# Help working emitter-follower design exercise.

#### Zazoo

Joined Jul 27, 2011
114
I'm working on exercise 2.2 in The Art of Electronics (2nd ed. Horowitz & Hill, pg. 67)

The exercise is:

"Use a follower with base driven from a voltage divider to provide a stiff source of +5V from an available regulated +15V supply. Load current (max) = 25mA. Choose your resistor values so that the output voltage doesn't drop more than 5% under full load."

Based on the problem description I have the following circuit: What I've got so far:

RL(min) = 200Ω (using the maximum load condition: 5V/25mA = 200Ω)

I need about 5.6V at the base (VB) so R1/R2 = 1.68

Output and input resistances:
Rout ≈ Rth/(hfe+1)
Rin ≈ (hfe+1)(RE//RL)

(note: Rth = R1//R2)

The book has been assuming a typical hfe of 100 in most of its examples, so I'm assuming that's the assumption used here as well.

I'm a bit confused about how to proceed from here. The variable load current and maximum voltage drop condition is really throwing me. I just need a nudge in the right direction.

Thanks

#### t_n_k

Joined Mar 6, 2009
5,455
A starting point might be to consider the required internal "source" resistance.

You know that the output voltage can fall from 5V to 4.75V (-5%) with a 25mA increase in Ie. That is to say, the source resistance should be no more than about 0.25V/25mA or 10Ω.

The source resistance will be about (R1||R2)/ (1+β). So R1||R2 would need to be no more than about 1kΩ.

I would suggest 0.6V for Vbe is too small - a value of ~0.7V would be more reasonable.

• Zazoo

#### t_n_k

Joined Mar 6, 2009
5,455
Keep in mind also that you have to consider the dynamic emitter resistance which is added to the referred base side resistance.

If Ie=25mA then you have an effective dynamic emitter resistance of ~1Ω which adds to the aforementioned referred value. So you may in fact have to keep R1||R2 as low as 900Ω with β=100.

• Zazoo

#### Hi-Z

Joined Jul 31, 2011
158
Don't forget RE needs a value too, in order to minimise the voltage rise for the no-load situation. A "nice" value would set the minimum emitter current to about 1mA, so maybe around 5k. (This means max emitter current is actually 26mA.)

• Zazoo

#### Hi-Z

Joined Jul 31, 2011
158
I realise this is only an academic exercise, but for anyone attempting to use such a circuit as a voltage regulator for a power-supply application, I have the following warning: emitter-followers driving capacitive loads will probably oscillate!

You may get round this by including a "base-stopper" resistor in the base circuit, but my advice would be to avoid the use of an emitter-follower altogether.

• Zazoo

#### Zazoo

Joined Jul 27, 2011
114
A starting point might be to consider the required internal "source" resistance.

You know that the output voltage can fall from 5V to 4.75V (-5%) with a 25mA increase in Ie. That is to say, the source resistance should be no more than about 0.25V/25mA or 10Ω.

The source resistance will be about (R1||R2)/ (1+β). So R1||R2 would need to be no more than about 1kΩ.

I would suggest 0.6V for Vbe is too small - a value of ~0.7V would be more reasonable.
I calculated R1 and R2 to be about 2660 and 1630 respectively (using Vbe = 0.7 and ignoring the dynamic resistance of the BE junction for now.)

Don't forget RE needs a value too, in order to minimise the voltage rise for the no-load situation. A "nice" value would set the minimum emitter current to about 1mA, so maybe around 5k. (This means max emitter current is actually 26mA.)
What do you mean by "nice" - is the choice really that arbitrary?

I don't mean to come across as dense, but I lack the experience to have any real "feel" for the rules-of-thumb or short-cuts that must become obvious with time and practice.

Edit: Also, one other question. On page 66 the following is written in reference to the emitter resistance RE:
"[RL] is in parallel with RE, but with RE dominating the parallel resistance"
They use this as a justification for the approximation: Rin = (β+1)RE.

This seems backwards to me, wouldn't RE typically be larger than RL, making Rin = (β+1)RL a better approximation?

