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1. Figure 1 shows the circuit diagram for a common-source amplifier. You may assume that B= 0.01 A/Vv², v, = 0.6 V, C is negligibly small and VA = 20 V for the transistor. All other values are shown beside the schematic. %3D a. Calculate the DC Ips for the circuit. b. Calculate gm and ro for the MOSFET. c. Draw the small-signal equivalent circuit for the amplifier. d. Derive formulas for each of the following: i. Voltage gain. ii. Input resistance. iii. Output resistance.

1. Figure 1 shows the circuit diagram for a common-source amplifier. You may assume that B= 0.01 A/Vv², v, = 0.6 V, C is negligibly small and VA = 20 V for the transistor. All other values are shown beside the schematic. %3D a. Calculate the DC Ips for the circuit. b. Calculate gm and ro for the MOSFET. c. Draw the small-signal equivalent circuit for the amplifier. d. Derive formulas for each of the following: i. Voltage gain. ii. Input resistance. iii. Output resistance.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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1. Figure 1 shows the circuit diagram for a common-source amplifier. You may assume that B = 0.01 A/N², V, = 0.6 V, Cs is negligibly small and VA = 20 V for the transistor. All other values are shown beside the schematic. %3! a. Calculate the DC Ips for the circuit. b. Calculate gm and ro for the MOSFET. c. Draw the small-signal equivalent circuit for the amplifier. d. Derive formulas for each of the following: i. Voltage gain. ii. Input resistance. iii. Output resistance. e. Calculate the numerical values for each of the attributes in Part (d). VDD = 3 V RI = 4k R2 = 5k RD = 5k VDD R1 3 2 RD C2 %3! C1 M1 out in HE RS = 6k Rext = 20k Rext cs R2 3 RSE ov Figure 1 VDD = 3 V R1 = 4k R2 = 6k RS = 1k VDD R1 C1 M1 in HE C2 out Rexd R2 3 RS 2 ov Figure 2 RD2 = 4k RS2 = 6k VDD = 2 V RD2 Vss = 0 V 2 RD1 R13 M2 RI = 4k M1 R2 = 4k RD1 = 4k V1 VOUT RSI = 4k R2 RS13 Cs1= RS23 CS2 VIN Figure 3 VDD = 2 V RS2 = 6k VDD Vss =0 V RD1 R1 = 4k S R1 HT M2 R2 = 4k RDI = 4k Vin M1 out RS1 = 4k %3D Vout R2 { RS1 Tcs1RS2 Vss = 0 V Figure 4
Transcribed Image Text:1. Figure 1 shows the circuit diagram for a common-source amplifier. You may assume that B = 0.01 A/N², V, = 0.6 V, Cs is negligibly small and VA = 20 V for the transistor. All other values are shown beside the schematic. %3! a. Calculate the DC Ips for the circuit. b. Calculate gm and ro for the MOSFET. c. Draw the small-signal equivalent circuit for the amplifier. d. Derive formulas for each of the following: i. Voltage gain. ii. Input resistance. iii. Output resistance. e. Calculate the numerical values for each of the attributes in Part (d). VDD = 3 V RI = 4k R2 = 5k RD = 5k VDD R1 3 2 RD C2 %3! C1 M1 out in HE RS = 6k Rext = 20k Rext cs R2 3 RSE ov Figure 1 VDD = 3 V R1 = 4k R2 = 6k RS = 1k VDD R1 C1 M1 in HE C2 out Rexd R2 3 RS 2 ov Figure 2 RD2 = 4k RS2 = 6k VDD = 2 V RD2 Vss = 0 V 2 RD1 R13 M2 RI = 4k M1 R2 = 4k RD1 = 4k V1 VOUT RSI = 4k R2 RS13 Cs1= RS23 CS2 VIN Figure 3 VDD = 2 V RS2 = 6k VDD Vss =0 V RD1 R1 = 4k S R1 HT M2 R2 = 4k RDI = 4k Vin M1 out RS1 = 4k %3D Vout R2 { RS1 Tcs1RS2 Vss = 0 V Figure 4
