HW 2 - NMOS Inverters (Resistor & Nonlinear Loads), PMOS Review

Chapter 6.5-6.6.3 (NMOS Inverters)

6.1

Consider 100,000 gates, with 1 $W$ on the entire chip.

a

We have:

P_{g a t e, o n} = \frac{1 W}{100, 000} = 10 μ W

So average power is:

P_{a v g} = \frac{1}{2} 10 μ W = 5 μ W

b

We know:

I = \frac{P}{V} = \frac{5 μ W}{2.5 V} = 2 μ A

6.2

Consider 100,000,000 gates, with 100 $W$ to dissipate:

a

Same as HW 2 - NMOS Inverters (Resistor & Nonlinear Loads), PMOS Review#6.1:

P_{o n} = \frac{100 W}{100, 000, 000} = 1 μ W

b

So then:

P_{a v g} = 0.25 P_{o n} = 0.25 μ W

Calculate the current:

I_{a v g} = \frac{P_{a v g}}{V} = \frac{0.25 μ W}{2.5 V} = 0.1 μ A

c

Just multiply the total number of gates by our current:

100, 000, 000 \cdot I_{a v g} = 10 A

6.39

Consider the following, where $V_{D D} = 2.5 V$ :

Pasted image 20240122120532.png

We can use the table for values:

K_{n}^{'} = 100 μ A / V^{2}, V_{t n} = 0.6 V

a

We know:

K_{n} = K_{n}^{'} \cdot 3 = 300 μ A / V^{2}

We know that $V_{o} = V_{L}$ so then $V_{I} = V_{H} = 2.5 V$ . We know for these types that:

V_{o} = V_{D S} = V_{D D} - I_{D} R = 2.5 - I_{D} R

Since $V_{G S} - V_{t n} \geq V_{D S}$ then we are in Triode so then:

I_{D} = K_{n} (V_{O V} - V_{D S} / 2) V_{D S} = K_{n} (2.5 - V_{t n} - V_{L} / 2) V_{L}

So then plug into our original equation and get:

V_{L} = 2.5 - 300 μ A / V^{2} \cdot 200 k Ω \cdot (2.5 - 0.6 - V_{L} / 2) V_{L}

Giving:

\begin{aligned} V_{L} & = 2.5 - 60 / V (1.9 - V_{L} / 2) V_{L} \\ = 2.5 - 60 (1.9 V_{L} - V_{L}^{2} / 2) \\ = 2.5 - 114 V_{L} + 30 V_{L}^{2} \\ \Rightarrow 0 = 30 V_{L}^{2} - 115 V_{L} + 2.5 \end{aligned}

So then we can solve and get $V_{L} = 3.811 V, 0.022 V$ so clearly $V_{L} = 0.022 V$ . We can get power dissipation by finding $I_{D}$ :

I_{D} = 12.5 μ A \Rightarrow P = I V = 2.5 V \cdot 12.5 μ A = 31.3 μ W

b

We repeat (a):

K_{n} = 6 K_{n}^{'} = 600 μ A / V^{2}

V_{L} = 2.5 - 600 μ A / V^{2} \cdot 400 k Ω \cdot (2.5 - 0.6 - V_{L} / 2) V_{L}

Thus:

\begin{aligned} V_{L} & = 2.5 - 240 / V^{2} (1.9 - V_{L} / 2) V_{L} \\ = 2.5 - 456 V_{L} + 120 V_{L}^{2} \\ \Rightarrow V_{L} = 0.00548 V, 3.803 V \end{aligned}

So clearly $V_{L} = 5.48 m V$ . Plug in as before:

I_{D} = 6.24 μ A \Rightarrow P = I V = 2.5 V \cdot 6.24 μ A = 15.6 μ W

6.43

We as the noise margins for the circuit in HW 2 - NMOS Inverters (Resistor & Nonlinear Loads), PMOS Review#6.39.

a

V_{I L} = \frac{1}{K_{n} R} + V_{t n} = \frac{1}{300 μ A / V^{2} \cdot 200 k Ω} + 0.6 V = 0.6167 V

And:

