# How to set emitter resistors in transistor emitter follower

I got this circuit out of an old model railroad electronics book from the 1970s. It's your basic common-collector (a.k.a. emitter follower). Control voltage goes into the base of Q1, and then Q1, Q2, Q3 amplify the current to drive a motor load off Q3's emitter. But I cannot for the life of me understand the theory of how the values for the resistors R3, R4, R5 were chosen. I think they are there to stabilize the transistor behavior - maybe keep them in their active region and keep the output voltage more constant as the load varies? Can anyone explain to me the purpose of those resistors?

• If im not wrong, I think they pre-polarize the transistors into the active zone to linearize the control and make it less abrupt
– tac
Commented Feb 5, 2023 at 4:15

simulate this circuit – Schematic created using CircuitLab

If $$\P\$$ is the potentiometer position where 0 means the position is at the bottom and 1 means it is at the top then $$\V_{_\text{P}}=P\cdot 12\:\text{V}\$$ and its Thevenin resistance is $$\P\left(1-P\right)\cdot 5\:\text{k}\Omega\$$.

So, from that find $$\0\:\text{V} \le V_1\le 11.83\:\text{V}\$$ and the Thevenin resistance seen at the base of $$\Q_1\$$ will be as little as $$\4.6\:\text{k}\Omega\$$ (when $$\P=0\$$ or $$\P=1\$$) and as much as $$\5.8\:\text{k}\Omega\$$ when $$\P=\frac12\$$.

The 2N5088 is a fairly high $$\\beta\$$ device with $$\300\le\beta\le 1200\$$. The 2N4922 is a medium power BJT with $$\30\le\beta\le 150\$$. And, of course, the 2n3055 is a higher power BJT with $$\20\le\beta\le 70\$$ around the maximum current this circuit indicates ($$\3\:\text{A}\$$.)

Taking the minimum $$\\beta\$$ values from above, this means $$\20\cdot 30\cdot 300=18\:\text{k}\$$. So the base current for the 2N5088 is likely to be well under $$\20\:\mu\text{A}\$$ and the voltage drop across the $$\Q_1\$$ base supply's Thevenin resistance will be under $$\10\:\text{mV}\$$.

In short, they are supplying the entire system with a relatively stiff (low impedance) divider. That's good. And it means that the Thevenin resistance to the base of $$\Q_1\$$ can likely be discounted.

simulate this circuit

I've left the load itself off of that schematic because it's important to realize that there may not even be a load (no train on the tracks.) So there are two situations that bound the possibilities.

Starting with the question about sizing $$\R_5\$$, note that it's already been determined that the recombination current required by the base of $$\Q_1\$$ will be under $$\20\:\mu\text{A}\$$ and that the voltage drop for that reason will be under $$\10\:\text{mV}\$$. Including it at all guarantees that there is a galvanic path and load on the potentiometer and so long as $$\R_5\$$ is sized such that doesn't require a lot more than $$\Q_1\$$ then it also won't much impact the voltage at the base of $$\Q_1\$$. In this case, they set it at about twice the worst-case expected base recombination current. There isn't a bright line about making that particular choice. But it is a comfortable choice to my eyes.

$$\R_3\$$ provides a galvanic connection that keeps $$\Q_1\$$ operating regardless of the load on the track and sets the collector/emitter current for $$\Q_1\$$ as a function of $$\V_1\$$. Lower values of $$\V_1\$$ will mean lower operating currents for $$\Q_1\$$. But that's fine as the following transistors will be supplying less current to the load (train), too.

A first concern will be about the maximum power dissipation in $$\Q_1\$$. And in this case that peaks at about $$\60\:\text{mW}\$$ when set half-way ($$\6\:\text{V}\$$.) This is well within the dissipation spec (about 10% of it) for the 2N5088 device. So on that score, this choice is fine.

