Temperature controller from The Art of Electronics - Page 1
Forum section: electronics / projects

author: tree
author_rank: Regular Contributor
timestamp: December 14, 2013, 04:28:36 am
content: I have attached an excerpt from the art of electronics. It's a temperature controller that uses a differential pair.I don't understand a few aspects of this circuit. The first being the output of the differential pair and how it's related to the inputs. Furthermore, how does that output control Q9?

author: IanB
author_rank: Super Contributor
timestamp: December 14, 2013, 04:40:26 am
content: A constant current flows through Q6. If the temperature is too cold this current flows into the base of Q9, turning it, and the heater, on. If the temperature is too warm the current flows through Q3 instead and Q9 turns off.

author: tree
author_rank: Regular Contributor
timestamp: December 14, 2013, 04:43:58 am
content: Quote from: IanB on December 14, 2013, 04:40:26 amHow? I tried simulating the set temp pot by using fixed voltage sources (at various voltages) and I do not get constant currents. The sum of currents through Q5 and Q6 has to be constant due to the current mirror at the emitter, but the individual currents are not constant.

author: IanB
author_rank: Super Contributor
timestamp: December 14, 2013, 04:50:57 am
content: I'd have to simulate it to understand better how it works, but the configuration looks like what is found in standard op amp or comparator circuits. I just recognize that I see that arrangement of transistors all the time and I accept it without questioning the details too much. I guess you've given me a project for the weekend...

author: tree
author_rank: Regular Contributor
timestamp: December 14, 2013, 04:54:13 am
content: Quote from: IanB on December 14, 2013, 04:50:57 amYeah, the input is a standard darlington differential pair, active load and a current mirror. I just don't understand how the output controls the heating element based on the input. That is not to say I don't understand how the actual heating and current limiting stages work. I'm trying to simulate the entire circuit to understand what happens. We'll see how far I get...

author: IanB
author_rank: Super Contributor
timestamp: December 14, 2013, 04:59:10 am
content: Quote from: tree on December 14, 2013, 04:43:58 amThe collector current of Q6 should mirror the collector current of Q5. So I guess that is not a constant current exactly, merely a mirrored current.

author: tree
author_rank: Regular Contributor
timestamp: December 14, 2013, 05:00:55 am
content: Never mind, solved it. This circuit is less analog than I thought. The output merely controls Q9 (on/off). So if the set voltage is higher than the voltage across the thermistor (i.e. warmer than it should be), Q3/Q4 are on while Q1/Q2 are off due to the reverse biased base emitter junctions of Q1/Q2 and No current flows into the base of Q9. Instead it flows through the darlington pair Q3/Q4. If the set voltage is lower, Q3/Q4 are off and the current out of Q5 flows into the base of Q9 turning it on and the heater on.

author: IanB
author_rank: Super Contributor
timestamp: December 14, 2013, 05:14:22 am
content: The key to the circuit apparently is that the voltage at the collector of Q8 floats up and down due to the constant current through Q8. If Q1/Q2 turn on then the voltage goes up and Q4/Q3 turn off. If Q1/Q2 turn off then the voltage goes down and Q4/Q3 turn on.

author: tree
author_rank: Regular Contributor
timestamp: December 14, 2013, 05:17:39 am
content: Quote from: IanB on December 14, 2013, 05:14:22 amPretty much. The voltage at the collector of Q8 is determined by the larger of the two voltages either across R3 (the base of Q1) or from R5 (base of Q4). As a result only one of the darlington differential pair transistors can be on at a time.

author: grumpydoc
author_rank: Super Contributor
timestamp: December 14, 2013, 12:46:29 pm
content: It's just a discreet op amp isn't it?

author: SeanB
author_rank: Super Contributor
timestamp: December 14, 2013, 01:00:21 pm
content: If you listened to the 555 video the operation of the comparators and the current steering logic was explained there.

