L7Part2P5.159 P5.163 with PSpice Output L6_2

L7Part2P5.159 P5.163 with PSpice Output L6_2 - Problem...

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Problem 5.159 With PSPICE Output Zubair Rehman
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2 I E = 0.3 mA β = 120 r o = 300 k r x = 50 f T = 700 MHz C μ = 1 pF g m = 12 mS C π = 1.728 pF A m = -16.12 f H = 1.79 MHz f L = 175 Hz V E = 1.17 V V B = 2 V V C = 3.59 V V BE = 0.797 V V BC = - 1.623 V V CE = 2.42 V I E = 0.3 mA I B = 2.5 μ A I C = 2.975 μ A
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3 DC Analysis V E = I E R E = 0.3 mA x 39 k = 1.17 V I B = I E / ( β + 1) = 0.3 mA / 120 = 2.5 μ A V BE = V CC x {R 2 / (R 1 + R 2 )} = 5 V x 22 k / (33 k + 22 k) = 2 V R B = R 1 || R 2 = 13.2 k V B = I B R B + V BE + I E R E V BE = V B - I B R B - I E R E = 2 V - (2.5 μ A x 13.2 k) - 1.17 V = 0.797 V I C = I E - I B = 300 μ A - 2.5 μ A = 297.5 μΑ SPICE Parameter I S I S = I C e -VBE / Vt = 297.5 μ A e - 0.797/25.85mV I S = 1.212 E -17 A = 0.01212 fA Omit Spaces IS=0.01212f SPICE Parameter β F β F = β = 120 Omit Spaces BF=120 SPICE Parameter VAF VAF = I c r o = 0.3 mA x 300 k = 90 V Omit Spaces VAF=90 SPICE Parameter RB RB = r
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L7Part2P5.159 P5.163 with PSpice Output L6_2 - Problem...

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