Deck 4: Building Blocks of Integrated-Circuit Amplifiers

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    Figure 8.1.1 For the circuit shown in Fig. 8.1.1, assume  Q_{1}  and  Q_{2}  are perfectly matched, and  Q_{3}, Q_{4} , and  Q_{5}  are perfectly matched. Also, assume all transistors have very high  \beta  and  V_{A} . Find  V_{1}, I , and  V_{2} .<div style=padding-top: 35px>

Figure 8.1.1
For the circuit shown in Fig. 8.1.1, assume Q1Q_{1} and Q2Q_{2} are perfectly matched, and Q3,Q4Q_{3}, Q_{4} , and Q5Q_{5} are perfectly matched. Also, assume all transistors have very high β\beta and VAV_{A} . Find V1,IV_{1}, I , and V2V_{2} .
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Given the availability of NMOS and PMOS transistors that are matched and have kn=kp=k_{n}=k_{p}= 5 mA/V25 \mathrm{~mA} / \mathrm{V}^{2} and VAn=VAp=5 VV_{A n}=-V_{A p}=5 \mathrm{~V} , give circuits for the following amplifiers. In each case, find the output resistance and the voltage gain. In all cases, operate each transistor at a dc bias current of 100μA100 \mu \mathrm{A} and give the value of VOV\left|V_{O V}\right| at which each transistor is operating.
(a) An NMOS common-source amplifier with a PMOS current-source load.
(b) An NMOS common-source amplifier with an NMOS cascode stage and a single-transistor PMOS current-source load.
(c) An NMOS common-source amplifier with an NMOS cascode stage and a cascode PMOS current-source load.
Question
    Figure 8.3.1 Design the double-cascode current source shown in Fig. 8.3.1 to provide  I=0.2 \mathrm{~mA}  and the largest possible signal swing at the output; that is, design for the minimum allowable voltage across each transistor. The CMOS fabrication process available has  \left|V_{t p}\right|=0.4 \mathrm{~V},\left|V_{A}^{\prime}\right|=6 \mathrm{~V} / \mu \mathrm{m} , and  \mu_{p} C_{o x}=100 \mu \mathrm{A} / \mathrm{V}^{2} . Use devices with  L=   0.4 \mu \mathrm{m} , and operate at  \left|V_{O V}\right|=0.2 \mathrm{~V} . (a) Specify  V_{G 1}, V_{G 2} , and  V_{G 3} . (b) Find the  W / L  ratios of the transistors. (c) What is the value of  R_{O}  achieved? (d) What is the maximum allowable voltage at the current-source output? (e) If this current source is used as the load of an NMOS double-cascode amplifier having a shortcircuit transconductance of  2 \mathrm{~mA} / \mathrm{V}  and an output resistance equal to  R_{O}  of the current source, what voltage gain is obtained?<div style=padding-top: 35px>

Figure 8.3.1
Design the double-cascode current source shown in Fig. 8.3.1 to provide I=0.2 mAI=0.2 \mathrm{~mA} and the largest possible signal swing at the output; that is, design for the minimum allowable voltage across each transistor. The CMOS fabrication process available has Vtp=0.4 V,VA=6 V/μm\left|V_{t p}\right|=0.4 \mathrm{~V},\left|V_{A}^{\prime}\right|=6 \mathrm{~V} / \mu \mathrm{m} , and μpCox=100μA/V2\mu_{p} C_{o x}=100 \mu \mathrm{A} / \mathrm{V}^{2} . Use devices with L=L= 0.4μm0.4 \mu \mathrm{m} , and operate at VOV=0.2 V\left|V_{O V}\right|=0.2 \mathrm{~V} .
(a) Specify VG1,VG2V_{G 1}, V_{G 2} , and VG3V_{G 3} .
(b) Find the W/LW / L ratios of the transistors.
(c) What is the value of ROR_{O} achieved?
(d) What is the maximum allowable voltage at the current-source output?
