capacitor for providing extra current on short period of time - Page 1
Forum section: electronics / projects

author: john23
author_rank: Frequent Contributor
timestamp: May 13, 2024, 06:47:45 pm
content: Hello,i need to drive 5 mosfets as shown below. Each mosfet draws 15A when the pulse opens the mosfet. My power supply is gives the 48V but only 3.3A,If I understand correctly I need to put very high decoupling capacitors at the voltage source so I will have in that short period of time the 5*15=75A .My V4 power supply in real life can supply only 4A as shown in the link and datasheet below. When i put a 10uF capacitor i cant see any higher amount of current. how do i properly simulate this capacitor phenomena In LTspiceI have created such current limited voltage source as shown below.I have added a 10uF capacitor as shown below.However i cant see that the capacitor creates a current source about the limited 4A.My mosfets get zero current.Where did i go wrong in the simulation?LTspice files are attached.Thanks.https://www.meanwell-web.com/content/files/pdfs/productPdfs/MW/HRPG-150/HRPG-150-spec.pdf

author: thm_w
author_rank: Super Contributor
timestamp: May 13, 2024, 11:52:49 pm
content: Your FETs are all shorted together, is that the intent, all of them in parallel?You are using UCC5304 as a high side driver, is it correctly wired to function that way? Maybe check or use it as a low side driver.

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 14, 2024, 02:04:20 am
content: If this helps: time and capacitance are connected as t=RC.

author: john23
author_rank: Frequent Contributor
timestamp: May 14, 2024, 08:20:28 am
content: Hello , i put a resistor in parralel to the capacitor as shown below.I get a pulse of viltage and current on the pulse .The higher the ressistor the higher the higher voltage i get on my load resistor.How the capacitor and resistor shown below play together .I want to have a 45V and 30A on my load resistor over 2u pulse when its on.Thanks.

author: john23
author_rank: Frequent Contributor
timestamp: May 14, 2024, 09:34:19 am
content: Currently my cuurent is at 27A is ther a way to increase it?Thanks.

author: shapirus
author_rank: Super Contributor
timestamp: May 14, 2024, 09:58:20 am
content: Quote from: john23 on May 14, 2024, 09:34:19 amYou're not going to get more than 30 A with a 1.5 Ohms resistor and a 45 V voltage source (45V/1.5Ohm=30A). The extra 3 A are lost probably because of the FET's drain-source resistance, not sure what it is exactly in this simulation.Another potential issue is that you're putting the load resistors between source and ground. What's the reason behind that? That may create issues if the gate driver's output is ground-referenced, because as the voltage drop on the load resistor increases, it will subtract from the gate-source voltage, preventing the FET from fully turning on as the current rises high enough.In the latest simulation you connect the gate driving voltage source between the gate and source pins, so that's not an issue, but keep this in mind for the real circuit.Additionally, this schematic is rather hard to read with all the wires ending nowhere and simulation-only primitives like current sources. I'd suggest drawing it in a more traditional way and use voltage sources as signal input.I am working on something similar myself right now: a device that can provide short pulses of high current to a DUT, and it's also powered from a low-power source and has a bulk capacitor storing charge that can be quickly released when a trigger signal arrives. It simulates quite fine in ngspice, and I can't understand what your difficulty really is, because I can hardly understand the schematic.

author: ArdWar
author_rank: Frequent Contributor
timestamp: May 14, 2024, 11:09:29 am
content: Why not just put a (or maybe even a bunch of) really big capacitor? You can control both ON and OFF time of the pulse anyway. Do you need the exponential slope of a specific time constant?Also most commercial PSU protections will simply shut down the unit when overloaded instead of reverting to constant current operation. Fortunately 2us overload usually isn't long enough to trip it, and with that fast pulse you're only discharging its output capacitor anyway.