Proper usage of a MOSFET to control USB

Question

On the RaspberryPi.SE it was suggested that one can use MOSFET BS170 to switch a USB device on and off by a GPIO pin (3.3V). I need to get 500mA to power a GSM dongle; the voltage drop should preferrably be minimal since USB is quite strict on voltage).

I'm not sure if I did the right thing. I cut the power wire of the USB cable, and connected the MOSFET to it in the following way:

^{simulate this circuit – Schematic created using CircuitLab}

(I use a speaker in the schematic to represent my USB device, and a switch to represent a pin of the GPIO.)

The problem is that:

1. I'm not sure that the two grounds are actually connected.
2. According to a scheme on RPi's elinux page, I should have cut the GND wire instead of the PWR one.
3. The official gate voltage is 3V and I supplied 3.3V to it.

I think I demaged the transistor I bought. I can buy a new one (even couple of them, they're really cheap), but I would prefer to know what to do before I experiment again or cut some more wires.

EDIT:

I did some more search, and I found a schematic by bretth on sparkfun.com, which seems to do what I want, using a P-MOSFET instead of N-MOSFET. Is this schematic angood way to go? It is a problem that GPIO provides 3.3V and maximal V_GS is 3V for N-FET and (minus) 2.4V for P-FET?

^{simulate this circuit}

Answer 1

So I decided I'd give it a try, I bought AOP605, wired everything, and it works! :) So the following scheme is what can be used to control USB power using 3.3V input voltage:

When the GPIO pin is off, the USB DEVICE is off, switching the GPIO on switches on the USB DEVICE as well. I use 7.5k as R1 (I don't have 10k by hand) and it works well, too. Credit to betth and his post on sparkfun.com, I only changed the transistors to something more suitable for my case that I can buy here.

Answer 2

Alternatives: There are dedicated USB power switch ICs, for example Maxim 1562 and lots of others.

Or, you might consider a small reed relay. If you want to avoid the holding current, then you might choose a latching reed relay.

Answer 3

Following advice from Dave Tweed it is apparent that switching a USB load via the ground rail is not good so, what remains of this answer is all about choosing a mosfet and checking the appropriate graphs in the spec for the device. In short, look at the graph of drain current versus drain-source voltage and choose a device that has the lowest volt drop for the gate voltage you are able to provide. In this particular case it's appropriate to use a p channel mosfet switching the positive rail to the load and to do this requires an NPN BJT to interface between the gpio and the mosfet's gate.

You should have inserted the N channel mosfet in the ground wire to the load with source to ground and drain to the positive rail connected load. Also you need to put a resistor across the gate and source to leak away charge once the switch has open circuited. The gate-source pins are effectively a small capacitor with very little ability to drain away charge / voltage fed to them when the switch is closed. Try 10kohm or 100kohm.

EDIT NOTE - originally the question showed an IRF530 in the circuit

3.3V may be a little too lightweight for the the MOSFET you've shown in the diagram. The IRF530 is a 100V rated device and it has quite a high gate-source threshold voltage (2V min to 4V max). This means at 3.3V it may be only just starting to turn the MOSFET on fully: -

I've marked in red what the volt drop across the MOSFET is at a current of 1A flowing when the gate-source voltage is 4.5V - they don't specify any less than 4.5V so this is also a big clue that you should look for a device with much smaller Vgs(threshold). As you can see, with 1A flowing you are losing 1V across the MOSFET and this isn't what you should be aiming for. Clearly if you could supply a gate drive voltage of (say) 10V then at 1A drain current the volt drop will be about 100mV.

Try finding a device that shows the same graph with gate voltages down to 3V and look for no more than 100mV lost across the FET when 1A is flowing.

A BS170 FET is going to be worse if you want 1A: -

At 1A you'll need at least 6V to drop maybe 1.8V. Miles worse than the IRF530. However, if your load current was more like 100mA, a 3V gate-source voltage might only "lose" 0.5volts across the device.

Answer 4

First, you don't need the 10k resistor, I guess it was inherited from a diagram involving NPN+PNP transistors. Second, there's another option which is sometimes even simpler, consisting in using an LDO regulator with a shutdown input pin. That's what is used on the BusPirate board and a number of other ones. Such regulators are able to show almost no voltage drop between input and output and come in a 5-pin SOT23 package and are small and very convenient. Otherwise there are USB power switches which include the two transistors as well (and sometimes current limiting and possibly fuse emulation).