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I am having a circuit which I have tried to simulate and understand the working.

Circuit parameters :

  • Zener Breakdown voltage = 6.2V

  • P-MOSFET Vgs threshold voltage = 1.5V

  • Inrush current limiter = 3.3ohms

Below are my simulation efforts :

MOSFET Parameters :

MOSFET

ZENER Parameters :

ZENER

Current Limiter Resistor :

Current Limiter

There are other aluminium and electrolytic capacitors which are present. But for simple simulation circuit analysis I have ignored them.

My questions :

I have set the load current as 19.43mA using a 500Ohm Resistor.

  1. I have a Zener diode whose Breakdown Voltage is 6.2V. So, as you can see the MOSFET Vgs Threshold Voltage is -1.5V. So, My question is, if the Vgs threshold voltage of the MOSFET is only 1.5V, the voltage drop between the MOSFET gate and source should only be 1.5V right? But from the simulation we can see that the voltage between the MOSFET gate and source is around 6.01V. Why is this?

From my understanding of BJTs, the voltage drop between the transistor base and emitter would always be around 0.6V-0.7V typically or whatever voltage they have mentioned in the transistor datasheet.

So, isn't the case similar to the MOSFET? Should the Voltage between the MOSFET gate and source be the maximum threshold Vgs voltage(if the voltage across it, exceed Maximum Vgs threshold).

  1. So, I have used a 3.3Ohms current limiter resistor. I understand that Current limiting resistors are chosen to be a small value so that it does not dissipate high power during normal operating conditions.

But during the rise time of the power supply, how does a small value of resistor can help to reduce the high inrush current? Since, low resistance values do not offer much resistance to the current, right? So, how does a small resistance can help to reduce considerable inrush current during the initial turn on condition?

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So, isn't the case similar to the MOSFET? Should the Voltage between the MOSFET gate and source be the maximum threshold Vgs voltage

No, the MOSFET gate-source region is very high impedance and the voltage will freely rise to any voltage you choose to place across it (until the insulation breaks). The threshold voltage is that gate-source voltage that (usually) produces a current of 1 uA (dependent on data sheet) into the drain terminal.

But during the rise time of the power supply, how does a small value of resistor can help to reduce the high inrush current?

If drain current starts to become a significant quantity (say 1 amp for example), there will be 3.3 volts dropped across a 3.3 ohm resistor and that (in very simple terms) has the effect of lowering the gate-source voltage and shutting off the MOSFET. In more complicated terms, it means that the MOSFET acts as a current limiter and does the job of inrush limiting.

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  • \$\begingroup\$ Thank you. I understood your answer for my second question. But in your first answer, I understand in BJT voltage drop between base and emitter will be only around 0.6V-0.8V. But that won't be the case of MOSFET? The MOSFET will take the entire voltage whatever that is applied irrespective of its threshold voltage? Is the understanding correct? And what detemines the voltage drop across the MOSFET and the 3.3ohm resistor - The Load Current? \$\endgroup\$ – Newbie Apr 23 '20 at 9:23
  • \$\begingroup\$ infineon.com/dgdl/… - This is my MOSFET Datasheet. On page 2, Electrical Characteristics, it is mentioned that the MOSFET Gate to Source threshold voltage is betweev -1V to -2V for a drain current of 370uA. So, if the gate source voltage is only -2V, maximum drain current is only 370uA? For higher drain current, in the order of 100mA, what would be the gate to source voltage? There is no graph provided Vgs Vs Id. Where to find this information of Vgs corresponding to drain current? \$\endgroup\$ – Newbie Apr 23 '20 at 9:28
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    \$\begingroup\$ @Newbie look at "5 Typ. output characteristic" in the data sheet. That graph shows typical responses for gate-source voltages up to 10 volts. In your first comment all your questions I answer yes. You have understood! \$\endgroup\$ – Andy aka Apr 23 '20 at 9:31
  • \$\begingroup\$ Thank you very much for your answer \$\endgroup\$ – Newbie Apr 23 '20 at 9:34
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    \$\begingroup\$ Yes, you have to interpolate a bit but, you also have to remember that as drain current rises, gate-source voltage reduces and therefore this circuit has negative feedback and acts like a current regulator hence. The very current through the drain that you want to rise in order to safely power the load is the same current that acts to shut-down the gate source voltage. It's not straight-forward to analyse!! \$\endgroup\$ – Andy aka Apr 23 '20 at 9:53

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