# Cheap alternative to limit inrush current (eg. relay bypassing a fixed resistor or MOSFET)

I'm editing the question because several (great) answers focus on the fact that the LTC3780 module (inside ajustable my power supply) can handle the inrush current (maybe, maybe not). I'm glad to know, but it doesn't answer my main question which is more global: how would you limit inrush current, in a cheap manner?
I apologized, my question wasn't clear.
(So the circuit I'm talking about below is just an example, let's say I use another power supply than the LTC3780 that deliver 12V-80W but can't handle the inrush current― but please keep the precious information provided in your answer about it).

The situation:
I want to power 5 fans in parallel (12V 0.80A, each) using an LTC 3780 (it's (DIY adjustable power supply)

Because my fan are brushless, I will have a high inrush current. My LTC 3780 can't handle more than 80W so I need to limit the current to about 6A (=80W/12V)

The possible solution:
I am looking for a simple/cheap solution (so no NTC thermistors*)

One cheap/simple solution would be to use a fixed resistor (R1) to limit inrush current to 6A, and a switch to bypass it. After a couple of second as the fans speed up, counter-electromotive force starts to limit the current. So the resistor R1 becomes useless and even inefficient as it consumes voltage/limits current that I need for my fans. The switch then allow to bypass R1.

The question is which switch : a relay, or even better, a MOSFET ? But how can I switch it? Could it work with a capacitor (or 555 timer ?)

Here is a working circuit with a relay (the fans are replaced by resistors with inductors).

How to choose the components ?
If this is a good solution, how can I determine the value for R1, R2, C1?
- If my understanding is right, R1 would need to be more than 2 ohms (= 12V/6A), so I guess a 4,7 Ohms would be right for R1.
- I guess I need a R2, but how to choose it's value.
- And what about C1? I guess the bigger it is the longer, it will take to switch on the relay.

And also how to choose the relay so that it doesn't switch on too early (before my fans reach their steady-state or C1 is full) : is there a way to know the minimum voltage/current needed to activate it? Would something like this MOFSET (or this relay) works ?

Better solutions ?
I also found this dirt cheap Time Delay Relay. If I use one with a fixed time like 5-10 seconds, would it work with my 12V/6A power supply and my five 12V, 0.8A fans?

Also is using a relay the best solution ? Couldn't I use a MOSFET as they consume less then a relay.
Here is my try at making a circuit with a MOSFET (but I can't find a way to use the capacitor to switch on the MOSFET):

(* Thermistors are expensive and I prefer to invest in components that I can re-use in other circuits.)

• Your link to the LTC3780 refers to a module. This module has a potmeter for setting the maximum current. I would simply set it to slightly above the nominal current of 4 A. I do not understand why the inrush current is a problem, the LTC3780 has current mode control so it will already limit the maximum current. In my opinion: The inrush current is not an issue but feel free to provide evidence that proves me wrong. Commented May 21, 2019 at 11:24
• Are you aware that the link you give to thermistors is the cost for a pack of ten? \$0.95 each is still not cheap, but I wouldn't consider it a dealbreaker level of expensive. And getting them from digikey or mouser can get you what you need for cheaper, anyway. Commented May 21, 2019 at 11:32
• @Hearth The cheapest thermistor I found is 9€ on aliexpress (I don't need 10 pieces, so it's like paying 9€ for a single thermistor) Digikey would cost me 18€ just for the shipping of a single piece, (free shipping in europe are over 50€), so 19€ for a single thermistor. I can get a relay for less than 2€ (free shipping), and I could use it in many other projects since they are rated for high amps (by the way, I've updated my question). Commented May 21, 2019 at 16:21
• It seems to me that what you need, is managing your stocks and supplies with a bit more anticipation. Because you can get a relay for much less than 2€ on mouser/digikey. Actually even less than 1€. So if you group your orders, and, say, buy a handful of resistors, 555 chips, relays, you'll be saving money on all components and be able to get your NTC for maybe 0.3€. Which is cheaper than all the circuits you suggested above.
– dim
Commented May 21, 2019 at 18:58

If the module you bought contains a real LTC3780, you do not need any downstream protection for it, as it is already short circuit protected.

If the chip on there is made of chinesium, and you do not trust it, you can try and hack the board to regulate the inrush current changing the cap connected to the SS pin. Limiting the inrush current will surely also limit the output current, saving the board.

• thanks! changing the cap connected to the SS pin if this is the case, do you mean I should increase the capacitor size ? To which value? And what this capacitor is used for ? (also I've updated my question). Commented May 21, 2019 at 16:21
• Did you check the LTC datasheet? And you should not edit your answer, if you need to ask a different question, please do so. And again, the LTC can handle being shorted to ground, so your phrasing "maybe, maybe not" is a bit off in my opinion. Commented May 22, 2019 at 8:40

If you look at the datasheet for the fan, you'll see that the maximum current is 0.80A, apparently limited by winding impedance, so your total maximum load shouldn't hit the limit of the power supply.

The back EMF generated by motion of the rotor brings down the running current to a value that produces a torque that balances the fan load at operating speed, but at standstill, when the back EMF is obviously zero, if there are no active current limiting mechanisms, the current is limited only by the resistance of the windings.

The term "impedance protected" comes from UL1004-2 describing motors that don't need thermal fuses or similar to prevent them from exceeding their insulation temperature limits when held in locked rotor condition, because the current determined by the impedance is low enough not to create an excessive temperature rise. For a DC supply, the 'impedance' is purely resistive, the inductance of the windings is not relevant.

• PhilG Would the "winding impedance" protect the power supply from the inrush current? I thought we had to wait for the brushless motor to reach their steady state for the winding impedance to start to resist the current. Or maybe I am confusing "counter-electromotive force" and "winding impedance", do I? If so, how concretely the impedance (the sum of the resistance and the reactance ― which is, if I got it right, a resistance that change with the frequency) could protect the motor against inrush current? Commented May 21, 2019 at 16:27
• Added some more detail. Commented May 21, 2019 at 18:24