Ouput Stage

Terminology¶

THD: Total Harmonic Distortion

Class A¶

Power delivered to load ( $P_{L}$ )

\begin{matrix} P_{L} = V_{O - r m s} \cdot I_{L - r m s} = \frac{1}{2} \frac{{V_{O - p}}^{2}}{R_{L}} \end{matrix}

Power drawn from supply ( $P_{S}$ )

\begin{aligned} P_{S} & = (P from M_{1}) + (P from M_{2}) \\ = V_{D D} \overset{―}{I_{D 1}} + V_{S S} I_{D 2} \\ = (V_{D D} + V_{S S}) I \end{aligned}

note

consider the original definition of (AC) power

\begin{matrix} P = \frac{1}{2 π} \int_{0}^{2 π} v (t) i (t) d t \end{matrix}

in the $P_{L}$ case, we have the vary $v_{O}$ , thus we write

\begin{matrix} P_{L} = \frac{1}{2} V_{p} I_{p} \cos (θ) = V_{r m s} I_{r m s} \end{matrix}

however, in the $P_{S}$ case, we have an constant source $V_{D D}$ and $V_{S S}$ , thus

\begin{aligned} P_{S} & = \frac{1}{2 π} \int_{0}^{2 π} V \cdot i (t) d t \\ = V \cdot \frac{1}{2 π} \int_{0}^{2 π} i (t) d t \\ = V \cdot I_{a v g} \end{aligned}

Class B¶

Power delivered to load ( $P_{L}$ )

ignoring crossover distortion

\begin{matrix} P_{L} = \frac{1}{2} \frac{{V_{O - p}}^{2}}{R_{L}} \end{matrix}

Power drawn from supply ( $P_{S}$ )

for each MOS, only conducting in half period

\begin{aligned} P_{S n} = P_{S p} & = V_{D D} \cdot \overset{―}{I_{D}} \\ = V_{D D} \cdot \frac{1}{2 π} \int_{0}^{2 π} \frac{V_{O - p}}{R_{L}} \sin θ d θ \\ = V_{D D} \cdot \frac{V_{O - p}}{π R_{L}} \end{aligned}

and for total power

\begin{matrix} P_{S} = P_{S p} + P_{S n} = \frac{2}{π} \frac{V_{O - p}}{R_{L}} V_{D D} \end{matrix}

efficiency $η$

\begin{matrix} η = \frac{P_{L}}{P_{S}} = \frac{π}{4} \frac{V_{o - p}}{V_{D D}} \end{matrix}

dissipated power $P_{D}$

\begin{matrix} P_{D} = P_{S} - P_{L} = \frac{2}{π} \frac{V_{O - p}}{R_{L}} V_{D D} - \frac{1}{2} \frac{V_{O - p}^{2}}{R_{L}} \end{matrix}

$P_{D - m a x}$ w.r.t $V_{o - p}$

\begin{matrix} \frac{d P_{D}}{d V_{o}} = \frac{2}{π} \frac{V_{D D}}{R_{L}} - \frac{V_{o}}{R_{L}} = 0 \\ ⟹ V_{o} = \frac{2}{π} V_{D D} \\ ⟹ P_{D - m a x} = 2 P_{D n - m a x} = 2 P_{D p - m a x} = \frac{2 V_{D D}^{2}}{π^{2} R_{L}} \end{matrix}

Class AB¶

CS Buffer¶

Output Resistance consider feedback (without $R_{L}$ ) $R_{o u t}$

\begin{matrix} R_{o u t - p} = \frac{r_{o}}{1 + μ g_{m p} r_{o}} \approx \frac{1}{μ g_{m p}} \\ R_{o u t - n} = \frac{r_{o}}{1 + μ g_{m n} r_{o}} \approx \frac{1}{μ g_{m n}} \\ ⟹ R_{o u t} = R_{o u t - p} | | R_{o u t - n} = \frac{1}{μ (g_{m p} + g_{m n})} \end{matrix}

Gain Error $G_{E}$

\begin{matrix} G_{E} \equiv \frac{v_{o} - v_{i}}{v_{i}} = \frac{A}{1 + A} - 1 = - \frac{1}{1 + A} \approx \frac{- 1}{A} = \frac{- 1}{2 μ g_{m} R_{L}} \end{matrix}

Class D¶

Theoretically power-conversion efficiency is $100 %$
since the input 1/0 pulse make transistors act as on-off switch, when the transistor on, there is no cross voltage, but current pass through. On the other hand, when transistor off, there is a large cross voltage but no current.