V_{O H} = V_{D D} - \frac{1}{2 K_{n} R} = 2.5 - \frac{1}{2 \cdot 300 μ A / V^{2} \cdot 200 k Ω} = 2.4917 V

And:

V_{I H} = V_{t n} - \frac{1}{K_{n} R} + 1.63 \sqrt{\frac{V_{D D}}{K_{n} R}} = 0.916 V

And:

V_{O L} = \sqrt{\frac{2 V_{D D}}{3 K_{n} R}} = 0.167 V

So then:

N M_{L} = V_{I L} - V_{O L} = 0.456 V

and:

N M_{H} = V_{O H} - V_{O L} = 1.5717 V

b

We can repeat the same thing as (a), except that $K_{n} = 600 μ A / V^{2}$ and $R = 400 k Ω$ are now changed using (b)'s values from before, giving:

V_{I L} =

6.59

Pasted image 20240122130006.png

First, some calculations:

K_{n, L} = 100 μ A / V^{2} \cdot 0.5 = 500 μ A / V^{2}, K_{n, S} = 4 K_{n}^{'} = 400 μ A / V^{2}

Since $γ = 0$ then:

V_{H} = V_{D D} - V_{t n} = 3.3 - 0.6 = 2.7 V

Solving for $V_{L}$ , we know that the currents are the same. $M_{L}$ is forced saturation, while $M_{S}$ is in triode when $V_{I} = V_{H} = 2.7 V$ (since we know $V_{I} >> V_{O}$ ):

\begin{aligned} I_{L, s a t} & = I_{S, t r i} \\ K_{n, L} / 2 \cdot (V_{G S, L} - V_{t n})^{2} & = K_{n, S} \cdot V_{L} (V_{H} - V_{t n} - V_{L} / 2) \\ 25 μ A / V^{2} \cdot (3.3 - V_{L} - 0.6)^{2} & = 400 μ A / V^{2} \cdot V_{L} (2.7 - 0.6 - 2.7 / 2) \\ (2.7 - V_{L})^{2} & = 16 V_{L} (2.1 - V_{L} / 2) \\ V_{L}^{2} - 5.4 V_{L} + 7.29 & = 33.6 V_{L} - 8 V_{L} \\ V_{L}^{2} - 31 V_{L} + 7.29 = 0 \\ V_{L} = 0.237 V \end{aligned}

Thus we can find $I_{D} = 152 μ A$ so then $P = 500 μ W$ .

6.60

We just need to find $V_{t n}$ and do the previous problem again. Notice:

V_{t n} = V_{T O} + γ (\sqrt{V_{B S} + 2 ϕ_{F}} - \sqrt{2 ϕ_{F}}) =

Where $V_{T O} = 0.6 V$ , $2 ϕ_{F} = 0.6 V$ :

V_{t n} = 0.6 + 0.6 (\sqrt{V_{H} + 0.6} - \sqrt{0.6})

So:

V_{H} = V_{D D} - V_{t n} = 3.3 - 0.6 - 0.6 (\sqrt{V_{H} + 0.6} - 0.775)

Just plug in $V_{H}$ in a numeric solver to get:

V_{H} = 2.17 V

Thus we do the same thing as prior, except with:

V_{t n, s i t} = 0.6 + 0.6 (\sqrt{2 V_{L} + 0.6} - \sqrt{0.6})

Thus:

\begin{aligned} I_{D L} & = I_{D S} \\ K_{n, L} / 2 \cdot (V_{G S, L} - V_{t n})^{2} & = K_{n, S} \cdot V_{L} (V_{H} - V_{t n} - V_{L} / 2) \end{aligned}

Plug into a numeric solver and we get:

V_{L} = 0.256 V

Chapter 4.3 (PMOS Review)

