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JFET amplifier

The diagram above shows a very simple JFET amplifier. Let's make some DC calculations (no input signal). T1 is a BF245A. The datasheet tells us that the yfs is 3mA/V if -1<VGS<0. So we choose VGS = -0.5V. At that voltage, ID = 2.5mA. We want VOUT to be 6V, so RD = 6V/2.5mA = 2.4k.

Since VG = 0V and VGS = -0.5V, VS must be 0.5V. So RS = 0.5V/2.5mA = 200Ω.

Next, we connect a 0.1Vt sinus signal to the IN termial. What will be the the amplitude of the output signal? A 0.1V change in VGS causes a 0.1V∙3mA/V = 0.3mA change in ID and thus a 0.3mA∙2.4k = 0.8V change in the output signal. So the voltage gain is 8 (or actually -8; it's an inverting amplifier).

Capacitor CS makes sure that VS remains constant, so that vGS = vIN (for AC signals).

RI is usually 1M or so. It guarantees that VIN = 0V (DC), while the input resistance remains very high.

Just like the current gain (hFE) of a bipolar transistor may vary over a wide range, so may the forward transfer admittance of a JFET. In case of the BF245A, yFS may vary between 3 and 6.5mS. Far worse is the fact that the pinch-off voltage of a BF245A ranges from -0.25 to -8V. That means that at VGS=-0.5V, ID can be much less or greater than the 2.5mA mentioned above; that was just a typical value. Resistor RS is therefore often replaced by a current source.