How to calculate gate resistance, Rg in the JFET amplifier circuit?
And What is the function of Rg?
Rg is a pull-down resistor and also determines the input impedance of our amplifier.
Because the input resistance of the FET itself is very high (many megohms) then the gate resistor is added to allow the coupling capcitor at the input to charge up and discharge in a reasonable time but not fast enough to cause loss of signal.
So, if the input was steady at zero volts and was suddenly increased to 3 volts, the capacitor will pass this change to the gate of the FET because it doesn't (or shouldn't) have time to charge up through the resistor Rg.
But if the input stayed at 3 volts, then the capacitor needs to charge up to 3 volts via Rg so that the voltage across Rg again became zero until any change came along.
The resistor value can be calculated by making it large compared with the reactance of the capacitor at the lowest frequency you want to pass to the gate of the FET.
For example, if the capacitor was 0.01 uF and you wanted to pass a frequency of 300 Hz, the capacitor would have a reactance of 53000 ohms. So, you might make Rg twenty times that, or 1 Megohm.
The capacitor could still charge up in 10 mS, but a 300 Hz signal would have changed polarity several times before that happens.
However, a more permanent change in the input voltage will cause the capacitor to charge up and the voltage on the gate of the FET to return to zero.
If Rg were not there, then the gate of the FET would be dragged to the DC input voltage for many minutes and would probably give no output due to incorrect bias conditions.
Thx vk6kro for a brief explanation and Jony130! ! :)
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