# Need help with limiting transformer inrush current.

#### AndrejaKo

Joined Sep 6, 2010
16
I have a 220 V 50 VA to 2* 12 V 2.1 A transformer. If my calculations are correct, maximum input current for it should be around 227 mA. I want to protect the circuit which it is powering with a fuse. Unfortunately, when I power on the transformer, fuse pops (at that time, I was using 315 mA fast 5x20 fuses). After doing some research, I believe that fuses are dieing because of inrush current of the transformer.

I'm looking for a way to stop inrush current form popping the fuse (and slow fuses don't help, I tried), so the most obvious solution would be NTC thermistor. I found some 100Ω @ 25°C rated at 450 mW which look like they may be useful to me. Would they be a good choice or should I look for some with higher power rating? Also, from what I can see, they heat up and then their resistance drops. How hot would they be running? Also, I read in some documents that it's a good idea to serially connect several thermistors in order to increase performance. Should I do that? If yes, how much thermistors would be enough?

Another way which was recommended to me was to use chokes to control input current. It looks interesting to me because that way, I'd be able to avoid waiting for thermistors to cool down before powering the circuit again. Where would I start with them? I do understand that the choke will induce current which will oppose current coming in from power source, but I don't have much information on what inductances should I look for when getting a choke.

#### Kermit2

Joined Feb 5, 2010
4,158
You could make a 'soft start' circuit.

The easest way is a momentary switch that is normally closed.

Wire that across a large 10 watt 2-5 ohm resistor, so that the switch bypasses the resistor unless it is depressed, putting the resistor in circuit.

You must hold the momentary switch 'open' while you start(switch on) the power supply. the current will have a path through the resistor, which will limit the inrush, and when you release the switch, the resistor is bypassed by the switch and is out of the circuit.

The automatic way to do this is with a douple pole relay set to latch itself on.

#### AndrejaKo

Joined Sep 6, 2010
16
Is there any more elegant way to solve this?

#### Kermit2

Joined Feb 5, 2010
4,158

You just need one switch for the OFF function, but you can put it anywhere in the circuit where it will break the relay latch up.

#### debe

Joined Sep 21, 2010
1,139
Generaly you use slowblow fuses in that aplication.

#### Kermit2

Joined Feb 5, 2010
4,158
That was the "catch". He told us he had tried slo-blo and they wouldn't work for him.

#### pjax

Joined Dec 21, 2010
2
use a 500 mA fuse

#### Audioguru

Joined Dec 20, 2007
11,251
A transformer does not produce an inrush current. It is the very high current charging the huge filter capacitor that produces the high inrush current.

#### marshallf3

Joined Jul 26, 2010
2,358
In most cases you shouldn't be concerned with this, just fuse the input for at least three times what the output equivalent wattage will be, in your case a 1A fuse should sufficient but in my opinion I'd use a 1-6/10 or 2A fuse. Should anything go wrong on the primary side that will catch it.

The primary side of a transformer in a power supply circuit is actually the least of your concerns, the secondary side (usually after rectification and filtering) is where to place your emphasis.

#### creakndale

Joined Mar 13, 2009
68
I'd probably use a 2.7 ohm 1/2 watt resistor in series with a 500mA fuse on the input side of the transformer.

IMO, the fuse value = nominal current * 2, then generally pick the next larger available value.

creakndale

#### gootee

Joined Apr 24, 2007
447
Attached is my schematic for a circuit that limits the inrush current of the filter capacitor(s), which limits the transformer current. This circuit goes between the rectifier bridge and the first filter capacitor. For dual-polarity supplies, I use two of them and it works equally well between positive and ground and between ground and negative. The version shown worked with both 25V and 30V AC RMS transformer secondary voltages. Your mileage may vary.

When power is applied, the capacitor currents are forced to flow through R2, which limits the current. The MOSFET turns on between 150 and 180 ms after power was applied (in the circuits shown at the link below), bypassing R2 with a very low on-resistance (Rdson) for normal circuit operation. The mosfet dissipates about 115 mW after startup, in the 22V supply shown at the link below, when the load is drawing 4 Amps.

I used a 5-Watt one Ohm resistor for R2 (594-AC05W1R000J at mouser.com). But a lower-rated one would probably be fine, since it only conducts significant current for a very short time (averaging about 41 Watts dissipation over 160 ms [with an initial 400-Watt peak for a few milliseconds] in the 22 VDC supply shown at the link below and 69 W in the 28V supply).

You can see examples of the inrush limiter circuit being used in a dual power supply schematic at

http://www.fullnet.com/~tomg/gooteesp.htm .

As configured, in the +/-22V power supply shown at the link, the inrush limiter cuts the peak inrush currents almost in half. The transformer primary current went from 8.4 Amps peak to 4.4 Amps peak and the first large electrolytic's peak startup charging current went from 12 Amps to about 6.1 Amps, which should keep it within its ripple current rating at up to 70 deg C. The transformer secondary's peak startup current went from about 38 Amps without the limiter to about 20 Amps with the limiter circuit.

You can download the LTspice model for the power supply circuit at the link above, and try different secondary voltages, current-limit resistor values, etc. (Hint: Use the "Alternate Solver" to speed up the simualations drastically.)

Perhaps interestingly, my original motivation for designing the inrush limiters was because with relatively large-value capacitors on the main 5-Amp regulators' adjust pins the supply had very low output ripple voltage, but then also had a slower-rising output voltage at startup. And in order to ensure that the main regulators' maximum input-to-output differential voltage specification was not exceeded during start-up, it was necessary to add the inrush-limiter circuit, in order to also slow down the rise of the regulator's input voltage.

Cheers,

Tom

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#### marshallf3

Joined Jul 26, 2010
2,358
Kind of a complicated way to deal with it when simple thermistor can do the same thing.

#### R!f@@

Joined Apr 2, 2009
9,647
I have heard of people over explaining things...and this thread is one good example.

OP's transformer is a mere 50VA... Current will be around 300mA at worst.
Any fuse close to this value might blow if connected to a transformer. So in practice most of us use three times the capacity.
In your case a slow blow 800mA will be the appropriate primary fuse. If the fuse is blowing, you are doing something wrong or your former is faulty

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#### gootee

Joined Apr 24, 2007
447
Kind of a complicated way to deal with it when simple thermistor can do the same thing.
Back when I looked at using PTC thermistors I couldn't figure out exactly what the "on" resistance would be. And don't the PTC inrush-limiter types have to run pretty hot?

I realize that the OP's current level is rather modest but for a high-current supply how can people tolerate a thermistor's resistance in line with their high-current path? How low does their resistance get, typically? I realize there are many different PTC devices but can they usually be selected to go below 0.1 Ohm or so when operating? (Assume a 3 to 8 amp 20-something volts supply, for example.)

#### marshallf3

Joined Jul 26, 2010
2,358
You would use an NTC thermistor in this application, in other words the resistance goes down as it heats up.

For transformers this small a simple 1A slo-blo fuse on the primary is going to work fine or, as mentioned above, your design needs some more attention.