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I tried to simulate a push-pull output stage in this configuration:

Push-Pull Switch opened

Here with the closed switch:

Push-Pull Switch closed As you can see, the output voltage is very low. I expected 3.3 V and 250 mA at the output.

Shouldn't the output be 3.3 V? Am I missing something here?

(The 13.2 ohm resistor simulates a load that draws 250 mA.)

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  • \$\begingroup\$ You will need pull-down for Q1 and pull-up for Q2 to make them turn off. \$\endgroup\$
    – winny
    Jun 2, 2022 at 9:27

5 Answers 5

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I expected 3.3V and 250mA at the output. Shouldn't the Output be 3.3V? Or am I missing something here?

Your "amplifier" is inverting because it uses transistors in a common source configuration hence, if you want 3.3 volts out you need to pull the input down to 0 volts (this activates Q2). Remember also that a MOSFET has a non-zero "on" resistance hence you won't quite achieve 3.3 volts at the output when loaded.

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The closed-switch operation is exactly as expected: Q1 is conducting because of the positive gate voltage compared to source at the ground rail.

The open-switch operation is the "kill me" mode of MOSFETs: leaving a gate floating not just gives you undefined behavior due to the very high gate isolation that makes the voltage across it a history of what went before. It also makes the transistor very susceptible to static charges that will kill it. Discrete MOSFETs may get delivered with a wire threaded through them that will protect them until they are soldered into the circuit.

You likely want a pull-down resistor on the common gates: otherwise once the switch has been closed once, not much will change because of the gate charge not being able to drain anywhere.

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  • \$\begingroup\$ Thanks for your Response. i've added a 10k Pull-Down Resistor to the Gates. The Result remained the same. 109 uA and 1.44 mV at the Output. \$\endgroup\$ Jun 2, 2022 at 12:57
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You don't state your chosen transistors. The current outside of the cutoff and saturation regions will be the amount of voltage in excess of the threshold voltage times the transconductance of the MOSFET. Whether 3.3V are sufficient for that purpose depends both on transistor model as well as specimen (there tends to be significant variation even given a particular model).

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S1 is not configured as it should be. You probably want to drive the gates of the FETs to 0 and to 3.3 V. When S1 is open, the gate V on the FETs is not defined (not controlled) -- you need a resistor (e.g. 1 MΩ) from the gates to GND to control the voltage in this condition.

Your FETs are not appropriately sized. For 99 % of circuit applications, there is no need to make individual FETs simultaneously both wider and longer than the minimum size. Your FETs have an on-resistance of about 30 kΩ which can't drive the 13 Ω load. You should make the FETs with L = 1 u (or smaller; depends on the the technology you are using). Even then you won't fully drive the load (you need FETs of about 1 Ω) -- so make the width about 300*100 = 30000 um.

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Thanks for all of your answers!

when i added the Pull-down resistor to the Gates, it still didn't work. Then i got suspicious about the MOSFET part. The MOSFET i was using, was a virtual MOSFET from Multisim. This virtual MOSFEt had no function. I resolved the problem by replacing the MOSFETs and adding the pull-down to the gates.

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