For this Art of Electronics circuit, why aren't the transistors specified? And what transistors do I use?

Question

If I were going to EE school, then there would be a lot of knowledge transferred just by going through all of the usual things -- labs, homework, exams, etc. In teaching myself, it's so hard when they just "Q1, Q2, Q3, ..." and I know that the choice of the individual transistors requires further decision making, and I can easily do it wrong. I wish they would place a more fully worked-out instance on a later page, sort of like a reference design.

Specifically, can I use jellybean 2N3904 and 2N3906 for Q1-Q8, the comparator-like subcircuit? Or do I have to use a dual (like PMBT3904YS,115) or a matched pair (like DMMT3904W-7-F)?

What follows are my semi-educated guesses. Though I'm not really completely sure which transistor to start with.

I would feel comfortable using the jellybean 2N3904 and 2N3906 for Q9 through Q12. Perhaps Q10 could be a 2N4401 -- it depends on how much we need to saturate Q11. Though I would put Q10 a distance from Q11 to separate it thermally and hopefully improve performance. And I would put a 2SA1908 at Q11, or some other power-transistor with a heat-sink.

Anyway, that's my current, probably somewhat divided view of what I would do, mixed with what I think might be required.

2.22 Temperature controller -- The schematic diagram in Figure 2.76 shows a temperature controller based on a thermistor sensing element, a device that changes resistance with temperature. Differential Darlington Q1-Q4 compares the voltage of the adjustable reference divider R4-R6 with the divider formed from the thermistor
and R2. (By comparing ratios from the same supply, the comparison becomes insensitive to supply variations; this particular configuration is called a Wheatstone bridge.) Current mirror Q5-Q6 provides an active load to raise the gain, and mirror Q7-Q8 provides emitter current. Q9 compares the differential amplifier output with a fixed voltage, saturating Darlington Q10-Q11, which supplies power to the heater, if the thermistor is too cold. R9 is a current-sensing resistor that turns on protection transistor Q12 if the output current exceeds about 6 amps; that removes base drive from Q10-Q11, preventing damage.

(Added emphasis mine.) The preceding excerpt is from page 105 of The Art of Electronics, Second Edition, by Horowitz, Paul; Hill, Winfield. Cambridge University Press. Kindle Edition of hardback (ISBN 978-0-521-37095-0):

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_{Note: I am trying to teach myself -- this is not homework or school work. A software developer trying to gain more knowledge of hardware so I will write better firmware.}

Answer 1

If you're building this at board level, you wouldn't buy individual transistors for any of the part that's powered by +15 V (except Q9). You'd just buy an op-amp IC. And if you built this in the last 20 years you'd probably power it from 5 V or 3.3 V instead of 15 V.

For the part powered by 50 V, you need to choose transistors that can withstand the 50 V that will be across them when they are "shut off", can handle the forward currents they will need to carry when they are "on", and can dissipate the heat they will be generating when they are at an intermediate state between "on" and "off". Since this circuit operates with very slow changes in voltage, there isn't any special need to worry about the frequency response of these transistors.

Answer 2

This circuit is not something you’d want to build in 2021. The performance will be poor compared to a few-penny op-amp because you can’t (inexpensively) buy transistors matched and on the same substrate for the two current mirrors and the input darlingtons- 8 transistors in total).

The output circuit has a current limit function (R9/Q12) that would cause Q11 to dissipate around 350W if the heater is shorted, but otherwise does nothing much. That kind of dissipation would require a huge heat sink and even a beefy TO-3 metal PNP transistor such as MJ2955 cannot survive for more than a couple hundred microseconds (SOA). It will also dissipate quite a bit of heat in normal operation (in the 5-10W range when ‘on’) so a substantial and expensive heat sink is required.

If you want to design a thermistor temperature controller, use a single-supply op-amp or comparator, and switch the load cleanly with a MOSFET that will not require a heat sink. Or use a relay if you don’t care about lifetime, and want more flexibility.

This is not something you should want to copy.

The AoE circuit may well be instructive (it is of help in understanding the inner workings of something like an LM393), but it’s less than useful in a practical sense.

If you want to build one to play around, SMT transistors from adjacent pockets in the tape are usually matched very well (at least from major makers) since they tend to come off of nearby areas on the same wafer in an automated line). If they are mounted to a board with little power dissipation they will tend to be at similar temperatures. In this case you have many watts being dissipated you’d tend to get the offset shifting wildly as the board heats and cools. Op-amps have a similar issue (even though it is mW and not watts) and use clever layout to place differential parts on isotherms to reduce offset induced distortion. Just as a rough number, a temperature difference of 0.05 degrees C will change the bias by around 10% if I guesstimated correctly.

Answer 3

Yes, you can use jellybean devices for all elements of this circuit except for Q11 as you mention. A 2N3055 would be suitable for that, although you need to ensure its gain (beta) is good (over 50) so that Q9 and Q10 can drive it. For the other devices, standard discrete transistors as you mention will be suitable -- their beta is high enough, their current-carrying capability is suitable and their voltage rating is enough.

You don't need matched devices for this application, which means that the opamp likely will have a small offset voltage (perhaps 10 mV). That is not critical.

Note that there is no AC stability compensation in this circuit and a breadboard or simple implementation may oscillate. While the thermal lag of the heater to the thermistor may be sufficient, this does depend on the physical construction. A capacitance from the base of Q9 to ground could compensate it. Try 1 uF to begin.

Answer 4

Why are you building this?

Are you actually trying to build this? The schematic looks like it came from a text book where it would be used to exemplify things like current sources, current mirrors, differential amplifiers, etc. In other words, it's a great schematic to use combined with "ideal" (i.e. "simplified") transistor models to teach concepts.

But as others have noted, this aint' the way to do it if you're trying to build a working circuit. If you're trying to build this because you actually need a working temperature controller for a 50W heating element, heck, you can buy stuff like that off the shelf at most hardware stores.

On the other hand, if you're just trying to build it for the fun of learning how, then replace the heating element with an 3.3vdc incandescent flashlight bulb (you can get them at any hardware store) and forget using the monster drivers the heating element would require. Do that and after scaling some of the resistors to maintain current levels, you can use vanilla transistors and a 5vdc source for all sources.

Or, better still, download a free SPICE simulator and simulate it. You'll learn everything other than the frustration of using a breadboard. (And be grateful you're not wire wrapping... which I had to do in college and I'm glad I never had to do it again.)