EE 230 NMOS examples - Iowa State University

NMOS example problems

off

ohmic (linear)

saturation

VGS < VT

vGS VT vDS < vGS ?VT

vGS VT vDS vGS ?VT

iD = 0 iD = Kn [2 (vGS - VT) vDS - vD2S]

5.0

Kn

=

1 2

nCox

W L

4.0

iD = Kn [vGS - VT]2

vGS = 5 V

drain current -- iDS (mA)

3-terminal D

symbol iD

+

G

+ vGS ?

vDS ?

S

G. Tuttle - 2022

3.0

2.0

1.0

0.0 0.0

vGS = 4 V

vGS = 3 V vGS = 2 V

2.0

4.0

6.0

8.0

drain voltage -- vDS (V)

vGS < VT

10.0

NMOS examples ? 1

Example 1

For the circuit shown, use the the NMOS equations to nd iD and vDS.

For the NMOS, VT = 1 V and Kn = 0.5 mA/V2. We see that

RD iD

1.0 k +

VGG

+ ?

+ vGS

vDS ?

?

5 V

vGS = VGG = 5 V > VT the NMOS is on.

Guess that the transistor is in saturation.

iD = Kn (vGS - VT)2 = (0.5 mA/V2)(5 V - 1 V)2 = 8 mA

vDS = VDD - iDRD = 15 V - (8.0 mA) (1 k) = 7 V vGS - VT = 4 V

vDS > vGS ? VT saturation con rmed. Q.E.D.

+ ?

VDD

15 V

G. Tuttle - 2022

NMOS examples ? 2

if if

Example 2

For the circuit shown, use the the NMOS equations to nd iD and vDS. (This looks a lot like Example 1, but the resistor connected to the source will change things.)

For the NMOS, VT = 1.0 V and Kn = 0.5 mA/V2.

First, is the transistor on? The problem is that vGS = vG ? vS, and vS = iDRS. Since we don't yet know iD, we don't know the value of vGS.

VDD

RD 1 k

15 V iD

+

VGG 5 V

+ vGS ?

vDS ?

RS 1 k

To help decide the issue, we could guess that the transistor is off, meaning that we are assuming vGS < VT. However, if the NMOS is off, then iD = 0. This makes vGS = VGG ? iDRS = VGG = 5 V. This is greater than VT, meaning that the transistor should be on -- in exact contradiction to the assumption of the transistor being off. So the transistor must be on.

Next, is the NMOS in saturation or ohmic? Here we can't even make a logical argument as we did in deciding on or off. So we must guess -- let's guess saturation and see what happens.

G. Tuttle - 2022

NMOS examples ? 3

if

Example 2 (cont.)

If the transistor is in saturation

iD = Kn (vGS - VT)2.

Also, we know that vGS = VGG ? iDRS. Inserting into the saturation current equation:

iD = Kn (VGG - iDRS - VT)2

This is a quadratic equation for iD. If we have the right type of calculator, we could type in this equation and hit "solve" to get the values (plural!) for iD. Or we can limber up our algebra muscles and solve the old fashioned way. First expand the square: (Treat VGG ? VT as as single constant.)

iD = Kn [(iDRS)2 - 2 (iDRS) (VGG - VT)2 - (VGG - VT)2]

Re-arranging and gathering terms:

G. Tuttle - 2022

iD2 -

2

(

VGG - RS

VT )

+

1 KnRS2

2

iD

+

(

VGG - RS

VT

)

=0

NMOS examples ? 4

Example 2 (cont.)

Now, using our old-fashioned calculator (Or slide rule. Or abacus. Or

ngers and toes.), we can plug in the numbers.

iD2 - (10 mA) iD + 16 mA2 = 0.

(Note the units on the third term.) Use the quadratic formula to nd

iD = 2 mA or iD = 8 mA. So another conundrum: Which is it? Or both?

ax2 + bx + c = 0

x = - b?

If a = 1:

x

=

-

b 2

?

b2 - 4ac 2a

(

b 2

2

)

-

c

The correct solution must be consistent with the NMOS being on and in saturation. Only one root will meet both conditions. Start with iD = 8 mA. In that case, vGS = VGG ? iDRS = 5 V ? (8 mA)(1k) = ?3 V. This is (a lot) less thanVT. The larger root fails the "on" test rather badly.

Now consider iD = 2 mA. In that case, vGS = 3 V > VT. So the NMOS is clearly on. Next, vDS = VDD ? iDRD ? iDRS = 15 V ? 2 V ? 2 V = 11 V, which is clearly bigger than vGS ? VT = 2 V. The saturation assumption is also con rmed. So the correct answers are:

G. Tuttle - 2022 iD = 2 mA and vDS = 11 V.

NMOS examples ? 5

if if if

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download