I'm using a mosfet for powering an Arduino mega 2560 on and off. The current drawn is less than 500mA with a 12V battery.

I chose the IRF7457Pbs (RS 543-0888), with these features:
VDSS - 20V
RDS(on) - 7.0mΩ
max ID - 15A

I measured the voltage drop on D-S:

  1. Vin=9.40, Vout=7.55
  2. Vin=12.15, Vout=10.24

This is too high for the mosfet specifications.

I'm using the mosfet driver UCC27517DBVT, (RS 774-1367), with Vdd=12V, IN+ TTL logic level from Arduino. The resistor at the gate is 5.1 ohm.

Why is the voltage drop too high?

  • 3
    \$\begingroup\$ What Vgs is that 7mohm gate resistance specified at? \$\endgroup\$
    – user16324
    Nov 20, 2014 at 12:08
  • \$\begingroup\$ Also - what VCC are you using on the UCC27517 gate driver? \$\endgroup\$ Nov 20, 2014 at 12:40
  • \$\begingroup\$ If you want to use an N-channel MOSFET as a high-side switch, you need to be able to drive its gate above the supply rail. Curiously, the data sheet never actually says "N-channel"; you need to infer it from the symbol they use. \$\endgroup\$
    – Dave Tweed
    Nov 20, 2014 at 13:54

1 Answer 1


You are using the MOSFET as a voltage follower. In other words you have drain connected to 12V and the source connected to the load. To switch on the device, the gate voltage needs to be a few volts higher than the source and your results demonstrate that i.e. Vin = 9.4 volts and Vout (source) = 7.55 volts i.e you have a volt drop of 1.85 volts trying to turn on the device.

Of course this isn't enough - if you read the data sheet, to turn the device on "properly" you need at least 2.7 volts (figure 1 page 3 of data sheet) and, to turn it on so that it produces 12 volts out means applying 14.7 volts to the gate.

I suspect you should be using a P channel device wired so that it can turn on hard with a gate voltage taken down to 0V: -

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