4.48

We know $K_{p}^{'} = μ_{p} C_{o x}^{″} = μ_{p} \cdot \frac{ϵ_{o x}}{t_{o x}}$ , so since $μ_{p} = 200 \frac{c m^{2}}{V s}$ , and $ϵ_{o x} = 345 f F c m^{- 1}$ for silicon oxide:

a) $K_{p}^{'} = 200 \frac{c m^{2}}{V s} \cdot \frac{345 F c m^{- 1}}{50 n m} = 13.8 \frac{μ A}{V^{2}}$
b) $K_{p}^{'} = 200 \frac{c m^{2}}{V s} \cdot \frac{345 F c m^{- 1}}{20 n m} = 34.5 \frac{μ A}{V^{2}}$
c) $K_{p}^{'} = 200 \frac{c m^{2}}{V s} \cdot \frac{345 F c m^{- 1}}{10 n m} = 69 \frac{μ A}{V^{2}}$
d) $K_{p}^{'} = 200 \frac{c m^{2}}{V s} \cdot \frac{345 F c m^{- 1}}{50 n m} = 138 \frac{μ A}{V^{2}}$

4.49

We are given:

Pasted image 20240123184522.png

We know that the pinchoff voltage is where the curve enters from triode into saturation. For say $V_{G S} = - 5 V \Rightarrow V_{S G} = 5 V$ we see that it levels off starting at about $V_{D S} = - 4.5 V \Rightarrow V_{S D} = 4.5 V$ , so then we have:

V_{t p} = - (V_{S G} - V_{S D}) = - (5 - 4.5) = - 0.5 V

Thus we can estimate the $K_{p}$ value by plugging in a saturation value and using the corresponding drain current:

I_{S D} = \frac{k_{p}}{2} (V_{S G} - | V_{t p} |)^{2} \Rightarrow 1250 μ A = \frac{k_{p}}{2} (3 V - 0.5 V)^{2} \Rightarrow K_{p} = 400 μ A / V^{2}

Since we know that $K_{p}^{'} = 40 μ A / V^{2}$ then $W / L = \frac{K_{p}}{K_{p}^{'}} = 10 / 1$ .

Typically enchancement MOSFETs turn "off" at voltages near 0, while depletion MOSFETs turn "off" at extreme voltages (high-low). In this case, the MOSFET turns "off" at a voltage of 0, so it is enhancement.

4.50

Recall that:

I_{D, s a t} = \frac{k_{p}}{2} (V_{S G} - | V_{t p} |)^{2}

At saturation $I_{D}$ stays relatively constant. We know what the values of $V_{t p}$ and everything else are, so for $V_{G S} = - 3.5, - 4.5 V$ we can see that:

I_{D, - 3.5 V} = \frac{400 μ A / V^{2}}{2} (3.5 - 0.5)^{2} = 1800 μ A / V^{2}

and

I_{D, - 4.5 V} = \frac{400 μ A / V^{2}}{2} (4.5 - 0.5)^{2} = 3200 μ A / V^{2}

Also, the voltage at which each becomes saturation is $V_{D S} = - 3 V$ and $V_{D S} = - 4 V$ respectively, which we use to plot the graph below:

4.51

We are given $W / L = 20$ so then $K_{p} = 40 μ A / V^{2} \cdot 20 = 800 μ A / V^{2}$ . Since $V_{B S} = 0 V$ then $V_{t p} = V_{T O} = - 0.75 V$ .

a

We are given $V_{G S} = - 1.1 V$ and $V_{D S} = - 0.2 V$ , so then notice that first $V_{G S} < V_{t p}$ so we are not in cutoff, but furthermore:

0.2 V = | V_{D S} | \leq | V_{G S} - V_{t p} | = 0.35 V

So then we are in triode, so:

I_{D} = 800 μ A / V^{2} (- 1.1 - (- 0.75) - \frac{- 0.2}{2}) (- 0.2) = 40 μ A

b

We are given $V_{G S} = - 1.3 V$ and $V_{D S} = - 0.2 V$ so then notice still $V_{G S} < V_{t p}$ but also:

0.2 V = | V_{D S} | \leq | V_{G S} - V_{t p} | = 0.55 V

So we are still in triode:

I_{D} = 800 μ A / V^{2} (- 1.3 - (- 0.75) - \frac{- 0.2}{2}) (- 0.2) = 72 μ A

c

We do the exact same process as before, except now since $V_{B S} = 1 V$ then:

V_{t p} = V_{T O} - γ (\sqrt{V_{B S} + 2 ϕ_{F}} - \sqrt{2 ϕ_{F}}) = - 0.995 V \approx - 1.0 V