(e) If this current source is used as the load of an NMOS double-cascode amplifier having a shortcircuit transconductance of 2 mA/V2 \mathrm{~mA} / \mathrm{V} and an output resistance equal to ROR_{O} of the current source, what voltage gain is obtained?
Question
    Figure 8.4.1 The cascode current source in Fig. 8.4.1 utilizes two identical PMOS transistors fabricated in a 0.18 -  \mu \mathrm{m}  CMOS process for which  V_{D D}=1.8 \mathrm{~V} ,  V_{t p}=-0.5 \mathrm{~V}, V_{A}^{\prime}=-6 \mathrm{~V} / \mu \mathrm{m} , and  \mu_{p} C_{o x}=   100 \mu \mathrm{A} / \mathrm{V}^{2} . Design the circuit to obtain  I=50 \mu \mathrm{A}  and  R_{O}=1 \mathrm{M} \Omega  and to allow for the maximum possible voltage swing at the output terminal of the current source. Utilize  \left|V_{O V}\right|=0.2 \mathrm{~V} . Specify the values of  L  and  W / L  for  Q_{1}  and  Q_{2} . As well, specify the required values of the dc bias voltages  V_{G 1}  and  V_{G 2} . What is the maximum allowable voltage at the output?<div style=padding-top: 35px>

Figure 8.4.1
The cascode current source in Fig. 8.4.1 utilizes two identical PMOS transistors fabricated in a 0.18 - μm\mu \mathrm{m} CMOS process for which VDD=1.8 VV_{D D}=1.8 \mathrm{~V} , Vtp=0.5 V,VA=6 V/μmV_{t p}=-0.5 \mathrm{~V}, V_{A}^{\prime}=-6 \mathrm{~V} / \mu \mathrm{m} , and μpCox=\mu_{p} C_{o x}= 100μA/V2100 \mu \mathrm{A} / \mathrm{V}^{2} .
Design the circuit to obtain I=50μAI=50 \mu \mathrm{A} and RO=1MΩR_{O}=1 \mathrm{M} \Omega and to allow for the maximum possible voltage swing at the output terminal of the current source. Utilize VOV=0.2 V\left|V_{O V}\right|=0.2 \mathrm{~V} . Specify the values of LL and W/LW / L for Q1Q_{1} and Q2Q_{2} . As well, specify the required values of the dc bias voltages VG1V_{G 1} and VG2V_{G 2} . What is the maximum allowable voltage at the output?
Question
    Figure 8.5.1 The modified Wilson current mirror of Fig. 8.5.1 utilizes four matched transistors for which  V_{t}=   0.4 \mathrm{~V}, \mu_{n} C_{o x}=400 \mu \mathrm{A} / \mathrm{V}^{2} , and  V_{A}=3 \mathrm{~V} . It is required to design the circuit so that  I_{O}=0.2 \mathrm{~mA}  with all transistors operating at  V_{O V}=0.2 \mathrm{~V} . (a) Find the required value of  R . (b) Find the required value of the  W / L  ratio for each of the four transistors. (c) Find the value of the output resistance  R_{O} . (d) What is the minimum voltage permitted at the output?<div style=padding-top: 35px>

Figure 8.5.1
The modified Wilson current mirror of Fig. 8.5.1 utilizes four matched transistors for which Vt=V_{t}= 0.4 V,μnCox=400μA/V20.4 \mathrm{~V}, \mu_{n} C_{o x}=400 \mu \mathrm{A} / \mathrm{V}^{2} , and VA=3 VV_{A}=3 \mathrm{~V} . It is required to design the circuit so that IO=0.2 mAI_{O}=0.2 \mathrm{~mA} with all transistors operating at VOV=0.2 VV_{O V}=0.2 \mathrm{~V} .
(a) Find the required value of RR .
(b) Find the required value of the W/LW / L ratio for each of the four transistors.
(c) Find the value of the output resistance ROR_{O} .
(d) What is the minimum voltage permitted at the output?
