Deck 3: Transistor Amplifiers

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    Figure 7.1.1 The NMOS transistor in the circuit in Fig. 7.1.1 has  \mu_{n} C_{o x}=0.4 \mathrm{~mA} / \mathrm{V}^{2}, W / L=25 , and  V_{t}=   0.4 \mathrm{~V}  (a) Find the value of  V_{G S}  that results in saturationmode operation with a dc current of  0.1 \mathrm{~mA} . Neglect the Early effect. (b) Find the value of  R_{D}  that results in a dc drain voltage of  0.5 \mathrm{~V} . (c) Find  g_{m}  and  r_{O}  at the dc operating point specified above. Assume  V_{A}=5 \mathrm{~V} . (d) Find the open-circuit voltage gain  A_{v o} . (e) If a sinusoidal signal with peak amplitude  V_{i}  is superimposed on the dc voltage  V_{G S} , find the maximum allowable value of  V_{i}  for which the transistor operates in saturation.<div style=padding-top: 35px>

Figure 7.1.1
The NMOS transistor in the circuit in Fig. 7.1.1 has μnCox=0.4 mA/V2,W/L=25\mu_{n} C_{o x}=0.4 \mathrm{~mA} / \mathrm{V}^{2}, W / L=25 , and Vt=V_{t}= 0.4 V0.4 \mathrm{~V}
(a) Find the value of VGSV_{G S} that results in saturationmode operation with a dc current of 0.1 mA0.1 \mathrm{~mA} . Neglect the Early effect.
(b) Find the value of RDR_{D} that results in a dc drain voltage of 0.5 V0.5 \mathrm{~V} .
(c) Find gmg_{m} and rOr_{O} at the dc operating point specified above. Assume VA=5 VV_{A}=5 \mathrm{~V} .
(d) Find the open-circuit voltage gain AvoA_{v o} .
(e) If a sinusoidal signal with peak amplitude ViV_{i} is superimposed on the dc voltage VGSV_{G S} , find the maximum allowable value of ViV_{i} for which the transistor operates in saturation.
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Question
An NMOS transistor fabricated in a 0.18μm0.18-\mu \mathrm{m} CMOS technology has L=0.54μmL=0.54 \mu \mathrm{m} and W=W= 10.8μm10.8 \mu \mathrm{m} . The technology is specified to have μnCox=400μA/V2,Vtn=0.5 V\mu_{n} C_{o x}=400 \mu \mathrm{A} / \mathrm{V}^{2}, V_{t n}=0.5 \mathrm{~V} , and VA=V_{A}^{\prime}= 5 V/μm5 \mathrm{~V} / \mu \mathrm{m} .
(a) If the device is operated in saturation with an overdrive voltage of 0.2 V0.2 \mathrm{~V} , find the required values of IDI_{D} and VGSV_{G S} , along with the resulting values of gmg_{m} and ror_{o} . (b) If the gate width is doubled but the value of VOVV_{O V} is maintained, what do the values of ID,gmI_{D}, g_{m} , and ror_{o} become?
Question
A 0.18μm0.18-\mu \mathrm{m} CMOS technology is specified to have μn=450 cm2/Vs,μp=100 cm2/Vs,Cox=\mu_{n}=450 \mathrm{~cm}^{2} / \mathrm{V} \cdot \mathrm{s}, \mu_{p}=100 \mathrm{~cm}^{2} / \mathrm{V} \cdot \mathrm{s}, C_{o x}= 8.6fF/μm2,Vtn=Vtp=0.5 V8.6 \mathrm{fF} / \mu \mathrm{m}^{2}, V_{t n}=-V_{t p}=0.5 \mathrm{~V} , and dc power supply of 1.8 V1.8 \mathrm{~V} .
(a) Find the transconductance parameters knk_{n}^{\prime} and kpk_{p}^{\prime} expressed in μA/V2\mu \mathrm{A} / \mathrm{V}^{2} .
(b) Find the W/LW / L ratios of matched NMOS and PMOS transistors that exhibit resistance rDSr_{D S} of 250Ω250 \Omega when operated in the triode region with an overdrive voltage of 0.3 V0.3 \mathrm{~V} . If twice-theminimum channel length is used, specify the width of the NMOS transistor and the PMOS transistor.
(c) If the devices in (b) are operated in saturation with VOV=0.2 V\left|V_{O V}\right|=0.2 \mathrm{~V} , what drain current results? If for each transistor the source is connected to ground, what should the gate voltages be? In each case, specify the range of voltages permitted at the drain for saturation-mode operation to be maintained.
(d) If the drain currents in (c) are to be reduced by a factor of 4 , what should the overdrive voltage be? If instead of changing VOV\left|V_{O V}\right| , the IC designer redesigns the widths of the transistors, what WW values would be required?
(e) If the devices described in (b) above are operated in saturation with VOV=0.2 V\left|V_{O V}\right|=0.2 \mathrm{~V} , find the resulting value of gmg_{m} .
(f) If an NMOS transistor as in (e) above is connected as a common-source amplifier with RD=R_{D}= 5kΩ5 \mathrm{k} \Omega and VDD=1.8 VV_{D D}=1.8 \mathrm{~V} , what dc voltage would appear at the drain? What small-signal voltage gain AvA_{v} would be obtained?
(g) Recalculate the voltage gain in (f) taking into account channel-length modulation. Assume the
Early voltage for the process technology is specified as 10 V/μm10 \mathrm{~V} / \mu \mathrm{m} .
Question
    Figure 7.4.1 The MOSFET in the circuit of Fig. 7.4.1 has  \mu_{n} C_{o x}(W / L)=4 \mathrm{~mA} / \mathrm{V}^{2} . (a) Find the value of  I  that causes the MOSFET to operate in saturation with an overdrive voltage of  0.25 \mathrm{~V} . (b) What value of  R_{D}  results in  V_{D}=0 \mathrm{~V}  ? (c) Find the value of  g_{m} . (d) Find the value of  r_{o}  given that  V_{A}=25 \mathrm{~V} . (e) If  R_{\text {in }}  is to be  1 \mathrm{M} \Omega , find the value that  R_{G}  must have. (f) For  R_{\text {sig }}=1 \mathrm{M} \Omega  and  R_{L}=40 \mathrm{k} \Omega , find the value of the overall voltage gain  v_{o} / v_{\text {sig }} . (g) What does the voltage gain become if  C_{S}  is removed?<div style=padding-top: 35px>

Figure 7.4.1
The MOSFET in the circuit of Fig. 7.4.1 has μnCox(W/L)=4 mA/V2\mu_{n} C_{o x}(W / L)=4 \mathrm{~mA} / \mathrm{V}^{2} .
(a) Find the value of II that causes the MOSFET to operate in saturation with an overdrive voltage of 0.25 V0.25 \mathrm{~V} .
(b) What value of RDR_{D} results in VD=0 VV_{D}=0 \mathrm{~V} ?
(c) Find the value of gmg_{m} .
(d) Find the value of ror_{o} given that VA=25 VV_{A}=25 \mathrm{~V} .
