While simulating a basic common source JFET amp I found that eliminating the input capacitor improved the gain of the circuit dramatically. This is because without the input capacitor, the gate voltage is held to ground by the VOFF = 0 of the VSIN source. See figure 1.

No Input Cap

Figure above shows the good biasing with no input capacitor and Vg = 0

When the capacitor is included, this DC Offset VOFF is blocked by the input capacitor. So the gate voltage is allowed to float. Figure 2 shows that with the capacitor in place, the gate voltage Vg ~ 730mV. This changes the Q point of the JFET, putting it into the linear region and killing all good amplifier behavior.

Signal and Cap

Figure above shows Vg = 730mV with signal and cap

No Signal and Cap

Figure above shows Vg = 730mV with no signal and cap

This made me curious as to where this ~730mV are coming from. Eliminating the VSIN source entirely did not change Vg = 730mV. So the presence of this gate voltage is caused by the supply voltage. Since the JFET is a depletion-mode device, it will conduct with no gate voltage applied. So when the supply voltage is applied across the drain and source of the JFET, the JFET will conduct. Since the drain of the FET will be much higher than the Gate voltage, Vdg > 0, This is like a reverse bias between the drain (connected to N-type channel) and gate (connected to P type well). Reverse biasing this p-n junction increases the voltage barrier across the junction. So the gate voltage present Vg is caused by the voltage drop Vdg due to the reverse biased PN junction. This does not sit well with me and I would like to better understand what is happening.

I tried a transient simulation using a VPULSE supply to see how the circuit behaves if the supply voltage is gradually increased instead of being an instantaneously present VDC supply. This experiment (figure 3 + 4) showed that the gate voltage only went to 1.8uV. Much less than the expected 730mV. What is going on here? Why is the DC bias so different when the supply voltage is pulsed from 0 to 3 instead of being a constant DC source?

Pulsed Supply and Cap

Figure above shows pulsed supply with input cap does not produce the Vg = 730mV

enter image description here

Figure above shows gate voltage plot where Vg goes to 1.8uV. Not close to expected 730mV.

What is causing this behavior? What causes the gate to be biased at 730mV? Can anyone explain the DC bias being different for the VPULSE vs VDC supply? Thank you

  • \$\begingroup\$ The positive voltage is from the (small) gate-drain leakage of the device. Try stretching out your transient simulation to the 10ms - 1s range and see if the voltage rises. Alternatively, consider adding a resistor from gate to ground in the MEG-GΩ range and see what that does. \$\endgroup\$
    – W5VO
    Oct 28, 2021 at 15:38
  • 1
    \$\begingroup\$ This doesn't answer the OP, but that amplifier design has no input bias control. The gate is basically floating, so leakages and such will greatly affect it. \$\endgroup\$
    – Aaron
    Oct 28, 2021 at 15:41
  • \$\begingroup\$ The reason why I excluded a gate bias resistor is because I was finding that my 2.5mVp signal that I am trying to amplify was presented as 1.5uVp signal at the gate of the JFET when I connected a 10Mohm resistor between gate and ground. So gate resistors greatly reduce the Vgate/Vsig ratio. 100Mohm would bring me up to Vg = 15uVp but I really need the entire 2.5mVp to be at the gate. @W5VO \$\endgroup\$
    – javamang
    Oct 28, 2021 at 16:52
  • \$\begingroup\$ @Aaron Is there a way to add a resistor between gate and ground that would not cause the signal voltage to be greatly reduced at the gate? \$\endgroup\$
    – javamang
    Oct 28, 2021 at 16:54

1 Answer 1


These simulations are not very reliable without including parasitic properties. In reality both the capacitor and reversed gate junction will leak. And it is not clear what will leak more.

If you include a capacitor leakage of eg. 1 GOhm it could bias correctly again. However a gate bias resistor is almost mandatory when ac coupling through a capacitor. Something is wrong if a 10M gate resistor diminishes your AC signal to almost zero: Either your signal frequency is way below the RC corner frequency or your source impedance is way higher than 1 MOhm.

But if your signal is ground referenced then you could indeed just DC couple it.. that is how I do it for my active JFET guitar cable for example.


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