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Deck 4: Building Blocks of Integrated-Circuit Amplifiers
1
    Figure 8.1.1 For the circuit shown in Fig. 8.1.1, assume  Q_{1}  and  Q_{2}  are perfectly matched, and  Q_{3}, Q_{4} , and  Q_{5}  are perfectly matched. Also, assume all transistors have very high  \beta  and  V_{A} . Find  V_{1}, I , and  V_{2} .

Figure 8.1.1
For the circuit shown in Fig. 8.1.1, assume Q1Q_{1} and Q2Q_{2} are perfectly matched, and Q3,Q4Q_{3}, Q_{4} , and Q5Q_{5} are perfectly matched. Also, assume all transistors have very high β\beta and VAV_{A} . Find V1,IV_{1}, I , and V2V_{2} .
   Figure 8.1.2 The analysis is shown on Fig. 8.1.2.

Figure 8.1.2
The analysis is shown on Fig. 8.1.2.
2
Given the availability of NMOS and PMOS transistors that are matched and have kn=kp=k_{n}=k_{p}= 5 mA/V25 \mathrm{~mA} / \mathrm{V}^{2} and VAn=VAp=5 VV_{A n}=-V_{A p}=5 \mathrm{~V} , give circuits for the following amplifiers. In each case, find the output resistance and the voltage gain. In all cases, operate each transistor at a dc bias current of 100μA100 \mu \mathrm{A} and give the value of VOV\left|V_{O V}\right| at which each transistor is operating.
(a) An NMOS common-source amplifier with a PMOS current-source load.
(b) An NMOS common-source amplifier with an NMOS cascode stage and a single-transistor PMOS current-source load.
(c) An NMOS common-source amplifier with an NMOS cascode stage and a cascode PMOS current-source load.
For each transistor,
kn=kp=5 mA/V2VAn=VAp=5 VID=100μA\begin{aligned}k_{n} & =k_{p}=5 \mathrm{~mA} / \mathrm{V}^{2} \\V_{A n} & =-V_{A p}=5 \mathrm{~V} \\I_{D} & =100 \mu \mathrm{A}\end{aligned}
ID=12kVOV20.1=12×5×VOV2VOV=0.2 Vgm=2IDVOV=2×0.10.2=1 mA/VrO=VAID=50.1=50kΩ\begin{aligned}I_{D} & =\frac{1}{2} k\left|V_{O V}\right|^{2} \\0.1 & =\frac{1}{2} \times 5 \times\left|V_{O V}\right|^{2} \\\Rightarrow\left|V_{O V}\right| & =0.2 \mathrm{~V} \\g_{m} & =\frac{2 I_{D}}{\left|V_{O V}\right|}=\frac{2 \times 0.1}{0.2}=1 \mathrm{~mA} / \mathrm{V} \\r_{O} & =\frac{\left|V_{A}\right|}{I_{D}}=\frac{5}{0.1}=50 \mathrm{k} \Omega\end{aligned}
(a)