(e) If Rin R_{\text {in }} is to be 1MΩ1 \mathrm{M} \Omega , find the value that RGR_{G} must have. (f) For Rsig =1MΩR_{\text {sig }}=1 \mathrm{M} \Omega and RL=40kΩR_{L}=40 \mathrm{k} \Omega , find the value of the overall voltage gain vo/vsig v_{o} / v_{\text {sig }} .
(g) What does the voltage gain become if CSC_{S} is removed?
Question
Figure 7.5.1 (refer to Figure below)
 Figure 7.5.1 (refer to Figure below)    shows a capacitively coupled amplifier. In the following, assume operation at midband frequencies where the coupling and bypass capacitors behave as short circuits. The BJT has  \beta=99  and the Early effect can be neglected. (a) If the de voltage at the base is to be  5 \mathrm{~V}  and the emitter current  1 \mathrm{~mA} , find the required value of  \left(R_{E 1}+R_{E 2}\right) . Assume  V_{B E}=0.7 \mathrm{~V} . (b) If the input resistance at the base,  R_{i b} , is to be  10 \mathrm{k} \Omega , find the required value of  R_{E 1}  and hence the value of  R_{E 2} . (c) If the dc current in  R_{B 2}  is to be  0.1 \mathrm{~mA} , find the values of  R_{B 1}  and  R_{B 2} . (d) Find the value of  R_{C}  that results in a de voltage of  +6 \mathrm{~V}  at the collector. (e) Find the input resistance  R_{\text {in }}  and the value of  v_{b} / v_{\text {sig. }} . (f) Find the value of  v_{o} / v_{b} . (g) Find the value of the overall voltage gain  v_{o} / v_{\text {sig }} . (h) If the peak amplitude of the signal between base and emitter  \left(v_{\pi}\right)  is to be limited to  5 \mathrm{mV} , what are the corresponding amplitudes of  v_{b}, v_{\mathrm{sig}} , and  v_{o}  ?<div style=padding-top: 35px>

shows a capacitively coupled amplifier. In the following, assume operation at midband frequencies where the coupling and bypass capacitors behave as short circuits. The BJT has β=99\beta=99 and the Early effect can be neglected.
(a) If the de voltage at the base is to be 5 V5 \mathrm{~V} and the emitter current 1 mA1 \mathrm{~mA} , find the required value of (RE1+RE2)\left(R_{E 1}+R_{E 2}\right) . Assume VBE=0.7 VV_{B E}=0.7 \mathrm{~V} .
(b) If the input resistance at the base, RibR_{i b} , is to be 10kΩ10 \mathrm{k} \Omega , find the required value of RE1R_{E 1} and hence the value of RE2R_{E 2} .
(c) If the dc current in RB2R_{B 2} is to be 0.1 mA0.1 \mathrm{~mA} , find the values of RB1R_{B 1} and RB2R_{B 2} .
(d) Find the value of RCR_{C} that results in a de voltage of +6 V+6 \mathrm{~V} at the collector.
(e) Find the input resistance Rin R_{\text {in }} and the value of vb/vsig. v_{b} / v_{\text {sig. }} .
(f) Find the value of vo/vbv_{o} / v_{b} .
(g) Find the value of the overall voltage gain vo/vsig v_{o} / v_{\text {sig }} .
(h) If the peak amplitude of the signal between base and emitter (vπ)\left(v_{\pi}\right) is to be limited to 5mV5 \mathrm{mV} , what are the corresponding amplitudes of vb,vsigv_{b}, v_{\mathrm{sig}} , and vov_{o} ?
Question
    Figure 7.6.1 (a) Perform a dc bias design for the amplifier in Fig. 7.6.1. For this purpose, assume  \beta  is very high and  V_{B E}=0.7 \mathrm{~V}  and neglect the Early effect. Design to obtain a dc base voltage of  V_{C C} / 3 , a dc emitter current of  1 \mathrm{~mA} , and a dc voltage at the collector that allows for  \pm 1-\mathrm{V}  signal swing at the collector with the minimum collector voltage no lower than  V_{B} . Use a de current in the base voltage divider of  I_{E} / 10 . What values are required for  R_{B 1}, R_{B 2}, R_{E} , and  R_{C}  ? (b) If the transistor has  \beta=100 , find the actual values obtained for  I_{E}, I_{C}, V_{B} , and  V_{C} . (c) What are the values of  g_{m}, r_{e} , and  r_{\pi}  at the dc bias point. (d) Assuming very large coupling and bypass capacitors, find the values of  R_{E 1}  and  R_{E 2}  that result in  R_{\text {in }}=10 \mathrm{k} \Omega . (e) Find the overall voltage gain  G_{v}=v_{o} / v_{\text {sig }} . (f) For  v_{o} , a sine wave with a  1-\mathrm{V}  peak amplitude, what is the peak amplitude required of the sine wave  v_{b e}  ? If this value is greater than  5 \mathrm{mV} , reduce  v_{\text {sig }}  to limit the peak amplitude of  v_{b e}  to  5 \mathrm{mV} . What is the resulting output voltage in this case?<div style=padding-top: 35px>

Figure 7.6.1
(a) Perform a dc bias design for the amplifier in Fig. 7.6.1. For this purpose, assume β\beta is very high and VBE=0.7 VV_{B E}=0.7 \mathrm{~V} and neglect the Early effect. Design to obtain a dc base voltage of VCC/3V_{C C} / 3 , a dc emitter current of 1 mA1 \mathrm{~mA} , and a dc voltage at the collector that allows for ±1V\pm 1-\mathrm{V} signal swing at the collector with the minimum collector voltage no lower than VBV_{B} . Use a de current in the base voltage divider of IE/10I_{E} / 10 . What values are required for RB1,RB2,RER_{B 1}, R_{B 2}, R_{E} , and RCR_{C} ?
(b) If the transistor has β=100\beta=100 , find the actual values obtained for IE,IC,VBI_{E}, I_{C}, V_{B} , and VCV_{C} .
(c) What are the values of gm,reg_{m}, r_{e} , and rπr_{\pi} at the dc bias point.
(d) Assuming very large coupling and bypass capacitors, find the values of RE1R_{E 1} and RE2R_{E 2} that result in Rin =10kΩR_{\text {in }}=10 \mathrm{k} \Omega .
(e) Find the overall voltage gain Gv=vo/vsig G_{v}=v_{o} / v_{\text {sig }} .
(f) For vov_{o} , a sine wave with a 1V1-\mathrm{V} peak amplitude, what is the peak amplitude required of the sine wave vbev_{b e} ? If this value is greater than 5mV5 \mathrm{mV} , reduce vsig v_{\text {sig }} to limit the peak amplitude of vbev_{b e} to 5mV5 \mathrm{mV} . What is the resulting output voltage in this case?
Question
The transistor in the emitter follower of Fig. 7.7.1 (refer to Figure below)
 The transistor in the emitter follower of Fig. 7.7.1 (refer to Figure below)    Figure 7.7.1 has  \beta=100 . Assume  V_{B E}=0.7 \mathrm{~V}  and neglect the Early effect. (a) Find the dc emitter current  I_{E} . (b) Find the value of the emitter resistance  r_{e} . (c) Find the input resistance  R_{\text {in }} . (d) Find the voltage gain from signal source to the transistor base,  v_{b} / v_{\text {sig }} . (e) Find the voltage gain from transistor base to the output,  v_{o} / v_{b} . (f) Find the overall voltage gain,  v_{o} / v_{\text {sig }} . (g) Find the output resistance  R_{\text {out }} .<div style=padding-top: 35px>

Figure 7.7.1
has β=100\beta=100 . Assume VBE=0.7 VV_{B E}=0.7 \mathrm{~V} and neglect the Early effect.
