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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?

enter image description here

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

2 Answers 2

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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.

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  • \$\begingroup\$ Do you think the MOSFET FDN360P will work? \$\endgroup\$
    – user27207
    Commented Aug 7, 2013 at 9:39
  • \$\begingroup\$ @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. \$\endgroup\$
    – radagast
    Commented Aug 7, 2013 at 9:43
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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: -

enter image description here

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.

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