# Suitable Half Bridge Driver to control capacitve load

I am looking for a method to reliably and quickly charge/discharge capacitors. Each capacitor should be individually charged/discharged. The charge/discharge rate should be in the microsecond range (if possible). Everything should be controlled by a micro-controller.

Capacitor: 40uF, DC

Power Source: 100V, DC

Switching time: ~us (originally nanoseconds)

Source Output impedance: order of milliohms ~20mOhm (originally 50 ohm)

I thought about using a half-bridge as a load switch, but so far I have been unsuccessful at finding a suitable IC. Therefore, I wanted to ask if anyone had done something similar in the past, or could advise me on which type of Load Switch/Half-Bridge to use.

Many thanks!

• It cannot be this: as fast as possible - so be reasonable and place numerical limits on things. Also state the source impedance of your power supply to avoid taking so much charge current that the MCU gets reset. Or, maybe the MCU is powered separately? – Andy aka Feb 28 at 9:58
• Without a purpose of high ripple current, this is like the job during the Depression for make work. Dig holes then fill 'em up and repeat. – Tony Stewart EE75 Feb 28 at 10:03
• So the capacitor is 40uF, and is charged from 0V to 100V, and dischared from 100V to 0V? If you want to do that in 1 microsecond, that means charge and discharge current has to be 4000 A. – Justme Feb 28 at 12:52
• What exactly are you trying to achieve with this device? – bobflux Feb 28 at 14:31
• You may need to do this with discrete FETs. And even then you're not getting 1 μs rise and fall times to 100 V. The calculation is $\frac{C·V}{t}$, so $\frac{40\ \mathrm{μF}·100\ \mathrm{V}}{1\ \mathrm{μs}}=4000\ \mathrm{A}$. – Hearth Feb 28 at 14:31

Capacitor: 40uF, DC

Power Source: 100V, DC

Switching time: ~us (originally nanoseconds)

Current in a capacitor can be calculated from the derivative vs time of voltage across it:

$$\ i = C dv/dt \$$

With dt=1µs, dv=100V, C=40µF we got i=4000 A. This is quite impractical.

If the charger does not use inductive energy storage, instead simply connecting a capacitor to the output of a (presumed) stiff low impedance power supply, then the same amount of energy that will be stored in the cap at the end of charge will also be dissipated in the switch and various resistances in the circuit.

$$\ E = \frac{1}{2} CV^2 \$$

Here E=0.2 Joules which is low enough to not make large things explode. MOSFET semiconductor die are not that large though.

Source Output impedance: order of milliohms ~20mOhm (originally 50 ohm)

Under high di/dt pulsed current, the inductance of power supply wires will be a larger problem than the output impedance.

But... to charge the cap, the current has to ramp up first, and preferably the ramp-up has to be shorter than the duration of the pulse. Current should also ramp down at the end. For current in an inductor to go from 0A to 4000A in 100ns with 100V across it, the inductor has to be:

$$\ e = L \frac{di}{dt} \rightarrow L = e \frac{dt}{di} = 2.5 nH\$$

This is ironically doable, barely, if the target capacitor has low enough inductance. But it will need a power supply of very low output inductance, which means lots of capacitors charged to a rather high voltage on the board... and capacitors don't discriminate, they'll discharge into anything that makes contact, including you.

Note current in wires creates magnetic fields, and a wire with current flowing through it in a magnetic field will be pushed by a force. With this kind of current you can expect things to jump around, and I wouldn't be surprised if the charged capacitor flies across the room. Whoever catches it will have a bad day.

Honestly this project is going to be expensive, complicated, and pretty lethal. If you say "I am very new to electrical engineering" I would recommend less dangerous stuff... or maybe the thing you actually want to do can be achieved by less "mad-scientist"-ey means.

Happy ending:

If the goal is to store energy, then a 1nF cap charged under 20kV also stores 0.2 Joules, and it can easily be charged in a microsecond with a current of only 20 Amps. Obviously, it is even more lethal.

• Thank you very much for your very thorough and nicely explained answer! 4000A is so much. I would never have thought that charging a small-cap will result in such big currents. I guess that I will need to lower my switching time. Also, do you know how I can make sure never to go above a certain threshold in current/voltage (power) on my PCB? Are there any kind of security ICs that I could buy? Thanks a lot. – henry Feb 28 at 21:23
• That kind of protection should be an integral part of the circuit, so you have to decide on a realistic charging time first then decide on what kind of circuit will do it. – bobflux Mar 1 at 8:20

Capacitor: 40uF, DC Power Source: 100V, DC Switching time: ~ns Source Output impedance: ~50Ohm

The charge/discharge rate should be in the nanosecond range

To charge a 40 μF capacitor from a source with 50 Ω impedance cannot be done in nano seconds. The time taken to charge the capacitor to approximately 63.2% of 100 volts will be: -

$$\text{50 Ω } \times \text{ 40 μF = 0.002 seconds}$$

This is nowhere near nanoseconds so maybe you should rethink your requirements. CR time constants graphically: -

Picture from here and red-lined by me.

To get to 99 volts, it will take about 10 ms; possibly 10,000 times longer than you hoped for. Discharging can be done quickly but, ultimately that depends on the ESR of the capacitor. For instance if the ESR is 1 ohm, then the capacitor discharge time constant is 40 μs. This means you can discharge it to about 36.8% in 40 μs with a perfect short circuit.

You do the math.

• Thanks a lot for your answer! Indeed, the specs don't seem to match up. This is very important, thanks for your feedback. I am quite new to this topic. Would you also know which type of Load Switch/Half-Bridge to use? Thanks. – henry Feb 28 at 11:21
• @henry you've been a member now for three years+, so you should know how thanks is properly given on this site!! – Andy aka Feb 28 at 11:35
• Why are you getting so worked up mate? I am asking a question. I gave it a +1, because the answer is useful, but so far this answer has not completely solved my problem, so why would I accept it? That would be against what stack exchange stands for. No hard feeling necessary. – henry Feb 28 at 11:47
• Yes, thanks is given by upvotes. Your load switch is immaterial if you can't meet your timing requirements of nanoseconds. If you want an answer to solve your problem, you should either improve the power source or reduce your expectations. – Andy aka Feb 28 at 11:51
• @henry how much peak current can the power supply produce into a short circuit? How long can it operate whilst shorted-out by a charging capacitor (for a short time). How much ESL (effective series inductance) does the supply and feed lines have. This is now important because you might get really bad ringing overshoot that puts approximately double the supply (200 volts) across the capacitor for a few microseconds (aka RLC 2nd order transient response). – Andy aka Feb 28 at 13:19