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L15Part2 BJT Cascode High Freq Open Circuit Time Constants_2

# L15Part2 BJT Cascode High Freq Open Circuit Time...

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6.8 Cascode Open-Circuit Time Constants Ref: L11 Part 2 Slide 9 R10 & R20 Zubair Rehman EE 311 Lecture 15 & EEO 311 Lecture 14 Part 2

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2 Cascode Amplifier
3 High Frequency Response Method of Open-Circuit Time Constants I C1 = I E2 = 10 mA β = 60 r x1 = 50 r x2 = 0 C π = 195 pF C μ = 5 pF R s = 50 R L = 1 k ω h = 2 π f h = 1 / Σ T jo Σ T jo = R 10 C π1 + R 20 C μ1 + R 30 C π2 + R 40 C μ2 R 10 = R π = r π || [(r x + R b + R e ) / (1 + g m R e )] e x b e b C e x b R r r R R r R r R R r r R R R 129 + + + + + + 129( + + + 129] + + + = = β β β π π π π μ ( ) ( ( [ || ) ( 20

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4 r e = V T / I E = 25 mV / 10 mA = 2.5 r π = ( β+129 r e = (60+1) * 2.5 = 152.5 V A = infinity r o = infinity V A = 200 V r o = V A / I C1 = 200 V / 10 mA = 20 k Q1 CE R e1 = 0 R b1 = R s = 50 R c = R in,Q2 = R in,CB = (r π2 ) / ( β 2 +129 = r e2 = 2.5 R 10 = R π1 = 152.5 || [(50+50+0) / 1] = 152.5 || 100 = 60.4 R 10 C π1 = 60.4 * 195 pF = 11778 ps = 11.8 ns R 20 = R μ1 = (50+50) || 152.5 + {2.5*[152.5 + 61(50+50+0)] / [50+50+152.5+0]} R 20 = 60.4 + 61.9 = 122.3 R 20 C μ1 = 122.3 * 5 pF = 611.5 ps = 0.6 ns Parameters I C1 = I E2 = 10 mA β = 60 r x1 = 50 r x2 = 0 C π = 195 pF C μ = 5 pF R s = 50 R L = 1 k
5 Q2 CB R b2 = 0 R c2 = R L = 1 kΩ R π = r π || [(r x + R b + R e ) / (1 + g m R e )] R e2 = R outQ1 = r 01 = infinity or 20 k R π2 = R 30 = 152.5 || [(0+0+infinity) / (1+g m2 *infinity)] = 152.5 || 1/g m2 R π2 =152.5 || r e2 = 152.5 || 2.5 = 2.46 r 01 = infinity r 01 = 20 k R π2 = R 30 = 152.5 || [(0+0+20k) / (1+g

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