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I'm trying to understand how this circuit operates. For this example lets say "From Battery Stack +" voltage is 42V. This circuit is designed to output about 4V at the source of the n-fet. I don't understand how that is possible.

I see 42V applied to both the drain of the n-fet and R1. R1 limits the current through the Zener so it can be rated relatively high resistance. The Zener is rated for 5.6V so 5.6V will be applied at the gate of the n-fet. 5.6V is above the threshold gate voltage of the fet which is about 2V so the n-fet will conduct.

Here is were my understanding is lost, if the n-fet is conducting and the voltage on the drain is 42V then wouldn't the source output of the n-fet be 42V as well? How could it be 4V?

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2 Answers 2

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Q1 outputs current to the load just as much as needed to lift the output voltage to 5.6V minus the treshold voltage of Q1 - more current would shut down the channel in Q1 because Vgs would be less than the treshold voltage.

The type of Q1 must be one which has treshold value of Vgs approximately 1.6V

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  • \$\begingroup\$ Thank you for your answer user287001 - "Q1 outputs current to the load just as much as needed to lift the output voltage to 5.6V minus the threshold voltage of Q1 " - how is this possible? Sorry but I must be missing some core fundamental here \$\endgroup\$
    – Tim51
    Nov 13, 2020 at 0:28
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    \$\begingroup\$ This is a search for balance situation - the circuit is a control system with feedback. If there were say 1kOhm load resistor between the cathode and the GND then at 4mA current the output voltage would be 4V. The voltage between g and s would be 1.6V If the current were say 5mA Vgs would be only 0.6V which copuldn't cause any current because opening the channel between d and s needs 1.6V. If the current were less than 4mA , say 3mA, Vgs would be 2.6V, the channel would be wide open and the current would be higher. The balance is found near Vgs=the treshold voltage which I assumed =1.6V \$\endgroup\$
    – user136077
    Nov 13, 2020 at 0:35
  • \$\begingroup\$ Ok that makes it crystal clear. I did not understand to apply a load value resistor between the source and ground (you said cathode, I assume that was a typo?) when looking at the behavior of the transistor. Seeing the output in terms of current makes it clear that with a low current at the source the voltage will be lower thereby closing the channel as the Vgs begins to drop below Vth. When the Vgs drops below the Vth the channel begins to close thereby raising Vgs which starts to re-open the channel. Fascinating. Thank you for your time @user287001! \$\endgroup\$
    – Tim51
    Nov 13, 2020 at 0:51
  • \$\begingroup\$ name cathode is a remnant, - a bad habit, in the datasheet there reads source.You can see it also elsewhere when someone who knows vacuum tubes writes something of a fet circuit. \$\endgroup\$
    – user136077
    Nov 13, 2020 at 0:54
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The transistor only knows about the voltages between its terminals - it doesn't know or care where you think "Zero Volts" is.

The transistor will only conduct when the gate is more than a couple of volts above the source, regardless of what the voltage is between source and ground.

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  • \$\begingroup\$ I understand the transistor will only conduct when the gate is more than a couple volts above the source. Right the transistor only cares about the voltages at its terminals so I shouldn't be referring to my "zero point". So how can I identify the voltages between its terminals? @peter bennett \$\endgroup\$
    – Tim51
    Nov 13, 2020 at 0:36
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    \$\begingroup\$ The Zener and R1 will set the voltage (relative to "Ground") at the gate. The characteristics of the transistor will then establish the voltage at its source. \$\endgroup\$ Nov 13, 2020 at 0:41

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