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S. RossEECS 40 Spring 2003 Lecture 20 Today we will Review NMOS and PMOS I-V characteristic Practice useful method for solving transistor circuits Build.

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Presentation on theme: "S. RossEECS 40 Spring 2003 Lecture 20 Today we will Review NMOS and PMOS I-V characteristic Practice useful method for solving transistor circuits Build."— Presentation transcript:

1 S. RossEECS 40 Spring 2003 Lecture 20 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!

2 S. RossEECS 40 Spring 2003 Lecture 20 G D S IDID IGIG - V DS + + V GS _ NMOS I-V CHARACTERISTIC Since the transistor is a 3-terminal device, there is no single I-V characteristic. Note that because of the gate insulator, I G = 0 A. We typically define the MOS I-V characteristic as I D vs. V DS for a fixed V GS. 3 modes of operation

3 S. RossEECS 40 Spring 2003 Lecture 20 triode mode cutoff mode saturation mode V DS IDID V GS = 3 V V GS = 2 V V GS = 1 V V DS = V GS - V TH(N) NMOS I-V CHARACTERISTIC (V GS ≤ V TH(N) )

4 S. RossEECS 40 Spring 2003 Lecture 20 NMOS I-V CHARACTERISTIC Cutoff Mode Occurs when V GS ≤ V TH(N) I D = 0 Triode Mode Occurs when V GS > V TH(N) and V DS < V GS - V TH(N) Saturation Mode Occurs when V GS > V TH(N) and V DS ≥ V GS - V TH(N)

5 S. RossEECS 40 Spring 2003 Lecture 20 G D S IDID IGIG - V DS + + V GS _ PMOS I-V CHARACTERISTIC Symbol has “dot” at gate. NMOS does not. I D, V GS, V DS, and V TH(P) are all negative for PMOS. These values are positive for NMOS. Channel formed when V GS < V TH(P). Opposite for NMOS. Saturation occurs when V DS ≤ V GS – V TH(P). Opposite for NMOS.

6 S. RossEECS 40 Spring 2003 Lecture 20 triode mode cutoff mode saturation mode V DS IDID V GS = -3 V V GS = -2 V V GS = -1 V V DS = V GS - V TH(P) PMOS I-V CHARACTERISTIC (V GS ≥ V TH(P) )

7 S. RossEECS 40 Spring 2003 Lecture 20 PMOS I-V CHARACTERISTIC Cutoff Mode Occurs when V GS ≥ V TH(P) I D = 0 Triode Mode Occurs when V GS V GS - V TH(P) Saturation Mode Occurs when V GS < V TH(P) and V DS ≤ V GS - V TH(P)

8 S. RossEECS 40 Spring 2003 Lecture 20 SATURATION CURRENT Since is small or zero, current I D is almost constant in saturation mode. We can call this current I DSAT :

9 S. RossEECS 40 Spring 2003 Lecture 20 LINEAR AND NONLINEAR ELEMENTS To solve a transistor circuit, obtain: 1)the nonlinear I D vs. V DS characteristic equation for the transistor 2)The linear relationship between I D vs. V DS as determined by the surrounding linear circuit Then simultaneously solve these two equations for I D and V DS. G D S IDID IGIG - V DS + + V GS _ Linear circuit We need to find out how transistors behave as part of a circuit.

10 S. RossEECS 40 Spring 2003 Lecture 20 1)Guess the mode of operation for the transistor. (We will learn how to make educated guesses). 2)Write the I D vs. V DS equation for this mode of operation. 3)Use KVL, KCL, etc. to come up with an equation relating I D and V DS based on the surrounding linear circuit. 4)Solve these equations for I D and V DS. 5)Check to see if the values for I D and V DS 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. SOLVING TRANSISTOR CIRCUITS: STEPS

11 S. RossEECS 40 Spring 2003 Lecture 20 CHECKING THE ANSWERS NMOS V DS must be positive I D must be positive PMOS V DS must be negative I D must be negative V DS < V GS – V T(N) in triode V DS ≥ V GS – V T(N) in saturation V GS > V T(N) in triode or saturation V GS ≤ V T(N) in cutoff V DS > V GS – V T(P) in triode V DS ≤ V GS – V T(P) in saturation V GS < V T(P) in triode or saturation V GS ≥ V T(P) in cutoff

