Question 4.6: The JFET shown in the circuit of Figure 4–15(a) has the tran...
The JFET shown in the circuit of Figure 4–15(a) has the transconductance curve shown in Figure 4–15(b). From the transconductance curve, determine V_{S} and I_{D} . Using this result, determine the value of V_{DS} .
Plot the line representing a 2.0 kΩ resistor by selecting the origin and any point on the resistor’s line. A convenient point is V_{GS} = – 4 V , I_{D} = 2 mA . The line between the origin and this point represents a 2.0 kΩ self-bias resistor as shown in Figure 4–15 (c)
The intersection of the 2.0 kΩ resistor and the transconductance curve represents V_{GS} and I_{D} for this case. Reading the plot, you can see that V_{GS} = -1.75 V and I_{D} is 0.85 mA. Since V_{G} = 0 V, V_{S} = + 1.75 V.The voltage across the drain resistor is found by Ohm’s law.
V_{R_{D} } = I_{D} R_{D} = (0.85 mA)(2.7 kΩ) = 2.3 V
To obtain the drain voltage, subtract the previous result from V_{DD} .
V_{D} = V_{DD} – V_{R_{D} } = 9.0 V – 2.3 V = 6.7 V
V_{DS} = V_{D} – V_{S} = 6.7 V – 1.75 V = 4.95 V
Interestingly, this result can be obtained graphically by load line analysis. The load line for the circuit is superimposed on the drain curves (from Figure 4–9 (b)) as shown in Figure 4–16 .