Power MOSFET¶

High $V_{t}$ : $2 V \sim 4 V$
In saturation region:
- for low $V_{G S}$ : $i_{D} \propto {V_{G S}}^{2}$
- for high $V_{G S}$ : $i_{D} \propto V_{G S}$
  - (the velocity saturation is due to the saturation of mobility $μ$ )

Temperature Effects¶

i_{D} = \frac{1}{2} μ C_{o x} \frac{W}{L} (V_{G S} - V_{t})^{2}

for $T ↑$ $⟹$ $μ ↓$ , $V_{T} ↓$

for low $V_{G S}$ , $T ↑ ⟹ i_{D} ↑$ ( $Δ [(V_{G S} - V_{t})^{2}]$ dominates)
for high $V_{G S}$ , $T ↑ ⟹ i_{D} ↓$ ( $Δ [μ]$ dominates)

Thermal Resistance¶

Junction temp. $T_{J}$
Ambient temp. $T_{A}$
Thermal resistance between junction and ambience. $θ_{J A}$
Temperature here acts as voltage, and the dissipated power acts as current. Thus by Ohm's law we have

\begin{matrix} T_{J m a x} - T_{A} = P_{D m a x} θ_{J A} \end{matrix}

Two Stage CMOS op-amp¶

Common-Mode¶

\begin{matrix} A_{c m} \approx \frac{- 1}{2 g_{m 3} R_{S S}} \\ C M R R \equiv \frac{| A_{d} |}{| A_{c m} |} = g_{m 1} (r_{o 2} | | r_{o 4}) \cdot 2 g_{m 3} R_{S S} \end{matrix}

Input Common-Mode Range¶

lower limit ( $M_{1}$ leaving saturation, when $V_{O V 1} = V_{D S 1}$ )

\begin{matrix} V_{I C M} + | V_{t p} | \geq V_{D 3} = - V_{S S} + V_{G S 3} \\ V_{I C M} \geq - V_{S S} + V_{G S 3} - | V_{t p} | \end{matrix}

upper limit ( $M_{5}$ leaving saturation)

\begin{matrix} V_{D D} - V_{O V 5} \geq V_{I C M} + V_{S G 1} \\ V_{I C M} \leq V_{D D} - V_{O V 5} - V_{S G 1} \end{matrix}

Output Swing¶

lower limit

\begin{matrix} v_{O} \leq - V_{S S} + V_{O V 6} \end{matrix}

upper limit

\begin{matrix} v_{O} \geq V_{D D} - V_{O V 7} \end{matrix}

PSRR¶

definition

\begin{matrix} {PSRR}^{+} \equiv \frac{A_{d}}{A^{+}} & where A^{+} \equiv \frac{v_{o}}{v_{d d}} \\ {PSRR}^{-} \equiv \frac{A_{d}}{A^{-}} & where A^{-} \equiv \frac{v_{o}}{v_{s s}} \end{matrix}

${PSRR}^{+} \to \infty$ :
just remember that $v_{O}$ via second stage and $v_{O}$ via first stage would cancel out each other.
${PSRR}^{-}$
- $v_{O}$ from first stage is $0$ (don't know fucking why)
- $v_{O}$ from second stage is

\begin{matrix} v_{o} = v_{s s} \frac{r_{o 7}}{r_{o 7} + r_{o 6}} \\ {PSRR}^{-} = \frac{A_{d}}{A^{-}} = \frac{g_{m 1} (r_{o 2} | | r_{o 4}) g_{m 6} (r_{o 6} | | r_{o 7})}{\frac{r_{o 7}}{r_{o 7} + r_{o 6}}} \end{matrix}

Slew Rate¶

when $| v_{i d} | > \sqrt{2} V_{O V}$ , the current would go through only one side of differential pair.

\begin{matrix} v_{o} (t) = \frac{Q_{C}}{C_{C}} = \frac{I}{C_{C}} t \\ SR = \frac{I}{C_{C}} = \frac{I}{G_{m 1} / ω_{t}} = \frac{I ω_{t}}{I / V_{O V}} = ω_{t} V_{O V} \end{matrix}

Folded-Cascode CMOS op-amp¶

Input Common-Mode Range¶

upper limit ( $M_{1}, M_{2}$ leave saturation)

\begin{matrix} V_{I C M} - V_{t n} \leq V_{D D} - V_{O V 9} \\ ⟹ V_{I C M} \leq V_{D D} - V_{O V 9} + V_{t n} \end{matrix}

lower limit ( $M_{11}$ leaves saturation)

\begin{matrix} V_{I C M} - V_{G S 1} \geq - V_{S S} + V_{O V 11} \\ ⟹ V_{I C M} \geq - V_{S S} + V_{O V 11} + V_{G S 1} \end{matrix}

Output Swing¶

upper limit ( $M_{4}$ leaves saturation)

\begin{matrix} v_{O} \leq V_{D D} - V_{O V 10} - V_{O V 4} \end{matrix}

lower limit ( $M_{6}$ leaves saturation)

\begin{matrix} v_{O} \geq - V_{S S} + V_{G S 7} + V_{G S 5} - V_{t n 6} \end{matrix}