12 S. RossEECS 40 Spring 2003 Lecture 20 EXAMPLE G D S IDID + V GS _ + V DS _ +_+_ +_+_ 4 V 3 V 1.5 k  V TH(N) = 1 V, ½ W/L  n C OX = 250  A/V 2, = 0 V -1. 1) Guess the mode: Since V GS > V TH(N), not in cutoff mode. Guess saturation mode. 2) Write transistor I D vs. V DS : 3)Write I D vs. V DS equation using KVL:

13 S. RossEECS 40 Spring 2003 Lecture 20 EXAMPLE G D S IDID + V GS _ + V DS _ +_+_ +_+_ 4 V 3 V 1.5 k  V TH(N) = 1 V, ½ W/L  n C OX = 250  A/V 2, = 0 V -1. 4) Solve: I D = 1mA V DS = 2.5 V 5) Check: I D and V DS are correct sign, and V DS ≥ V GS -V T(N) as required in saturation mode.

14 S. RossEECS 40 Spring 2003 Lecture 20 G D S IDID + V GS _ + V DS _ +_+_ +_+_ 4 V 3 V 1.5 k  V TH(N) = 1 V, ½ W/L  n C OX = 250  A/V 2, = 0 V -1. WHAT IF WE GUESSED THE MODE WRONG? 1) Guess the mode: Since V GS > V TH(N), not in cutoff mode. Guess triode mode. 2) Write transistor I D vs. V DS : 3)Write I D vs. V DS equation using KVL: I D = 2·250·10 -6 (3 – 1 – V DS /2)V DS

15 S. RossEECS 40 Spring 2003 Lecture 20 G D S IDID + V GS _ + V DS _ +_+_ +_+_ 4 V 3 V 1.5 k  V TH(N) = 1 V, ½ W/L  n C OX = 250  A/V 2, = 0 V -1. 4) Solve for V DS with quadratic equation: V DS = {4 V, 2.67 V} 5) Check: Neither value valid in triode mode! V DS > V GS – V T(N) not allowed in triode mode. WHAT IF WE GUESSED THE MODE WRONG?

16 S. RossEECS 40 Spring 2003 Lecture 20 GUESSING RIGHT How do you guess the right mode? Often, the key is the value of V GS. (We can often find V GS directly without solving the whole circuit.) V GS ≤ V T(N) V GS = V T(N) +  definitely cutoff V DS IDID probably saturation V DS IDID V GS - V T(N) = 

17 S. RossEECS 40 Spring 2003 Lecture 20 triode mode saturation mode V DS V GS - V TH(N) GUESSING RIGHT When V GS >> V TH(N), it’s harder to guess the mode. IDID If I D is small, probably triode mode

18 S. RossEECS 40 Spring 2003 Lecture 20 G D S IDID + V GS _ + V DS _ +_+_ +_+_ 4 V 3 V 1.5 k  In this circuit, the transistor delivered a constant current I DSAT to the 1.5 k  resistor. This circuit acts like a constant current source, as long as the transistor remains in saturation mode. I DSAT does not depend on the attached resistance if saturation is maintained. I DSAT 1.5 k  A CLOSER LOOK

19 S. RossEECS 40 Spring 2003 Lecture 20 G D S IDID + V GS _ + V DS _ +_+_ +_+_ V DD V GS RLRL I DSAT does depend on V GS ; one can adjust the current supplied by adjusting V GS. The circuit will go out of saturation mode if V GS < V T(N) or V DS < V GS – V T(N) This can happen if V GS is too large or too small, or if the load resistance is too large. I DSAT RLRL A CLOSER LOOK

20 S. RossEECS 40 Spring 2003 Lecture 20 ANOTHER EXAMPLE G D S IDID + V GS _ + V DS _ +_+_ 4 V 1.5 k  V TH(N) = 1 V, ½ W/L  n C OX = 250  A/V 2, = 0 V -1. 1) Guess the mode: What is VGS? No current goes into/out gate. V GS = 3 V by voltage division. Guess saturation (randomly). 2) Write transistor I D vs. V DS : 3)Write I D vs. V DS equation using KVL: 2 k  6 k  Effectively the same circuit as previous example: only 1 source.

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