In the circuit shown below, the transistor has very large $β$ . Some measurements have been made for this circuit . The results are indicated in the figure. The value of voltage $V_{2}$ = ___V.

-2

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Solution

The correct option is A -2
For pnp transistor if $β$ is very large then

$I_{E} = I_{C}$

$I_{C} = \frac{- 4 - (- 10)}{24 k} = 2.5 m A = I_{E}$

Applying KVL,

$- 12 + (5.6 \times 2.5) + V_{2} = 0$

$V_{2} = - 2 V$

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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 .

V_{T} = 1 V a n d μ_{n} C_{o x} (\frac{ω}{L}) = 2 m A / V^{2} .

(V_{D S})

V_{D_{1}} = V_{D_{2}} = V_{D}

V_{B E (o n)} .

β

The value of voltage gain \frac{v_{0}}{v_{i n}} for the circuit is equal to ______V/V.

(Assume V_{T} = 25 mV and β to be very large)

| \frac{V_{0}}{V_{s}} |

V_{T} = 25 m V

β

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