Q. Design a Resistive load inverter with R = 47.5 K $\Omega$ such that $V_{OL} = 0.25V$. The E-type nMOS driver transistor has following parameter $V_{DD} = 5V, V_{TO} = 1 V, \mu C_{OX} = 25 \mu A/V^2$

a) Determine W/L

b) Determine $V_{IL}$ and $V_{IH}$

c) Determine $NM_L , NM_H$

Power consumption in CMOS digital circuit

1. The dynamic (Switiching) power consumption

2. The short - circuit Power consumption

3. The leakage power consumption

1) Basic of Series - parallel nmos & pmos.

2) Series parallel equivalency rule

3) CMOS - 2 i/p NOR GATE Analysis & Design

4) CMOS - 2 i/p NAND GATE Analysis & Design

5) Numerical based on series parallel equivalency rule.

Q. Design a Resistive load inverter with R = 47.5 K $\Omega$ such that $V_{OL} = 0.25V$. The E-type nMOS driver transistor has following parameter $V_{DD} = 5V, V_{TO} = 1 V, \mu C_{OX} = 25 \mu A/V^2$

a) Determine W/L

b) Determine $V_{IL}$ and $V_{IH}$

c) Determine $NM_L , NM_H$

Power consumption in CMOS digital circuit

1. The dynamic (Switiching) power consumption

2. The short - circuit Power consumption

3. The leakage power consumption

1) Basic of Series - parallel nmos & pmos.

2) Series parallel equivalency rule

3) CMOS - 2 i/p NOR GATE Analysis & Design

4) CMOS - 2 i/p NAND GATE Analysis & Design

5) Numerical based on series parallel equivalency rule

Resistance $\alpha$ $\frac{1}{Width}$

Series - Parallel Equivalence Rule

3) NOR

Assumptions : i) Both i/p switch simultaneously i.e $V_A = V_B$

ii) Derive sizes in each block are identical, $(W/L)_{n,A} = (W/L)_{n,B}$ And $(W/L)_{p,A} = (W/L)_{p,B}$

iii) Substract bias effect for pmos transistor is neglected for simplicity

$M_1$ = saturation

$M_2$ = Saturation

$M_3$ = Linear

$M_4$ = Saturation

Combined brain current of noms transistors ($M_1, M_2$)

$I_D = K_n(V_{th} - V_{Th})^2$

$I_D = K_n[V_{th}^2 - 2V_{th}.V_{Th} - V_{TD}^2]$

$V_{th} = V_{TD}(\frac{I_D}{K_n})^{1/2}$ -----------(1)

Drain Current for $M_3$ & $M_4$

$I_{D3} = \frac{K_i}{2}[2(V_{DD - V_{Th} - |V_{TP}|}V_{SD3} - V_{SD3}^2)]$

$I_{D4} = \frac{K_P}{2}[2(V_{DD - V_{Th} - |V_{TP}|} - V_{SD4}^2)]$

=> $I_{D3} = I_{D4}$

we get,

$V_{DD} - V_{Th} - |V_{TP}| = 2(\frac{I_D}{K_P})^{1/2}$ -----------(2)

Combining 1 & 2 , we get

[$V_{th}(NOR 2) = \frac{V_{T,n} + \frac{1}{2}(\frac{K_P}{K_n})^{1/2}(V_{DD} - |V_{T,P}|)}{1 + \frac{1}{2}(\frac{K_P}{K_n})^{1/2}}$]

If $K_P = K_n$ & $V_{T,n} = V_{T,P}$

$V_{Th}(NOR) = \frac{V_{DD} + V_{T,n}}{3}$