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EECS 40 Spring 2003 Lecture 20 S. Ross Today we will • Review NMOS and PMOS I-V characteristic • Practice useful method for solving transistor circuits • Build a familiar circuit element using a transistor Good luck to Marquette in the NCAA Final Four! EECS 40 Spring 2003 Lecture 20 S. Ross NMOS I-V CHARACTERISTIC G IG ID S - VDS + D • Since the transistor is a 3-terminal device, there is no single I-V characteristic. • Note that because of the gate insulator, IG = 0 A. • We typically define the MOS I-V characteristic as ID vs. VDS for a fixed VGS. • 3 modes of operation EECS 40 Spring 2003 Lecture 20 S. Ross NMOS I-V CHARACTERISTIC ID triode mode saturation mode VGS = 3 V VDS = VGS - VTH(N) VGS = 2 V VGS = 1 V VDS cutoff mode (VGS ≤ VTH(N)) EECS 40 Spring 2003 Lecture 20 S. Ross NMOS I-V CHARACTERISTIC Cutoff Mode • Occurs when VGS ≤ VTH(N) ID = 0 Triode Mode • Occurs when VGS > VTH(N) and VDS < VGS - VTH(N) W ID μnCOX VGS VTH(N) VDS /2 VDS L Saturation Mode • Occurs when VGS > VTH(N) and VDS ≥ VGS - VTH(N) W 1 2 ID μnCOX VGS VTH(N) 1 λn VDS L 2 EECS 40 Spring 2003 Lecture 20 S. Ross PMOS I-V CHARACTERISTIC G IG ID S - VDS + D Symbol has “dot” at gate. NMOS does not. ID, VGS, VDS, and VTH(P) are all negative for PMOS. These values are positive for NMOS. Channel formed when VGS < VTH(P). Opposite for NMOS. Saturation occurs when VDS ≤ VGS – VTH(P). Opposite for NMOS. EECS 40 Spring 2003 Lecture 20 S. Ross PMOS I-V CHARACTERISTIC VDS cutoff mode (VGS ≥ VTH(P)) VGS = -1 V VGS = -2 V VDS = VGS - VTH(P) ID VGS = -3 V saturation mode triode mode EECS 40 Spring 2003 Lecture 20 S. Ross PMOS I-V CHARACTERISTIC Cutoff Mode • Occurs when VGS ≥ VTH(P) ID = 0 Triode Mode • Occurs when VGS < VTH(P) and VDS > VGS - VTH(P) W ID μpCOX VGS VTH(P) VDS /2 VDS L Saturation Mode • Occurs when VGS < VTH(P) and VDS ≤ VGS - VTH(P) W 1 2 ID μpCOX VGS VTH(P) 1 λp VDS L 2 EECS 40 Spring 2003 Lecture 20 S. Ross SATURATION CURRENT Since l is small or zero, current ID is almost constant in saturation mode. We can call this current IDSAT: W 1 IDSAT μnCOX VGS VTH(N) 2 L 2 W 1 IDSAT μpCOX VGS VTH(P) 2 L 2 for NMOS for PMOS EECS 40 Spring 2003 Lecture 20 S. Ross LINEAR AND NONLINEAR ELEMENTS We need to find out how transistors behave as part of a circuit. To solve a transistor circuit, obtain: 1) the nonlinear ID vs. VDS characteristic equation for the transistor Linear circuit 2) The linear relationship between ID vs. VDS as determined by the surrounding linear circuit G IG ID S - VDS + D Then simultaneously solve these two equations for ID and VDS. EECS 40 Spring 2003 Lecture 20 S. Ross SOLVING TRANSISTOR CIRCUITS: STEPS 1) Guess the mode of operation for the transistor. (We will learn how to make educated guesses). 2) Write the ID vs. VDS equation for this mode of operation. 3) Use KVL, KCL, etc. to come up with an equation relating ID and VDS based on the surrounding linear circuit. 