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? commonCollector schematic

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

1 Answer 1


Start with:


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.

  • \$\begingroup\$ 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. \$\endgroup\$
    – user318003
    Feb 6 at 5:16
  • \$\begingroup\$ 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. \$\endgroup\$
    – user318003
    Feb 6 at 5:26
  • \$\begingroup\$ 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. \$\endgroup\$
    – user318003
    Feb 6 at 22:34
  • \$\begingroup\$ @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. \$\endgroup\$ Feb 6 at 23:53

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