 For each transistor,  \begin{aligned} k_{n} & =k_{p}=5 \mathrm{~mA} / \mathrm{V}^{2} \\ V_{A n} & =-V_{A p}=5 \mathrm{~V} \\ I_{D} & =100 \mu \mathrm{A} \end{aligned}   \begin{aligned} I_{D} & =\frac{1}{2} k\left|V_{O V}\right|^{2} \\ 0.1 & =\frac{1}{2} \times 5 \times\left|V_{O V}\right|^{2} \\ \Rightarrow\left|V_{O V}\right| & =0.2 \mathrm{~V} \\ g_{m} & =\frac{2 I_{D}}{\left|V_{O V}\right|}=\frac{2 \times 0.1}{0.2}=1 \mathrm{~mA} / \mathrm{V} \\ r_{O} & =\frac{\left|V_{A}\right|}{I_{D}}=\frac{5}{0.1}=50 \mathrm{k} \Omega \end{aligned}  (a)   Figure 8.2.1 See Figure 8.2.1.  \begin{aligned} & R_{o}=r_{o n}\left\|r_{o p}=50\right\| 50=25 \mathrm{k} \Omega \\ & A_{V}=-g_{m} R_{O}=-1 \times 25=-25 \mathrm{~V} / \mathrm{V} \end{aligned}  (b)    Figure 8.2.2  \begin{aligned} R_{O 2} & \simeq\left(g_{m 2} r_{o 2}\right) r_{o 1} \\ & =(1 \times 50) \times 50=2500 \mathrm{k} \Omega \\ R_{O 3} & =r_{o 3}=50 \mathrm{k} \Omega \\ R_{O} & =R_{O 2}\left\|R_{O 3}=50\right\| 2500=49 \mathrm{k} \Omega \\ A_{v} & =-g_{m 1} R_{O}=-1 \times 49=-49 \mathrm{~V} / \mathrm{V} \end{aligned}  (c)    Figure 8.2.3  \begin{aligned} R_{o 2} \simeq\left(g_{m 2} r_{o 2}\right) r_{o 1} & =(1 \times 50) \times 50=2500 \mathrm{k} \Omega \\ R_{O 3} \simeq\left(g_{m 3} r_{o 3}\right) r_{o 4} & =(1 \times 50) \times 50=2500 \mathrm{k} \Omega \\ R_{O}=R_{O 2} \| R_{O 3} & =2500 \| 2500=1250 \mathrm{k} \Omega \\ A_{v}=-g_{m 1} R_{O} & =-1 \times 1250=-1250 \mathrm{~V} / \mathrm{V} \end{aligned}
Figure 8.2.1
See Figure 8.2.1.
Ro=ronrop=5050=25kΩAV=gmRO=1×25=25 V/V\begin{aligned}& R_{o}=r_{o n}\left\|r_{o p}=50\right\| 50=25 \mathrm{k} \Omega \\& A_{V}=-g_{m} R_{O}=-1 \times 25=-25 \mathrm{~V} / \mathrm{V}\end{aligned}
(b)
 For each transistor,  \begin{aligned} k_{n} & =k_{p}=5 \mathrm{~mA} / \mathrm{V}^{2} \\ V_{A n} & =-V_{A p}=5 \mathrm{~V} \\ I_{D} & =100 \mu \mathrm{A} \end{aligned}   \begin{aligned} I_{D} & =\frac{1}{2} k\left|V_{O V}\right|^{2} \\ 0.1 & =\frac{1}{2} \times 5 \times\left|V_{O V}\right|^{2} \\ \Rightarrow\left|V_{O V}\right| & =0.2 \mathrm{~V} \\ g_{m} & =\frac{2 I_{D}}{\left|V_{O V}\right|}=\frac{2 \times 0.1}{0.2}=1 \mathrm{~mA} / \mathrm{V} \\ r_{O} & =\frac{\left|V_{A}\right|}{I_{D}}=\frac{5}{0.1}=50 \mathrm{k} \Omega \end{aligned}  (a)   Figure 8.2.1 See Figure 8.2.1.  \begin{aligned} & R_{o}=r_{o n}\left\|r_{o p}=50\right\| 50=25 \mathrm{k} \Omega \\ & A_{V}=-g_{m} R_{O}=-1 \times 25=-25 \mathrm{~V} / \mathrm{V} \end{aligned}  (b)    Figure 8.2.2  \begin{aligned} R_{O 2} & \simeq\left(g_{m 2} r_{o 2}\right) r_{o 1} \\ & =(1 \times 50) \times 50=2500 \mathrm{k} \Omega \\ R_{O 3} & =r_{o 3}=50 \mathrm{k} \Omega \\ R_{O} & =R_{O 2}\left\|R_{O 3}=50\right\| 2500=49 \mathrm{k} \Omega \\ A_{v} & =-g_{m 1} R_{O}=-1 \times 49=-49 \mathrm{~V} / \mathrm{V} \end{aligned}  (c)    Figure 8.2.3  \begin{aligned} R_{o 2} \simeq\left(g_{m 2} r_{o 2}\right) r_{o 1} & =(1 \times 50) \times 50=2500 \mathrm{k} \Omega \\ R_{O 3} \simeq\left(g_{m 3} r_{o 3}\right) r_{o 4} & =(1 \times 50) \times 50=2500 \mathrm{k} \Omega \\ R_{O}=R_{O 2} \| R_{O 3} & =2500 \| 2500=1250 \mathrm{k} \Omega \\ A_{v}=-g_{m 1} R_{O} & =-1 \times 1250=-1250 \mathrm{~V} / \mathrm{V} \end{aligned}

Figure 8.2.2
RO2(gm2ro2)ro1=(1×50)×50=2500kΩRO3=ro3=50kΩRO=RO2RO3=502500=49kΩAv=gm1RO=1×49=49 V/V\begin{aligned}R_{O 2} & \simeq\left(g_{m 2} r_{o 2}\right) r_{o 1} \\& =(1 \times 50) \times 50=2500 \mathrm{k} \Omega \\R_{O 3} & =r_{o 3}=50 \mathrm{k} \Omega \\R_{O} & =R_{O 2}\left\|R_{O 3}=50\right\| 2500=49 \mathrm{k} \Omega \\A_{v} & =-g_{m 1} R_{O}=-1 \times 49=-49 \mathrm{~V} / \mathrm{V}\end{aligned}
(c)
 For each transistor,  \begin{aligned} k_{n} & =k_{p}=5 \mathrm{~mA} / \mathrm{V}^{2} \\ V_{A n} & =-V_{A p}=5 \mathrm{~V} \\ I_{D} & =100 \mu \mathrm{A} \end{aligned}   \begin{aligned} I_{D} & =\frac{1}{2} k\left|V_{O V}\right|^{2} \\ 0.1 & =\frac{1}{2} \times 5 \times\left|V_{O V}\right|^{2} \\ \Rightarrow\left|V_{O V}\right| & =0.2 \mathrm{~V} \\ g_{m} & =\frac{2 I_{D}}{\left|V_{O V}\right|}=\frac{2 \times 0.1}{0.2}=1 \mathrm{~mA} / \mathrm{V} \\ r_{O} & =\frac{\left|V_{A}\right|}{I_{D}}=\frac{5}{0.1}=50 \mathrm{k} \Omega \end{aligned}  (a)   Figure 8.2.1 See Figure 8.2.1.  \begin{aligned} & R_{o}=r_{o n}\left\|r_{o p}=50\right\| 50=25 \mathrm{k} \Omega \\ & A_{V}=-g_{m} R_{O}=-1 \times 25=-25 \mathrm{~V} / \mathrm{V} \end{aligned}  (b)    Figure 8.2.2  \begin{aligned} R_{O 2} & \simeq\left(g_{m 2} r_{o 2}\right) r_{o 1} \\ & =(1 \times 50) \times 50=2500 \mathrm{k} \Omega \\ R_{O 3} & =r_{o 3}=50 \mathrm{k} \Omega \\ R_{O} & =R_{O 2}\left\|R_{O 3}=50\right\| 2500=49 \mathrm{k} \Omega \\ A_{v} & =-g_{m 1} R_{O}=-1 \times 49=-49 \mathrm{~V} / \mathrm{V} \end{aligned}  (c)    Figure 8.2.3  \begin{aligned} R_{o 2} \simeq\left(g_{m 2} r_{o 2}\right) r_{o 1} & =(1 \times 50) \times 50=2500 \mathrm{k} \Omega \\ R_{O 3} \simeq\left(g_{m 3} r_{o 3}\right) r_{o 4} & =(1 \times 50) \times 50=2500 \mathrm{k} \Omega \\ R_{O}=R_{O 2} \| R_{O 3} & =2500 \| 2500=1250 \mathrm{k} \Omega \\ A_{v}=-g_{m 1} R_{O} & =-1 \times 1250=-1250 \mathrm{~V} / \mathrm{V} \end{aligned}

Figure 8.2.3
