem right -- 5V on that LED should result in several amps of current; you wouldn't be getting that from the signal generator. If you set the signal generator at 2.5V, the LED would "see" a 50 ohm load with 5V behind it -- that should result in about 30mA and 3.5V at the LED. If you looked at the signal generator end and the LED end simultaneously, you'd see the signal generator end go to 2.5V, then the LED would go to 3.5V about \$0.7 c\$ times the length of the cable later, then about the same time after that the voltage at the signal generator would go to 3.5V. \$\endgroup\$
    – TimWescott
    Commented Sep 22, 2020 at 3:03
  • \$\begingroup\$ Thanks, that is a helpful way of looking at things. \$\endgroup\$
    – Will
    Commented Sep 23, 2020 at 19:34

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