FET basics

Yes, more negative. Junction fets are normally on devices that turn off as the gate side of the gate to channel junction becomes more reverse biased.

(neglecting any base current to TR3)

Yes, though the current is not so clearly 1 mA. If the base to emitter drop is about .6 volts, then there is .3 volts left across the 220 ohm emitter resistor, for a current of about 1.4 mA. But the exact value is not so important to the concept.

The source voltage will be pulled down by TR3 until it is just far enough more positive than the gate voltage, to allow TR1 to carry the 1mA. If the source voltage were more positive than that required amount, there would not be a mA passing through the 220 ohm emitter resistor, so the base current for TR3 would increase, and the collector voltage would be pulled down to a less positive voltage, turning TR1 on more. If the source voltage of TR1 were less positive than what would regulate the drain current to 1 mA, there would be more current passing through the 220 ohm emitter resistor and the base current to TR1 would decrease, causing its collector voltage to go more positive. So the circuit finds whatever positive (with respect to the gate) source voltage it takes to force TR1 to carry all the current passing through TR3. To calculate what voltage is, you would need a set of curves that show the drain current for many combinations of drain to source voltage at various gate to source voltages.

Assuming it is possible for the fet to carry the regulated current, the source voltage must fall somewhere between .9 and 3.1 volts.

In this case, the average current is regulated by TR3, while variations around that average are controlled by TR1.

OK, so the variations of Vs are applied on capacitor C5, which causes a current proportional to it's capacity, and this current in turn causes the voltage across R1 to vary. Is this correct ?

Thanks all, I think I understand better now :)

applied on capacitor C5, which causes a

It helps to think about AC (signal) and DC (average / bias) conditions seperately. Capacitors are effectively open-circuit to DC, and low-impedance to AC. Think of the signal as a small AC component superimposed on top of the steady-state DC average.

If it were not for C5, Vgs and Id would be constant, and the input signal would appear on Vs.

The amount of charge stored on C5 is related to the voltage across it by:

Q = CV

To change this voltage, you have to move charge. Current is the rate of change of charge:

I = dQ/dt

You need a big current for a short time, or a small current for a long time. At signal freqeuncies, we do not have a long time, and the voltage cannot change fast enough in response to cycles of input freqeuncy, so C5 keeps Vs fairly constant. This means the AC input signal appears across Vgs, and this forces Id to vary in sympathy.

Make that the variations of Is...

which cause very small voltage changes, inverse to its capacity (I=C*(dv/dt) so dv or change in voltage = dt*I/C or time * current / capacitance) across the capacitor

If the capacitor is large enough, it will look like a voltage supply at signal frequencies. In other words, the signal current, during a half cycle of the signal, will pass current through the capacitor without changing its voltage significantly. The next half cycle will pass the same current in the opposite direction, again not changing the voltage, significantly. And since AC averages zero DC current, those alternating insignificant changes will not add up to any net, long term change of capacitor voltage. During this signal current detour through the capacitor, TR3 holds to a steady current that is the average current through TR1.

The capacitor voltage it is initially defined by the DC bias point that TR3 and TR1 settle to. So, for signal analysis, just replace TR3 and the capacitor with a battery that has the DC voltage you measure (or calculate) at the source of TR1.

Sorry I misunderstood your explanation. What I did not get was that the capacitor acts as a secondary path for the current, and at the same time fixes Vs. I finally see how the current can vary.

Thanks. vic.