Ro2(gm2ro2)ro1=(1×50)×50=2500kΩRO3(gm3ro3)ro4=(1×50)×50=2500kΩRO=RO2RO3=25002500=1250kΩAv=gm1RO=1×1250=1250 V/V\begin{aligned}R_{o 2} \simeq\left(g_{m 2} r_{o 2}\right) r_{o 1} & =(1 \times 50) \times 50=2500 \mathrm{k} \Omega \\R_{O 3} \simeq\left(g_{m 3} r_{o 3}\right) r_{o 4} & =(1 \times 50) \times 50=2500 \mathrm{k} \Omega \\R_{O}=R_{O 2} \| R_{O 3} & =2500 \| 2500=1250 \mathrm{k} \Omega \\A_{v}=-g_{m 1} R_{O} & =-1 \times 1250=-1250 \mathrm{~V} / \mathrm{V}\end{aligned}
3
    Figure 8.3.1 Design the double-cascode current source shown in Fig. 8.3.1 to provide  I=0.2 \mathrm{~mA}  and the largest possible signal swing at the output; that is, design for the minimum allowable voltage across each transistor. The CMOS fabrication process available has  \left|V_{t p}\right|=0.4 \mathrm{~V},\left|V_{A}^{\prime}\right|=6 \mathrm{~V} / \mu \mathrm{m} , and  \mu_{p} C_{o x}=100 \mu \mathrm{A} / \mathrm{V}^{2} . Use devices with  L=   0.4 \mu \mathrm{m} , and operate at  \left|V_{O V}\right|=0.2 \mathrm{~V} . (a) Specify  V_{G 1}, V_{G 2} , and  V_{G 3} . (b) Find the  W / L  ratios of the transistors. (c) What is the value of  R_{O}  achieved? (d) What is the maximum allowable voltage at the current-source output? (e) If this current source is used as the load of an NMOS double-cascode amplifier having a shortcircuit transconductance of  2 \mathrm{~mA} / \mathrm{V}  and an output resistance equal to  R_{O}  of the current source, what voltage gain is obtained?

Figure 8.3.1
Design the double-cascode current source shown in Fig. 8.3.1 to provide I=0.2 mAI=0.2 \mathrm{~mA} and the largest possible signal swing at the output; that is, design for the minimum allowable voltage across each transistor. The CMOS fabrication process available has Vtp=0.4 V,VA=6 V/μm\left|V_{t p}\right|=0.4 \mathrm{~V},\left|V_{A}^{\prime}\right|=6 \mathrm{~V} / \mu \mathrm{m} , and μpCox=100μA/V2\mu_{p} C_{o x}=100 \mu \mathrm{A} / \mathrm{V}^{2} . Use devices with L=L= 0.4μm0.4 \mu \mathrm{m} , and operate at VOV=0.2 V\left|V_{O V}\right|=0.2 \mathrm{~V} .
(a) Specify VG1,VG2V_{G 1}, V_{G 2} , and VG3V_{G 3} .
(b) Find the W/LW / L ratios of the transistors.
(c) What is the value of ROR_{O} achieved?
(d) What is the maximum allowable voltage at the current-source output?
(e) If this current source is used as the load of an NMOS double-cascode amplifier having a shortcircuit transconductance of 2 mA/V2 \mathrm{~mA} / \mathrm{V} and an output resistance equal to ROR_{O} of the current source, what voltage gain is obtained?
