Using the circuit below, I have measured Vd and Vs for two different MPF102 FETs. The supply voltage was 9.54V and the results are:

MPF102 #1, Vd = 4.64V and Vs = 1.80V
MPF102 #2, Vd = 5.78V and Vs = 1.38V

Is it possible to calculate what these values would be if the supply voltage was exactly 9.0V? If so, please show the math.

Thanks.

 

The source voltage would be the same because it is set by the idle current of the transistor. The drain voltage would decrease by .54 volts because the voltage drop across R3 will stay the same.

1.8/560 = .0032143 amps
.0032143 x 1500 = 4.8214V
9.54-9 = .54

 

The source voltage would be the same because it is set by the idle current of the transistor. The drain voltage would decrease by .54 volts because the voltage drop across R3 will stay the same.

1.8/560 = .0032143 amps
.0032143 x 1500 = 4.8214V
9.54-9 = .54

That's true for an ideal transistor. In reality the current will change very slightly due to the finite output impedance of the transistor. But to simplify the calculation for practical purposes the output impedance is often assumed to be infinite.

According to this data sheet the output conductance is 200μmhos (5kΩ resistance) at zero Vgs and 15V Vds.

 

If you showed your math I would be educated by it.

 

That's true for an ideal transistor. In reality the current will change very slightly due to the finite output impedance of the transistor. But to simplify the calculation for practical purposes the output impedance is often assumed to be infinite.

According to this data sheet the output conductance is 200μmhos (5kΩ resistance) at zero Vgs and 15V Vds.

If you showed your math I would be educated by it.

Me too. I am trying to understand JFETs and would be very interested. Thanks to you both.

 

jfets are basically transistors that are born "on". This idle current is the major hurdle to designing with them. Each jfet will have different idle current, even in the same part number. You can buy jfets according to their idle current range from less than a milliamp to hundreds of milliamps. The control terminal (gate) basically leads to a diode junction which will come on if you apply the wrong polarity. The right way to use jfets in an analog way is to apply voltage to the gate that makes them use less than their idle current. They are also used as digital switches.

There. That's a start.

 

If you showed your math I would be educated by it.

The math is not simple since the actual current is a function of both the transistor gain and the output conductance. If I had to solve the problem, I would just simulate it and let Spice do the calculations.

But an approximate calculation can be done by noting that, due to the inherent negative feedback from the source to the grounded gate, the source resistor effective value (as seem by any current though the transistor conductance) is increased by the transistor gain (the proof of this is left as a exercise for the reader since my brain isn't up to it. ). Thus for the minimum subject transistor gain (Id/Vgs) of 2000μmho, for example, the effective resistance of the 560Ω source resistor would be 560/2000u = 280kΩ. This gives a change in transistor current for the subject 0.56V change in battery voltage of 0.56/(280k + 5k + 1.5k) = 1.95μA, a very small increase.

 

That explains why I've gotten away with not knowing that math for all these years.
Tracecom can probably ignore it, too.
The difference amounts to less than 1/10th of 1%
However, when you get the drain voltage less than 2 volts from the source voltage, you start to seriously starve the jfet and the current comes tumbling down.
There is probably a big word for that but I don't know what it is.