# 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
Feb 5 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. Feb 6 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. Feb 6 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. Feb 6 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. Feb 6 at 23:53