Does an emitter follower really improve a zener regulator circuit?

The Art of Electronics (1st ed., page 56) has confused me!

On page 12, this circuit is shown as a simple zener regulator. I added the load, a 1k resistor.



When I simulate it, the regulator draws 49.5 mA, the vast majority of which runs through the 300 ohm resistor. The load gets approx 5 mA, which we would expect. So I guess it's normal for a regulator like this to be very inefficient. I was content to move on because the regulator does seem to do its job and keep voltage steady even if the load changes to a much lower resistance; even at 100 ohms its voltage is basically 5 volts.

But later, the authors claim that:

By using an emitter follower to isolate the zener [regulator circuit], you get the improved circuit shown in figure 2.11. Now the situation is better. Zener current can be made relatively independent of the load current, since the transistor base current is small, and far lower zener power dissipation is possible (reduced by as much as 1/beta).[p. 56]



Here's where they lose me. Yes, the zener draws less current. But now the voltage is loewr by a diode drop and the transistor and its collector resistor draw even more power.

In the original, the load dissipates 26mW and the whole circuit dissipates 990 mW (barely 3% efficiency). But in the "improved" circuit, the entire circuit dissipates 1.1 W and the load only 19mW, an efficiency of not even 2%. Plus now, both the transistor and collector resistor consume enough current to worry about their maximum dissipation.

Which begs the question: Is the improved dissipation of a zener regulator by an emitter follower a sound reason to increase the dissipation of a resistor and transistor? It seems that whatever was gained in terms of temperature stability wrt the zener has been shifted to the follower circuit.

What am I missing here? What kind of efficiency should I expect from either regulator circuit, and under what conditions is the follower an improvement?

Let's re-draw your circuits with a more conventional layout and reference designators, so we can talk about them more clearly.



simulate this circuit – Schematic created using CircuitLab

Since Vout is fixed by D1 at 5.1 V, we can directly compute the current through both resistors:

The difference, 44.57 mA, is the current flowing through D1, and it is dissipating 227.3 mW. But more to the point, the current through R1 is constant, which means that the circuit always draws 49.67 mA × 20 V = 1 Watt from the power source.

This puts some limits on the circuit. If the load current increases, the zener current decreases, and it ceases to regulate if the zener current drops to zero. So the maximum current this circuit can supply is 49.67 mA.

So let's add the emitter follower:



simulate this circuit

I've decreased the value of R2 to 100 Ω to represent the maximum 50 mA output current of the previous circuit. Because of the gain of the transistor (we'll assume it's 100), the load current for the zener, Ib only varies from 0 to 0.5 mA as the load current varies from 0 to 50 mA. This means that we can design the zener circuit for a much lower maximum current. I've increased the value of R1 to 7500 Ω which reduces the current through it to about 2 mA. This in turn means that as Ib varies from 0 to 0.5 mA, D1's current varies from 2 down to 1.5 mA, and it dissipates a maximum of about 10 mW.

Furthermore, when the load current drops to zero, this 2 mA is the only current drawn from the 20 V source by the circuit, representing a standby power of just 40 mW.

And yes, I've increased D1's voltage slightly to account for the VBE drop of Q1.

Finally, I've marked R3 as "optional". Its only purpose is to dissipate some of the power that would otherwise be dissipated by Q1.

In the zener-only shunt regulator, the diode is required to pass current that is not being drawn by the load. If the heaviest load is, say 100mA, then the diode will be required to pass at least some current in that state (to stay regulating), say 10mA. With the load removed, though, no-load diode current will be 110mA.

By adding an emitter-follower, we now rely on the transistor to pass 99% of the load (emitter) current directly from the collector, bypassing the shunt altogether. Consequently, we can drop diode current to a small fraction of its former value. Here are the two scenarios side-by-side:



simulate this circuit – Schematic created using CircuitLab

To compare how efficient these are, we can plot the sum of power dissipation in all elements, in each case. That is, plot the sum of power in R1 and D1 on the left, and the sum of power in R2, D2 and Q1 on the right, while sweeping load current from near zero to the maximum of 100mA:



Your argument that efficiency is similar under full-load conditions is true, but you didn't consider efficiency for smaller loads. The emitter-follower design is always more efficient, but it becomes less so as current approaches the maximum.

In the no-load condition, D1 is always conducting 110mA, and always has 5V across it. R1 is similarly always conducting 110mA, and has 7V across it. That's a lot of power for something that's not actually delivering any of it to the load:

$$ P_{SHUNT} \approx 7V \times 0.11A + 5V \times 0.11A = 1.3W $$

By contrast, for the emitter-follower, R2 is much larger, so current in R2 and D2 is greatly reduced, along with wasted power. Most importantly, though, Q1 conducts exactly as much as is necessary to produce \$V_{OUT2}=5V\$, regardless of the load. Under negligible load, it is almost completely off, having very high effective resistance, and dissipates negligible power, unlike R1 which is always 60Ω, and always getting hot.

Let's compare regulation quality, \$V_{OUT1}\$ vs. \$V_{OUT2}\$ as the load varies from 1mA to 100mA:



Not much difference there, they both suffer a drop in \$V_{OUT}\$ as load increases, but both regulate well enough for a lot of applications.

