I'm trying to design a power supply for a GSM modem with 2A instantaneous current, but only 210mA operating current.

I believe my power supply does not have to be rated at 2A does it?

I'm wondering if an inrush current limiter like a thermistor will help protect the power supply, while still ensuring the modem works fine within its limits.

Is this ok?


It does. Peak loading like you describe can be dealt with in one of two ways - rate the power supply for the worst-case load continuously, or temporarily. Either way, the power train needs to be able to deliver that power.

Since the worst-case load is only 2A, it is a reasonable design point to make the power supply work at 2A continuous output - then you'll never have an issue.

If you're talking about a 100A power supply that needs to deliver 200A temporarily, rating the power supply for 200A continuous is most likely overkill. In this case you can put a timer on the overload protection to allow the overload to persist for the short interval that's needed. The power train components in this case have to be rated for the surge, not the higher sustained steady-state - often the 100A power train parts could handle 200A for a few milliseconds as long as the thermals are managed well.

  • \$\begingroup\$ When you say "the power train needs to be able to deliver that power" does it mean that i cannot use an off the shelf power supply rated at say 500mA to power this modem, even with an inrush current limiter to protect the power supply? \$\endgroup\$ – TiOLUWA Jun 19 '12 at 11:18
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    \$\begingroup\$ @TiOLUWA, Inrush limiters often choke the output voltage to keep the current lower, this means you will stop the device from operating. \$\endgroup\$ – Kortuk Jun 19 '12 at 12:17
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    \$\begingroup\$ You CAN use a lower spec supply if its capacity exceeds the mean power drain and if you provide reservoir capacity as per my answer, BUT you may have to take special startup precautions. eg if you design to supply 2A peaks from capacitors you need to charge them before the peaks occur. If the load draws 2A at startup for say 200 mS you need the supply to charge a capacitor to this level BEFORE the load is applied. This is easily enough done BUT may need to be a specific design decision. \$\endgroup\$ – Russell McMahon Jun 19 '12 at 17:27
  • \$\begingroup\$ @RussellMcMahon This is what I mean by 'temporarily'. \$\endgroup\$ – Adam Lawrence Jun 19 '12 at 19:23
  • \$\begingroup\$ @Madmanguruman - Not to be rude, but you were unclear and it is not what will be taken from yiour answer. You say it CANNOT use a lower current suppply, but it can. A 2A 12V supply is often far dearer than a 500 mA or 200 mA supply (even though it should not be). You essentially say "Just get the big one". I say "with care the little one is fine. Your answer does not match mine. \$\endgroup\$ – Russell McMahon Jun 20 '12 at 4:44

If the load is rated at a genuine 2A peak then you need to provide a means of supplying it. If you want to avoid a brute force approach then knowing the worst case power profile would be of great assistance.

Placing a maximum current limited on the input to your power supply is OK if it is useful to do so, as long as you can obtain the average power level required and can service the peak output load. Placing a current limiter between power supply and modem guarantees failure.

The easy and obvious way to provide brief current peaks which are well in excess of suspply capability is to provide output capacitors. 1 farad will drop by one volt in one second when supplying one amp.
Or, rearranging in various ways:

dV = I.t/C | C = i.t/dV || t = C.dV/i || i = C.dV/t

These formulae (all the same formula rearranged), can help you make decisions about output filtering.

If the 2A peak lasts for 1 uS you need C ~= i.t/dV = 2 uF for 1 Volr drop = not much.

If the 2A peak lasts for 1 S you need C ~= i.t/dV = 2F !!! - largely supercap country.

What they probably have in mind is peaks in the uS to perhaps 10's of mS range that can be supplied by large conventional capacitors. For say 10 ms duration & 1 V droop.

C = i.t/dV = 2. 0.01 / 1= 0.020 F = 20 mF = 20,000 uF.

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    \$\begingroup\$ it might be worth learning the mathjax engine for how often you use math and equations, it can make your equations MUCH more clear. \$\endgroup\$ – Kortuk Jun 19 '12 at 12:18
  • \$\begingroup\$ @Kortuk - Aye .................. \$\endgroup\$ – Russell McMahon Jun 19 '12 at 17:13

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