(a) Find the dc emitter current IEI_{E} .
(b) Find the value of the emitter resistance rer_{e} .
(c) Find the input resistance Rin R_{\text {in }} .
(d) Find the voltage gain from signal source to the transistor base, vb/vsig v_{b} / v_{\text {sig }} .
(e) Find the voltage gain from transistor base to the output, vo/vbv_{o} / v_{b} .
(f) Find the overall voltage gain, vo/vsig v_{o} / v_{\text {sig }} .
(g) Find the output resistance Rout R_{\text {out }} .
Question
   Figure 7.8.1 The transistors in the amplifier shown in Fig. 7.8.1 have  \beta=100, V_{B E}=0.7 \mathrm{~V} , and negligible Early effect. (a) Design the circuit to obtain the following dc operating parameters:  V_{B 1}=+3 \mathrm{~V}, I_{E 1}=1 \mathrm{~mA} ,  V_{E 2}=+4 \mathrm{~V} , and  I_{E 2}=4 \mathrm{~mA} . Use  I=0.1 \mathrm{~mA} . Find the values of  R_{1}, R_{2}, R_{3},\left(R_{4}+R_{5}\right) , and  R_{6} . (b) Find the value of  R_{4}  (and hence  R_{5}  ) that results in  R_{\text {in }}=10 \mathrm{k} \Omega . (c) Find  R_{i 2} . (d) Find the voltage gain  v_{c 1} / v_{i} . (e) Find the voltage gain  v_{O} / v_{c 1} . (f) Find the overall voltage gain  v_{o} / v_{i} . (g) Find the output resistance  R_{\text {out }} . (h) In order to minimize nonlinear distortion, it is required to keep the maximum signal across the base-emitter junction of each of  Q_{1}  and  Q_{2}  to  5 \mathrm{mV} . Under this constraint, what is the maximum peak-to-peak sine wave signal that can be obtained at the output?<div style=padding-top: 35px>
Figure 7.8.1
The transistors in the amplifier shown in Fig. 7.8.1
have β=100,VBE=0.7 V\beta=100, V_{B E}=0.7 \mathrm{~V} , and negligible Early effect.
(a) Design the circuit to obtain the following dc operating parameters: VB1=+3 V,IE1=1 mAV_{B 1}=+3 \mathrm{~V}, I_{E 1}=1 \mathrm{~mA} , VE2=+4 VV_{E 2}=+4 \mathrm{~V} , and IE2=4 mAI_{E 2}=4 \mathrm{~mA} . Use I=0.1 mAI=0.1 \mathrm{~mA} . Find the values of R1,R2,R3,(R4+R5)R_{1}, R_{2}, R_{3},\left(R_{4}+R_{5}\right) , and R6R_{6} . (b) Find the value of R4R_{4} (and hence R5R_{5} ) that results in Rin =10kΩR_{\text {in }}=10 \mathrm{k} \Omega .
(c) Find Ri2R_{i 2} .
(d) Find the voltage gain vc1/viv_{c 1} / v_{i} .
(e) Find the voltage gain vO/vc1v_{O} / v_{c 1} .
(f) Find the overall voltage gain vo/viv_{o} / v_{i} .
(g) Find the output resistance Rout R_{\text {out }} .
(h) In order to minimize nonlinear distortion, it is required to keep the maximum signal across the base-emitter junction of each of Q1Q_{1} and Q2Q_{2} to 5mV5 \mathrm{mV} . Under this constraint, what is the maximum peak-to-peak sine wave signal that can be obtained at the output?
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Deck 3: Transistor Amplifiers
1
    Figure 7.1.1 The NMOS transistor in the circuit in Fig. 7.1.1 has  \mu_{n} C_{o x}=0.4 \mathrm{~mA} / \mathrm{V}^{2}, W / L=25 , and  V_{t}=   0.4 \mathrm{~V}  (a) Find the value of  V_{G S}  that results in saturationmode operation with a dc current of  0.1 \mathrm{~mA} . Neglect the Early effect. (b) Find the value of  R_{D}  that results in a dc drain voltage of  0.5 \mathrm{~V} . (c) Find  g_{m}  and  r_{O}  at the dc operating point specified above. Assume  V_{A}=5 \mathrm{~V} . (d) Find the open-circuit voltage gain  A_{v o} . (e) If a sinusoidal signal with peak amplitude  V_{i}  is superimposed on the dc voltage  V_{G S} , find the maximum allowable value of  V_{i}  for which the transistor operates in saturation.

Figure 7.1.1
The NMOS transistor in the circuit in Fig. 7.1.1 has μnCox=0.4 mA/V2,W/L=25\mu_{n} C_{o x}=0.4 \mathrm{~mA} / \mathrm{V}^{2}, W / L=25 , and Vt=V_{t}= 0.4 V0.4 \mathrm{~V}
(a) Find the value of VGSV_{G S} that results in saturationmode operation with a dc current of 0.1 mA0.1 \mathrm{~mA} . Neglect the Early effect.
(b) Find the value of RDR_{D} that results in a dc drain voltage of 0.5 V0.5 \mathrm{~V} .
(c) Find gmg_{m} and rOr_{O} at the dc operating point specified above. Assume VA=5 VV_{A}=5 \mathrm{~V} .
(d) Find the open-circuit voltage gain AvoA_{v o} .
(e) If a sinusoidal signal with peak amplitude ViV_{i} is superimposed on the dc voltage VGSV_{G S} , find the maximum allowable value of ViV_{i} for which the transistor operates in saturation.
