In the transistor amplifier circuit shown in the figure below- the transistor has the following parameters - - dc - 60- VBE - 0-7 Vhis- - hfc- -The capacitance CC can be assumed to be infiniteUnder the dc conditions- the collector-to-emitter voltage drop is V-

Applying KVL in base emitter loop, 12 I_ER_C I_BR_f 0.7 = 0 12 0.7 = I_E. 1 k+I_E(β +1).53 k I_E = 11.3 ( 1+5361 )k = 11.31.87k = 6.046 mA ≃ nbsp;6 mA V ...

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Byju's Answer

Standard XII

Physics

Transistor as an Amplifier

In the transi...

Question

In the transistor amplifier circuit shown in the figure below, the transistor has the following parameters :

$β_{d c} = 60,$ $V_{B E} = 0.7 V$
$h_{i s} \to \infty,$ $h_{f c} \to \infty,$

The capacitance $C_{C}$ can be assumed to be infinite

Under the dc conditions, the collector-to-emitter voltage drop is ___V.
6

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Solution

The correct option is A 6
Applying KVL in base - emitter loop,

$12 - I_{E} R_{C} - I_{B} R_{f} - 0.7 = 0$

$12 - 0.7 = I_{E} . 1 k + \frac{I_{E}}{(β + 1)} .53 k$

$I_{E} = \frac{11.3}{(1 + \frac{53}{61}) k} = \frac{11.3}{1.87 k} = 6.046 m A ≃ 6 m A$

$V_{C E} = V_{C C} - I_{E} R_{C}$

$= 12 - 6 m A \times 1 k = 6 V$

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In the transistor amplifier circuit shown in the figure below, the transistor has the following parameters : βdc=60, VBE=0.7V his→∞, hfc→∞, The capacitance CC can be assumed to be infinite Under the dc conditions, the collector-to-emitter voltage drop is ___V.6

The correct option is A 6Applying KVL in base - emitter loop, 12−IERC−IBRf−0.7=0 12−0.7=IE.1k+IE(β+1).53k IE=11.3(1+5361)k=11.31.87k=6.046mA≃6mA VCE=VCC−IERC =12−6mA×1k=6V