Problem 5.18

A particular enhancement MOSFET for which V_t = 0.5V and k'_n(W/L) = 0.1mA/V² is to be operated in the saturation region. If I_D is to be 12.5 uA, find the required V_GS and the minimum required V_DS. Repeat for I_D = 50 uA.

V_OV = V_DS at the edge of the saturation region.

For I_D = 12.5 uA

To find V_OV,

I_D = 1/2 k'_n (W/L) (V_OV)

.0125 = 1/2 (0.1) (V_OV)²

V_OV = 0.5 V

V_OV = V_DS = 0.5 V

V_OV = V_GS - V_t

0.5 V = V_GS - 0.5 V

V_GS = 1 V

For I_D = 50 uA

To find V_OV,

ID = 1/2 k'_n (W/L) (V_OV)²

.050 = 1/2 (0.1) (V_OV)²

V_OV = 1 V

V_OV = V_DS = 1 V

V_OV = V_GS - V_t

1 V = V_GS - 0.5 V

V_GS = 1.5 V

Problem 5.11

A p-channel MOSFET with a threshold voltage V_tp = -0.7 V has its source connected to ground.

(a) What should the gate voltage be for the device to operate with an overdrive voltage |V_OV| = 0.5V?

V_OV = V_SG - |V_t|

V_SG = V_S - V_G

V_S = 0 V

V_SG = -V_G

V_OV = -V_G - |V_t|

0.5 V = -V_G - 0.7 V

V_G = 1.2 V

(b)What is the highest Voltage allowed at the drain while the device operates in saturation region?

At the limit, V_OV = V_SD

V_OV = V_S - V_D

V_OV = -V_G - |V_t| = -1.2 V - 0.7 V

-1.9 V = 0 V - V_D

V_D = 1.9 V

(c) If the drain current obtained above is 1 mA, what would the current be for V_D = -10 mV and for V_D = -2 V?

 First lets find k_P

I_D = 1/2 k_P (V_OV)²

1 mA = 1/2 k_P (1.9)²

 k_P = 0.554 mA/V²

Now let's use the formula to find I_D for both V_D

For V_D = 10 mV

 I_D = 1/2 k_P (V_OV)²

I_D = 1/2 (.552 mA/V²) (.010)²

I_D = 0 mA

For V_D = 10 mV

 I_D = 1/2 k_P (V_OV)²

I_D = 1/2 (.552 mA/V²) (2)²

I_D = 1.104 mA

Problem 5.9

An NMOS transistor with a k_n = 1 mA/V² and V_t = 1V is operated with V_GS = 2.5 V. At what value of V_DS does the transistor enter the saturation region? What value of I_D is obtained in saturation?

As you can see from the graph from page 249 the transistor enters the saturation region when the V_OV = V_DS.

V_OV = V_GS - V_t

V_OV = 2.5 V - 1 V

V_OV = 1.5 V

V_DS = 1.5 V

To find I_D we use the formula:

I_D = 1/2 k_N(V_OV)²

I_D = 1/2 (1 mA/V²)(1.5V)²

I_D = 1.125 mA