In this Part 3, the optimal parallel-segment amplifier voltages are derived so that for either resistive or constant-current loads, the power dissipation in the parallel transistors is the same.

From High-Efficiency Analog Amplifiers, Part 1: The Efficiency-Linearity Tradeoff, we have the ||-segmented amplifier scheme, repeated below.

The question to be answered is what the *V*_{gn} voltages should be for equal maximum power dissipation, P̂_{Q}, in each of the parallel-segment output BJTs. For a total number of *N* segments, then the segment numbers, *n*, range from *n* = 1, 2, ...., *N*. An unsegmented amplifier has *N* = 1 and a maximum power loss at midrange, where the output power is also maximum;

The power dissipation in the BJT of the *n*th segment is designated as *P*_{Qn} and its supply voltage is *V*_{gn}. For a resistive load, P̂_{Q} occurs at the center voltage of the segment. The power loss of the transistor is

Let segments each have equal P̂_{Q}. Then

Solving for *V*_{gn},

Substituting for *V*_{o},

Solving this recursion equation with initial value, *V*_{g0} = 0 V,

For *n* = *N*, *V*_{gn} = *V*_{g}. Then *V*_{g} = 2 x P̂_{Q} x
√
*N* and

The graph of BJT power loss over the output voltage range is plotted below.

These derivations result in the following design formulas for ||-segmented amplifiers:

The *segment power fraction* is defined as

Then the amplifier efficiency is

As an example, a parallel-segmented amplifier with *N* = 3 segments has

Then the segment voltages are optimally

The efficiency is
η
= 85.7 %. With only three segments, the analog amplifier has efficiency comparable to low-end switching converters.

For *N* = 4,

The following table results from applying the design formulas for *N* = 4.

From this exercise in algebra, we find that we need supply voltages that differ from each other in ratio by *n*^{1/2}. As the upper end of the voltage range is approached, more segments are needed because power loss increases by the voltage-squared.

Sometimes an amplifier load is not resistive (or even reactive) but is a constant current load. For these kinds of loads, the advantage of segmentation is not as extensive, as can be seen from the design formulas:

More segments - about twice as many - are required for the same efficiency for a constant-current load relative to a resistive load.

I used two parallel segments for an amplifier in the Floating Differential Source instrument, which is the topic of a prior series of articles, The Floating Differential Source, Part 2: Specifications and Capabilities. The circuit diagram shown below, taken from the FDS Manual, is a working example of amplifier segmentation. The segmenting circuitry is in the upper-left. The hysteretic comparator consists of Q21, Q22. Its output switches the upper-segment Darlington, Q16, Q17, on or off. Darlington Q13, Q15 is the lower-segment output transistors.

Click here for larger image

What this foray into amplifier segmentation teaches is that with only three segments, the analog amplifier enters the realm of efficiency of switching converters, and that with four, it is competing with switching efficiency. There is no technical reason to stop at four.

What segmentation does is to displace the problem of analog amplifier efficiency to that of multiple-output power supplies. If the supplies are those of a switching converter, it is only a modest increase in cost and complexity to tap the secondary winding and add rectifiers and storage capacitors. The supply efficiency is decreased negligibly by the taps. The voltages of *V*_{gn} can usually have a wide tolerance, and ripple on them has no greater effect than on any other amplifiers. The parts count is indeed higher than for unsegmented or switching amplifiers, but when both analog precision and efficiency are required, it provides a solution path to an optimal design.