Transformer inrush current - Page 1
Forum section: electronics / beginners

author: IvoS
author_rank: Frequent Contributor
timestamp: March 22, 2015, 07:21:52 pm
content: Hello folks,I have two 300VA toroidal transformers switching both ON and OFF at the same time using SSR. I, however, want to incorporate in series resistor on primary side shunted by a regular relay for a slow start (and to limit inrush current) because both transformers have rectified secondaries with large capacitors. Is there like a rule of thumb what the resistor value should be? I was thinking maybe 20-30 ohm resistor shunted after 100-200ms . ?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 22, 2015, 07:36:52 pm
content: It needs to deliver at least enough current to reach steady state; which for a nice high inductance transformer, probably doesn't need to be very small. It also needs to precharge fast enough for desired results. Maximum value is whatever your desired inrush is; if you just want to take the edge off (say, 10A instead of 100A+ surge), 10 ohms or so might be fine (which is also typical of NTC inrush resistors, which don't need to be switched back out... if you don't mind an extra watt or two of loss in the process).Tim

author: SeanB
author_rank: Super Contributor
timestamp: March 22, 2015, 07:52:00 pm
content: 10R NTC inrush limiter and if you want a faster cool down simply bypass it with a relay ( low current coil) driven direct by the rectified secondary side. Current rating of the contacts can be low, just a little more than the primary current, as it only has to short out in a steady state current, and thus does not see any large turn on currents or even turn off currents.

author: IvoS
author_rank: Frequent Contributor
timestamp: March 22, 2015, 08:23:52 pm
content: Thanks guys. I hate thermistors. I will rather have relay contacts in the way, so, it means I will go with regular high power resistor and a SPST relay to shunt it. I am not sure which solution would be more reliable from turn ON turn OFF point of view - cycle reliability. How do I calculate inrush current? Is there a website I can have a look?

author: Cliff Matthews
author_rank: Unknown
timestamp: March 22, 2015, 09:14:57 pm
content: http://en.tdk.eu/tdk-en/180518/design-support/design-tools/ntc-thermistors/ntc-r-t-calculation-5-0With the resistor, you'd also need a thermal cut if the relay fails.

author: Simon
author_rank: Unknown
timestamp: March 23, 2015, 07:55:06 am
content: Why do you need to do this anyway ? primary side it's under 5A. You could say use a mosfet with a low pass filter on the gate so the gate voltage "slowly" ramps on the secondary.

author: MF-jockey
author_rank: Regular Contributor
timestamp: March 23, 2015, 08:42:12 am
content: The transformer itself will have a bigger inrush current of magnetization, depending on size and saturation.The load resistor must have enough short time power in accordance to the load capacity. A resistor from small wire may burn like a fuse during load the capacitors.

author: Simon
author_rank: Unknown
timestamp: March 23, 2015, 08:55:05 am
content: What about an inductor, would that slow the inrush down ?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 23, 2015, 01:06:25 pm
content: Yes, but only proportionally to its impact on load regulation as well, and it has to withstand nearly 120VAC at whatever current it passes, which means a choke as big as the transformer most likely.To put it another way: you're trying to dampen a current surge, but the desired current flow is itself alternating. If you attenuate a big peak one way, you can't avoid attenuating the smaller peaks in either polarity. Not with a linear element.It almost works on DC, since DC is DC, and transients are AC. Trouble is, since the load is capacitive, attempting to use an inductor will only increase overshoot. It needs to be heavily damped, meaning you need ESR in the loop. With low inductance and large, lossy capacitors, the ordinary short-circuit event is destructive (100s of amps) but otherwise fairly well behaved (little overshoot), because ESR > sqrt(L/C).Say for a typical example, there's a 1m twisted lead from automotive power / a 12V car battery, to a modest power amplifier that draws a few amperes. The supply is maybe 1uH, and 1000uF would be reasonable bypass. Such a capacitor might have ESR ~ 0.05 ohm, but sqrt(1u/1000u) ~= 0.03 ohm, so the ESR is dominant and it will go "thud" when connected. It'll also draw a peak of about 12V / 0.05 ohm = 240A, which will result in a snappy spark and that's all.If you use a lot of inductance, you do reduce the peak current. Suppose we put a "hash filter" on here that's 100uH. Now sqrt(101u/1000u) ~= 0.3 ohms and it will ring and overshoot. But the peak current has dropped to about 12V / sqrt(L/C) = 50A, so it is true that inrush will be improved. It's just not a real attractive way to do it. Oh, notice that also assumes the hash choke is itself still linear at 50A -- unlikely given it only needs to handle a few amps DC. It most likely won't be linear, and therefore the same exact problem that affects the OP's toroidial transformer will pop up here: spiky inrush due to saturation.Tim