4) Solve these equations for ID and VDS. 5) Check to see if the values for ID and VDS are possible for the mode you guessed for the transistor. If the values are possible for the mode guessed, stop, problem solved. If the values are impossible, go back to Step 1. EECS 40 Spring 2003 Lecture 20 S. Ross CHECKING THE ANSWERS NMOS • VDS must be positive • ID must be positive • VDS < VGS – VT(N) in triode VDS ≥ VGS – VT(N) in saturation • VGS > VT(N) in triode or saturation VGS ≤ VT(N) in cutoff PMOS • VDS must be negative • ID must be negative • VDS > VGS – VT(P) in triode VDS ≤ VGS – VT(P) in saturation • VGS < VT(P) in triode or saturation VGS ≥ VT(P) in cutoff EECS 40 Spring 2003 Lecture 20 S. Ross EXAMPLE 1) Guess the mode: 1.5 kW D 4V ID + _ 3V G + _ Since VGS > VTH(N), not in cutoff mode. Guess saturation mode. + 2) Write transistor ID vs. VDS: VDS _ ID IDSAT ID 250 106 A / V 2 3 V 1 V 2 S VTH(N) = 1 V, ½ W/L mnCOX = 250 m A/V2, l = 0 V-1. ID 1 mA 3) Write ID vs. VDS equation using KVL: - VDS - 1.5 kW ID 4V 0 EECS 40 Spring 2003 Lecture 20 S. Ross EXAMPLE 4) Solve: 1.5 kW D 4V ID + _ 3V G + _ + VDS _ S VTH(N) = 1 V, ½ W/L mnCOX = 250 m A/V2, l = 0 V-1. ID = 1mA VDS = 2.5 V 5) Check: ID and VDS are correct sign, and VDS ≥ VGS-VT(N) as required in saturation mode. EECS 40 Spring 2003 Lecture 20 S. Ross WHAT IF WE GUESSED THE MODE WRONG? 1) Guess the mode: Since VGS > VTH(N), not in cutoff mode. Guess triode mode. 1.5 kW D 4V ID + _ 3V G + _ 2) Write transistor ID vs. VDS: + VDS _ S VTH(N) = 1 V, ½ W/L mnCOX = 250 m A/V2, l = 0 V-1. ID = 2·250·10-6(3 – 1 – VDS/2)VDS 3) Write ID vs. VDS equation using KVL: - VDS - 1.5 kW ID 4V 0 EECS 40 Spring 2003 Lecture 20 S. Ross WHAT IF WE GUESSED THE MODE WRONG? 4) Solve for VDS with quadratic equation: 1.5 kW D 4V ID + _ 3V G + _ + VDS _ S VTH(N) = 1 V, ½ W/L mnCOX = 250 m A/V2, l = 0 V-1. VDS = {4 V, 2.67 V} 5) Check: Neither value valid in triode mode! VDS > VGS – VT(N) not allowed in triode mode. EECS 40 Spring 2003 Lecture 20 S. Ross GUESSING RIGHT How do you guess the right mode? Often, the key is the value of VGS. (We can often find VGS directly without solving the whole circuit.) ID ID VGS ≤ VT(N) VGS = VT(N) + e probably saturation definitely cutoff VDS VGS - VT(N) = e VDS EECS 40 Spring 2003 Lecture 20 S. Ross GUESSING RIGHT When VGS >> VTH(N), it’s harder to guess the mode. ID triode mode saturation mode VGS - VTH(N) If ID is small, probably triode mode VDS EECS 40 Spring 2003 Lecture 20 S. Ross A CLOSER LOOK In this circuit, the transistor delivered a constant current IDSAT to the 1.5 kW resistor. 1.5 kW D 4V ID + _ 3V G + _ This circuit acts like a constant current source, as long as the transistor remains in saturation mode. IDSAT does not depend on the attached resistance if saturation is maintained. + VDS _ S IDSAT 1.5 kW EECS 40 Spring 2003 Lecture 20 S. Ross A CLOSER LOOK The circuit will go out of saturation mode if • VGS < VT(N) or • VDS < VGS – VT(N) IDSAT does depend on VGS; one can adjust the current supplied by adjusting VGS. RL D VDD ID + _ VGS G + _ This can happen if VGS is too large or too small, or if the load resistance is too large. + VDS _ S IDSAT RL EECS 40 Spring 2003 Lecture 20 S. Ross ANOTHER EXAMPLE 1) Guess the mode: 1.5 kW 2 kW 4V ID + _ D G + 6 kW VDS _ What is VGS? No current goes into/out gate. VGS = 3 V by voltage division. Guess saturation (randomly). 2) Write transistor ID vs. VDS: ID IDSAT ID 250 106 A / V 2 3 V 1 V 2 ID 1 mA S VTH(N) = 1 V, ½ W/L mnCOX = 250 m A/V2, l = 0 V-1. 3) Write ID vs. VDS equation using KVL: - VDS - 1.5 kW ID 4V 0 Effectively the same circuit as previous example: only 1 source.