Current Through a Capacitor - Page 1
Forum section: electronics / projects

author: nicknails
author_rank: Regular Contributor
timestamp: December 20, 2016, 02:57:08 pm
content: I'll start by saying that, yes, this is for work. I don't know if that matters on this forum, and if it does, please feel free to delete the post, but I'm stumped.This is all mathematical, I don't currently have the means to do a lab test. I have a DC/DC converter with +/-95V out. It's connected to a high power opamp that's switching a 2nF cap from the +95 to the -95 at 40kHz. I need to determine the current draw of the capacitor to determine the power rating of the DC/DC converter that is needed.Formula wise, I'm doing i = C (dv/dt), so i = 2nF (190V/(40000)E-1)) = 15.2mA. Therefore, power of DC/DC converter is 190V * (15.2mA + 26mA (quiescent current of op-amp)) = 7.8W.However, doing a spice model, nothing more than a source (190V AC @ 40kHz) and a cap, I'm getting +/-95mA or 190mA total, meaning the power is 41W! Considerably higher.Any help would be appreciated.

author: orolo
author_rank: Frequent Contributor
timestamp: December 20, 2016, 03:21:26 pm
content: In your spice model, you're using 190V of amplitude, not 190V peak-to-peak. So you are overestimating the simulated current by a factor of two.Also, you can do better with your theoretical computation of the current. I = C · dv / dt. Since the excitation is sinusoidal, v = 95·sin(w·t) , so dv/dt = 95·w·cos(wt). From there, I_max = C · 95 · w = 95 · C · 2 · Pi · f. So you were underestimating your peak current by a factor of Pi (Since 95·2 = 190).So:Correct model current, about 100/2 = 50mA peak.Correct theoretical current, about 15.2 · 3.1416 = 47.75mA peak.

author: snarkysparky
author_rank: Frequent Contributor
timestamp: December 20, 2016, 03:25:24 pm
content: energy stored in a capacitor is 0.5*C*V^2 in joules.E = 0.5*(2e-9)*190^2= 3.61e-5 joules.times two for charging and discharging.40000 times per second makes2.88 wattsequation 4 in this paperhttp://www.ti.com/lit/an/scaa035b/scaa035b.pdf

author: StillTrying
author_rank: Super Contributor
timestamp: December 20, 2016, 03:34:16 pm
content: You can plot the instantaneous power in a component by holding down 'Alt' while hovering over the component.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: December 20, 2016, 04:03:18 pm
content: Quote from: snarkysparky on December 20, 2016, 03:25:24 pmYou made the same mistake the OP did in his calculation - that's the average power over the entire waveform. The power supply needs to be sized to handle the peak current, not the average current. The peak occurs when dv/dt is maximum, which is when the sinusoid is crossing through 0. orolo has the correct calculation for the peak current requirement.

author: snarkysparky
author_rank: Frequent Contributor
timestamp: December 20, 2016, 06:42:32 pm
content: Well the OP is asking about switching a square voltage wave into a capacitor. I think....Which for a perfect capacitor is impossible to do. The calculation will depend on the slew rate of the op-amp and thetotal resistance in the circuit including the capacitor ESR.And as he asked for the power required I would assume the op-amp supply is sufficiently bypassed with a large enough capacitor to average out the switching pulses leaving only the average current needing to be supplied.Its a common issue with CMOS digital circuit design as mostly what is going on there is switching gate capacitances.

author: Zero999
author_rank: Super Contributor
timestamp: December 20, 2016, 07:11:41 pm
content: The DC-DC converter only needs to be rated to compensate for the losses in the system. Add an inductor (7.9mH) in parallel with the capacitor to bring the power factor close to unity. The supply current will drop dramatically.

author: nicknails
author_rank: Regular Contributor
timestamp: December 20, 2016, 07:28:08 pm
content: Thanks everyone! I got the math to match the spice model and I'm happy with that. This forum is awesome.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 20, 2016, 09:48:14 pm
content: Why would you not use an inductor for that job?An op-amp will dissipate all the VARs as watts, divided by efficiency.Tim

author: nicknails
author_rank: Regular Contributor
timestamp: December 20, 2016, 10:27:05 pm
content: I am only determining the spec for the power supply. The customer is the one with the circuit. I can't change or do anything beyond the power supply.

author: Zero999
author_rank: Super Contributor
timestamp: December 20, 2016, 11:33:54 pm
content: Quote from: T3sl4co1l on December 20, 2016, 09:48:14 pmBecause:QuoteIn other words, the inductor pushes the power back into the capacitor, so it doesn't need to be dissipated by the op-amp or whatever is driving it.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 21, 2016, 12:14:33 am
content: Quote from: nicknails on December 20, 2016, 10:27:05 pmThis is really weird. Why doesn't the customer measure their circuit, and give you a spec?How can they expect you to design a power supply when they withhold critical design information?!Even if they tell you the efficiency, they might not know about quirks under high frequency or reactive loads, or if they later decide to increase the frequency (because, customers, amirite?), and suddenly the design falls apart. (Or you give them a double sized power supply, and then they complain that it's too big and expensive. You can't win like this.)Tim