5 MOS Field-Effect Transistors (MOSFETs)

1 5 MOS Field-Effect Transistors (MOSFETs). Section : Device Structure and iD for vGS changing from V by +10 mV and Physical Operation by 10 mV. Comment. An NMOS transistor is fabricated in a m CMOS process with L = and W = m. The process technology is specified to have tox = nm, n = 400 cm2 /V s, and Vtn = V. (a) Find Cox , k n , and kn . (b) Find the overdrive voltage VOV and the mini- I. mum value of VDS required to operate the transis- tor in saturation at a current ID = 100 A. What gate-to-source voltage is required? +. (c) If vDS is very small, what values of VOV and VGS are required to operate the MOSFET as a VDS =VGS. 2-k resistance? If VGS is doubled, what rDS . results? If VGS is reduced, at what value does rDS.

2 Become infinite? Figure Section : Current Voltage An NMOS transistor fabricated in a process for Characteristics which the process transconductance parameter is 400 A/V2 has its gate and drain connected together. The resulting two-terminal device is fed An NMOS transistor fabricated in a m with a current source I as shown in Fig. process has L = m and W = 2 m. With I = 40 A, the voltage across the device The process technology has Cox = fF/ m2 , is measured to be V. When I is increased to n = 450 cm2 /V s, and Vtn = V. Neglect the 90 A, the voltage increases to V. Find Vt channel-length modulation effect . and W/L of the transistor . Ignore channel-length (a) If the transistor is to operate at the edge of modulation.

3 The saturation region with ID = 100 A, find the values required of VGS and VDS . (b) If VGS is kept constant at the value found in (a) while VDS is changed, find ID that results at VDS An NMOS transistor for which kn = 4 mA/V2. equal to half the value in (a) and at VDS equal to and Vt = V is operated with VGS = VDS =. the value in (a). V. What current results? To what value can (c) To investigate the operation of the MOSFET VDS be reduced while maintaining the current as a linear amplifier, let the operating point be at unchanged? If the transistor is replaced with VGS = V and VDS = V. Find the change in another fabricated in the same technology but 5-1. 5-2. with twice the width, what current results? For Section : MOSFET Circuits at DC.

4 Each of the two Transistors when operated at small VDS , what is the range of linear resis- tance rDS obtained when VGS is varied over VDD = +1 V. the range V to 1 V? Neglect channel-length modulation. ID RD. VD. An NMOS transistor is fabricated in a m process having k n = 500 A/V2 , and V A =. 5 V/ m. (a) If L = m and W = m, find VA and . (b) If the device is operated at VOV = V and VS. VDS = V, find ID . (c) Find ro at the operating point specified in (b). ID RS. (d) If VDS is increased to V, what is the corre- sponding change in ID ? Do this two ways: using the expression for ID and using ro . Compare the results obtained. VSS = 1 V. Figure VDD = +1 V. VDD = V. ID RD. VD. vG. ID. vD. VS. Figure I. The PMOS transistor in Fig.

5 Has V tp = VSS = 1 V. V, k p = 100 A/V2 , and W/L = 10. (a) Find the range of vG for which the transistor Figure conducts. (b) In terms of vG , find the range of vD for which The NMOS transistor in the circuit in Fig. the transistor operates in the triode region. has Vtn = V, k n = 400 A/V2 , W/L = 10, and (c) In terms of vG , find the range of vD for which = 0. the transistor operates in saturation. (a) Design the circuit ( , find the required val- (d) Find the value of vG and the range of vD for ues for RS and RD ) to obtain ID = 180 A and which the transistor operates in saturation with VD = + V. Find the voltage VS that results. ID = 20 A. Assume = 0. (b) If RS is replaced with a constant-current source (e) If | | = V 1 , find ro at the operating point I, as shown in Fig.

6 , what must the value of in (d). I be to obtain the same operating conditions as in (f) For VOV equal to that in (d) and | | = V 1 , (a)? find the value of ID at VD = 1 V and at VD = 0 V. (c) What is the largest value to which RD. Use these values to calculate the output resistance can be increased while the transistor remains in ro and compare the result to that found in (e). saturation? 5-3. VDD = +1 V. VDD = +5 V. RS. RG1 RD1 RS2. VS 3 M k k . Q2. VD. Q1. RD. RG2 RS1 RD2. 2 M k 8 k . VSS = 1 V. Figure Figure The PMOS transistor in the circuit in Fig. has Vtp = V, k p = 100 A/V2 , W/L = 20, For the circuit in Fig. , the NMOS transis- and = 0. tor has Vtn = V, kn = 10 mA/V2 , and n = 0, (a) Find RS and RD to obtain ID = mA and and the PMOS transistor has Vtp = V, kp =.