The low-load power argument alone may be sufficient justification for the added cost and board space of transistor Q1, and perhaps a collector resistor to offload some of the power dissipation from Q1. If load is expected to vary (as most loads do, as LEDs switch on or off, for example), then the emitter follower offers significantly better efficiency as load diminishes.

Is the improved dissipation of a zener regulator by an emitter follower a sound reason to increase the dissipation of a resistor and transistor?

Yes. As already stated in the question, by using an emitter follower the zener current (and hence zener power dissipation) is now independent of load current, so zener voltage will be more stable over the full load range, and the zener can be selected to suit the available components and the application. A low-power zener can be used for a higher-power load. Of course, to realise this benefit, the zener current must be reduced, so the 300ohm resistor between power supply and zener must be increased to, say, 3k ohm (Iz~5mA, rather than 50mA).

The extra voltage drop of the BJT B-E junction can be compensated by putting a forward-biased diode in series with the zener.

It seems that whatever was gained in terms of temperature stability wrt the zener has been shifted to the follower circuit.

Correct, however there is an important difference regarding how the two circuits behave as load current varies from zero to full-load. For the zener-only circuit, the power dissipation in the zener decreases as load current increases. For the zener-plus-transistor circuit, the power in the transistor increases as load current increases (while the power in the zener is constant).

What am I missing here? What kind of efficiency should I expect from either regulator circuit, and under what conditions is the follower an improvement?

Regarding efficiency: the zener-plus-follower circuit has much better efficiency at light load. This can be explained by referring to the OPs original circuits, and the following assumptions:

Let's consider how both circuits behave when the load current is at 10% of the design maximum, ie: load current = 5mA.

The power delivered to the load is the same in both cases: 5.1V * 5mA = 25.5mW.

For the zener-only circuit, the power drawn from the 20V supply is: 20V * 50mA = 1W. Efficiency = 25.5mW / 1W = 2.55%.

For the zener-plus-follower circuit, the power drawn from the 20V supply is: 20V * (5mA + 5mA) = 20V * 10mA = 200mW. Efficiency = 25.5mW / 200mW = 12.75%. This is a big improvement over the zener-only circuit.

Let's new consider how both circuits behave when the load current is at 100% of the design maximum, ie: load current = 50mA. Let's assume there is still some current in the zener that it is still regulating Vout.

The power delivered to the load is the same in both cases: 5.1V * 50mA = 255mW.

For the zener-only circuit, the power drawn from the 20V supply is unchanged: 20V * 50mA = 1W. Efficiency = 255mW / 1.0W = 25.5%.

For the zener-plus-follower circuit, the power drawn from the 20V supply is: 20V * (50mA + 5mA) = 20V * 55mA = 1.1W. Efficiency = 255mW / 1.1W = 23.2%. This is only slightly worse than the zener-only circuit.

We can conclude that the zener-plus-follower circuit will be more efficient than the zener-only circuit for load currents from 0mA up to about 45mA. At 45mA load current, both circuits draw the same current (50mA) from the 20V power supply. Between 45mA and 50mA the zener-only circuit will be slightly more efficient than the zener-plus-follower.

Note 1: The zener-only will deliver a maximum load current of ~50mA (=14.9V / 300 ohm), beyond this the zener current falls to zero so the zener voltage will no longer be at 5.1V.

The follower draws very little current from zener leg maintaining the zener current constant during load changes. Thus constant zener current ensures constant output voltage.

As this is more of a homework question, rather than giving away the answers exactly, we prefer a more didactic approach to answering it.

You say it's "improved". Improved in what sense?

(The "improved" circuit might even be impaired in some respects. Can you find which?)

You say the transistor and resistor "draw even more power", but is this just an assertion? Under what condition(s) would this be true? -- Have you evaluated it under equal conditions?

I believe on nearby pages, you will find an explanation of the design process, and thus, where the above points fit in. (It has been a long time since I looked in AoE 1st ed., admittedly!)

Or, more fundamentally: start with the assumption that the zener is conducting in breakdown, as intended, and calculate its current for all combinations / the full range of input and load conditions. If the current flow direction is correct (zener has positive conventional current from cathode to anode) and is within acceptable bounds, the assumption is confirmed.

What resistor value(s) are necessary to meet this condition, for the two circuits?

What is the total power dissipation for both circuits, under the same design parameters and operating conditions?

Alternately, we can take a "critical reading" approach:

I think what you are missing is what has been improved.

The authors did not claim that the circuit is more efficient. They also did not claim the output voltage stays the same.

Obviously when the output voltage drops, you get less current and power into the load, for comparing the results, assume you use 5.6V zener to get 5.0V to load.

The voltage is an easy fix. Use a 5.6V Zener and you get about 5.0V regulated output from the emitter.

The circuit will have better regulation.

And it will then allow for further optimizations.

So this is not about optimizing efficiency. All linear regulators making 5V from 20V to a 1000 ohm load have equal efficiency when the supply current of the regulator itself is ignored. And if the supply current of regulator itself is made smaller, you have better efficiency.

In the original circuit the Zener needs to handle all the 50mA or be able to dissipate 250mW when there is no load. With the transistor handling voltage buffering and current gain, you can easily use 3kohm resistor so Zener needs to only handle 5mA at most, dissipating only roughly 25mW. Assume the circuit will be adjusted as pure emitter follower with no collector resistance.