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#### t_n_k

Joined Mar 6, 2009
5,455
I calculated R1 and R2 to be about 2660 and 1630 respectively (using Vbe = 0.7 and ignoring the dynamic resistance of the BE junction for now.)
I now don't think that one needs to include the dynamic emitter resistance since the assumption of a suitable (higher) Vbe value includes the variation in Vbe with emitter current. If we had assumed Vbe=0.6 say then it might be a valid point with Ie varying from no-load to full load current.

With your selected values we would get

Vth for the R1:R2 voltage divider as

Vth=15*1630/(2660+1630)=5.7 V

and a Thevenin resistance

Rth=2660||1630=1011 Ω

With 26mA in the emitter leg at full load (i.e. a no load current =1mA) this would translate to a

Vb=5.7-1011*Ib

With β=100 and Ie=26mA this would give Ib=257.4uA

This makes Vb=5.7-1011*257.4uA=5.7-0.26V=5.44V

With Vbe=0.7 this gives Ve=4.74V which is just outside of spec by 0.01V and probably OK for the purposes of the design.

Slight variations in β can make or break the design goal in this situation. This is a rather contrived problem to the degree that a real implementation could have a wide spread of regulation performance.

With your selected R1:R2 values I simulated the circuit using a transistor (BD139) with a β value of about 95 and Re=5k. The no load voltage was 5.08V and the full load voltage (25mA load current) was 4.73V which doesn't meet the 4.75V minimum at full load. The difference is quite small in the overall picture - 5.4% down on a nominal 5V. Although it is down 6.9% on the no load voltage of 5.08V. At no load Vbe=611mV and at full load Vbe=702mV. These values are subject to the "as supplied" simulator device model.

I agree with Hi-Z - It's all rather academic and of little practical value.

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• Zazoo

#### Hi-Z

Joined Jul 31, 2011
158
...What do you mean by "nice" - is the choice really that arbitrary?

I don't mean to come across as dense, but I lack the experience to have any real "feel" for the rules-of-thumb or short-cuts that must become obvious with time and practice.

Edit: Also, one other question. On page 66 the following is written in reference to the emitter resistance RE:
"[RL] is in parallel with RE, but with RE dominating the parallel resistance"
They use this as a justification for the approximation: Rin = (β+1)RE.

This seems backwards to me, wouldn't RE typically be larger than RL, making Rin = (β+1)RL a better approximation?
Well, if you make RE small enough, you can make the output voltage pretty independent of load current - but that's a pretty ridiculous way of implementing a voltage regulator.

Variation in output voltage due to variation in emitter current is mostly due to 2 factors:

1. The effect of base current variation on base voltage, due to loading of R1, R2 (as discussed at length above).
2. The effect of base current variation on Vbe.

You can minimise (1) by making R1, R2 small. You can only minimise Vbe variation by limiting the variation in emitter current. If I remember correctly (from about 25 years ago), Vbe will increase by about 18mV per octave increase (i.e. doubling in) Ib, or about 60mV per decade (i.e. tenfold increase).

As you can see, it's a case of diminishing returns here. Limiting Ie variation to a factor of 2 will limit Vbe variation to 18mV, but this would be rather extreme. Limit it to a factor of 10, and you get a Vbe variation of 60mV; to a factor of 20, 78mV. I happened to choose a factor of 25, so we'd expect a Vbe variation of just over 80mV. (Actually, in practice, with a real transistor, it would probably be somewhat worse than this.)

If you list the various sources of output voltage error (including the above and Vbe variation due to sample variation and temperature variation), you'll see why emitter followers wouldn't be my favourite means of implementing a voltage regulator (even if they didn't oscillate!).

• Zazoo

#### Zazoo

Joined Jul 27, 2011
114
I'm starting to get the picture now.
For some reason transistor circuits have really been a roadblock for me in my studies. I just can't get my head around them sometimes - this really helps.

Many thanks to you and t_n_k for your input. It has helped me immensely in understanding this circuit.

#### Hi-Z

Joined Jul 31, 2011
158
I'm very pleased that we've helped in your understanding of transistor circuits (you've used the word "immensely" - that just makes it all the more rewarding for us).

If you have any further questions about transistor circuits, don't be afraid to ask!

• anhnha