    Figure 8.3.2 (a) Refer to Figure 8.3.2. For  \left|V_{t p}\right|=0.4 \mathrm{~V}  and  \left|V_{O V}\right|=0.2 \mathrm{~V}, V_{S G}=0.6 \mathrm{~V}  and the minimum required  V_{S D}  for each transistor is  0.2 \mathrm{~V} . Thus,  V_{G 1}=+1.2 \mathrm{~V}, \quad V_{G 2}=+1.0 \mathrm{~V}, \quad  and  \quad V_{G S}=+0.8 \mathrm{~V}  (b)  \begin{aligned} I_{D} & =\frac{1}{2} \mu_{p} C_{O x}\left(\frac{W}{L}\right)\left|V_{O V}\right|^{2} \\ 0.2 & =\frac{1}{2} \times 0.1 \times \frac{W}{L} \times 0.2^{2} \\ \Rightarrow \frac{W}{L} & =100 \end{aligned}  (c)  \begin{aligned} r_{o 1} & =\frac{\left|V_{A}\right|}{I_{D}}=\frac{\left|V_{A}^{\prime}\right| L}{I_{D}}=\frac{6 \times 0.4}{0.2}=12 \mathrm{k} \Omega \\ r_{o 2} & =r_{o 3}=r_{o 1}=12 \mathrm{k} \Omega \\ g_{m 2} & =g_{m 3}=\frac{2 I_{D}}{\left|V_{O V}\right|}=\frac{2 \times 0.2}{0.2}=2 \mathrm{~mA} / \mathrm{V} \\ R_{o 2} & =\left(g_{m 2} r_{o 2}\right) r_{o 1}=(2 \times 12) \times 12=288 \mathrm{k} \Omega \\ R_{o 3} & =\left(g_{m 3} r_{o 3}\right) R_{o 2}=(2 \times 12) \times 288=6912 \mathrm{k} \Omega \end{aligned}  (d)  v_{O \max }=0.8+0.4=+1.2 \mathrm{~V}  (e)  \begin{aligned} A_{v} & =-G_{m}\left(R_{O} \| R_{O}\right) \\ & =-2(6912 \| 6912)=-6912 \mathrm{~V} / \mathrm{V} \end{aligned}

Figure 8.3.2
(a) Refer to Figure 8.3.2. For Vtp=0.4 V\left|V_{t p}\right|=0.4 \mathrm{~V} and VOV=0.2 V,VSG=0.6 V\left|V_{O V}\right|=0.2 \mathrm{~V}, V_{S G}=0.6 \mathrm{~V} and the minimum required VSDV_{S D} for each transistor is 0.2 V0.2 \mathrm{~V} . Thus,
VG1=+1.2 V,VG2=+1.0 V,V_{G 1}=+1.2 \mathrm{~V}, \quad V_{G 2}=+1.0 \mathrm{~V}, \quad and VGS=+0.8 V\quad V_{G S}=+0.8 \mathrm{~V}
(b)
ID=12μpCOx(WL)VOV20.2=12×0.1×WL×0.22WL=100\begin{aligned}I_{D} & =\frac{1}{2} \mu_{p} C_{O x}\left(\frac{W}{L}\right)\left|V_{O V}\right|^{2} \\0.2 & =\frac{1}{2} \times 0.1 \times \frac{W}{L} \times 0.2^{2} \\\Rightarrow \frac{W}{L} & =100\end{aligned}
(c)
ro1=VAID=VALID=6×0.40.2=12kΩro2=ro3=ro1=12kΩgm2=gm3=2IDVOV=2×0.20.2=2 mA/VRo2=(gm2ro2)ro1=(2×12)×12=288kΩRo3=(gm3ro3)Ro2=(2×12)×288=6912kΩ\begin{aligned}r_{o 1} & =\frac{\left|V_{A}\right|}{I_{D}}=\frac{\left|V_{A}^{\prime}\right| L}{I_{D}}=\frac{6 \times 0.4}{0.2}=12 \mathrm{k} \Omega \\r_{o 2} & =r_{o 3}=r_{o 1}=12 \mathrm{k} \Omega \\g_{m 2} & =g_{m 3}=\frac{2 I_{D}}{\left|V_{O V}\right|}=\frac{2 \times 0.2}{0.2}=2 \mathrm{~mA} / \mathrm{V} \\R_{o 2} & =\left(g_{m 2} r_{o 2}\right) r_{o 1}=(2 \times 12) \times 12=288 \mathrm{k} \Omega \\R_{o 3} & =\left(g_{m 3} r_{o 3}\right) R_{o 2}=(2 \times 12) \times 288=6912 \mathrm{k} \Omega\end{aligned}
(d)
vOmax=0.8+0.4=+1.2 Vv_{O \max }=0.8+0.4=+1.2 \mathrm{~V}
(e)
Av=Gm(RORO)=2(69126912)=6912 V/V\begin{aligned}A_{v} & =-G_{m}\left(R_{O} \| R_{O}\right) \\& =-2(6912 \| 6912)=-6912 \mathrm{~V} / \mathrm{V}\end{aligned}
4
