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I am replacing a 2N2222 BJT inverter circuit with a MOSFET. The circuit needs to feed into the CW and CCW direction pins on a motor driver board, setting one high and then the other at 8-10kHz. So the switching needs to be clean and not introduce much of a delay in inverted half of the waveform that completes the cycle. But I worry the wrong MOSFET with a high RDS will do the same thing as the transistor with its .7V drop and switching delay.

The input signal is a digital pulse train and I would like clean 50us and 100us pulses where the one part of the pulse goes to the non-inverted input and then hopefully an equal pulse goes to the other completing each cycle. The input is a 5V signal from a microcontroller and the output is logic level into another high impedance circuit in the motor driver. Not being familiar with the popular and cheap MOSFETS, I looked at the 2N7000 and BS170, FQP30N06L, I wondered about switching time and anything else I need to think about for a good design. Do I need a GS resistor? Here is my sample circuit. In the simulator, I made the voltage 2.5V with a 2.5V DC offset to get a 0-5V pulse:

MOSFET Inverter

And the modeled waveform looks like this. Orange is the input pulse and blue is the inverted pulse. The switching time seems to be just 1 or 2 microseconds, but there it a small clipped off corner in the rising edge of the inverted pulse. What might cause that?:

Switching waveform

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  • \$\begingroup\$ If the 2N2222 circuit isn’t working then I’d suggest you look at resolving that first as it should work. I don’t see how going to a mosfet is going to make it any better. \$\endgroup\$
    – Kartman
    Mar 2, 2022 at 6:00
  • \$\begingroup\$ A general-purpose transistor doesn't have enough delay to matter to any normal motor control function. \$\endgroup\$
    – Whit3rd
    Mar 2, 2022 at 6:59
  • \$\begingroup\$ Can you edit your question to show the existing circuit with transistor input drivers and output loads. As others have said, that should work so let's understand its failings. Only then can we understand what any modifications or replacement would have to solve. \$\endgroup\$
    – TonyM
    Mar 2, 2022 at 9:28
  • \$\begingroup\$ The transistor circuit does work just fine. I was wanting to replace it with a MOSFET. I know it works using an IC like a 74LS04 as well. Thank you. \$\endgroup\$
    – FDecker
    Mar 5, 2022 at 1:46

1 Answer 1

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The MOSFETS pulls the output down with an RDS(on) of about 2 Ω; the resistor pulls the output up with 1 kΩ. There are parasitic capacitances at the output (the transistor itself, and the cable) that will take longer to be (dis)charged, so the rising edge will be slower.

The corner looks clipped because the simulation did not use smaller time steps. The actual waveform would be more rounded.

For a 10 kHz signal, the slower edge probably does not matter. But if you care, you can replace R with a P-channel MOSFET to get a CMOS inverter that makes both edges equally fast. In practice, you would not build such an inverter from discrete transistors but simply use an integrated logic gate like the (SN)74AHC1G14.

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