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D2 is a 18V Zener, LM3Z18VT1, diode. M1 is a NCE01P30K. Node DO is 8V by default and 0V when the circuit activates.

If M2 in the circuit below opens, what is the voltage at the gate of M1? Now because there is no current flow through M2 I would assume that M1's gate would 18V due to the Zener. But something doesn't seem right here. Any help is greatly appreciated.

Thank you for your time.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ DO has a pull down resistor, yet you state that 8V is the default, which would require a pull up resistor. Plus if DO is ZERO volts, that means that your GATE = SOURCE, ie it's shutdown. What are you trying to do here? Is this all simulation? Or do yo have a physical circuit? \$\endgroup\$
    – GB - AE7OO
    Mar 24, 2020 at 0:22
  • \$\begingroup\$ I have a physical circuit that I just wanted to understand a bit better. DO is actually the output an IC. That output in default is 8V and when active is 0V. So normal operation has M2 closed. \$\endgroup\$
    – Tim51
    Mar 24, 2020 at 17:16

2 Answers 2

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The voltage would be 42V because it is a dead-end so no current flows which means no current flows through R1 resistor which means no voltage drop across a resistor which means voltage on both sides are the same.

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  • \$\begingroup\$ Do you mean 42V? @DKNguyen \$\endgroup\$
    – Tim51
    Mar 23, 2020 at 21:19
  • \$\begingroup\$ @Tim51 Corrected. \$\endgroup\$
    – DKNguyen
    Mar 23, 2020 at 21:19
  • \$\begingroup\$ Thank you for the clarification. @DKNguyen \$\endgroup\$
    – Tim51
    Mar 23, 2020 at 21:21
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Now because there is no current flow through M2 I would assume that M1's gate would 18V due to the Zener.

When the current through M2 is nearly zero, the current through the zener cannot be larger, because (neglecting leakage currents): I(M2DS)=I(R2)=I(R1)+I(D2)

At (almost) zero current through the zener, the voltage across the zener is not 18V but (almost) 0V.
Check the graph below, 4th quadrant.

enter image description here

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