7 VD = 0 V. mA/V2 , and | p | = 0. Observe that Q1 and its (b) What is the largest RD for which the Transistors surrounding circuit is the same as the circuit ana- remains in saturation. At this value of RD , what is lyzed in Problem (Fig. ), and you may use the voltage at the drain, VD ? the results found in the solution to that problem here. Analyze the circuit to determine the currents in all branches and the voltages at all nodes. VDD = +5 V. I ID RD = k . VDD = +10 V. RG1 = 3 M . VD I. RG1 ID RD. VG. VD. VS VG Q. VS. RG2 = 2 M ID RS = k RG2 ID RS. Figure Figure The NMOS transistor in the circuit in Fig. has Vt = V, kn = 10 mA/V2 , and = 0. Ana- Design the circuit in Fig. to obtain I = 1 A, lyze the circuit to determine the currents through ID = mA, VS = 2 V, and VD = 5 V.

8 The NMOS. all branches and to find the voltages at all nodes. transistor has Vt = V, kn = 4 mA/V2 , and = 0. 5-4. + V + V. +. vI Q1 V CS I.. vO vO. +. V CS I vI Q2.. V V. (a) (b). + V + V. + +. V CS I V CS I.. vI Q1 vO Q2. vI Q1. Q2 vO. + +. V CS I V CS I.. V V. (c) (d). Figure The Transistors in the circuits of Fig. have in saturation. In each case find the allowable |Vt | = V, kn = kp = 20 mA/V2 , and = 0. Also, ranges of vO and vI . Assume that the minimum I = mA. For each circuit find vO as a func- voltage VCS required across each current source tion of vI assuming the Transistors are operating is V. 5 MOS Field-Effect Transistors (MOSFETs). To operate in saturation, VDS must at least be equal (a) to VOV , thus L = = = m ox VDSmin = V.

9 Cox =. tox The gate-to-source voltage is where ox = 0 = 10 12 VGS = Vtn + VOV = + = V. = 10 11 F/m (c) When vDS is small, 10 11 F/m Cox = iD kn VOV vDS. 10 9 m = 10 2 F/m2 and v = 10 2 1015 10 12 fF/ m2 rDS DS = 1/kn VOV. iD. = fF/ m2. Thus, for rDS = 2 k , k n = n Cox 1. = 400 (cm2 /V s) (fF/ m2 ) 2 103 =. 10 3 VOV. = 400 108 ( m2 /V s) 10 15 (F/ m2 ). VOV = V. = 512 10 6 (F/V s) = 512 10 6 (A/V2 ). and, correspondingly, = 512 A/V2. kn = k n (W/L) VGS = + = V. = 512 = 3413 A/V2 If VGS is doubled, we obtain kn = mA/V2 VGS = 2 = V. (b) When the MOSFET operates in saturation, we and have 1 2 VOV = = V. ID = kn VOV. 2. Thus, Thus, correspondingly, rDS becomes 1 1 1. 100 = 2. 3413 VOV rDS = =. 2 kn VOV 10 3 VOV = V =.

10 5-5. 5-6. As VGS is reduced, rDS increases, becoming infi- For nite when the channel disappears, which occurs as VOV reaches zero or, correspondingly, vDS = VDSmin = = V. VGS = Vtn = V we get iD = 576 10 ( ). (a) When the transistor operates in saturation, we = A. obtain . 1 W 2 (c) For VGS = V ( , VOV = V) and VDS =. ID = n Cox VOV. 2 L V, the MOSFET will be operating in saturation where with . n Cox = 450 (cm2 /V s) (fF/ m2 ) 1 W 2. ID = n Cox VOV. 2 L. = 450 108 ( m2 /V s) 10 15 (F/ m2 ). 1 2. = 576 10 6 (F/V s) = 576 2 = 576 A/V2 = A. To obtain ID = 100 A, VOV can be found from Now, if vGS is increased by a 10-mV increment, then vGS = + = V. 1 2 m 2. 100 = 576 VOV and the current becomes 2 m VOV = V 1 2. iD = 576 ( )2.