(a)
 (a)    Figure 7.1.1  \begin{aligned} I_{D} & =\frac{1}{2}\left(\mu_{n} C_{O x}\right)(W / L)\left(V_{G S}-V_{t}\right)^{2} \\ 0.1 & =\frac{1}{2} \times 0.4 \times 25\left(V_{G S}-0.4\right)^{2} \\ \Rightarrow V_{G S} & =0.541 \mathrm{~V} \end{aligned}  (b)  R_{D}=\frac{1.5-V_{D}}{I_{D}}=\frac{1.5-0.5}{0.1}=10 \mathrm{k} \Omega  (c)  g_{m}=\frac{2 I_{D}}{V_{O V}}  where  \begin{aligned} V_{O V} & =V_{G S}-V_{t}=0.541-0.4=0.141 \mathrm{~V} \\ g_{m} & =\frac{2 \times 0.1}{0.141}=1.42 \mathrm{~mA} / \mathrm{V} \\ r_{o} & =\frac{V_{A}}{I_{D}}=\frac{5}{0.1}=50 \mathrm{k} \Omega \end{aligned}  (d)  \begin{aligned} A_{v o} & =-g_{m}\left(R_{D} \| r_{o}\right) \\ & =-1.42(10 \| 50) \\ & =-11.8 \mathrm{~V} / \mathrm{V} \end{aligned}  (e) At the edge of saturation,  v_{G \max }-v_{D \min }=V_{t}  Here,  \begin{aligned} v_{G \max } & =V_{G S}+\hat{V}_{i}=0.541+\hat{V}_{i} \\ v_{D \min } & =V_{D}-\left|A_{v o}\right| \hat{V}_{i} \\ & =0.5-11.8 \hat{V}_{i} \end{aligned}  Thus,  \begin{gathered} 0.541 +\hat{V}_{i}-0.5+11.8 \hat{V}_{i}=0.4 \\ \Rightarrow \hat{V}_{i}=28 \mathrm{mV} \end{gathered}

Figure 7.1.1
ID=12(μnCOx)(W/L)(VGSVt)20.1=12×0.4×25(VGS0.4)2VGS=0.541 V\begin{aligned}I_{D} & =\frac{1}{2}\left(\mu_{n} C_{O x}\right)(W / L)\left(V_{G S}-V_{t}\right)^{2} \\0.1 & =\frac{1}{2} \times 0.4 \times 25\left(V_{G S}-0.4\right)^{2} \\\Rightarrow V_{G S} & =0.541 \mathrm{~V}\end{aligned}
(b)
RD=1.5VDID=1.50.50.1=10kΩR_{D}=\frac{1.5-V_{D}}{I_{D}}=\frac{1.5-0.5}{0.1}=10 \mathrm{k} \Omega
(c)
gm=2IDVOVg_{m}=\frac{2 I_{D}}{V_{O V}}
where
VOV=VGSVt=0.5410.4=0.141 Vgm=2×0.10.141=1.42 mA/Vro=VAID=50.1=50kΩ\begin{aligned}V_{O V} & =V_{G S}-V_{t}=0.541-0.4=0.141 \mathrm{~V} \\g_{m} & =\frac{2 \times 0.1}{0.141}=1.42 \mathrm{~mA} / \mathrm{V} \\r_{o} & =\frac{V_{A}}{I_{D}}=\frac{5}{0.1}=50 \mathrm{k} \Omega\end{aligned}
(d)
Avo=gm(RDro)=1.42(1050)=11.8 V/V\begin{aligned}A_{v o} & =-g_{m}\left(R_{D} \| r_{o}\right) \\& =-1.42(10 \| 50) \\& =-11.8 \mathrm{~V} / \mathrm{V}\end{aligned}
(e) At the edge of saturation,
vGmaxvDmin=Vtv_{G \max }-v_{D \min }=V_{t}
Here,
vGmax=VGS+V^i=0.541+V^ivDmin=VDAvoV^i=0.511.8V^i\begin{aligned}v_{G \max } & =V_{G S}+\hat{V}_{i}=0.541+\hat{V}_{i} \\v_{D \min } & =V_{D}-\left|A_{v o}\right| \hat{V}_{i} \\& =0.5-11.8 \hat{V}_{i}\end{aligned}
Thus,
0.541+V^i0.5+11.8V^i=0.4V^i=28mV\begin{gathered}0.541+\hat{V}_{i}-0.5+11.8 \hat{V}_{i}=0.4 \\\Rightarrow \hat{V}_{i}=28 \mathrm{mV}\end{gathered}
2
An NMOS transistor fabricated in a 0.18μm0.18-\mu \mathrm{m} CMOS technology has L=0.54μmL=0.54 \mu \mathrm{m} and W=W= 10.8μm10.8 \mu \mathrm{m} . The technology is specified to have μnCox=400μA/V2,Vtn=0.5 V\mu_{n} C_{o x}=400 \mu \mathrm{A} / \mathrm{V}^{2}, V_{t n}=0.5 \mathrm{~V} , and VA=V_{A}^{\prime}= 5 V/μm5 \mathrm{~V} / \mu \mathrm{m} .
(a) If the device is operated in saturation with an overdrive voltage of 0.2 V0.2 \mathrm{~V} , find the required values of IDI_{D} and VGSV_{G S} , along with the resulting values of gmg_{m} and ror_{o} . (b) If the gate width is doubled but the value of VOVV_{O V} is maintained, what do the values of ID,gmI_{D}, g_{m} , and ror_{o} become?
(a) Saturation with VOV=0.2 VV_{O V}=0.2 \mathrm{~V} ,
ID=12(μnCOx)(W/L)VOV2=12×0.4×(10.8/0.54)×0.22=0.16 mAVGS=Vtn+VOV=0.5+0.2=0.7 V\begin{aligned}I_{D} & =\frac{1}{2}\left(\mu_{n} C_{O x}\right)(W / L) V_{O V}^{2} \\& =\frac{1}{2} \times 0.4 \times(10.8 / 0.54) \times 0.2^{2} \\& =0.16 \mathrm{~mA} \\V_{G S} & =V_{t n}+V_{O V}=0.5+0.2=0.7 \mathrm{~V}\end{aligned}
gm=2IDVOV=2×0.160.2=1.6 mA/Vro=VAID=VALID=5×0.540.16=16.9kΩ\begin{aligned}g_{m} & =\frac{2 I_{D}}{V_{O V}}=\frac{2 \times 0.16}{0.2}=1.6 \mathrm{~mA} / \mathrm{V} \\r_{o} & =\frac{V_{A}}{I_{D}}=\frac{V_{A}^{\prime} L}{I_{D}}=\frac{5 \times 0.54}{0.16}=16.9 \mathrm{k} \Omega\end{aligned}
(b) If WW is doubled while VOVV_{O V} is maintained, we see from Eq. (1) that IDI_{D} doubles to
ID=0.32 mAI_{D}=0.32 \mathrm{~mA}
From
gm=2IDVOVg_{m}=\frac{2 I_{D}}{V_{O V}}
we see that gmg_{m} doubles to
gm=3.2 mA/Vg_{m}=3.2 \mathrm{~mA} / \mathrm{V}
Finally, from
ro=VAIDr_{o}=\frac{V_{A}}{I_{D}}
we see that rOr_{O} is halved to
ro=8.45kΩr_{o}=8.45 \mathrm{k} \Omega
3
A 0.18μm0.18-\mu \mathrm{m} CMOS technology is specified to have μn=450 cm2/Vs,μp=100 cm2/Vs,Cox=\mu_{n}=450 \mathrm{~cm}^{2} / \mathrm{V} \cdot \mathrm{s}, \mu_{p}=100 \mathrm{~cm}^{2} / \mathrm{V} \cdot \mathrm{s}, C_{o x}= 8.6fF/μm2,Vtn=Vtp=0.5 V8.6 \mathrm{fF} / \mu \mathrm{m}^{2}, V_{t n}=-V_{t p}=0.5 \mathrm{~V} , and dc power supply of 1.8 V1.8 \mathrm{~V} .
(a) Find the transconductance parameters knk_{n}^{\prime} and kpk_{p}^{\prime} expressed in μA/V2\mu \mathrm{A} / \mathrm{V}^{2} .
(b) Find the W/LW / L ratios of matched NMOS and PMOS transistors that exhibit resistance rDSr_{D S} of 250Ω250 \Omega when operated in the triode region with an overdrive voltage of 0.3 V0.3 \mathrm{~V} . If twice-theminimum channel length is used, specify the width of the NMOS transistor and the PMOS transistor.
