# Would this inrush current circuit work in practice?

I am building a USB hub circuit, which, because of the 120uF capacitance requirements on upstream USB ports, means that I need to charge those large capacitors in my circuit. However, my hub will be bus-powered over USB-C, which I believe means I need to be careful of the USB spec's inrush current requirements. Many discrete current limited switches I've seen have a "hot-start" function for this purpose, but most don't have the capability to pass 3A of current like I need in my application.

No problem- just build the switch out of discretes. I came up with this pretty simple circuit in Falstad circuit simulator. Operation is pretty simple- the 1uF capacitor is charged through the resistor until the voltage across it (which is Vgs) reaches the threshold voltage, the P-Channel MOSFET starts conducting in the saturation region, where the Rdson limits the current. Then as the Vgs continues to increase the switch turns fully on.

I know that operating a MOSFET in the saturation region is a nice way to release the magic smoke through P=I^2R heating, but how long is too long? Looking at the datasheet of, for example, the AO3401A, does Figure 10 say that it is okay to dissipate ~10W for a 10ms pulse? If so, I imagine it would be okay, since, according to the "Power consumed" graph in Falstad, the peak power consumption is 1.7W and the whole inrush event lasts for less than 5ms.

Also, though I think it goes without saying- am I missing something that can't show up in the simulator, but will ruin my day in the real world?

• I think Andy hit the key points. An alternative solution is to use a simple inrush limiting resistor which is bypassed by a mosfet. Some kind of delay circuit keeps the mosfet off for a fixed delay which gives the capacitor time to charge up. Commented Feb 6, 2022 at 18:33
• My rule of thumb when it comes to circuits which require dissipation of energy (where dissipating energy is the purpose of the circuit) is that it is always cheaper and safer to dissipate energy in a resistor rather than in silicon. Dissipating energy in silicon should be a last resort. The reason is that silicon is expensive and cannot tolerate high temperatures. Resistors are less expensive and often can tolerate higher temperatures than silicon without failing. Commented Feb 6, 2022 at 18:36

It will work -- the first time you plug something in.

Consider what happens when your 5 V connection has an intermittent connection -- when the 5V becomes open circuit, there is nothing to discharge the 1 uF capacitor. The FET will still be on, so the 1 uF will tend to remain charged by the 300 uF (even as it discharges). If you happen to reconnect when the 300 uF is not fully discharged (but still above a threshold voltage), you won't get inrush limiting the 2nd time.

Note also that while the 300 uF remains charged, the '5V' line will remain at 5 V when disconnected. This makes detecting the removal of the connector difficult.

While the current won't be any larger than it reached during the initial plug-in, the rate of rise of current will be and ultimately the small arcing this causes can damage the connector.

There isn't a solution to this issue without some additional components and complexity.

• This answer unfortunately seems correct- I see that some nasty current spikes occur if 5V is momentarily disrupted, since that 1uF capacitor stays charged. Guess I'll have to go back to the drawing board on this one :/ I do wonder what a good solution would be. I like the sound of @mkeith 's idea with the bypassed resistor, though I worry if it could be susceptible to the same problem. Might just look for an IC... Commented Feb 6, 2022 at 20:45
• To the OP: you have to work through the cases. The delay circuit could potentially be designed to be very bounce resistant or could even use an IO pin from the processor with the delay controlled by firmware. Commented Feb 8, 2022 at 21:14

Would this inrush current circuit work in practice?

The MOSFET might spend too long in its linear region and fail. It's a tricky one to get right unless you are using an over-sized MOSFET or one that is designed to run without significant thermal runaway in its linear region. It can "fry" in significantly less time than a millisecond. It's also known as the Spirito effect (see this article from Analog Devices for further info.

Having said the above, the thermal runaway characteristic of the MOSFET you have chosen looks about as good as it gets for avoiding significant thermal runaway: -

So, if the gate source voltage is about -1.5 volts, the MOSFET may rapidly warm from 25°C to 125°C in very little time but, I suspect that this MOSFET will survive that - the difference between the two curves is quite small. However, the MOSFET chosen is also quite small. BTW, you can't rely on heatsinking copper for avoiding this. Try and and do some decent testing.

• If there's room for it in the device, I would suggest using a larger part with a higher thermal mass, perhaps something in SOT-223 or even DPAK. It's thermal mass that matters here; the time scale is too short for heatsinking to make any difference. This is what you were getting at, I think, when you mentioned an oversized FET. Commented Feb 6, 2022 at 17:07
• @Hearth yes, it's the thermal mass of the silicon die that will save the day and, as I said and you said, external heatsinking won't prevent thermal runaway. Commented Feb 6, 2022 at 17:09
• Yeah--I know I repeated some of what you'd already said, but I wanted to expand on it a bit. Commented Feb 6, 2022 at 17:16
• Thank you for this answer- that article from Analog is helpful reading. If I do end up using a discrete MOSFET, I think I'll try and look for a beefier one than SOT-23. Haven't quite got the room for SOT-223, but a nice power 3.3x3.3 or 5x6 DFN package would fit the bill I think. Commented Feb 6, 2022 at 20:52

Andy's answer already addresses whether or not this circuit would work, but I'd like to draw your attention to something you may not know existed: the inrush current limiter. It's a part designed for exactly the purpose you need, and they don't cost very much money (this one for instance is \$0.40 in single quantity).

What it is is just an NTC thermistor designed for relatively high continuous current (I'm not certain, but I suspect they may have some thermal insulation too); you put it in series with your load, and at room temperature it has a resistance of a few ohms, which prevents too much current from flowing, but as it heats itself up, its resistance drops.

This solution will result in a little more wasted power than a MOSFET with sufficiently low Rds,on, but it's also more foolproof, so it's up to you which way you want to go. It may also result in too much voltage drop for a 5 V circuit; ICLs are more often used on mains-voltage devices.