Question 11.4: Consider the following situation in the FET series gate, Vs ......
Consider the following situation in the FET series gate,\ V_{s} = ± 2 V,\ R_{s} = 100 Ω,\ R_{L} = 10 kΩ. The FET has the following parameters:
\ V_{GS(OFF)max} = −10 V and\ R_{D(ON)} = 20 Ω
Calculate the voltage levels of the control signal,\ I_{D}, error due to\ R_{S} and error due to\ R_{D(ON)}.
The control signal should have a value\ V_{1} for\ Q_{1} to be ON.
Therefore,\ V_{1} = V_{s(peak)} = 2 V
For\ Q_{1} to be OFF, the control signal should be\ –V_{2}:
\ −V_{2} = −(V_{s(peak)} + V_{GS(OFF)max} + 1V)
\ −V_{2} = −(2 + 10 + 1) = −13 V
\ I_{D} = \frac{V_{S}}{R_{S} + R_{D(ON)}+ R_{L}} = \frac{2 V}{100 + 20 + 10000} = 197.4 μA
\ I_{D}R_{S} = 197 × 10^{−6} × 0.1 × 10^{3} = 19.74 mV
Error due to\ R_{S} = \frac{19.74 mV}{2V} × 100 % = 0.987 %
\ I_{D}R_{D(ON)} = 197.4 μA × 20 Ω = 3.948 mV
Error due to\ R_{D(ON)} = \frac{3.948 mV}{2} × 100 % = 0.197 %