Consider the circuit shown in the figure below.

The above silicon transistor is biased at operating point Q(7V, 1 mA) and with $V_{c c} = 12 V .$ If the value of $β = 100$ , then the value of stability factor $S (I_{c o})$ is _________

56.75

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Solution

The correct option is A 56.75

Given, $V_{C E} = 7 V$ and $I_{C} = 1 m A$
$∴ R_{C} = \frac{V_{c c} - V_{C E}}{I_{C}} = \frac{12 - 7}{1 \times 10^{- 3}} = 5 k Ω$
now, $I_{B} = \frac{I_{c}}{β} = \frac{1 \times 10^{- 3}}{100} = 10 μ A$
$R_{B} = \frac{V_{c c} - V_{B E} - I_{C} R_{C}}{I_{B}} = \frac{12 - 0.7 - 1 \times 10^{- 3} \times 5 \times 10^{3}}{10 \times 10^{- 6}} = 630 k Ω$
$S = \frac{1 + β}{1 + β [\frac{R_{C}}{R_{C} + R_{B}}]} = \frac{1 + 100}{1 + 100 [\frac{5 \times 10^{3}}{(5 + 630) \times 10^{3}}]} = 56.50$

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V_{B E (O N)}

S (I_{C O})

C o n s i d e r t h e c i r c u i t s h o w n i n t h e f i g u r e b e l o w .

T h e t r a n s i s t o r i s b i a s e d s u c h t h a t t h e v a l u e o f g_{m} = 2 m A / V, t h e n t h e v a l u e o f o u t p u t r e s i s t a n c e R_{0} i s e q u a l t o . . . K Ω . [A s s u m e v a l u e o f λ = 0.1 V^{- 1}]

β = β_{1} = β_{2} = 100, β_{3} 99

V_{B E (o n)} = 0.7

I_{0} = 100 m A

R_{1}

Ω

R_{B} (in k Ω)

I_{C} = 1 mA, V_{C E} = 3 V, V_{B E} = 0.5 V, V_{C C} = 12 V

β = 100

c o n s i d e r t h e c i r c u i t s h o w n i n t h e f i g u r e b e l o w . t h e v a l u e o f C_{1}, C_{2} a n d C_{c} i s v e r y l a r g e a n d t h e t r a n s i s t o r i s b i a s e d s u c h t h a t t h e v a l u e o f g_{m} = 20 m A / V λ = 0.01 V^{- 1} I f t h e M O S t r a n s i s t o r h a s t h e n t h e v a l u e o f v o l t a g e g a i n \frac{V_{0}}{V_{s}} i s e q u a l t o . . . . . V / V .

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