# 3.3v high side switch

I want to replace my relay with a high side MOSFET switch. I am a new to all this and I don't know how to choose the components.

The load needs around 200mA at 3.3v with the control coming from a open collector pin on a microcontroller. After some searching I found the following circuit. How do I modify this to run at 3.3v instead of 12v?

• I'd love to know how the value of R3 was chosen. 4 ohms? Commented Aug 7, 2013 at 9:24

You can regard R1 and T1 as the microcontroller's open-collector (actually it's open-drain) output, then you only need the pull-up resistor R2 and the P-channel MOSFET. Since you're working at low voltages you need a logic-level gate FET.

• Do you think the MOSFET FDN360P will work? Commented Aug 7, 2013 at 9:39
• @user27207: Yes. Look at figure 5. That shows that at 3 V you can have up to 2 A drain current, so it works for your low voltage. Commented Aug 7, 2013 at 9:43

The problem is the MOSFET that is shown will not turn as as expected with the 3.3V supply rail. It's all down to its $V_{GS(threshold)}$ being too high: -

The top diagram is for the device shown in the question - note that it doesn't give you an "on resistance" curve for gate voltages less than 4.5V and even if you were running from a 5V supply you'd barely be able to supply 0.33A without dropping 0.5V.

The lower device (chosen at random from googling low Vgs FETs) is a different story. For a start it specifies "on" resistance down to a gate voltage of 1.5V and it's easy to see that at this gate voltage, the FET would be able to conduct 2A and drop 0.5V. On a 3.3V supply, the likely gate drive voltage would be 3V and clearly with a 0.5V drop you could conduct over 6A.

The 2nd device is much superior and you should be able to find plenty that are similar.

In all other respects (apart from the 4$\Omega$ resistor) the circuit is good. You'll need a pull-up from the open collector output so directly connect the O/C to the base of T1 and use the 10k as a pull up from this point. You should be able to drive the FET directly from where T1's collector is providing the MCU's supply is the same as the FET supply.