15 V Dmeas VDS 2N4416 GS 115 kJ2 Rsmeas b meas Figure 133

15 V D(meas) = VDS 2N4416 GS+ 1.15 kJ2 Rs(meas) =-b (meas) = Figure 13-3 Self-bias circuit.

Solution

    Since no gate current flows through the reverse-biased gate-source, the gate current I_G = 0 and, therefore:

V_G = I_G R_G = 0

    With a drain current I_D the voltages at S and D are:
V_S = I_D R_S = I_D (1.2k)

V_{D =} V_DD – I_D (R_D) = 15 – I_D (1k)

    The gate-source voltage is then:

V_GS = V_G – V_S = 0 – I_D R_S = – I_D R_S = – I_D (1.2k)

    So voltage drop across resistance R_S provides the biasing voltage V_Gg and no external source is required for biasing and this is the reason that it is called self-biasing.

    The operating point (that is zero signal I_D and V_DS) can easily be determined from equation and equation given below :

V_{DS =} V_DD – I_D (R_{D +} R_S) = 15 – I_D (1k ₊ 1.2k) = 15 – I_D (2.2k)