My question is related to the question here. I need very fast rise and fall time (below 2ns) and pulse width of 50ns. Load is 50ohm and voltage is 250V. I have achieved the rise time with a mercury wetted relay (700ps). But problem here is that the fall time is in range of 1.6 - 3ns.

I connected every part as directly as possible to eliminate paths which slow down the pulse.

I can see 1.6ns fall time. But it doesn't happen all the time. Is there a way to decrease the fall time or stabilise it to have constant 1.6 ns? Or should I completely try a different relay, transmission line or maybe another way?

Transmission line is RG58 Coaxial cable.


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ The transmission line has capacitance and inductance, so it is storing energy. You want to discharge this energy as fast as possible into a 50Ohm load. Do the math. Are you trying to violate physics? \$\endgroup\$
    – Kartman
    Dec 14, 2020 at 6:14
  • \$\begingroup\$ @Kartman Thank you for your response. Actually I couldn't find any researches about calculating rise and fall times. If anyone has experience about this type of pulsers, they might suggest some solutions like changing the relay type or cable type. Or adding extra components. Or maybe completely different solution. \$\endgroup\$
    – Ismail
    Dec 14, 2020 at 8:48
  • \$\begingroup\$ Can you explain what you are trying to do here and what the t-line brings to the party. I think you are trying to use the inverted reflection to pulse the 50 ohm load but details are needed, especially about the t-line. \$\endgroup\$
    – Andy aka
    Dec 14, 2020 at 10:23
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    \$\begingroup\$ The relay should certainly form part of a transmission line, not simply be 'connected ... as directly as possible'. \$\endgroup\$
    – Neil_UK
    Dec 14, 2020 at 13:53
  • 3
    \$\begingroup\$ Not sure how your short-circuited Tline supports 250 V across it. Fill in the other components on your schematic. RG58 does not have zero loss or infinite bandwidth. Have you simulated what its loss can do? Have you simulated what its limited bandwidth can do. The risetime just sees the end of the transmission line. The fall time sees twice the length of the transmission line. I'd expect the imperfections of the line to show up on the falling edge, not the leading edge. \$\endgroup\$
    – Neil_UK
    Dec 14, 2020 at 16:17

2 Answers 2


Unfortunately this isn't really a mechanical problem and has little to do with the relay itself.

A mercury-wetted relay is certainly the right choice, but their contact closure speeds are a couple orders of magnitude faster than the rise times you're seeing. They typically achieve contact closure times of 3-5 picoseconds, and the contact break time (at least, in the galvanic sense) is going to be on a similar scale.

The rise time as well as the longer fall time are hundreds of times slower than the actual physical contact closure/release times due to parasitics.

1. It's all about energy

Any transmission line (or any conductor for that matter) will have some inductance. Inductance is a measure of how much energy for a particular current something will store in a magnetic field. During the rise time, the rise in current will not occur at the speed of the relay contact closure, but rather the speed at which the parasitic inductance allows. The pulse at the relay will have much sharper rise times, but the pulse as seen at the other end of the transmission line will be slower, caused by the slowed current rise time. The current can't rise as fast as the voltage at the relay contacts would otherwise cause because the inductance represents reactance/imaginary component of impedance, which is just like resistance but represents stored (as opposed to dissipated) energy. It is still impedance, it is still correctly measured in Ω, and it causes a voltage drop. Of course, the contribution to the impedance from inductance (assuming a constant voltage across the inductive element) is temporary and falls to zero once the current (and the resulting magnetic field it creates) reaches a steady state.

2. You don't need much

And while a good transmission line will have very low inductance, it is important to understand just how little inductance is needed to cause something like a 700ps rise time: 10nH would do it at 250V.

On a circuit board that is 0.78mm thick with a solid ground plane on one side, any trace on the opposite side that is 1mm wide will have roughly 200pH of inductance per millimeter. And again, that is with a ground plane right next to the trace, helping to cancel out most of the magnetic field and inductance. Even in that case, a 5cm/2inch trace will have enough inductance to yield the rise times you're seeing.

Now look at your setup, and think about any areas anywhere in the signal path, like the internal path in the relay itself, where the signal and return paths are probably further apart and thus will create a more substantial magnetic field and have more inductance.

My point is that 10nH is a very small amount and unless there is some obvious low-hanging fruit in your signal path that is likely contributing a good portion of this parasitic inductance that you can eliminate, be prepared for a potentially costly up-hill battle to reduce the inductance enough to meet your requirements.

3. But what about the fall time?

I know 700ps is within spec, but everything above was really to provide context for the longer fall time.

