Question

(a) Figure Q5(a) is a BJT amplifier circuit which has an infinite value of AC collector resistance, $r_o$ (or $r_c$) with $r_e = 28.48 Omega$ and $A_v = -72.91$; 14 V $C_{wi} = 5 ext{ pF}$ $C_{bc} = 12 ext{ pF}$ $C_{wo} = 8 ext{ pF}$ $C_{be} = 40 ext{ pF}$ $C_{ce} = 8 ext{ pF}$ $V_i$ $5.6 ext{ k}Omega$ $68 ext{ k}Omega$ $0.47 mu F$ $eta = 120$ $0.47 mu F$ $3.3 ext{ k}Omega$ $V_i$ $10 ext{ k}Omega$ $1.2 ext{ k}Omega$ $20 mu F$ $0.47 mu F$ Figure Q5(a) (i) Determine the high cut-off frequencies, $f_{Hi}$ and $f_{Ho}$. (8 marks) (ii) The internal capacitors influenced the high frequency response. In your opinion, how to minimize the effect of internal capacitors?

          (a) Figure Q5(a) is a BJT amplifier circuit which has an infinite value of AC collector
resistance, $r_o$ (or $r_c$) with $r_e = 28.48 Omega$ and $A_v = -72.91$;
14 V
$C_{wi} = 5 	ext{ pF}$  $C_{bc} = 12 	ext{ pF}$
$C_{wo} = 8 	ext{ pF}$  $C_{be} = 40 	ext{ pF}$
$C_{ce} = 8 	ext{ pF}$
$V_i$
$5.6 	ext{ k}Omega$
$68 	ext{ k}Omega$
$0.47 mu F$
$eta = 120$
$0.47 mu F$
$3.3 	ext{ k}Omega$
$V_i$
$10 	ext{ k}Omega$
$1.2 	ext{ k}Omega$
$20 mu F$
$0.47 mu F$
Figure Q5(a)
(i) Determine the high cut-off frequencies, $f_{Hi}$ and $f_{Ho}$.
(8 marks)
(ii) The internal capacitors influenced the high frequency response. In your
opinion, how to minimize the effect of internal capacitors?

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(a) Figure Q5(a) is a BJT amplifier circuit which has an infinite value of AC collector
resistance, ro (or rc) with re = 28.48 Omega and Av = -72.91;
14 V
Cwi = 5 ext pF Cbc = 12 ext pF
Cwo = 8 ext pF Cbe = 40 ext pF
Cce = 8 ext pF
Vi
5.6 ext kOmega
68 ext kOmega
0.47 mu F
eta = 120
0.47 mu F
3.3 ext kOmega
Vi
10 ext kOmega
1.2 ext kOmega
20 mu F
0.47 mu F
Figure Q5(a)
(i) Determine the high cut-off frequencies, fHi and fHo.
(8 marks)
(ii) The internal capacitors influenced the high frequency response. In your
opinion, how to minimize the effect of internal capacitors?

Added by Cynthia S.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

04/14/2023

Step 1

The Miller capacitance is given by: $C_{M} = C(1 - A_v)$ For the input capacitance, we have: $C_{Mi} = C_{wa}(1 - A_v) = 5\,\text{pF}(1 - (-72.91)) = 365.55\,\text{pF}$ For the output capacitance, we have: $C_{Mo} = C_{wo}(1 - A_v) = 8\,\text{pF}(1 - Show more…

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Figure Q5(a) is a BJT amplifier circuit which has an infinite value of AC collector resistance, ro (or rc) with re = 28.48 ̩ and Av = -72.91; Cwi = 5 pF Cbc = 12 pF Cwo = 8 pF Cbe = 40 pF Cce = 8 pF ̢ = 120 Figure Q5(a) (i) Determine the high cut-off frequencies, fHi and fHo. (ii) The internal capacitors influenced the high frequency response. In your opinion, how to minimize the effect of internal capacitors?

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Transcript

00:01 Hello students we are given the following figure.

00:04 The high frequency cutoff can be given as fhi is equal to 1 upon 2 pi rhi into ci.

00:15 Let's say this is equation 1.

00:18 Now this ci will be equal to cwi plus cbe plus 1 minus av into cce.

00:31 Substituting values 5 plus 40 plus 2956 .4 pico farad or ci is equal to 3000 pico farad.

00:46 Now rhi is equal to r1 in parallel with r2 in parallel with re into beta plus 1 which is equal to 68 kilo ohm is in parallel with 10 kilo ohm is in parallel with 28 .48 ohms into 121 or 68 kilo ohms is in parallel with 10 kilo ohms is in parallel with 3 .44 kilo ohms.

01:25 This is equal to 68 into 10 upon 68 plus 10 in parallel with 3 .44 which is equal to 8 .71 in parallel with 3 .44.

01:40 Calculating this we get the final value of rhi equals to 2 .46 kelvin.

01:48 Thus substituting values in 1 we get fhi equals to 1 upon 2 pi into 2 .46 into 10 to the power 3 multiplied by 3 into 10 to the power minus 9.

02:07 Solving this we get fhi equals to 21 .57 kilo hertz.

02:20 Now to find the low cutoff frequency we use the formula fho equals to 1 upon 2 pi rho co.

02:31 Let's say this is equation 2.

02:34 Here again co will be equal to cwo plus cce plus 1 minus 1 upon av into cbc which is equal to 8 plus 8 plus 1 minus 1 upon minus 72 .91 into 12 picofarad.

03:01 Solving this we get co is equal to 28 .164 picofarad...

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