load line AB is constructed as shown in Fig. This locates the second point A (OA = 3mA) of the load line on the collector current axis. When V CE = 0, I C = V CC/R C = 6V/2 kΩ = 3 mA. This locates the first point B (OB = 6V) of the load line on collector-emitter voltage axis as shown in Fig. What is the stability factor ? Solution : load line and determine the operating point. 3 (i) shows that a silicon transistor with β = 100 is biased by base resistor method. (ii) When R B is made equal to 50 kΩ, then it is easy to see that base current is doubled i.e. (i) Referring to Fig.2 (ii) and applying Kirchhoff ’s voltage law to the circuit ABEN, we get, It may be noted that negative terminals of the power supplies are grounded to get a complete path of current. Here, we need show only the supply voltages, + 2V and +9V. The same circuit is shown in a simplified way in Fig. 2 (i), biasing is provided by a battery V BB (= 2V) in the base circuit which is separate from the battery V CC (= 9V) used in the output circuit. (ii) If R B in this circuit is changed to 50 kΩ, find the new operating point. (i) Determine the collector current I C and collector-emitter voltage V CE . 2 (i) shows biasing with base resistor method. ∴ Base voltage (signal voltage) = Collector current / (5 mA/V )= 3 mA /( 5 mA/V) = 600 mV Q3. Fig. Now Collector current / Base voltage (signal voltage) = 5 mA/V allowed collector current, i C =12 V /R C = 12 V/ 4 KΩ = 3 mA allowed voltage across R C = 13 − 1 = 12 V What is the maximum input signal if β = 100 ? Given V knee = 1V and a change of 1V in V BE causes a change of 5mA in collector current. During the positive peak of the signal, i C = 1 + 1 = 2mAĪnd during the negative peak (point B), i C = 1 − 1 = 0 mA Q2. A transistor employs a 4 kΩ load and V CC = 13V.
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