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 :
ZENER Parameters :
Current Limiter Resistor :
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.
- 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).
- 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?