Question 4.7: Assume you are troubleshooting the JFET shown in Figure 4–18...
Assume you are troubleshooting the JFET shown in Figure 4–18 . You do not know the transconductance of the transistor, but you need to fi nd out if the circuit is working properly.
(a) Estimate the expected V_{G} and V_{S} .
(b) Assume you measured the source voltage and found it was +5.4 V. Is the circuit functioning as expected? Based on this measurement, what is the drain voltage?
(c) Assume you replace the transistor. The measured source voltage for the new transistor is +4.0 V. From this measurement, what is the expected drain voltage?
(a) Start with V_{G} because this value can be computed accurately and quickly with the voltage-divider rule. The gate voltage is about one-fourth of V_{DD} as shown in the following equation:
V_{G} = \left(\frac{R_{2} }{R_{1} + R_{2} } \right) V_{DD} = \left(\frac{330 k\Omega }{1.0 M\Omega + 330 k\Omega } \right) 12 V = 2.98 V
You know immediately that if the circuit is operating properly, the source voltage must be more positive than this value. Your estimate of the source voltage should be about +4 V.
(b) The measured value of +5.4 V may indicate a problem. This is larger than the expected 4 V and is nearly half of V_{DD} . Since R_{D} is even larger than R_{S} , it should drop even more of the total voltage. A quick check of V_{DS} shows that it is 0 V! This confirms a problem with the circuit; there is probably a drainto-source short in the transistor, causing V_{D} to also be 5.4 V.
(c) The drain current for the new transistor can be found by applying Ohm’s law to the source resistor.
I_{D} = \frac{4.0 V}{1.8 k\Omega } = 2.2 mA
Subtracting V_{RD} from V_{DD} gives V_{D} .
V_{D} = V_{DD} – I_{D} R_{D} = 12 V – (2.2 mA)(2.2 kΩ) = 7.16 V