author: dannyf
author_rank: Super Contributor
timestamp: December 14, 2013, 01:14:29 pm
content: QuoteThe CCS Q7/Q8 holds the current steady. Q5/Q6 is an active load for higher gain.The book does a very good job explaining how it works.QuoteOr a comparator.QuoteThere are two comparators here: one formed by the differential input stage + associated resistors. Another is formed by Q9 + R7/R8.

author: GK
author_rank: Super Contributor
timestamp: December 14, 2013, 01:34:01 pm
content: This is not a very good circuit. There is potential for the b-e junction of Q9 to get avalanched whenever the control loop is temporarily saturated (temp regulation not in equilibrium / out of range).

author: c4757p
author_rank: Super Contributor
timestamp: December 14, 2013, 01:44:37 pm
content: What do you know - I was just eying that area trying to figure out if that would happen. This particular problem could be solved with a clamp diode from emitter to base, I believe.I'm personally a bit nervous about the lack of hysteresis, so that the circuit could possibly oscillate around the set point. It could get quite warm in places not marked "heater" if that were to happen quickly...

author: GK
author_rank: Super Contributor
timestamp: December 14, 2013, 01:50:35 pm
content: The best way to fix the circuit would be to invert the differential long tail pair (ie PNP Darlingtons, ground-referenced NPN current mirror) and then Q9 can simply have its emitter tied to ground. Limit Ic Q9 with a resistor in series with its collector.

author: SeanB
author_rank: Super Contributor
timestamp: December 14, 2013, 02:28:13 pm
content: True, though it will likely only see a reverse bias of about -6V, so most transistors will be happy with it. Most cases will have the set point voltage at about 6-9V and the thermistor selected to give about the same range. Q9 will likely be a 140V or higher transistor as well, so it will have a reverse base limit of more than 10V. At least leakage in Q12 will offset the leakage in Q9 somewhat, though I would have been happier with a 10k resistor across Q12 CE to limit it at high temperature, as the power stage is likely to run rather hot in use.

author: dannyf
author_rank: Super Contributor
timestamp: December 14, 2013, 03:11:55 pm
content: QuoteUse a pnp for Q9 and let it drive a npn switch - the load is on the npn's collector to Vcc. The rest remains the same.

author: IanB
author_rank: Super Contributor
timestamp: December 14, 2013, 05:56:49 pm
content: Quote from: SeanB on December 14, 2013, 01:00:21 pmI did watch that video and Dave didn't really explain how the comparator worked, he just said "that [arrangement of current mirrors and darlington pairs] is not too dissimilar to a regular comparator chip that you would buy off the shelf" and then moved on.If you are not familiar with it (as I am not) then that arrangement of circuit building blocks looks somewhat funky and it is not at all obvious how it does what it does.

author: GK
author_rank: Super Contributor
timestamp: December 15, 2013, 12:07:03 am
content: Quote from: SeanB on December 14, 2013, 02:28:13 pmWith the wiper of the pot at the far ground end, the b-e junction will see a reverse bias of about 12V.

author: SeanB
author_rank: Super Contributor
timestamp: December 15, 2013, 08:31:45 am
content: If you are worried put a diode in series with the base of Q9 and it will then not stress the transistor.I made a quick and nasty crystal heater just using a TO220 transistor soldered to the crystal and used the bias to heat it to a setpoint where the NTC thermistor would finally starve the base bias enough to stop it heating. Was not fast and needed a regulated 12V supply ( was a power rail anyhow) along with SOT resistors to heat the transistor and the power resistor on the other side to equal power during startup. But it did work, regulated to around 70C, was only a little bigger than the crystal and fitted inside a small insulating case around the crystal.

author: GK
author_rank: Super Contributor
timestamp: December 15, 2013, 11:03:31 am
content: If you put a Si diode in series with Q9's base the emitter potential will also have to be lowered otherwise the current mirror will not have enough headroom to turn Q9 on.

author: SeanB
author_rank: Super Contributor
timestamp: December 15, 2013, 11:36:52 am
content: R7 to 1k5 will solve that.