    Figure 8.4.1 The cascode current source in Fig. 8.4.1 utilizes two identical PMOS transistors fabricated in a 0.18 -  \mu \mathrm{m}  CMOS process for which  V_{D D}=1.8 \mathrm{~V} ,  V_{t p}=-0.5 \mathrm{~V}, V_{A}^{\prime}=-6 \mathrm{~V} / \mu \mathrm{m} , and  \mu_{p} C_{o x}=   100 \mu \mathrm{A} / \mathrm{V}^{2} . Design the circuit to obtain  I=50 \mu \mathrm{A}  and  R_{O}=1 \mathrm{M} \Omega  and to allow for the maximum possible voltage swing at the output terminal of the current source. Utilize  \left|V_{O V}\right|=0.2 \mathrm{~V} . Specify the values of  L  and  W / L  for  Q_{1}  and  Q_{2} . As well, specify the required values of the dc bias voltages  V_{G 1}  and  V_{G 2} . What is the maximum allowable voltage at the output?

Figure 8.4.1
The cascode current source in Fig. 8.4.1 utilizes two identical PMOS transistors fabricated in a 0.18 - μm\mu \mathrm{m} CMOS process for which VDD=1.8 VV_{D D}=1.8 \mathrm{~V} , Vtp=0.5 V,VA=6 V/μmV_{t p}=-0.5 \mathrm{~V}, V_{A}^{\prime}=-6 \mathrm{~V} / \mu \mathrm{m} , and μpCox=\mu_{p} C_{o x}= 100μA/V2100 \mu \mathrm{A} / \mathrm{V}^{2} .
Design the circuit to obtain I=50μAI=50 \mu \mathrm{A} and RO=1MΩR_{O}=1 \mathrm{M} \Omega and to allow for the maximum possible voltage swing at the output terminal of the current source. Utilize VOV=0.2 V\left|V_{O V}\right|=0.2 \mathrm{~V} . Specify the values of LL and W/LW / L for Q1Q_{1} and Q2Q_{2} . As well, specify the required values of the dc bias voltages VG1V_{G 1} and VG2V_{G 2} . What is the maximum allowable voltage at the output?
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5
    Figure 8.5.1 The modified Wilson current mirror of Fig. 8.5.1 utilizes four matched transistors for which  V_{t}=   0.4 \mathrm{~V}, \mu_{n} C_{o x}=400 \mu \mathrm{A} / \mathrm{V}^{2} , and  V_{A}=3 \mathrm{~V} . It is required to design the circuit so that  I_{O}=0.2 \mathrm{~mA}  with all transistors operating at  V_{O V}=0.2 \mathrm{~V} . (a) Find the required value of  R . (b) Find the required value of the  W / L  ratio for each of the four transistors. (c) Find the value of the output resistance  R_{O} . (d) What is the minimum voltage permitted at the output?

Figure 8.5.1
The modified Wilson current mirror of Fig. 8.5.1 utilizes four matched transistors for which Vt=V_{t}= 0.4 V,μnCox=400μA/V20.4 \mathrm{~V}, \mu_{n} C_{o x}=400 \mu \mathrm{A} / \mathrm{V}^{2} , and VA=3 VV_{A}=3 \mathrm{~V} . It is required to design the circuit so that IO=0.2 mAI_{O}=0.2 \mathrm{~mA} with all transistors operating at VOV=0.2 VV_{O V}=0.2 \mathrm{~V} .
(a) Find the required value of RR .
(b) Find the required value of the W/LW / L ratio for each of the four transistors.
(c) Find the value of the output resistance ROR_{O} .
(d) What is the minimum voltage permitted at the output?
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