When you have energy stored in a magnetic field, it doesn't just disappear. If you abruptly stop or interrupt the current that had been generating those magnetic fields, those fields collapse and induce current to flow in the conductor (due to plain old Faraday's law of induction - collapsing magnetic fields are definitely changing magnetic fields, and thus they induce a voltage). Unfortunately, the voltage these collapsing fields induce will far exceed 250V. You've tried to very rapidly interrupt the current flow, so the voltage (also called back EMF) will rise as high as is necessary. The energy has to go somewhere, and the only other option is an electric field. Storing energy in an electric field requires capacitance.

So guess what the recently opened contacts of your relay are now providing? Yep, capacitance. The contacts are two plates separated by a dielectric. A 3ps break time for a 50ns pulse into 50Ω at 250V could produce a back EMF pulse as high as ~27kV depending on conditions.

Regardless, the voltage will be plenty high enough to arc across the initially small gap between the relay contacts, and once that ion channel is formed, it doesn't need a high voltage to sustain it. Current keeps flowing even after the contacts have opened, which means there is not just the initial energy stored from the pulse rise time to deal with, but also on-going energy from the current still flowing from the 250V through the relatively low-resistance arc that has formed between the contacts.

Do note that this voltage spike will be a voltage across the relay contacts, meaning it is also dropping across them. So even while all this is occurring inside the relay, the spike will not show up in your pulse at the other end of the transmission line, but will be across the switch itself only. So even if you don't see some awful kV transient on your 50Ω load, trust me, it's there. But only across the relay.

This is why relays will have different make and break ratings, because the act of breaking a flowing current compared to making contact when there is obviously no current flowing are not symmetric at all.

4. Conclusion

The answer here is, unfortunately, probably not the one you had hoped to hear. The longer fall time compared to the rise time is both normal and expected. And there isn't much you can do about it except figure out how to reduce your parasitic inductance.

The mercury-wetted relay is a hundred times faster than your rise times and your rise and fall times are being caused by line inductance (or parasitic inductance due to the relay geometry itself, or likely small contributions of many different parasitic sources throughout your signal path) and you'll need to lower that inductance to get faster rise and fall times.

  • \$\begingroup\$ 50 ns pulsed using a relay is the best choice? I don't think that's what you meant. \$\endgroup\$ Dec 19, 2020 at 0:10
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 I have done this before. Sub ns rise time is very simple and easy with 30ns pulse width. With a simple mercury relay. The problem here is fall time. \$\endgroup\$
    – Ismail
    Dec 21, 2020 at 6:37
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 Actually I don't know if it is the best way but I have done this before. metacollin Thanks a lot for informative answer. So in that case I have nothing else to do but trying to reduce parasitic inductances. And maybe changing charge line, and trying different relays. \$\endgroup\$
    – Ismail
    Dec 21, 2020 at 6:58
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    \$\begingroup\$ Parasitic reactances involving the switch should be addressed. I can't imagine a switch physically small enough, or having 50 ohm transmission-line characteristics. I've managed ~1ns rise/fall with a small avalanche transistor switch...same circuit. \$\endgroup\$
    – glen_geek
    Dec 21, 2020 at 21:47

You're on the right track with this. In the old days people were pulsing radar tubes with a similar circuit, except instead of using a relay they'd use an arc tube such as a gas thyratron in conjunction with a transmission line ( your circuit doesn't show it but the far end of the transmission line should be open circuited). The beauty of this approach is you only need the switch to close rapidly. The trailing edge of the pulse as it uses up the transmission line provides the turn-off. As long as the load impedance is the same as the transmission line impedance the turn-off should be abrupt. You might see some echoes after the pulse on your oscilloscope if the match isn't good. I can think of a few other things that can mess this up:

  • the mercury contacts are conducting 5 amps at the start of the pulse. This is a lot and there might be some arc formation adding some series impedance.
  • the circuit through the relay contacts probably doesn't maintain a good 50 ohm transmission line geometry. The contacts will look a little inductive. Try to get a return conductor as close as you can parallel to the contact circuit, from transmission line outer conductor to the load.
  • the load might also have some inductance.

Things to experiment with might also be:

-fiddle with the load impedance

-use a good piece of coax, teflon insulator or maybe air insulated for more bandwidth. You say you're going for 50 ns pulse width. That's probably 15 or so feet of coax, enough that the GHz components of you pulse might be attenuated by a few dB.

-try a different style relay. You can experiment with some flat pieces of copper and short them by hand with a plastic rod.


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