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voltage follower for high voltage pulses

What I’m doing:

I have some unknown resistive loads that I want to apply an exponential decay pulse to.

Regardless of the resistance of the load, I want to be able to specify the pulse peak voltage and time constant of the pulse delivered to the load without measuring the load.

In a typical application, I want to specify a pulse with peak voltage between 50V and 350V and with a time constant between 1ms and 20ms.

Problem:

The desired pulse time constant and pulse peak voltage result across the load if the time constant is long enough. So the circuit produces the desired result for longer time constants. However, for shorter time constants the pulse peak voltage drops below the supply voltage, and the dropped voltage depends on the supply voltage.

For example, for a desired pulse of 5ms with a peak of 100V, the pulse seen by the load is 5ms with a 90V peak. For a desired pulse of 5ms with a peak of 200V, the pulse seen by the load is 5ms with a 172V peak.

The higher the supply voltage the higher the percentage of dropped voltage at a given time constant. If I increase the time constant, the percentage of dropped voltage decreases for a given supply voltage.

Circuit information:

I am using 50Ω for Rcharge, 1kΩ for Rfollow, 100Ω for Rg, and Rload can be anywhere between 200Ω and 10kΩ. The capacitance of the Carray is chosen to specify the time constant of the pulse based on the resistance of Rfollow (i.e. the time constant of the pulse across Rload will be τ=Rfollow*Carray). The switches are PAA140L optical switches, and the transistors are BUJ303A NPN BJT transistors.

The capacitor array charges when S1 is closed and S2 is open. When S1 is opened and S2 is closed, the capacitor array discharges across Rfollow. This also turns on the transistors and allows voltage from the supply to drop across the load according to the voltage dropped across Rfollow.

Questions:

  1. What can I change/add to my circuit to get the same voltage across the load that I set the supply to (less the voltage drops across the transistor junctions) for any time constant between 1ms and 20ms?

  2. Any thoughts on why my circuit is not able to supply the supply voltage to the load when the time constant is short, but it is able to do so when the time constant is long?

  3. I’m not married to my circuit, so I’m open to alternative circuit approaches.

Other information:

When I simulate this circuit in multisim, I do not get the behavior that I see with my physical circuit.

When I simulate this circuit in multisim and add in the 13Ω switch resistance, then I do get a little bit of a drop in output voltage. By increasing Rfollow to 10kΩ in Multisim I get the same output voltage as I did before adding the switch resistance.

Based on this I increased Rfollow in my actual circuit to 10kΩ and this resulted in a reduced gap between the peak pulse voltage seen by the load and my supply voltage. Increasing Rfollow to 100kΩ did not improve the situation further.

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    \$\begingroup\$ I take it, the switches are alternate (exclusive)? How much overlap do they have, if any -- IRL vs. sim? Try varying Rfollow instead of Carray? Have you taken waveforms, say on Carray, Rfollow, Q2-B, Q1-E? \$\endgroup\$ Sep 14, 2023 at 1:24
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    \$\begingroup\$ Since the typical button bounces about 5ms after press you are not able to generate a too short pulses (shorter then about 10ms). \$\endgroup\$ Sep 14, 2023 at 1:49
  • \$\begingroup\$ To get a full supply voltage on the output you need to drive the base with higher voltage then Vsupp. \$\endgroup\$ Sep 14, 2023 at 1:53
  • \$\begingroup\$ @Tim, yes when one switch is open the other is closed. There is likely some overlap but I'm not sure how much. Increasing Rfollow helped to a limited degree as I described in my post. I have a 4 channel o-scope that I use to monitor the supply, Carray, and Rload. That is how I know that the voltage is lower than it should be when the time constant is short. \$\endgroup\$
    – WRG
    Sep 14, 2023 at 19:56
  • \$\begingroup\$ @ Michal, as I mentioned in my post, I know I will not get the same voltage on my load as my supply but I am dropping way more voltage than can be explained by the transistor BE junctions. \$\endgroup\$
    – WRG
    Sep 14, 2023 at 19:58

1 Answer 1

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The turn-on/off time of the sold state relay (SSR) is the cause of the lower than expected output voltage.

Switching to a faster SSR (AC30 as mentioned in a comment) fixed the issue and now the pulse peak voltage at the load is roughly equal to to supply voltage minus the losses from the transistor junctions.

This circuit was tested with pulse peak voltages up to 360V and pulse time constants down to 1ms.

I also found an application note from Vishay called Solid State Relays Input Resistor Selection. At the end is a small section called Faster Turn-On/Off Times. It describes adding a resistor and capacitor in parallel to the SSR input resistor to speed up with switching speed if even faster speeds are needed in the future. Here is a link

https://www.vishay.com/docs/83861/appnote59.pdf

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  • \$\begingroup\$ SSR? Where is there an SSR in the circuit in the question? If you had modified the circuit from what's shown in the question, it'd be nice to edit the question so that the answer and the question are congruous. \$\endgroup\$ Sep 15, 2023 at 19:02
  • \$\begingroup\$ @Kubahasn'tforgottenMonica "The switches are PAA140L optical switches", and from comments. The datasheet link might've made this more obvious, I suppose. This outcome seemed likely to me, but I wanted to see waveforms to be sure, as there could be other things omitted. Perhaps that was enough of a nudge to solve the problem -- hence this answer :) \$\endgroup\$ Sep 15, 2023 at 19:31

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