PCB 0.2mm as power resistor - Page 1
Forum section: electronics / beginners

author: Flunze
author_rank: Regular Contributor
timestamp: June 30, 2020, 04:29:36 pm
content: Hello, I want to implement a 200mOhm resistor for my Boost converter for the initial capacitor charge up current.- The pulse starts at 17A; after 3ms it is at 5A. and after 10ms 1A- After 10ms I gradually engange the MOSFET to fully on, to feather out any remaining voltage difference.My plan was 80mm of 0.2mm track in a 0.2mm clearance zig-zag, which is almost spot on. However I do not know anything of pulsed power handling capabilities of PCBs. I do not think the PCB resistor would radiate much EMI at all, since it is on the filtered DC side (Battery side) and after the delay the Mosfet would basically completely short it out.Any ideas? Will it burn through. Will it stay absolutely cool because 10ms is to short?Thanks for your advice.Oh and why? I want to save one component.Keeping BOM low.

author: exe
author_rank: Unknown
timestamp: June 30, 2020, 04:49:45 pm
content: How did you calculate it will be 200mOhm? My calculation for 35oz copper layer it's going to be 0.00566Ohm. Apart from that, it may have quite some inductance, which may cause voltage spikes.

author: bob91343
author_rank: Super Contributor
timestamp: June 30, 2020, 05:01:15 pm
content: Don't ignore the temperature coefficient of resistance of copper, about 0.393% per degree C.

author: fcb
author_rank: Super Contributor
timestamp: June 30, 2020, 05:10:52 pm
content: Probably work if your maths check's out.You have just created a fuse though, would expect 10ms to be on the long side as there isn't much copper to absorb the pulse - but give it ago.When I've done similar things before (soft-start on boost converters) I've either softswitched a big FET or had a dedicated charge resistor to bring the caps up to rail voltage BEFORE engaging a bypass MOSFET and beginning switching, 1-2 seconds timescale (this is what our USB connected instruments do when plugged into PC).

author: max_torque
author_rank: Super Contributor
timestamp: June 30, 2020, 06:59:06 pm
content: Off the top of my head, for 1 oZ copper (35uM thickness, ~0.6 mOhm per square)80mm track 0.2mm wide is 400 "squares" and hence has a resistance of around 240 mOhm and consists of a total of 0.005g of copper (0.56mm3 of copper (0.2 X 35uM x 80) at 8.96g/cm3)Average current say 5A, means 6 watts into that track ((5^2)*0.24 ohm), say for 10 ms pulse, means 0.06 joules of energyWorst case is ZERO heat loss to ambient during the current flow ie adiabatic heating which is unlikely for a thin track on a 10ms long pulse which is a reasonable amount of time, but gives us a worst case because all the energy goes into the copper of the track and is not radiated or conducted away during the pulse.Specific heat of copper is 0.385 joules per gram per degC.We have 0.06 joules and 0.005g of copper, so assuming it is evenly distributed (no hot sports or zones of high resistance) then that means an upheat during the pulse of 31 degC!So design the track without sharp turns or zones of non conformability and it looks ok to me, as long as the pulse isn't repetative!you might want to check my numbers, i've only quickly done the calcs, so i could be wrong, but it"feels" kinda right by my thumb.........

author: bob91343
author_rank: Super Contributor
timestamp: June 30, 2020, 11:57:07 pm
content: One more consideration is the connections made to the trace. This is a trouble spot. Some builders use eyelets or rivets, some use connectors, and so on. The problem arises due to current paths which easily can concentrate current in a small area of trace and consequent overheating. It's the downfall of many tube circuits which utilize traces for filament power.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 01, 2020, 02:29:26 am
content: Is it really cheaper to use board area than to plop down a $0.10 resistor?Or, it sounds like you're switching it with a FET (or several!) anyway, why not control it so it just doesn't blow itself up in the process? (There are ICs for this function: load drivers and wired-OR controllers; you may prefer a discrete solution if cost is absolutely critical.)Quote from: max_torque on June 30, 2020, 06:59:06 pm17A down to 5A is about one time constant (30% vs. 37%), which implies C = (3ms) / (200mΩ) = 15mF. 17A into 200mΩ implies Vin = 3.4V.The capacitance is suspiciously large, but the voltage is quite easily handled, and furthermore this implies E = 87mJ (not far from your estimate).(If there's a load, which is on immediately, the energy will be higher due to load current plus charging current, of course.)QuoteFR-4 is a pretty awful thru-plane conductor, so it's actually a pretty good estimate.35°C rise is pretty tolerable for a PCB at modest ambient temperatures, so I wouldn't mind this as a possible solution. I just suspect there are better ways to solve it...Tim

author: Flunze
author_rank: Regular Contributor
timestamp: July 01, 2020, 02:49:34 am
content: Well, pretty much correct. But 4.2V and a 3 way filter with 2x3mF, 0.5uH, 2x3mF. This gives me a a Voltage drop of less than 100mV, if I instantly load down the regulator from 0 to 3 Amps.Yeah, it does likely not matter that much, but I overconstruct my converters all the time. 3-4.2V to 12V 3A at around 90-91% efficiency without any ICs.I love the challenge to create such circuits just with standard components and 0 chips. It uses 1uH. This means, that the current could ramp to saturation during power on.But it has a current control. The advantage of such small inductances is the quick response and very low ripple at low loads (without a precise PWM it would overshoot everytime the load drops to 0. But aith 1uH even at 20A there is almost no energy stored at all, so it cannot overshoot a lot)I will use a 1 Ohm resistor for the inrush. I think it is better to have some safety margin.The circuit is pretty complicated (compared to IC solutions), but it is for the fun. I love to set myself a goal and than try to design a product that works "good enought for the market" (wich would use IC, I know).Thanks for the advice. Was very helpful.