Question

\(r_o = \infty, V_{cc} = 10V, R_{B1} = R_{B2} = 200k\Omega, R_E = 1k\Omega, R_C = R_L = 2k\Omega, v_s = 0.1V\) \* r_e = \frac{V_T}{I_E} = 25mV/I_E \quad r_\pi = \frac{V_T}{I_B} \quad r_\pi = (\beta + 1)r_e \quad g_m = \frac{I_C}{V_T}

          \(r_o = \infty, V_{cc} = 10V, R_{B1} = R_{B2} = 200k\Omega, R_E = 1k\Omega, R_C = R_L = 2k\Omega, v_s = 0.1V\)
\* r_e = \frac{V_T}{I_E} = 25mV/I_E \quad r_\pi = \frac{V_T}{I_B} \quad r_\pi = (\beta + 1)r_e \quad g_m = \frac{I_C}{V_T}

ro = ∞, Vcc = 10V, RB1 = RB2 = 200kΩ, RE = 1kΩ, RC = RL = 2kΩ, vs = 0.1V
re = (VT)/(IE) = 25mV/IE rπ= (VT)/(IB) rπ= (β+ 1)re gm = (IC)/(VT)

Added by Dustin R.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sri K

01/17/2023

Step 1

Step 1: The circuit is a common-emitter amplifier with an emitter resistor. Show more…

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Thanks for your feedback!

(r = ∞, VCC = 10 V, RB1 = RB2 = 200 kΩ, RE = 1 kΩ, RC = RL = 2 kΩ, vs = 0.1 V) * re = VT/IE = 25 mV/IE rπ = VT/IB rπ = (β + 1)re gm = IC/VT RB1 VCC Rout RC RL CB VC VB CC CE VE RB2 VS RE Rin