(c) If the devices in (b) are operated in saturation with VOV=0.2 V\left|V_{O V}\right|=0.2 \mathrm{~V} , what drain current results? If for each transistor the source is connected to ground, what should the gate voltages be? In each case, specify the range of voltages permitted at the drain for saturation-mode operation to be maintained.
(d) If the drain currents in (c) are to be reduced by a factor of 4 , what should the overdrive voltage be? If instead of changing VOV\left|V_{O V}\right| , the IC designer redesigns the widths of the transistors, what WW values would be required?
(e) If the devices described in (b) above are operated in saturation with VOV=0.2 V\left|V_{O V}\right|=0.2 \mathrm{~V} , find the resulting value of gmg_{m} .
(f) If an NMOS transistor as in (e) above is connected as a common-source amplifier with RD=R_{D}= 5kΩ5 \mathrm{k} \Omega and VDD=1.8 VV_{D D}=1.8 \mathrm{~V} , what dc voltage would appear at the drain? What small-signal voltage gain AvA_{v} would be obtained?
(g) Recalculate the voltage gain in (f) taking into account channel-length modulation. Assume the
Early voltage for the process technology is specified as 10 V/μm10 \mathrm{~V} / \mu \mathrm{m} .
(a)
kn=μnCox=450cm2 Vs×8.6×1015Fμm2=450×108μm2 Vs×8.6×1015Fμm2=387×106FVs=387μA/V2kp=kn×μpμn=387×100450=86μA/V2\begin{aligned}k_{n}^{\prime} & =\mu_{n} C_{o x} \\& =450 \frac{\mathrm{cm}^{2}}{\mathrm{~V} \cdot \mathrm{s}} \times 8.6 \times 10^{-15} \frac{\mathrm{F}}{\mu \mathrm{m}^{2}} \\& =450 \times 10^{8} \frac{\mu \mathrm{m}^{2}}{\mathrm{~V} \cdot \mathrm{s}} \times 8.6 \times 10^{-15} \frac{\mathrm{F}}{\mu \mathrm{m}^{2}} \\& =387 \times 10^{-6} \frac{\mathrm{F}}{\mathrm{V} \cdot \mathrm{s}} \\& =387 \mu \mathrm{A} / \mathrm{V}^{2} \\k_{p}^{\prime} & =k_{n}^{\prime} \times \frac{\mu_{p}}{\mu_{n}}=387 \times \frac{100}{450} \\& =86 \mu \mathrm{A} / \mathrm{V}^{2}\end{aligned}
(b) For an NMOS transistor operating in a triode region,
rDS=1kn(W/L)NVOV250=1387×106×(W/L)N×0.3(W/L)N=34.5\begin{aligned}r_{D S} & =\frac{1}{k_{n}^{\prime}(W / L)_{N} V_{O V}} \\250 & =\frac{1}{387 \times 10^{-6} \times(W / L)_{N} \times 0.3} \\\Rightarrow(W / L)_{N} & =34.5\end{aligned}
For a matched PMOS transistor,
(W/L)P=(W/L)N×knkp=34.5×38786=155\begin{aligned}(W / L)_{P} & =(W / L)_{N} \times \frac{k_{n}^{\prime}}{k_{p}^{\prime}} \\& =34.5 \times \frac{387}{86}=155\end{aligned}
Thus,
L=2Lmin=2×0.18=0.36μmWN=34.5×0.36=12.4μmWP=155×0.36=55.8μm\begin{aligned}L & =2 L_{\min }=2 \times 0.18=0.36 \mu \mathrm{m} \\W_{N} & =34.5 \times 0.36=12.4 \mu \mathrm{m} \\W_{P} & =155 \times 0.36=55.8 \mu \mathrm{m}\end{aligned}
(c)
IDP=IDN=12kn(W/L)NVOV2=12×387×34.5×0.22=267μA\begin{aligned}I_{D P} & =I_{D N}=\frac{1}{2} k_{n}^{\prime}(W / L)_{N}\left|V_{O V}\right|^{2} \\& =\frac{1}{2} \times 387 \times 34.5 \times 0.2^{2} \\& =267 \mu \mathrm{A}\end{aligned}
For the NMOS transistor,
VG=VGS=Vtn+VOV=0.5+0.2=0.7 VV_{G}=V_{G S}=V_{t n}+V_{O V}=0.5+0.2=0.7 \mathrm{~V}
For the PMOS transistor,
 (a)  \begin{aligned} k_{n}^{\prime} & =\mu_{n} C_{o x} \\ & =450 \frac{\mathrm{cm}^{2}}{\mathrm{~V} \cdot \mathrm{s}} \times 8.6 \times 10^{-15} \frac{\mathrm{F}}{\mu \mathrm{m}^{2}} \\ & =450 \times 10^{8} \frac{\mu \mathrm{m}^{2}}{\mathrm{~V} \cdot \mathrm{s}} \times 8.6 \times 10^{-15} \frac{\mathrm{F}}{\mu \mathrm{m}^{2}} \\ & =387 \times 10^{-6} \frac{\mathrm{F}}{\mathrm{V} \cdot \mathrm{s}} \\ & =387 \mu \mathrm{A} / \mathrm{V}^{2} \\ k_{p}^{\prime} & =k_{n}^{\prime} \times \frac{\mu_{p}}{\mu_{n}}=387 \times \frac{100}{450} \\ & =86 \mu \mathrm{A} / \mathrm{V}^{2} \end{aligned}  (b) For an NMOS transistor operating in a triode region,  \begin{aligned} r_{D S} & =\frac{1}{k_{n}^{\prime}(W / L)_{N} V_{O V}} \\ 250 & =\frac{1}{387 \times 10^{-6} \times(W / L)_{N} \times 0.3} \\ \Rightarrow(W / L)_{N} & =34.5 \end{aligned}  For a matched PMOS transistor,  \begin{aligned} (W / L)_{P} & =(W / L)_{N} \times \frac{k_{n}^{\prime}}{k_{p}^{\prime}} \\ & =34.5 \times \frac{387}{86}=155 \end{aligned}  Thus,  \begin{aligned} L & =2 L_{\min }=2 \times 0.18=0.36 \mu \mathrm{m} \\ W_{N} & =34.5 \times 0.36=12.4 \mu \mathrm{m} \\ W_{P} & =155 \times 0.36=55.8 \mu \mathrm{m} \end{aligned}  (c)  \begin{aligned} I_{D P} & =I_{D N}=\frac{1}{2} k_{n}^{\prime}(W / L)_{N}\left|V_{O V}\right|^{2} \\ & =\frac{1}{2} \times 387 \times 34.5 \times 0.2^{2} \\ & =267 \mu \mathrm{A} \end{aligned}  For the NMOS transistor,  V_{G}=V_{G S}=V_{t n}+V_{O V}=0.5+0.2=0.7 \mathrm{~V}  For the PMOS transistor,   Figure 7.3.1 From Fig. 7.3.1, we see that for the NMOS transistor to operate in saturation,  V_{D} \geq V_{O V}  that is,  V_{D} \geq 0.2 \mathrm{~V}  For the PMOS transistor to operate in saturation,  V_{D} \leq-\left|V_{O V}\right|  that is,  V_{D} \leq-0.2 \mathrm{~V}  (d) Since  I_{D}  is proportional to  \left|V_{O V}\right|^{2} , the drain current can be reduced by a factor of 4 by reducing  \left|V_{O V}\right|  by a factor of 2 , that is, to  0.1 \mathrm{~V} . Alternatively,  \left|V_{O V}\right|  can be kept constant and  W  reduced by a factor of 4 , resulting in  W_{N}=\frac{34.5}{4}=8.63 \mu \mathrm{m}  and  W_{P}=\frac{155}{4}=38.8 \mu \mathrm{m}  (e)  g_{m}=\frac{2 I_{D}}{\left|V_{O V}\right|}=\frac{2 \times 0.267}{0.2}=2.67 \mathrm{~mA} / \mathrm{V}  (f)   Figure 7.3.2 Refer to Fig. 7.3.2.  \begin{aligned} V_{D} & =V_{D D}-I_{D} R_{D} \\ & =1.8-0.267 \times 5=0.465 \mathrm{~V} \\ A_{v} & =-g_{m} R_{D}=-2.67 \times 5=-13.35 \mathrm{~V} / \mathrm{V} \end{aligned}   (\mathrm{g})   \begin{aligned} V_{A} & =V_{A}^{\prime} \times L \\ & =10 \times 0.36=3.6 \mathrm{~V} \\ r_{O} & =\frac{V_{A}}{I_{D}}=\frac{3.6}{0.267}=13.5 \mathrm{k} \Omega \\ A_{v} & =-g_{m}\left(R_{D} \| r_{O}\right) \\ & =-2.67(5 \| 13.5) \\ & =-9.74 \mathrm{~V} / \mathrm{V} \end{aligned}
Figure 7.3.1
From Fig. 7.3.1, we see that for the NMOS transistor to operate in saturation,
VDVOVV_{D} \geq V_{O V}
that is,
VD0.2 VV_{D} \geq 0.2 \mathrm{~V}
For the PMOS transistor to operate in saturation,
VDVOVV_{D} \leq-\left|V_{O V}\right|
that is,
VD0.2 VV_{D} \leq-0.2 \mathrm{~V}
(d) Since IDI_{D} is proportional to VOV2\left|V_{O V}\right|^{2} , the drain current can be reduced by a factor of 4 by reducing VOV\left|V_{O V}\right| by a factor of 2 , that is, to 0.1 V0.1 \mathrm{~V} . Alternatively, VOV\left|V_{O V}\right| can be kept constant and WW reduced by a factor of 4 , resulting in
WN=34.54=8.63μmW_{N}=\frac{34.5}{4}=8.63 \mu \mathrm{m}
and
WP=1554=38.8μmW_{P}=\frac{155}{4}=38.8 \mu \mathrm{m}
(e)
gm=2IDVOV=2×0.2670.2=2.67 mA/Vg_{m}=\frac{2 I_{D}}{\left|V_{O V}\right|}=\frac{2 \times 0.267}{0.2}=2.67 \mathrm{~mA} / \mathrm{V}
(f)
 (a)  \begin{aligned} k_{n}^{\prime} & =\mu_{n} C_{o x} \\ & =450 \frac{\mathrm{cm}^{2}}{\mathrm{~V} \cdot \mathrm{s}} \times 8.6 \times 10^{-15} \frac{\mathrm{F}}{\mu \mathrm{m}^{2}} \\ & =450 \times 10^{8} \frac{\mu \mathrm{m}^{2}}{\mathrm{~V} \cdot \mathrm{s}} \times 8.6 \times 10^{-15} \frac{\mathrm{F}}{\mu \mathrm{m}^{2}} \\ & =387 \times 10^{-6} \frac{\mathrm{F}}{\mathrm{V} \cdot \mathrm{s}} \\ & =387 \mu \mathrm{A} / \mathrm{V}^{2} \\ k_{p}^{\prime} & =k_{n}^{\prime} \times \frac{\mu_{p}}{\mu_{n}}=387 \times \frac{100}{450} \\ & =86 \mu \mathrm{A} / \mathrm{V}^{2} \end{aligned}  (b) For an NMOS transistor operating in a triode region,  \begin{aligned} r_{D S} & =\frac{1}{k_{n}^{\prime}(W / L)_{N} V_{O V}} \\ 250 & =\frac{1}{387 \times 10^{-6} \times(W / L)_{N} \times 0.3} \\ \Rightarrow(W / L)_{N} & =34.5 \end{aligned}  For a matched PMOS transistor,  \begin{aligned} (W / L)_{P} & =(W / L)_{N} \times \frac{k_{n}^{\prime}}{k_{p}^{\prime}} \\ & =34.5 \times \frac{387}{86}=155 \end{aligned}  Thus,  \begin{aligned} L & =2 L_{\min }=2 \times 0.18=0.36 \mu \mathrm{m} \\ W_{N} & =34.5 \times 0.36=12.4 \mu \mathrm{m} \\ W_{P} & =155 \times 0.36=55.8 \mu \mathrm{m} \end{aligned}  (c)  \begin{aligned} I_{D P} & =I_{D N}=\frac{1}{2} k_{n}^{\prime}(W / L)_{N}\left|V_{O V}\right|^{2} \\ & =\frac{1}{2} \times 387 \times 34.5 \times 0.2^{2} \\ & =267 \mu \mathrm{A} \end{aligned}  For the NMOS transistor,  V_{G}=V_{G S}=V_{t n}+V_{O V}=0.5+0.2=0.7 \mathrm{~V}  For the PMOS transistor,   Figure 7.3.1 From Fig. 7.3.1, we see that for the NMOS transistor to operate in saturation,  V_{D} \geq V_{O V}  that is,  V_{D} \geq 0.2 \mathrm{~V}  For the PMOS transistor to operate in saturation,  V_{D} \leq-\left|V_{O V}\right|  that is,  V_{D} \leq-0.2 \mathrm{~V}  (d) Since  I_{D}  is proportional to  \left|V_{O V}\right|^{2} , the drain current can be reduced by a factor of 4 by reducing  \left|V_{O V}\right|  by a factor of 2 , that is, to  0.1 \mathrm{~V} . Alternatively,  \left|V_{O V}\right|  can be kept constant and  W  reduced by a factor of 4 , resulting in  W_{N}=\frac{34.5}{4}=8.63 \mu \mathrm{m}  and  W_{P}=\frac{155}{4}=38.8 \mu \mathrm{m}  (e)  g_{m}=\frac{2 I_{D}}{\left|V_{O V}\right|}=\frac{2 \times 0.267}{0.2}=2.67 \mathrm{~mA} / \mathrm{V}  (f)   Figure 7.3.2 Refer to Fig. 7.3.2.  \begin{aligned} V_{D} & =V_{D D}-I_{D} R_{D} \\ & =1.8-0.267 \times 5=0.465 \mathrm{~V} \\ A_{v} & =-g_{m} R_{D}=-2.67 \times 5=-13.35 \mathrm{~V} / \mathrm{V} \end{aligned}   (\mathrm{g})   \begin{aligned} V_{A} & =V_{A}^{\prime} \times L \\ & =10 \times 0.36=3.6 \mathrm{~V} \\ r_{O} & =\frac{V_{A}}{I_{D}}=\frac{3.6}{0.267}=13.5 \mathrm{k} \Omega \\ A_{v} & =-g_{m}\left(R_{D} \| r_{O}\right) \\ & =-2.67(5 \| 13.5) \\ & =-9.74 \mathrm{~V} / \mathrm{V} \end{aligned}
Figure 7.3.2
Refer to Fig. 7.3.2.
