Inrush limiter also provides short-circuit protection
MOSFET keeps inrush currents due to capacitive loads under control.
By Ryan Brownlee, Process Automation Division, Pepperl & Fuchs Inc, Twinsburg, OH; Edited by Brad Thompson and Fran Granville -- EDN, June 9, 2005
For containing large amounts of bulk capacitance, controlling inrush currents poses problems. The simplest approach involves placing an inrush-limiting resistor in series with the capacitor bank, but a resistor wastes power and adds a voltage drop. The circuit in Figure 1 addresses these problems and provides an additional benefit. At start-up, bipolar PNP transistor Q2 holds N-channel power MOSFET transistor Q1 off until the voltage across capacitor Cl reaches a high enough level to turn off Q2. During this interval, resistor R1 supplies C1 and the rest of the circuit with start-up current. When Q2 turns off, Q1 turns on and provides a low-resistance path across R1. When you shut off external power, the circuit resets as C1 discharges.
As an additional benefit, this circuit provides protection against short-circuited loads. As current through Q1 increases, the voltage drop across Q1 increases due to Q1's internal on-resistance. When the voltage drop across Q1 reaches approximately 0.6V (Q2's VBE(ON)voltage), Q2 turns on, turning off Q1 and forcing load current through R1. Removing the short circuit restores normal operation, allowing Q2 to turn off and Q1 to turn on. Note that, because Q1's on-resistance acts as a current-sense resistor for this function, the short-circuit trip point may vary depending on ambient temperature and Q1's characteristics. You can adjust Q1's turn-on and -off threshold by selecting R1 and Q1's on-resistance characteristic. Adding a conventional or zener diode in series with Q2's emitter increases the short-circuit trip current.
The components and values for constructing this circuit depend on the application. Depending on the design requirements, you may need to select a high-power resistor for R1 or add a heat dissipater to Q1, but, for many applications, the circuit saves power over a conventional approach.
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If one doesn't need the short-circuit current limiting, there is a cheaper and much more sure-fire way to do inrush limiting, using the fundamental properties of the MOSFET. It only takes the addition of two resistors and a capacitor, of the smallest size you'd want to use, and works >exactly< as advertised. The concept is described in a number of app notes from various semi vendors. The best one I've seen is the one authored by Mitter at Motorola Semiconductor, application note AN1542, before that chunk of Mot got spun off as On Semiconductor. Googling on "AN1542" pulls up a number of places where the app note entitled "Active Inrush Current Limiting Using MOSFETs" is kept by different folks. I have used this method on a number of consumer and automotive products, and have found it to work beautifully in a deployed base of >1M units.
Dan Wagner - 2005-8-7 09:51:00 PDT -
The circuit is just unworkable. Period. Even if a resistor is added at the base of Q2, the MOSFET would go on instantly on power ON and defeat the inrush-limiting feature. This is because at power ON the output voltage is zero and Q2 base-emitter current would flow causing Q1 to conduct without a delay; this would cause Q1 to go ON immediately. The external Drain-Source resistor does nothing since C1 gets charged through Q1 starting from power ON.
N. A. Gandhi - 2005-17-6 01:03:00 PDT -
Well, my DI certainly has received plenty of attention, unfortunately for me; it is not quite the attention I would prefer. However, as the author of this DI, I would like to clarify a few things so that there is no confusion. I would also like to take this opportunity to apologize to those readers I may have confused. I'm sure that we all can agree that sometimes, the simple, obvious items are the ones most often overlooked.
Issue 1. The text refers to an N-channel MOSFET, but the schematic shows a P-channel. The text should have read P-channel.
Issue 2. The base of Q2 does need a resistor, if you would like the circuit to function.
Issue 3. The circuit can be used for supply voltages over the gate-drain breakdown voltage (usually about 20V) if protected with a zener from gate to drain.
Thanks to those of you who first noticed the errors, and posted comments.
Ryan Brownlee - 2005-16-6 09:39:00 PDT -
Having esablished, by a large consensus of opinions, that Q1 should be a P-channel device and that Q2 should be provided with a resistance in series with the base, I think one should also mention that the maximum source-voltage should not exceed 20V, in order to avoid destroying the gate-source isolation barrier of Q1.
Mario Pazzini - 2005-16-6 06:00:00 PDT -
There is something wrong with this design idea: Q1 is described as an N channel device, but it is drawn with the intrinsic diode upside down or with source and drain interchanged.The second alternative should be the correct one, i.e. the current enters the N-channel drain and exits the source, but....in that case transistor Q2 won't work because it's emitter would be connected to Q1's drain and not to Q1's source. There are two possible alternatives: either Q1 is a P-channel device or, if it has to be an N-channel device, it should be placed in the current's return path....plus other obvious modifications to the circuit.
Mario Pazzini - 2005-15-6 06:38:00 PDT
