Question

7. For the FET amplifier shown, given gm = 1.18 mS a) Calculate the mid-band gain, Av = Vo/Vi. b) Determine Zi. c) Determine upper cut-off frequencies, f_Hi and f_Ho. d) Determine the approximate upper cutoff frequency of the amplifier. 18 V C_wi = 3 pF C_gd = 4 pF C_wo = 5 pF C_gs = 6 pF C_ds = 1 pF 3 k? (drain resistor) 4.7 ?F (coupling cap to output) 1 k? (source of input series resistor) 0.1 ?F (input coupling cap) I_DSS = 6 mA V_P = -6 V, r_d = ? 3.9 k? (load resistor at output) 1 M? (gate bias resistor to ground) 1.2 k? (source resistor) 10 ?F (source bypass cap)

          7. For the FET amplifier shown, given gm = 1.18 mS
a) Calculate the mid-band gain, Av = Vo/Vi.
b) Determine Zi.
c) Determine upper cut-off frequencies, f_Hi and f_Ho.
d) Determine the approximate upper cutoff frequency of the amplifier.

18 V
C_wi = 3 pF
C_gd = 4 pF
C_wo = 5 pF
C_gs = 6 pF
C_ds = 1 pF
3 k? (drain resistor)
4.7 ?F (coupling cap to output)
1 k? (source of input series resistor)
0.1 ?F (input coupling cap)
I_DSS = 6 mA
V_P = -6 V, r_d = ?
3.9 k? (load resistor at output)
1 M? (gate bias resistor to ground)
1.2 k? (source resistor)
10 ?F (source bypass cap)

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7. For the FET amplifier shown, given gm = 1.18 mS
a) Calculate the mid-band gain, Av = Vo/Vi.
b) Determine Zi.
c) Determine upper cut-off frequencies, fHi and fHo.
d) Determine the approximate upper cutoff frequency of the amplifier.

18 V
Cwi = 3 pF
Cgd = 4 pF
Cwo = 5 pF
Cgs = 6 pF
Cds = 1 pF
3 k? (drain resistor)
4.7 ?F (coupling cap to output)
1 k? (source of input series resistor)
0.1 ?F (input coupling cap)
IDSS = 6 mA
VP = -6 V, rd = ?
3.9 k? (load resistor at output)
1 M? (gate bias resistor to ground)
1.2 k? (source resistor)
10 ?F (source bypass cap)

Added by Jennifer Y.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sri K

06/17/2023

Step 1

The mid-band gain for a common-source FET amplifier is given by: Av = -gm * Rd Where: - gm = 1.18 mS = 1.18 x 10^-3 S - Rd is the parallel combination of the drain resistor and the load resistor. Rd = (3 kΩ || 3.9 kΩ) Calculate Rd: 1/Rd = 1/3000 + 1/3900 Rd ≈ Show more…

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Transcript

00:01 So here we are given i m this is equal to 1 .18 m a by divided by v.

00:10 So small signal mid base ac equivalent.

00:15 This are capacitors shortened and if we talk about a v this is equal to minus i m multiplied by 3k multiplied by 3 .9 k vgs divided by vgs multiplied by 10 raised to the power 6 divided by 10 raised to the power 6 plus 10 raised to the power 3.

00:51 So a v this is equal to minus 1 .18 into 10 raised to the power minus 3 multiplied by 1 .695 multiplied by 10 raised to the power 3.

01:09 So from here we get a v as minus 2 .0008 which is basically equal to minus 2.

01:20 So a v this is equal to minus 2 and then we have z i this is equal to 10 raised to the power 6 plus 10 raised to the power 3.

01:32 So this is going to be equal to 1 0 0 1 k ohms.

01:40 Now talking about the high frequency.

01:45 So by using the miller theorem so we have c m this is equal to c gd 1 minus a v.

01:58 So substituting we have 4 pf 1 plus 2.

02:07 This is equal to 4 multiplied by 3 is 12.

02:12 So it will be 12 pf.

02:15 Similarly c n this is equal to c gd 1 minus 1 divided by a v and we have 4 pf 1 plus 1 divided by 2.

02:30 So we have this as 6 pf.

02:34 Now c in this is equal to cwi plus cgs plus c m.

02:47 So this is nothing but 3 plus 6 plus 12...

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