# Open Circuit Voltage with Open FET and Zener?

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.

simulate this circuit – Schematic created using CircuitLab

• 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? Mar 24, 2020 at 0:22
• 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. Mar 24, 2020 at 17:16