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I'm designing a device powered from USB. The device uses FTDI FT2232 chip for USB connection. Upon a command from a computer FT2232 chip should enable power via a MOSFET switch to rest of the circuit. This additional circuit has a capacitance of 50uF (FPGA + aux stuff) and is powered from the same USB port. After the switch is turned on, this additional 50uF capacitance will sink a huge current until it's charged.

How to limit this inrush current 1) to avoid voltage drop on power rails and 2) to avoid USB PTC from disconnecting power to the device?

Is it enough to put a ferrite bead in series with MOSFET switch to limit the inrush current? Or should I use a special chips, like chips for limiting current or chips for slew rate control?

Note: all devices are powered from 3.3V. So a small drop on 5V rail should not be a problem if it does not prevent an LDO to output stable 3.3V.

jpc
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ibiza
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    50 uF across the USB 5V should not be an issue. PTCs have a time constant, so a short duration current spike will not trip it. Anyways, I have some project with several hundred uF across the USB rails, to no ill effect. – Connor Wolf Jan 22 '11 at 12:43
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    @Fake Name. PTCs are slow and, agreed, they do not trip because of short duration spike. However, such a spike can create a sufficient drop on voltage rail, which I would like to avoid. The USB spec says the inrush equivalent load should not exceed 10uF paralleled with 44 Ohms. – ibiza Jan 22 '11 at 14:32
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    My laptop has flashed overcurrent errors and not worked with devices with 47 uF across VBUS. Stick with the USB spec of <10 uF. – endolith Jan 23 '11 at 02:28
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    I guess USB current sensing is pretty implementation/motherboard specific. If you want to be safe, put a powered USB hub in series with your device. That way it's much harder to blow anything on your computer, even if something does go wrong. – Connor Wolf Jan 23 '11 at 10:49
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    I've had similar problems with the USB terminals on a Beckhoff industrial PLC, which was really finicky and would actually flash a warning onscreen when I plugged in an unpowered USB hub with nothing connected (the hub itself had too much capacitance). FakeName is right, it's very implementation/motherboard specific. – Kevin Vermeer Mar 07 '12 at 19:45
  • @ConnorWolf The inrush limit isn't for preventing damage to the computer; it's for preventing voltage sag that resets other devices on the same bus, and for increasing connector contact life because of reduced arcing. – endolith Aug 29 '16 at 21:03
  • @endolith - Good point. – Connor Wolf Aug 29 '16 at 21:49

3 Answers3

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Use an RC circuit in the MOSFET gate to slow down the turn-on.

One of the FTDI app notes has this example of a soft-start circuit on USB Vbus:

enter image description here

boardbite
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mikeselectricstuff
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    Agreed. An RC circuit will limit the current going to the capacitor. Care must be taken to ensure that minimum rise times on the devices connected to the rail are met, else there may be weird start-up logic glitches which are hard to diagnose. Also, a fast turn-off diode across the resistor may be needed (to quickly shut off the MOSFET). – Adam Lawrence Jan 22 '11 at 12:47
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    This seems the best solution. Indeed MOSFET based soft-start circuit is shown in FTDI guidelines http://www.ftdichip.com/Documents/AppNotes/AN_146_USB_Hardware_Design_Guidelines_for_FTDI_ICs.pdf. Thank you. – ibiza Jan 22 '11 at 14:21
  • When I simulate this circuit it doesn't do very much; I still get an 18 A current spike. The IRLML6402 typical threshold is -0.55 V, so it turns on pretty quickly. Am I missing something? – endolith Aug 29 '16 at 21:01
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    There seems to be a better version with an extra cap to the input explained here https://semianalog.com/articles/fet-inrush/fet-inrush.pdf (but I haven't tried it myself) – patstew Feb 20 '17 at 15:49
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The USB specification chapter 11, Interoperability and Power Delivery, places rather stringent limits on power draw. The cited capacitance there is only 10µF, to avoid too much voltage drop. There are specialized ICs (like LM3525) to do both current limiting and power switching, which might help, but make sure the circuitry behind it handles the slow voltage rise correctly. A brown-out detector may be enough, but a few devices require many voltages in specific orders.

Rev
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Yann Vernier
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  • Similar specialized ICs for USB current-limiting: [AP2337](http://diodes.com/catalog/single_channel_usb_switch/ap2337.html), [BD82034FVJ-GE2](http://www.rohm.com/web/global/products/-/product/BD82034FVJ), etc. – davidcary Sep 18 '14 at 15:44
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The other answers are good but if you prefer a one component solution there exists Inrush Current Limiters.

I've used them before to prevent fuses blowing when hot-plugging the power supply. Their operation is really simple. Basically they have a resistance at room temperature, say 5Ω. When you plug in a 5V power source the surge current is now limited to 1A even if there is a direct short on the other side of the ICL. (5V/5Ω = 1A) As soon as current flows through the ICL it starts to warm up and its resistance then drops very close to 0Ω (check the datasheet) and it's like the component is no longer in the circuit.

I like these because they're usually easy to rework into an existing designs and it's only one component.

ACD
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  • This is certainly a novel way to limit current, but the OP wanted a solution that prevented an voltage sagging. – sherrellbc Aug 01 '14 at 12:13
  • Oops, I thought I was sorting by newest. Didn't realize this was years old. OP wanted something that wouldn't lead to a voltage drop on the power rails. He said nothing about sagging. Plus the "sag" would only happen for microseconds(more of a soft start). Once steady state is reached the 5V rail wouldn't be off by more than a couple of millivolts, and the 3V3 rail would be unaffected. – ACD Aug 01 '14 at 12:33
  • I interpreted what the OP was saying about limiting voltage drop as something that was a direct consequence of the current limiting technique. That is, if the ICL you suggested was passing 1A of current, 5V would be dropped across it (i.e. voltage *sag* momentarily as the source charges the capacitance). I could be completely wrong here though. Perhaps I am throwing terms around that are not interchangeable. Either way, I like your solution. – sherrellbc Aug 01 '14 at 13:51