Complementary Symmetry Power Amplifiers

In most applications where AC power is driving a load, a complementary symmetry (push-pull) power amplifier is employed. This amplifier usually has a voltage gain of one, and a large current gain. It is the most efficient configuration for transforming DC power from the power supply to the AC power driving the load.

A basic complementary symmetry BJT power amplifier is shown in Fig. 3.1

  

Figure 3.1: Basic Complementary Symmetry Power Amp

The BJT's provide a current gain of $\beta$, and a low output resistance. When the input is positive and greater than 0.7V, the npn BJT is on and the pnp is off. Under these conditions, the npn acts like an emitter follower, taking the input current from the base, multiplying it by $\beta$, to provide current and a low source resistance to drive the load. Of course the extra current derived from the $\beta$ multiplication has its origin at the positive DC power supply.

When the input voltage is less the -0.7V, the npn is off and the pnp is on. Under these conditions, the pnp acts like an emitter follower, taking the input current from the base, multiplying it by $\beta$, to sink current and provide a low source resistance to sink the load. Here, the extra current required by $\beta$ multiplication has its origin at the negative DC power supply.

The output resistance R_out is given by the parallel output resistance of the npn and the pnp (R_on||R_op). For the positive side of the cycle, since the pnp is off, $R_{op}=\infty$. So under these conditions the output resistance is just that of the npn which is

$$R_{out}=R_{on} \approx \frac{1}{g_{mn}} + \frac{R_S}{\beta_n}~~~~~~~v_{in} \gt 0.7$$ (59)

During the negative cycle $R_{on} \approx \infty$, which yields the following for the negative cycle output resistance:

$$R_{out}=R_{op} \approx \frac{1}{g_{mp}} + \frac{R_S}{\beta_p}~~~~~~~v_{in} < - 0.7$$ (60)

Where g_mn, g_mp, $\beta_n$ and $\beta_p$ are the small signal transconductances and current gains of the the npn and pnp BJT's respectively. (Note the approximation sign. This is because, strictly speaking, we are not dealing with small signals so using a single value of g_m is only an approximation.)

To appreciate how power is converted from supply by the circuit, consider the following. Each power supply provides current during only half the cycle. During the positive cycle, the positive DC supply is sourcing current, while during the negative part of the cycle the negative DC supply is sinking current. Therefore, the average current provided by the positive supply is

$$I_{+supply}=\frac{1}{T}\int_0^{T/2}I_c(t)dt$$ (61)

which for a sinusoidal waveform gives

$$I_{+supply}=\frac{I_o}{\pi}=\frac{V_o}{\pi R_L}$$ (62)

During the negative part of the cycle, a totally analogous analysis applies, but for the negative supply current I_-supply. to give the same result.

The average power for the total cycle is then obtained by adding the power from each half cycle:

P_supply=V_CCI_+supply+V_EEI_-supply

Since we take the magnitudes of the DC supplies to be equal, we have

$$P_{supply}=2V_{CC}I_{+supply}=\frac{2V_{CC}V_o}{\pi R_L}$$ (64)

The power conversion efficiency $\eta$, which gives a measure of how well power is transferred from the supply to the load, is given by the ratio of the power to the load P_L to the power provided by the supplies P_supply:

$$\eta= \frac{P_L}{P_{supply}}=\frac{\pi V_o}{4V_{CC}}$$ (65)

Thus, for full swing $V_o \approx V_{CC}$, the power conversion efficiency of the complementary symmetry reaches is maximum value of approximately 75%. This number is much greater than that of the single ended emitter follower that we discussed earlier.

You may have noticed that the complementary symmetry configuration has a small problem which is that no signal can propagate when -0.7 < V_s < 0.7 because both BJT's are off at this time. This region is often referred to as the cross-over region between the npn and pnp cycles. Hence, the accompanied distortion that results is often referred to as cross-over distortion. To avoid cross-over distortion, circuit designers often bias the two transistors slightly into the forward active region with diodes. For example, look at the circuit in Fig. 3.2.

  

Figure 3.2: Complementary Symmetry Amp Biased to Eliminate Cross-Over Distortion

The diodes, in conjunction with resistors R₁ and R₂, constitute the biasing network to make sure these is always a $V_{BE} \approx 0.7V$ across both transistors. That way one BJT is always on, and the cross-over region is eliminated.