HW 5 - CMOS Inverter Dynamic and Power Energy Characteristics

7.3: CMOS Inverter Dynamic Characteristics

1

We can find the $K_{n}^{'}$ and $K_{p}^{'}$ via:

K_{n}^{'} = μ_{n} C_{o x} = μ_{n} \cdot \frac{ε_{o x}}{t_{o x}}

and similar for the PMOS. Here $ε_{o x} = 3.9 \cdot 8.854 \cdot 10^{- 14} F / c m = 345.3 f F / c m$ , and we have $t_{o x} = 100 Angstoroms = 10 nm$ :

K_{n}^{'} = 500 \frac{{cm}^{2}}{Vs} \cdot \frac{345.3 fF / cm}{10^{- 6} cm} = 173 μ A/V^{2}

and:

K_{p}^{'} = 200 \frac{{cm}^{2}}{Vs} \cdot \frac{345.3 fF / cm}{10^{- 6} cm} = 69.1 μ A/V^{2}

24

a

We know that:

τ_{P H L} = 1.2 C \cdot \frac{1}{K_{n} (V_{H} - V_{t n})}

τ_{P L H} = 1.2 C \cdot \frac{1}{K_{p} (V_{H} + V_{t p})}

and:

t_{f} = 3 τ_{P H L}, t_{r} = 3 τ_{P L H}

where here $V_{D D} = V_{H} = 2.5 V$ . Notice that:

K_{n} = K_{n}^{'} \cdot (W / L)_{N} = 100 μ {A/V}^{2} \cdot 2 = 200 μ {A/V}^{2}

and:

K_{p} = K_{p}^{'} \cdot (W / L)_{P} = 40 μ {A/V}^{2} \cdot 5 = 200 μ {A/V}^{2}

Thus:

τ_{P H L} = 1.2 \cdot 0.20 \cdot 10^{- 12} F \cdot \frac{1}{200 μ {A/V}^{2} \cdot (2.5 V - 0.6 V)} = 0.632 ns

τ_{P L H} = 1.2 \cdot 0.20 \cdot 10^{- 12} F \cdot \frac{1}{200 μ {A/V}^{2} \cdot (2.5 V - 0.6 V)} = 0.632 ns

therefore $τ_{P} = 0.632 ns$ by symmetry. Furthermore:

t_{f} = 3 τ_{P H L} = t_{r} = 3 τ_{P L H} = 1.896 ns

b and c

The same calculations apply, just change $V_{D D} = 2.0 V$ and $V_{D D} = 1.8 V$ respectively:

τ_{P} = 1.2 \cdot 0.20 \cdot 10^{- 12} F \cdot \frac{1}{200 μ {A/V}^{2} \cdot (2.0 V - 0.6 V)} = 0.857 ns

then:

t_{r} = t_{f} = 2.57 ns

and for c:

τ_{P} = 1 ns

then:

t_{r} = t_{f} = 3 ns

27

Use the same formula as prior:

K_{n} = 2 \cdot 100 μ {A/V}^{2} = 200 μ {A/V}^{2}

K_{p} = 2 \cdot 40 μ {A/V}^{2} = 80 μ {A/V}^{2}

thus:

τ_{P H L} = 1.2 \cdot 0.4 \cdot 10^{- 12} F \cdot \frac{1}{200 μ {A/V}^{2} \cdot (2.5 V - 0.6 V)} = 1.264 ns

τ_{P L H} = 1.2 \cdot 0.4 \cdot 10^{- 12} F \cdot \frac{1}{80 μ {A/V}^{2} \cdot (2.5 V - 0.6 V)} = 3.158 ns

so:

τ_{P} = \frac{τ_{P H L} + τ_{P L H}}{2} = 2.211 ns

and:

t_{f} = 3 τ_{P H L} = 3.792 ns

t_{r} = 3 τ_{P L H} = 9.474 ns

29

Again just use the same formula:

K_{n} = 2 \cdot 100 μ {A/V}^{2} = 200 μ {A/V}^{2}

K_{p} = 5 \cdot 40 μ {A/V}^{2} = 200 μ {A/V}^{2}

by the symmetry of $K_{n}, K_{p}$ as well as $V_{t n}, V_{t p}$ then we'll have $τ_{P} = τ_{P H L} = τ_{P L H}$ :