Thus for (a) we have $V_{G S} < V_{t p}$ and $| V_{D S} | \geq | V_{G S} - V_{t p} |$ so we are in saturation:

I_{D} = 400 μ A / V^{2} (- 1.1 - (- 1.0))^{2} = 4 μ A

and for (b) we have $V_{G S} < V_{t p}$ and $| V_{D S} | \leq | V_{G S} - V_{t p} |$ so we are in triode:

I_{D} = 800 μ A / V^{2} (- 1.3 - (- 1.0) - \frac{- 0.2}{2}) (- 0.2) = 32 μ A

4.54

a

The on-resistance is given by:

R_{o n} = \frac{1}{K_{n} (| V_{G S} - V_{t p} |)}

We are given that:

K_{n} = 200 \cdot 40 μ A / V^{2} = 8000 μ A / V^{2}

Furthermore, we know $V_{G S} = - 5 V, V_{t p} = - 0.75 V$ , so $| V_{G S} - V_{t p} | = 4.25 V$ , thus:

R_{o n} = \frac{1}{8000 μ A / V^{2} \cdot 4.25 V} = 29.41 Ω

b

If we repeat where $V_{G S} = 5 V$ and $V_{t n} = 0.75 V$ for an NMOS transistor, we get the same exact finding since $V_{G S} - V_{t n} = 4.25 V$ , except now $K_{n} = 100 μ A / V^{2} \cdot 200 = 20000 μ A / V^{2}$ so:

R_{o n} = \frac{1}{20000 μ A / V^{2} \cdot 4.25 V} = 11.76 Ω

c

We require:

R_{o n} = 11.76 Ω = \frac{1}{40 μ A / V^{2} \cdot W / L \cdot 4.25 V} \Rightarrow W / L = 500 / 1

4.57

The maximum on resistance comes from when $V_{D S} = 0.1 V$ (as small as it can be) while we consider it's current $I_{D} = 0.5 A$ , so then:

R_{o n, m a x} = {[\frac{\partial I_{D}}{\partial V_{G S}}]}^{- 1} |_{Q - p t} = \frac{0.1 V}{0.5 A} = 0.2 Ω

So then:

R_{o n} \leq 0.2 Ω \Rightarrow \frac{1}{K_{n} | V_{G S} - V_{t p} |} \leq 0.2 Ω \Rightarrow K_{p} \geq \frac{1}{0.2 Ω \cdot | - 10 V - (- 2 V) |} = 0.625 A / V^{2}

is the maximum value for $K_{p}$ .

4.59

Since $V_{B S} = 4 V$ then using our table values:

V_{t p} = V_{T O} - γ (\sqrt{V_{B S} + 2 ϕ_{F}} - \sqrt{2 ϕ_{F}}) = - 1.435 V

So since $- 1.5 V = V_{G S} < V_{t p}$ we are not in cutoff. Further, $| V_{D S} | \geq | V_{G S} - V_{t p} |$ so we are in saturation. Using $K_{p} = 40 μ A / V^{2} \cdot 25 = 1000 μ A / V^{2}$ :

I_{D} = 500 μ A / V^{2} \cdot (- 1.5 V - (- 1.435 V))^{2} = 2.113 μ A

7.1

Note that:

K_{n}^{'} = μ_{n} \cdot C_{o x}^{″} = μ_{n} \cdot \frac{e_{o x}}{t_{o x}} = 500 c m^{2} / (V s) \cdot \frac{3.9 \cdot 8.854 \cdot 10^{- 14} F / c m}{100 \cdot 10^{- 8} c m} = 172.7 \frac{μ A}{V^{2}}

And for $K_{p}^{'}$ it's very similar:

K_{p}^{'} = μ_{p} \cdot \frac{3.9 \cdot ϵ_{o x}}{t_{o x}} = 200 c m^{2} / (V s) \cdot \frac{3.9 \cdot 8.854 \cdot 10^{- 14} F / c m}{100 \cdot 10^{- 10} m \cdot 100 c m / m} = 69.1 \frac{μ A}{V^{2}}