22-Elec-B5 Advanced Electronics · May 2015

Question 3 of 5: Complementary MOS Class B Output Stage

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 3: Complementary MOS Class B Output Stage (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A complementary source-follower pair: the n-channel $M_1$ conducts on positive half-cycles from $+V_{DD}$, the p-channel $M_2$ conducts on negative half-cycles from $-V_{SS}$, and both drive the common source node into $R_L$. There is no bias between the gates, so the stage is true class B and each device conducts for exactly half of every cycle.

Find. (a) the maximum power delivered to $R_L$, (b) the average power dissipated in $M_1$ at that operating condition, and (c) the resulting power-conversion efficiency.

Approach. The peak output is not the rail: it is the point at which the conducting device reaches the edge of saturation, so solve that condition first, then apply the standard class B half-sine energy integrals for load power, device dissipation and efficiency.

1. Part (a) — find the peak output the stage can actually reach. On a positive half-cycle $M_1$ carries $i_D=\hat V_o/R_L$ and needs an overdrive of $V_{OV}=\sqrt{2i_D/K}$ to do so. Its drain sits at $V_{DD}$ and its source at the output, so $V_{DS}=V_{DD}-\hat V_o$; saturation requires $V_{DS}\ge V_{OV}$. At the limit the two are equal:$$V_{DD}-\hat V_o=\sqrt{\frac{2\hat V_o}{K R_L}}$$Squaring and collecting terms gives a quadratic in $\hat V_o$:$$\begin{aligned}\hat V_o^{\,2}-\left(2V_{DD}+\frac{2}{KR_L}\right)\hat V_o+V_{DD}^{\,2}&=0\\\hat V_o^{\,2}-20.5\,\hat V_o+100&=0\end{aligned}$$whose roots are $8.0\ \text{V}$ and $12.5\ \text{V}$. The second exceeds the rail and is rejected, so$$\boxed{\hat V_o=8.00\ \text{V}}$$Checking the limit closes exactly: $\hat i_D=8/8=1.00\ \text{A}$, $V_{OV}=\sqrt{2\times1.00/0.5}=2.00\ \text{V}$, and $V_{DD}-\hat V_o=2.00\ \text{V}$ as required.
2. Maximum RMS output power. The output is a full sine of amplitude $\hat V_o$, so its RMS value is $\hat V_o/\sqrt{2}=5.66\ \text{V}$ and$$P_{L,\max}=\frac{\hat V_o^{\,2}}{2R_L}=\frac{(8.00\ \text{V})^{2}}{2\times8\ \Omega}=\boxed{4.00\ \text{W}}$$Had the stage been able to swing all the way to the rail the answer would have been $6.25\ \text{W}$; the finite gate drive available costs the design more than a third of its power.
3. Part (b) — average power dissipated in $M_1$. $M_1$ conducts only over the positive half-cycle, during which it drops $v_{DS}=V_{DD}-\hat V_o\sin\theta$ while passing $i_D=(\hat V_o/R_L)\sin\theta$. Averaging over the whole period,$$P_{D1}=\frac{1}{2\pi}\int_0^{\pi}\left(V_{DD}-\hat V_o\sin\theta\right)\frac{\hat V_o}{R_L}\sin\theta\ d\theta=\frac{V_{DD}\hat V_o}{\pi R_L}-\frac{\hat V_o^{\,2}}{4R_L}$$Substituting the numbers,$$P_{D1}=\frac{10\times8.00}{\pi\times8}-\frac{(8.00)^{2}}{4\times8}=3.183-2.000=\boxed{1.18\ \text{W}}$$By symmetry $M_2$ dissipates the same, so the pair burns $2.37\ \text{W}$ between them at full output.
4. Part (c) — efficiency. Each supply delivers the average of a half-sine of peak $\hat V_o/R_L$, so the total power drawn from the two rails is$$P_S=\frac{2V_{DD}\hat V_o}{\pi R_L}=\frac{2\times10\times8.00}{\pi\times8}=6.37\ \text{W}$$$$\eta=\frac{P_{L,\max}}{P_S}=\frac{4.00}{6.37}=\boxed{62.8\ \%}$$The books’ result follows immediately: $\eta=(\pi/4)(\hat V_o/V_{DD})=0.785\times0.80=62.8\ \%$, so the stage reaches only 80 % of the theoretical class B ceiling of $\pi/4=78.5\ \%$ precisely because it cannot swing to the rail. The power balance also closes: $P_S-P_{L}=6.37-4.00=2.37\ \text{W}=2P_{D1}$.