Inrush current - should it be dealt with?

Thread Starter

jgrv

Joined Mar 2, 2018
36
Hi,

I'm pretty new to electronics, and I hope to get some help and advise from you guys :)

I’m in the design process of the supply section of a PCB. It will have a 12VDC supply and will be powered by either a 12V solar battery, or from a 12V wall adapter. Here is the circuit:


First is a P channel MOSFET, for protection, that turns off if the connections (+/-) are reversed. After that I have a 2200uF capacitor (will be electrolytic).
I have a question regarding inrush current caused by the capacitor. I did a simulation of the circuit and saw current go all the way up to 1500A!


The current stabilizes after approximately 75 microseconds. This simulations contains almost no resistance to the capacitor, so in real life, the current would probably not go this high. This circuit will normally draw very small currents (<50mA), but could draw up to 3A at most. But anyway; could this inrush current cause the diode, MOSFET and/or power supply to break? If so, is there a good way to suppress this current without reducing the efficiency of the circuit in steady state? Or does it last too short to be able to affect any components at all?

Thanks :)
 

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dendad

Joined Feb 20, 2016
4,452
Your real world supplies cannot source that current so it will be limited already by the internal resistance. There will not be any real problems.
 

crutschow

Joined Mar 14, 2008
34,285
Why do you have a diode in parallel with the MOSFET? :confused:
It's redundant.
The MOSFET carries the current at a very low voltage drop, so the diode does nothing.
 

ebp

Joined Feb 8, 2018
2,332
Diodes like the 1N400x series usually have a current rating for a single (some times two) half cycle at 50 or 60 Hz. This rating is included specifically to address inrush current in typical applications with capacitive filtering. The Vishay data sheet actually has two numbers - one for a half cycle at 60 Hz (30 A) and one for a 1 ms rectangular pulse (40 A).

You don't need the 1N4001 in any case. All power MOSFETs have a "body diode" in parallel with the actual transistor. Often the diode is rated for the same current as the actual FET. For the FET you have chosen the current rating is less - 2.1 A vs. 7.6 A at 25°C case temperature. But it is rated for 80 A for a 100 µs pulse. If you do the arithmetic for power in the FET at 7.6 A versus the power in the diode at 2.1 A, you'll see why the continuous current ratings are different. In reality, if you operate at the max for either for more than a very brief time the part would reach at least 150°C with the standard "footprint."

In your circuit, the FET will turn on and conduct the current virtually instantly, so even the body diode won't ever conduct current. If the input voltage rose slowly (which of course removes your original concern) the diode could conduct until the input voltage reached the gate-source voltage required to turn the FET on to a voltage less than the diode voltage. Your FET will start to turn on somewhere between 1.2 & 2.5 volts and should be well turned on at 3-4 volts.
 

Thread Starter

jgrv

Joined Mar 2, 2018
36
Thanks, guys, for all your feedback!

I feel a little bit wiser now :D

I realized after i posted this thread that the extra diode parallell to the MOSFET was unnecessary after seeing the symbol in the datasheet. But I did not realize all MOSFETs came with a "body diode". All the different symbols used for MOSFETs confuse me o_O
 

ebp

Joined Feb 8, 2018
2,332
They confuse me too. I like the diode on the symbol so I don't get the blasted things wrong way up. Ida thunk after several years of designing with FETs I get it right, but noooo...

The diode characteristics have been improved over the years. In some applications the diode behavior is detrimental. In years gone by, in switching power supplies you would sometimes see a schottky diode in series with the FET and another across it, just to prevent the body diode from conducting.
The diode behaves like a power zener to excessively high (relative to FET rating) voltage. In many modern FETs this is actually well specified and it is acceptable practice to deliberately let the diode act like a zener. FETs like that have an "avalanche" rating (for diodes that you buy as zeners, only those at the low end of the voltage range are true zeners; above about 5 volts they are avalanche diodes - the actual mechanisms of reverse conduction are different). I think I saw a repetitive avalanche rating it in the data sheet for the FET you are using.
 

crutschow

Joined Mar 14, 2008
34,285
All the different symbols used for MOSFETs confuse me
That's understandable.
But most MOSFET symbols come with an arrow showing the direction of the substrate diode (which all MOSFETs have). Two common symbols are below.
The normal drain-source bias would be such as to reverse-bias this diode, thus the N-channel has the drain more positive than the source, and the opposite for the P-channel.

One notable exception to this is when a MOSFET is used as an ideal diode or as an active flyback diode in a switching regulator. In that case the MOSFET is turned on to conduct in the same direction as the diode, and turned off to block in the normal direction (substrate diode reverse biased).
This works because MOSFETs conduct equally well in either direction when biased on.

upload_2018-3-4_10-58-8.png
 

Thread Starter

jgrv

Joined Mar 2, 2018
36
Thanks!

By the way. In the circuit i showed in the first post. I removed the diode 1N4001, but noticed that at 12V supply I measure 262uV between drain and source. If I increase the supply to 15V I measure 600mV between drain and source. Even if I increase the load to 10kOhm, I still measure 600mV. According to the datasheet the drain-source resistance should be very little, and 262uV and 600mV seems a little too much. Can anyone see why this is, if not an error in the spice model?
 
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