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I'm trying to understand why the above circuit is designed the way it is, especially with regards to the Zener diode. I understand that the Zener diode is needed to act as a clamp to help keep the Vgs of the MOSFET within the typical 20 V maximum, but I'm not sure how exactly it accomplishes that. I am also confused on the need for the resistors and also whether the Vds of the MOSFET will stay at around 50 V since it is tied in parallel to the Zener. Would really appreciate any insights or clarifications. Thank you!

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  • \$\begingroup\$ One immediate thought that comes to mind is transient responses. Diodes have non-zero switching time. \$\endgroup\$
    – TLW
    Commented Oct 15, 2022 at 23:10
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    \$\begingroup\$ Do you know what a zener diode does? \$\endgroup\$
    – Hearth
    Commented Oct 15, 2022 at 23:11
  • \$\begingroup\$ Have you tried to build a test bench for Zener diodes in your circuit simulator of choice? \$\endgroup\$
    – winny
    Commented Oct 16, 2022 at 10:24

3 Answers 3

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Consider if MOSFET M2 is 'on' such that it is basically shorting R2 to ground.

R1 and R2 will form a voltage divider across the 50 V supply, Vin. Since the resistors are equal in value the voltage across each would be half of the supply, or 25 V.

The MOSFET M1 has a \$V_{gs}\$ rating of 20 V, so 25 V is going to exceed that rating.

The Zener is across R1. When the voltage across a Zener exceeds it's rated voltage it starts to conduct, and as long as there is a series resistance large enough to limit the current to within the Zener's ratings, the voltage across it will be roughly it's rated voltage. With a 16 V Zener M1's \$V_{gs}\$ will be kept within it's rating.

As for \$V_{ds}\$, that will be the voltage from \$V_{in}\$ to \$V_{out}\$ (it is NOT in parallel with the Zener). As drawn there is no DC path from the drain to ground, so it would be 0 V. We would assume that the external load will provide this path though.

So assuming a load on \$V_{out}\$, with M1 off (not conducting) M1 will act like a very high resistance, nearly an open cirtcuit, so \$V_{ds}\$ would be the supply voltage, 50 V. With M1 on (fully conducting) M1 acts like a very low resistance, nearly a short, so \$V_{ds}\$ will be close to 0 V.

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  • \$\begingroup\$ Your answer definitely cleared things up for me! Just one more confusion though: why is R1 needed for keeping M1 off? Since a 16V Zener is there and the voltage of the battery is 50V, when the NMOS is on, it will conduct and switch on the PMOS so I don't see how R1 is useful here. \$\endgroup\$
    – calvin g
    Commented Nov 3, 2022 at 1:39
  • \$\begingroup\$ The zener conducting is not what turns the M1 on, M1 is on when the gate is pulled low by M2. When M2 is off there's no current through the zener so it doesn't conduct, leaving the gate of M1 basically floating. The resistor is there to pull up the gate of M1 and turn it off. \$\endgroup\$
    – GodJihyo
    Commented Nov 3, 2022 at 13:31
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While your drawing does not show inductors, even the wires that connect components have some inductance. The present of inductance results in voltage spikes when the MOSFET opens and attempts to block the flow of current. These spikes can appear across the gate-to-source junction and, if they are large enough, they can damage the transistor's gate. This generally renders the MOSFET unusable.

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I understand that the Zener diode is needed to act as a clamp to help keep the Vgs of the Mosfet within the typical 20V maximum, but I'm not sure how exactly it accomplishes that.

When M2 is conducting, R2 will dissipate the extra voltage.

For instance, if M2 is turned on, the Zener will clamp the gate of M1 to 50-16 = 34V and there will be 34V across R2.

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  • \$\begingroup\$ Your answer definitely cleared things up for me! Just one more confusion though: why is R1 needed for keeping M1 off? Since a 16V Zener is there and the voltage of the battery is 50V, when the NMOS is on, it will conduct and switch on the PMOS so I don't see how R1 is useful here. \$\endgroup\$
    – calvin g
    Commented Nov 3, 2022 at 1:39
  • \$\begingroup\$ Because you want V_GS to go to zero when M2 is not conducting. WIthin M1 there is a small capacitor between the gate and source and this will hold V_GS unless it is drained. So R1 can be thought of as a bleeder resistor for this capacitance. It also weakly pulls up the gate to the source making V_GS = 0 unless M2 is conducting. \$\endgroup\$
    – ErikR
    Commented Nov 3, 2022 at 1:51

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