VD=VDDIDRD=1.80.267×5=0.465 VAv=gmRD=2.67×5=13.35 V/V\begin{aligned}V_{D} & =V_{D D}-I_{D} R_{D} \\& =1.8-0.267 \times 5=0.465 \mathrm{~V} \\A_{v} & =-g_{m} R_{D}=-2.67 \times 5=-13.35 \mathrm{~V} / \mathrm{V}\end{aligned}
(g)(\mathrm{g})
VA=VA×L=10×0.36=3.6 VrO=VAID=3.60.267=13.5kΩAv=gm(RDrO)=2.67(513.5)=9.74 V/V\begin{aligned}V_{A} & =V_{A}^{\prime} \times L \\& =10 \times 0.36=3.6 \mathrm{~V} \\r_{O} & =\frac{V_{A}}{I_{D}}=\frac{3.6}{0.267}=13.5 \mathrm{k} \Omega \\A_{v} & =-g_{m}\left(R_{D} \| r_{O}\right) \\& =-2.67(5 \| 13.5) \\& =-9.74 \mathrm{~V} / \mathrm{V}\end{aligned}
4
    Figure 7.4.1 The MOSFET in the circuit of Fig. 7.4.1 has  \mu_{n} C_{o x}(W / L)=4 \mathrm{~mA} / \mathrm{V}^{2} . (a) Find the value of  I  that causes the MOSFET to operate in saturation with an overdrive voltage of  0.25 \mathrm{~V} . (b) What value of  R_{D}  results in  V_{D}=0 \mathrm{~V}  ? (c) Find the value of  g_{m} . (d) Find the value of  r_{o}  given that  V_{A}=25 \mathrm{~V} . (e) If  R_{\text {in }}  is to be  1 \mathrm{M} \Omega , find the value that  R_{G}  must have. (f) For  R_{\text {sig }}=1 \mathrm{M} \Omega  and  R_{L}=40 \mathrm{k} \Omega , find the value of the overall voltage gain  v_{o} / v_{\text {sig }} . (g) What does the voltage gain become if  C_{S}  is removed?

Figure 7.4.1
The MOSFET in the circuit of Fig. 7.4.1 has μnCox(W/L)=4 mA/V2\mu_{n} C_{o x}(W / L)=4 \mathrm{~mA} / \mathrm{V}^{2} .
(a) Find the value of II that causes the MOSFET to operate in saturation with an overdrive voltage of 0.25 V0.25 \mathrm{~V} .
(b) What value of RDR_{D} results in VD=0 VV_{D}=0 \mathrm{~V} ?
(c) Find the value of gmg_{m} .
(d) Find the value of ror_{o} given that VA=25 VV_{A}=25 \mathrm{~V} .
(e) If Rin R_{\text {in }} is to be 1MΩ1 \mathrm{M} \Omega , find the value that RGR_{G} must have. (f) For Rsig =1MΩR_{\text {sig }}=1 \mathrm{M} \Omega and RL=40kΩR_{L}=40 \mathrm{k} \Omega , find the value of the overall voltage gain vo/vsig v_{o} / v_{\text {sig }} .
(g) What does the voltage gain become if CSC_{S} is removed?
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Figure 7.5.1 (refer to Figure below)
 Figure 7.5.1 (refer to Figure below)    shows a capacitively coupled amplifier. In the following, assume operation at midband frequencies where the coupling and bypass capacitors behave as short circuits. The BJT has  \beta=99  and the Early effect can be neglected. (a) If the de voltage at the base is to be  5 \mathrm{~V}  and the emitter current  1 \mathrm{~mA} , find the required value of  \left(R_{E 1}+R_{E 2}\right) . Assume  V_{B E}=0.7 \mathrm{~V} . (b) If the input resistance at the base,  R_{i b} , is to be  10 \mathrm{k} \Omega , find the required value of  R_{E 1}  and hence the value of  R_{E 2} . (c) If the dc current in  R_{B 2}  is to be  0.1 \mathrm{~mA} , find the values of  R_{B 1}  and  R_{B 2} . (d) Find the value of  R_{C}  that results in a de voltage of  +6 \mathrm{~V}  at the collector. (e) Find the input resistance  R_{\text {in }}  and the value of  v_{b} / v_{\text {sig. }} . (f) Find the value of  v_{o} / v_{b} . (g) Find the value of the overall voltage gain  v_{o} / v_{\text {sig }} . (h) If the peak amplitude of the signal between base and emitter  \left(v_{\pi}\right)  is to be limited to  5 \mathrm{mV} , what are the corresponding amplitudes of  v_{b}, v_{\mathrm{sig}} , and  v_{o}  ?

shows a capacitively coupled amplifier. In the following, assume operation at midband frequencies where the coupling and bypass capacitors behave as short circuits. The BJT has β=99\beta=99 and the Early effect can be neglected.
(a) If the de voltage at the base is to be 5 V5 \mathrm{~V} and the emitter current 1 mA1 \mathrm{~mA} , find the required value of (RE1+RE2)\left(R_{E 1}+R_{E 2}\right) . Assume VBE=0.7 VV_{B E}=0.7 \mathrm{~V} .
(b) If the input resistance at the base, RibR_{i b} , is to be 10kΩ10 \mathrm{k} \Omega , find the required value of RE1R_{E 1} and hence the value of RE2R_{E 2} .
(c) If the dc current in RB2R_{B 2} is to be 0.1 mA0.1 \mathrm{~mA} , find the values of RB1R_{B 1} and RB2R_{B 2} .
(d) Find the value of RCR_{C} that results in a de voltage of +6 V+6 \mathrm{~V} at the collector.
(e) Find the input resistance Rin R_{\text {in }} and the value of vb/vsig. v_{b} / v_{\text {sig. }} .
(f) Find the value of vo/vbv_{o} / v_{b} .
(g) Find the value of the overall voltage gain vo/vsig v_{o} / v_{\text {sig }} .
(h) If the peak amplitude of the signal between base and emitter (vπ)\left(v_{\pi}\right) is to be limited to 5mV5 \mathrm{mV} , what are the corresponding amplitudes of vb,vsigv_{b}, v_{\mathrm{sig}} , and vov_{o} ?
