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The goal

I would like to know how to connect a transistor (MOSFET) to a capacitive load. I would like to rapidly charge and discharge a capacitor.

How do I do it correctly? I basically want the transistor to work as a mechanical switch.

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What I have tried so far

I have tried the following circuit, but for some reason, VDS (voltage, drain-source) is always zero. Here VDS = V_cap (Voltage over the capacitor)

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Here is a transient analysis:

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The issue:

As you can see VDS is always zero. I expect it to be alternating.

Additional Material

EDIT:

As suggested, putting a small Resistor (here I put 10Ohms) in parallel solves the issue:

enter image description here

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... but I don't understand why? Wouldn't we have a floating node if the transistor is open? enter image description here

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  • \$\begingroup\$ @JRE Thanks for the edit! \$\endgroup\$
    – henry
    Feb 24, 2021 at 12:46
  • \$\begingroup\$ The second circuit is not the same as the first. do you want to switch the high side or the low side? \$\endgroup\$
    – Parker
    Feb 24, 2021 at 14:32
  • \$\begingroup\$ Regarding your edited question: capacitor and bleeding resistor form a closed circuit. What is floating is the charging part of the circuit. \$\endgroup\$ Feb 24, 2021 at 15:57
  • \$\begingroup\$ One example of a circuit specialized for driving capacitive loads is what's called a gate driver, so called because its main purpose is to quickly and decisively drive the gate of a FET or IGBT (which is capacitive in nature). \$\endgroup\$
    – Hearth
    Feb 24, 2021 at 16:28
  • \$\begingroup\$ @Hearth Many thanks for your comment. I don't have much experience with that. Do you have a reference where this is used to drive a capacitive load? Thanks \$\endgroup\$
    – henry
    Feb 24, 2021 at 18:04

1 Answer 1

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To observe a VDS of 5V you must have some way of reducing the voltage across the capacitor to zero. Your circuit has no path for current flow to accomplish that. Try putting a resistor in parallel with the capacitor and you should see VDS change.

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  • \$\begingroup\$ Thanks a lot for your help! Very much appreciated! However, I don't fully understand what happens if the MOSFET is in the open state. Wouldn't that produce a floating node? (see EDIT in the question). \$\endgroup\$
    – henry
    Feb 24, 2021 at 15:39
  • \$\begingroup\$ Yes, it is floating so its voltage does not change. The first time you closed the switch/mosfet you brought one end of the capacitor to ground, which meant that the voltage across the capacitor was 5V. When you open the switch/mosfet one end of the capacitor is floating, so the voltage across the capacitor never changes...it is still 5V...so the voltage at the left end of the capacitor is still 0. \$\endgroup\$ Feb 24, 2021 at 15:46
  • \$\begingroup\$ By the way, the usual way this site works is that the person who asks a question will "accept" an answer that most helps them or solves their problem. The people who write answers really appreciate that feedback. \$\endgroup\$ Feb 24, 2021 at 15:47
  • \$\begingroup\$ Thanks for your comment. Yes, I will accept the answer as soon as I have understood it. So far I still have troubles with the floating node. Correct me if I am wrong, but you are essentially saying that the voltage across the capacitor -once charged- will always stay 5V. Why is it then that I see those charge/discharge cycles in the simulation (see the plot in the edit)? Thanks a lot for your help. \$\endgroup\$
    – henry
    Feb 24, 2021 at 16:06

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