The peak current will be about $$\23\:\text{mA}\$$ when $$\V_1\$$ is at its maximum. While that's less than half of what's given in the Absolute Maximum Ratings, it's a lot. Especially given that the expectation is that $$\Q_2\$$'s required recombination current will be $$\\le\frac13\:\text{mA}\$$ -- about 70 times less. Even 10 times that, or $$\4\:\text{mA}\$$, would probably be over-kill and certainly stiff enough.

Another consideration is the dissipation of $$\R_3\$$. In this case, it gets as bad as a little over $$\\frac14 \:\text{W}\$$. Okay. So that means sizing for at least $$\\frac12\:\text{W}\$$ and maybe better still at $$\1\:\text{W}\$$.

So, I probably would use something larger for $$\R_3\$$. I'd want to set the maximum collector current to at least 5 times less than the Absolute Maximum, or $$\10\:\text{mA}\$$. (Though far less would still be comfortable to me.) So I'd suggest $$\R_3\ge 1.2\:\text{k}\Omega\$$. At this value, its worst case dissipation is $$\100\:\text{mW}\$$, which means a $$\\frac14 \:\text{W}\$$ resistor would be acceptable here. But I think it could be still larger, without difficulties.

I can't fully explain their particular choice for $$\R_3\$$ except that back then dissipating excessive power was considered more acceptable and larger resistors were about as easy to use as smaller ones. (No SMT back then!)

The logic for $$\R_4\$$ is similar to the above. It also is a galvanic connection to keep $$\Q_2\$$ active. They set the peak collector/emitter current for $$\Q_2\$$ to about $$\180\:\text{mA}\$$ and worst case dissipation for $$\Q_2\$$ is then about $$\\frac12\:\text{W}\$$ and for $$\R_4\$$ it is nearing $$\2\:\text{W}\$$ (so use a $$\5\:\text{W}\$$ resistor for this one.) As the base of $$\Q_3\$$ will require as much as $$\100\:\text{mA}\$$, I think this choice is about where I'd have also gone, trying to balance things. This one is harder to argue with.

$$\Q_3\$$ itself may not be active if there's no load on the track. That's fine. But to use a voltmeter to check everything out with a predictable measurement then perhaps adding a small load in parallel (between $$\1\:\text{k}\Omega\$$ and $$\10\:\text{k}\Omega\$$) would help.

The circuit is "open loop" in the sense that there's no measurement at the tracks that is used to feed back to control $$\V_1\$$. It's also not a switcher, which is likely a more reasonable approach today because of availability of options, the lower dissipation, and the number of added safety features that are available, as well.

• Your calculations are about spot-on with what I get in a circuit simulator. I did kind of figure R3 and R4 were in there to make sure Q1 and Q2 are in their active region. Because I didn't like the higher power dissipations on Q1 and Q2, I upped R3 and R4 to be 2.2k and 220 ohm, respectively. This still provides plenty of drive to Q3's base. I don't see a negative side effect to doing this. I don't want a switcher. This supply can be used to control "DCC equipped" motors, and the DCC decoder can be confused by the switcher pulses. Commented Feb 6, 2023 at 5:16
• Q3's gain is more like 60 in this circuit, even at heavy load, so its base current doesn't get near 100mA, more like 30. I can't use a switcher. This supply must be linear because it can be used to control "DCC equipped" motors, and the DCC decoder can be confused by the switcher pulses. Commented Feb 6, 2023 at 5:26
• Looking back at what you wrote, I think I see something wrong. It is minor, but nonetheless... you say "If P is the potentiometer position where 0 means the position is at the bottom and 1 means it is at the top then V1=P⋅12V", but V1 is between the divider, R3 and R5. Wouldn't V1 be = P * 12v * (330k/334.7k) ? That seems to be consistent with what follows when you say V1 will be between 0 and 11.83v. Commented Feb 6, 2023 at 22:34
• @user318003 Yes. But if you look at the next paragraph, you will see the correct range given for $V_1$. I probably should have given them two different labels. Sorry about that. I made a slight correction, which I hope you find okay. Commented Feb 6, 2023 at 23:53