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6
    Figure 7.6.1 (a) Perform a dc bias design for the amplifier in Fig. 7.6.1. For this purpose, assume  \beta  is very high and  V_{B E}=0.7 \mathrm{~V}  and neglect the Early effect. Design to obtain a dc base voltage of  V_{C C} / 3 , a dc emitter current of  1 \mathrm{~mA} , and a dc voltage at the collector that allows for  \pm 1-\mathrm{V}  signal swing at the collector with the minimum collector voltage no lower than  V_{B} . Use a de current in the base voltage divider of  I_{E} / 10 . What values are required for  R_{B 1}, R_{B 2}, R_{E} , and  R_{C}  ? (b) If the transistor has  \beta=100 , find the actual values obtained for  I_{E}, I_{C}, V_{B} , and  V_{C} . (c) What are the values of  g_{m}, r_{e} , and  r_{\pi}  at the dc bias point. (d) Assuming very large coupling and bypass capacitors, find the values of  R_{E 1}  and  R_{E 2}  that result in  R_{\text {in }}=10 \mathrm{k} \Omega . (e) Find the overall voltage gain  G_{v}=v_{o} / v_{\text {sig }} . (f) For  v_{o} , a sine wave with a  1-\mathrm{V}  peak amplitude, what is the peak amplitude required of the sine wave  v_{b e}  ? If this value is greater than  5 \mathrm{mV} , reduce  v_{\text {sig }}  to limit the peak amplitude of  v_{b e}  to  5 \mathrm{mV} . What is the resulting output voltage in this case?

Figure 7.6.1
(a) Perform a dc bias design for the amplifier in Fig. 7.6.1. For this purpose, assume β\beta is very high and VBE=0.7 VV_{B E}=0.7 \mathrm{~V} and neglect the Early effect. Design to obtain a dc base voltage of VCC/3V_{C C} / 3 , a dc emitter current of 1 mA1 \mathrm{~mA} , and a dc voltage at the collector that allows for ±1V\pm 1-\mathrm{V} signal swing at the collector with the minimum collector voltage no lower than VBV_{B} . Use a de current in the base voltage divider of IE/10I_{E} / 10 . What values are required for RB1,RB2,RER_{B 1}, R_{B 2}, R_{E} , and RCR_{C} ?
(b) If the transistor has β=100\beta=100 , find the actual values obtained for IE,IC,VBI_{E}, I_{C}, V_{B} , and VCV_{C} .
(c) What are the values of gm,reg_{m}, r_{e} , and rπr_{\pi} at the dc bias point.
(d) Assuming very large coupling and bypass capacitors, find the values of RE1R_{E 1} and RE2R_{E 2} that result in Rin =10kΩR_{\text {in }}=10 \mathrm{k} \Omega .
(e) Find the overall voltage gain Gv=vo/vsig G_{v}=v_{o} / v_{\text {sig }} .
(f) For vov_{o} , a sine wave with a 1V1-\mathrm{V} peak amplitude, what is the peak amplitude required of the sine wave vbev_{b e} ? If this value is greater than 5mV5 \mathrm{mV} , reduce vsig v_{\text {sig }} to limit the peak amplitude of vbev_{b e} to 5mV5 \mathrm{mV} . What is the resulting output voltage in this case?
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7
The transistor in the emitter follower of Fig. 7.7.1 (refer to Figure below)
 The transistor in the emitter follower of Fig. 7.7.1 (refer to Figure below)    Figure 7.7.1 has  \beta=100 . Assume  V_{B E}=0.7 \mathrm{~V}  and neglect the Early effect. (a) Find the dc emitter current  I_{E} . (b) Find the value of the emitter resistance  r_{e} . (c) Find the input resistance  R_{\text {in }} . (d) Find the voltage gain from signal source to the transistor base,  v_{b} / v_{\text {sig }} . (e) Find the voltage gain from transistor base to the output,  v_{o} / v_{b} . (f) Find the overall voltage gain,  v_{o} / v_{\text {sig }} . (g) Find the output resistance  R_{\text {out }} .

Figure 7.7.1
has β=100\beta=100 . Assume VBE=0.7 VV_{B E}=0.7 \mathrm{~V} and neglect the Early effect.
(a) Find the dc emitter current IEI_{E} .
(b) Find the value of the emitter resistance rer_{e} .
(c) Find the input resistance Rin R_{\text {in }} .
(d) Find the voltage gain from signal source to the transistor base, vb/vsig v_{b} / v_{\text {sig }} .
(e) Find the voltage gain from transistor base to the output, vo/vbv_{o} / v_{b} .
(f) Find the overall voltage gain, vo/vsig v_{o} / v_{\text {sig }} .
(g) Find the output resistance Rout R_{\text {out }} .
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8
   Figure 7.8.1 The transistors in the amplifier shown in Fig. 7.8.1 have  \beta=100, V_{B E}=0.7 \mathrm{~V} , and negligible Early effect. (a) Design the circuit to obtain the following dc operating parameters:  V_{B 1}=+3 \mathrm{~V}, I_{E 1}=1 \mathrm{~mA} ,  V_{E 2}=+4 \mathrm{~V} , and  I_{E 2}=4 \mathrm{~mA} . Use  I=0.1 \mathrm{~mA} . Find the values of  R_{1}, R_{2}, R_{3},\left(R_{4}+R_{5}\right) , and  R_{6} . (b) Find the value of  R_{4}  (and hence  R_{5}  ) that results in  R_{\text {in }}=10 \mathrm{k} \Omega . (c) Find  R_{i 2} . (d) Find the voltage gain  v_{c 1} / v_{i} . (e) Find the voltage gain  v_{O} / v_{c 1} . (f) Find the overall voltage gain  v_{o} / v_{i} . (g) Find the output resistance  R_{\text {out }} . (h) In order to minimize nonlinear distortion, it is required to keep the maximum signal across the base-emitter junction of each of  Q_{1}  and  Q_{2}  to  5 \mathrm{mV} . Under this constraint, what is the maximum peak-to-peak sine wave signal that can be obtained at the output?
Figure 7.8.1
The transistors in the amplifier shown in Fig. 7.8.1
have β=100,VBE=0.7 V\beta=100, V_{B E}=0.7 \mathrm{~V} , and negligible Early effect.
(a) Design the circuit to obtain the following dc operating parameters: VB1=+3 V,IE1=1 mAV_{B 1}=+3 \mathrm{~V}, I_{E 1}=1 \mathrm{~mA} , VE2=+4 VV_{E 2}=+4 \mathrm{~V} , and IE2=4 mAI_{E 2}=4 \mathrm{~mA} . Use I=0.1 mAI=0.1 \mathrm{~mA} . Find the values of R1,R2,R3,(R4+R5)R_{1}, R_{2}, R_{3},\left(R_{4}+R_{5}\right) , and R6R_{6} . (b) Find the value of R4R_{4} (and hence R5R_{5} ) that results in Rin =10kΩR_{\text {in }}=10 \mathrm{k} \Omega .
(c) Find Ri2R_{i 2} .
(d) Find the voltage gain vc1/viv_{c 1} / v_{i} .
(e) Find the voltage gain vO/vc1v_{O} / v_{c 1} .
(f) Find the overall voltage gain vo/viv_{o} / v_{i} .
(g) Find the output resistance Rout R_{\text {out }} .
(h) In order to minimize nonlinear distortion, it is required to keep the maximum signal across the base-emitter junction of each of Q1Q_{1} and Q2Q_{2} to 5mV5 \mathrm{mV} . Under this constraint, what is the maximum peak-to-peak sine wave signal that can be obtained at the output?
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