Question

For the amplifier of given figure, $I_E = 1.3 mA$, $R_E = 2k\Omega$, $R_C = 500\Omega$, $V_T = 26mV$, $\beta = 100$, $V_{CC} = 15V$, $V_S = 0.01 \sin(\omega t)$ and $C_E = C_C = 10\mu F$ a- what is the small-signal voltage gain $A_V = \frac{V_O}{V_S}$? b. What is the approximate $A_V$ if $C_E$ is removed? c. what will $V_O$ be if $C_C$ is short circuited?

          For the amplifier of given figure,
$I_E = 1.3 mA$, $R_E = 2k\Omega$, $R_C = 500\Omega$,
$V_T = 26mV$, $\beta = 100$, $V_{CC} = 15V$,
$V_S = 0.01 \sin(\omega t)$ and $C_E = C_C = 10\mu F$
a- what is the small-signal voltage
gain $A_V = \frac{V_O}{V_S}$?
b. What is the approximate $A_V$ if
$C_E$ is removed?
c. what will $V_O$ be if $C_C$ is short
circuited?

For the amplifier of given figure,
IE = 1.3 mA, RE = 2kΩ, RC = 500Ω,
VT = 26mV, β = 100, VCC = 15V,
VS = 0.01 sin(ω t) and CE = CC = 10μ F
a- what is the small-signal voltage
gain AV = (VO)/(VS)?
b. What is the approximate AV if
CE is removed?
c. what will VO be if CC is short
circuited?

Added by James H.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Darshan Maheshwari

Step 1

To find the small-signal voltage gain, we can use the formula: Av = -gm * (R2 || Rp) Where R2 || Rp is the parallel combination of R2 and Rp. Show more…

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For the amplifier of the given figure, I = 1.3mA, R2 = 2kÎ©, Rp = 500Î©, Vbe = 26mV, Vcc = 15V, Î² = 100, gm = 3mS, and f0 = 500Hz. a- What is the small-signal voltage gain? b- What is the approximate output resistance? c- What will be the phase shift at 45Â° for a frequency of 9kHz?