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Review of exponential charging and discharging in RC Circuits

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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 – PowerPoint PPT presentation

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Title: Review of exponential charging and discharging in RC Circuits


1
  • 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

2
NMOS I-V CHARACTERISTIC
G
IG
VGS _
ID
S
D
- VDS
  • 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

3
NMOS I-V CHARACTERISTIC
ID
triode mode
saturation mode
VGS 3 V
VDS VGS - VTH(N)
VGS 2 V
VGS 1 V
VDS
(VGS VTH(N))
cutoff mode
4
NMOS I-V CHARACTERISTIC
  • Cutoff Mode
  • Occurs when VGS VTH(N)
  • ID 0
  • Triode Mode
  • Occurs when VGS gt VTH(N) and VDS lt VGS -
    VTH(N)
  • Saturation Mode
  • Occurs when VGS gt VTH(N) and VDS VGS -
    VTH(N)

5
PMOS I-V CHARACTERISTIC
G
IG
VGS _
ID
S
D
- VDS
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 lt VTH(P).
Opposite for NMOS. Saturation occurs when VDS
VGS VTH(P). Opposite for NMOS.
6
PMOS I-V CHARACTERISTIC
(VGS VTH(P))
cutoff mode
VDS
VGS -1 V
VGS -2 V
VDS VGS - VTH(P)
ID
VGS -3 V
triode mode
saturation mode
7
PMOS I-V CHARACTERISTIC
  • Cutoff Mode
  • Occurs when VGS VTH(P)
  • ID 0
  • Triode Mode
  • Occurs when VGS lt VTH(P) and VDS gt VGS -
    VTH(P)
  • Saturation Mode
  • Occurs when VGS lt VTH(P) and VDS VGS - VTH(P)

8
SATURATION CURRENT
Since l is small or zero, current ID is almost
constant in saturation mode.
We can call this current IDSAT
9
LINEAR AND NONLINEAR ELEMENTS
We need to find out how transistors behave as
part of a circuit.
  • To solve a transistor circuit, obtain
  • the nonlinear ID vs. VDS characteristic equation
    for the transistor
  • The linear relationship between ID vs. VDS as
    determined by the surrounding linear circuit

Linear circuit
G
IG
VGS _
ID
Then simultaneously solve these two equations for
ID and VDS.
D
S
- VDS
10
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.

11
CHECKING THE ANSWERS
  • NMOS
  • VDS must be positive
  • ID must be positive
  • VDS lt VGS VT(N) in triode
  • VDS VGS VT(N) in saturation
  • VGS gt VT(N) in triode or saturation
  • VGS VT(N) in cutoff
  • PMOS
  • VDS must be negative
  • ID must be negative
  • VDS gt VGS VT(P) in triode
  • VDS VGS VT(P) in saturation
  • VGS lt VT(P) in triode or saturation
  • VGS VT(P) in cutoff

12
EXAMPLE
  • Guess the mode
  • Since VGS gt VTH(N), not in cutoff mode. Guess
    saturation mode.

2) Write transistor ID vs. VDS
  • Write ID vs. VDS equation using KVL

VTH(N) 1 V, ½ W/L mnCOX 250 m A/V2, l 0
V-1.
13
EXAMPLE
  • Solve
  • ID 1mA VDS 2.5 V
  • Check
  • ID and VDS are correct sign, and VDS VGS-VT(N)
    as required in saturation mode.

VTH(N) 1 V, ½ W/L mnCOX 250 m A/V2, l 0
V-1.
14
WHAT IF WE GUESSED THE MODE WRONG?
  • Guess the mode
  • Since VGS gt VTH(N), not in cutoff mode. Guess
    triode mode.

2) Write transistor ID vs. VDS
ID 225010-6(3 1 VDS/2)VDS
  • Write ID vs. VDS equation using KVL

VTH(N) 1 V, ½ W/L mnCOX 250 m A/V2, l 0
V-1.
15
WHAT IF WE GUESSED THE MODE WRONG?
  • Solve for VDS with quadratic
  • equation
  • VDS 4 V, 2.67 V
  • 5) Check
  • Neither value valid in triode mode! VDS gt VGS
    VT(N) not allowed in triode mode.

VTH(N) 1 V, ½ W/L mnCOX 250 m A/V2, l 0
V-1.
16
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) e
VGS VT(N)
probably saturation
definitely cutoff
VDS
VDS
VGS - VT(N) e
17
GUESSING RIGHT
When VGS gtgt VTH(N), its harder to guess the mode.
ID
triode mode
saturation mode
VGS - VTH(N)
If ID is small, probably triode mode
VDS
18
A CLOSER LOOK
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.
In this circuit, the transistor delivered a
constant current IDSAT to the 1.5 kW resistor.
1.5 kW
D
ID
VDS _
G
4 V
VGS _
1.5 kW
IDSAT
3 V
S
19
A CLOSER LOOK
  • The circuit will go out of saturation mode if
  • VGS lt VT(N) or
  • VDS lt VGS VT(N)
  • This can happen if VGS is too large or too small,
    or if the load resistance is too large.

IDSAT does depend on VGS one can adjust the
current supplied by adjusting VGS.
RL
D
ID
VDS _
G
VDD
VGS _
RL
IDSAT
VGS
S
20
ANOTHER EXAMPLE
  • Guess the mode
  • What is VGS?
  • No current goes into/out gate.
  • VGS 3 V by voltage division.
  • Guess saturation (randomly).

1.5 kW
2 kW
D
ID
G
4 V
VDS _
2) Write transistor ID vs. VDS
VGS _
6 kW
S
  • Write ID vs. VDS equation using KVL

VTH(N) 1 V, ½ W/L mnCOX 250 m A/V2, l 0
V-1.
Effectively the same circuit as previous example
only 1 source.
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