1. B= 0.01 A/V. v, = 0.6 V.C. =0 (so far as we care) and VA = 20 V. a Calculate the DC Iuso for the circuit. Ipsg Is the QUIESCENT los je, the DC cument in the circuit when there is no applica sigiai Tuhe circuit is resting, if you like). It is also known as the bias point or the operating point. There is no lo. therefore V, =V R. -=3*(5/(5+4)) = 1.67 V o R +R. We also have Vs losRs and assuming the MOSFET is in saturation las = where Ves = Va - V Va-V.) This can be rearranged to give I =V.-„R, -V,) %3D Substituting in the values and solving the quadratic equation for les gives Las - 149 LA or les = 212 HA One of these solutions is wrong. The source voltage Vs = Iusks hence Vs =0.89 V ar Vy = 1.27 V Therefore Vas = 0.77 V or Vos = 0.39 V The second solution gives Vas - V, <0 therefore it cannot be true Check: for los = 149 HA we get Ves - V,= 0.17 V. and Vos = Vp - Vs = Veo - losRo - Vs =3 - 149u*5k -0.89 = 1.36 V So Vps > Vas - V, hence device is in saturation. b. Calculate g, and r, for the MOSFET. 5. =2 =2 0.01*149u =1.72 ms . = VAlles = 20/149u = 134 k2. c. Draw the small-signal equivalent circuit for the amplitier. out In Omgs SRD Rext Vout ROS Cgs to Vin d. We can write down the following hy looking a the circuit and using Ohm's and Kirchult's laws: L=- So i. Voltage gain. - - G -.r, N, --(r, IR, IR) * v. ii. Input resistance, If the frequency is not very high then we can assume that there is very little current in C. hence i, = 0, therefore V. R. R, = R, MR, R+ R, R, + R; iii. Output resistance. The output circuit for this amplifier is a pretty simple parallel combination of resistors. R is ignored - it is outside the amplifier, hence R=, IR,- R, , (134k) is much higger than either Rp (Sk) or Reu (20). We can then approximate to R,R. i Voltage gain R +R G, =-,r, IR, IR)--R.(R, NR)= 2-1.72m"(5k*20ISK+20k) = -6.88 i R. R. 4kU5k = 2 2k i. R -R. since Ra sCL. therefore R 5k +R.
Transcribed Image Text:1. B= 0.01 A/V. v, = 0.6 V.C. =0 (so far as we care) and VA = 20 V. a Calculate the DC Iuso for the circuit. Ipsg Is the QUIESCENT los je, the DC cument in the circuit when there is no applica sigiai Tuhe circuit is resting, if you like). It is also known as the bias point or the operating point. There is no lo. therefore V, =V R. -=3*(5/(5+4)) = 1.67 V o R +R. We also have Vs losRs and assuming the MOSFET is in saturation las = where Ves = Va - V Va-V.) This can be rearranged to give I =V.-„R, -V,) %3D Substituting in the values and solving the quadratic equation for les gives Las - 149 LA or les = 212 HA One of these solutions is wrong. The source voltage Vs = Iusks hence Vs =0.89 V ar Vy = 1.27 V Therefore Vas = 0.77 V or Vos = 0.39 V The second solution gives Vas - V, <0 therefore it cannot be true Check: for los = 149 HA we get Ves - V,= 0.17 V. and Vos = Vp - Vs = Veo - losRo - Vs =3 - 149u*5k -0.89 = 1.36 V So Vps > Vas - V, hence device is in saturation. b. Calculate g, and r, for the MOSFET. 5. =2 =2 0.01*149u =1.72 ms . = VAlles = 20/149u = 134 k2. c. Draw the small-signal equivalent circuit for the amplitier. out In Omgs SRD Rext Vout ROS Cgs to Vin d. We can write down the following hy looking a the circuit and using Ohm's and Kirchult's laws: L=- So i. Voltage gain. - - G -.r, N, --(r, IR, IR) * v. ii. Input resistance, If the frequency is not very high then we can assume that there is very little current in C. hence i, = 0, therefore V. R. R, = R, MR, R+ R, R, + R; iii. Output resistance. The output circuit for this amplifier is a pretty simple parallel combination of resistors. R is ignored - it is outside the amplifier, hence R=, IR,- R, , (134k) is much higger than either Rp (Sk) or Reu (20). We can then approximate to R,R. i Voltage gain R +R G, =-,r, IR, IR)--R.(R, NR)= 2-1.72m"(5k*20ISK+20k) = -6.88 i R. R. 4kU5k = 2 2k i. R -R. since Ra sCL. therefore R 5k +R.
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