Solutions to Homework Chapter 4-2

(Due 03/12/2004)

4.100(a)

Since V_DG = 0 V , the channel at the drain end is pinched off and transistor is in saturation mode

and V_DS = V_GS

Applying KVL we get

I_DS = (12 – V_GS) / 10⁵ ohm

But from the MOS equation for saturation mode

Equating two equations for I_DS , we get the following quadratic in V_GS

12.5V_GS² – 17.8 V_GS – 4.97 = 0

Solving this equation we get

V_GS = -0.24 V or V_GS = 1.66 V

Since negative V_GS implies transistor in cut off mode

Therefore, V_GS = 1.66 V = V_DS

Substituting in any of the equations we get

I_DS = (12 – 1.66) V / 10⁵ ohm = 103 uA

4.100(b)

Applying KVL we get

V_DS = 10⁷ I_G + V_GS

But I_G = 0

Therefore, V_GS = V_DS

Hence transistor in saturation mode because V_GD = 0 V and hence the channel is pinched off at drain end

From the MOS equation for saturation mode

I_DS = (12 – V_DS) / 330 kohm = (12 – V_GS) / 330 kohm

Equating two equations for I_DS , we get the following quadratic in V_GS

41.25V_GS² – 60.88 V_GS + 11.2 = 0

Solving this equation we get

V_GS = 0.215 V or V_GS = 1.26 V

Since V_GS = 0.215 V which is less than V_TN , that implies transistor in cut off mode

Therefore, V_GS = 1.26 V = V_DS

Substituting in any of the equations we get

I_DS = (12 – 1.26) V / 330 kohm = 32.5 uA

4.84

By voltage divider rule we get

V_G = 100 kohm * 12 V / (100 kohm + 220 kohm) = 3.75 V

Applying KVL we get

I_DS = (3.75 – V_GS) / 24 kohm

Equating two equations for I_DS , we get the following quadratic in V_GS

1.5V_GS² – 2 V_GS – 2.25 = 0

Solving this equation we get

V_GS = 2.061 V

Substituting in any of the equations we get

I_DS = (3.75 – 2.061) V / 24 kohm = 70.4 uA

Applying KVL we get

V_DS = 12 – 36 kohm * I_DS = 12 – 36 kohm * 70.4 uA

V_DS = 9.467 V