τ_{p} = 1.2 \cdot 0.20 \cdot 10^{- 12} F \cdot \frac{1}{200 μ {A/V}^{2} \cdot (3.3 V - 0.75 V)} = 0.471 ns

then:

t_{r} = t_{f} = 3 τ_{P} = 1.412 ns

31

For simplicity, let's design a symmetrical CMOS inverter, so we want $τ_{P H L} = τ_{P L H} = τ_{P} = 3 ns$ . We also require this since we want $t_{f} = t_{r}$ . Then:

\begin{aligned} 3 ns & = 1.2 \cdot 1 \cdot 10^{- 12} F \cdot \frac{1}{K_{t} (2.5 V - 0.6 V)} \\ K_{t} & = 211 μ {A/V}^{2} \end{aligned}

where $K_{t}$ is a placeholder for either $K_{n}, K_{p}$ as the process is the same for both. Thus:

(W / L)_{N} = \frac{K_{n}}{K_{n}^{'}} = \frac{211}{100} = 2.11

(W / L)_{P} = \frac{K_{p}}{K_{p}^{'}} = \frac{211}{40} = 5.28

7.4: CMOS Power Characteristics

43

a

Each gate can get:

P = \frac{100 W for all gates}{500 million gates} = 200 nW per gate

b

If $V_{D D} = 1.8 V$ then:

P_{t o t a l} = 100 W = 1.8 V \cdot I_{D D} \Rightarrow I_{D D} = 55.56 A

47

a

Here:

K_{n} = 15 \cdot 100 μ {A/V}^{2} = 1500 μ {A/V}^{2}

K_{p} = 15 \cdot 40 μ {A/V}^{2} = 600 μ {A/V}^{2}

so then $K_{R} = \frac{K_{n}}{K_{p}} = 2.5$ . At $V_{M}$ (peak current) we have both $M$ 's in saturation:

1500 / 2 \cdot μ {A/V}^{2} \cdot (V_{M} - 0.6 V)^{2} = 600 / 2 \cdot μ {A/V}^{2} \cdot (V_{M} - 3.3 V + 0.6 V)^{2}

Giving:

V_{M} \approx 1.414 V

Plugging back in:

I_{D, m a x} \approx 497 μ A

b

Doing the same process:

1500 / 2 \cdot μ {A/V}^{2} \cdot (V_{M} - 0.6 V)^{2} = 600 / 2 \cdot μ {A/V}^{2} \cdot (V_{M} - 2.5 V + 0.6 V)^{2}

Giving:

V_{M} \approx 1.104 V

Plugging back in:

I_{D, m a x} \approx 190 μ A

51

Here $C = 0.20 pF$ and $V_{D D} = 2.5 V$ . We solved this problem in HW 5 - CMOS Inverter Dynamic and Power Energy Characteristics#24. By the symmetry of the $K_{t}$ 's and the $V_{t}$ 's then our rise and fall times and similar were similar. Here:

P D P (V_{D D} = 2.5 V) \geq \frac{C V_{D D}^{2}}{5} = \frac{0.2 \cdot 10^{- 12} F \cdot (2.5 V)^{2}}{5} = 250 fJ

P D P (V_{D D} = 2.0 V) = 160 fJ

P D P (V_{D D} = 1.8 V) = 130 fJ

If $f = 100 Mhz$ , using $τ_{P} = 0.632 ns$ for the first, $τ_{P} = 0.857 ns$ for the second, and $τ = 1 ns$ for the latter:

P (V_{D D} = 2.5 V) = C V_{D D}^{2} \cdot f = 5 P D P \cdot f = 125 μ W

P (V_{D D} = 2.0 V) = 80 μ W

P (V_{D D} = 1.8 V) = 65 μ W

52

a

Again, just use the value from HW 5 - CMOS Inverter Dynamic and Power Energy Characteristics#29:

P D P = \frac{C V_{D D}^{2}}{5} = \frac{0.20 \cdot 10^{- 12} F \cdot (3.3 V)^{2}}{5} = 436 fJ

b

The highest switching frequency is when $t_{a}, t_{b} \to 0$ , so then:

T \geq \frac{2 t_{r}}{0.8} = 5 τ_{P}

so since $τ_{P} = 0.471 ns$ then $T = 2.355 ns$ so then $f_{m a x} = 425 Ghz$ .

c

At this highest frequency:

P_{m a x} = 5 P D P \cdot f_{m a x} = 926 μ W