From Q1 the gain at low frequencies is -6.88. Therefore CM1 = (1 + 6.88) * 100f = 788 fF and Cy2 = 100f * (1 + 6.88)/6.88 = 114 fF At the input the overall capacitance is C1 = Cgs + CM1 1 RG RG jaC 1 Ro + R, || jac, 1+ j@R,C RG Ro + 1+ j@R,C, Vg =Vin Ro + Ro + j@R,R,C, 1 RG =V in Ro + R 1+ j@(R, || R,)C, Therefore 1 r, II R, V our =-8mV gs (r, I| R, I| R.) |I- joCM 24 =-8m' s 1+ j@(r, I| R, || R)CM 2 Vg Vout the overall voltage gain is V out Since G, V in %3D Vin Vgs 1 RG G,- 8m(r, I| R, || R) Ro + R. (1+ j@(R, || R,)C)(1+ j@(r, I| R, I| R)CM 2) Tutorial 1 Solutions: Page O D. R. S. Cumming & B. Choubey, University of Glasgow 13 e. Poles The high frequency response has two poles at 1 fpi = 2T(R, || R.)C and fr2 27(r, I| R, I|R,)CM 2 RG = 2.2k and r,llRpl|Rext = 134kl|5k||20k = 134k||4k = 4k C1 = 100f + 788f = 888 fF. Therefore: fpl = 260 MHz and fp2 = 279 MHz. (NB - would hope for transistors with better B, but this would have the effect of reducing fo1).

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Explain maths in both sections
From Q1 the gain at low frequencies is -6.88.
Therefore CM1 = (1 + 6.88) * 100f = 788 fF and Cy2 = 100f * (1 + 6.88)/6.88 = 114 fF
At the input the overall capacitance is C¡ = Cgs + CM1
1
RG
RG
jaC,
1
R. + R ||
joc,
1+ j@R,C
RG
Ro +
1+ j@R,C,
=V in
Ves = V in
Ro + Rg + j@R,R,C,
RG
1
=V in
Ro + R, 1+ j@(R, || R,)C,
Therefore
1
r, IIR,
V out =-8mV gs (r, I| R, I| R,) |I-
joCM2.
=-8m' s 1+ j@(r, I| R, || R)CM 2
g5
V out
Since G,
Vg Vout the overall voltage gain is
%3D
V in
V in Vgs
1
RG
G,- 8m(r, I| R, || R)
Ro + R. (1+ j@(R, || R,)C)(1+ j@(r, I| R, ||R)CM 2)
Tutorial 1 Solutions: Page
O D. R. S. Cumming & B. Choubey, University of Glasgow
13
e. Poles
The high frequency response has two poles at
1
fpi =
2T(R, || R.)C
and fr2
27(r, I| R, I|R,)CM 2
RG = 2.2k and r„|Rpl|Rext = 134k||5k||20k = 134k||4k = 4k
C1 = 100f + 788f = 888 fF.
Therefore:
fpl = 260 MHz and fp2 = 279 MHz.
(NB – would hope for transistors with better B, but this would have the effect of reducing fp1).
Transcribed Image Text:From Q1 the gain at low frequencies is -6.88. Therefore CM1 = (1 + 6.88) * 100f = 788 fF and Cy2 = 100f * (1 + 6.88)/6.88 = 114 fF At the input the overall capacitance is C¡ = Cgs + CM1 1 RG RG jaC, 1 R. + R || joc, 1+ j@R,C RG Ro + 1+ j@R,C, =V in Ves = V in Ro + Rg + j@R,R,C, RG 1 =V in Ro + R, 1+ j@(R, || R,)C, Therefore 1 r, IIR, V out =-8mV gs (r, I| R, I| R,) |I- joCM2. =-8m' s 1+ j@(r, I| R, || R)CM 2 g5 V out Since G, Vg Vout the overall voltage gain is %3D V in V in Vgs 1 RG G,- 8m(r, I| R, || R) Ro + R. (1+ j@(R, || R,)C)(1+ j@(r, I| R, ||R)CM 2) Tutorial 1 Solutions: Page O D. R. S. Cumming & B. Choubey, University of Glasgow 13 e. Poles The high frequency response has two poles at 1 fpi = 2T(R, || R.)C and fr2 27(r, I| R, I|R,)CM 2 RG = 2.2k and r„|Rpl|Rext = 134k||5k||20k = 134k||4k = 4k C1 = 100f + 788f = 888 fF. Therefore: fpl = 260 MHz and fp2 = 279 MHz. (NB – would hope for transistors with better B, but this would have the effect of reducing fp1).
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