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I've a problem with the attached voltage regulator. It works fine at a room temperature, but when the ambient temperature drops so is the output voltage.

What is the reason for that? Is there a way to fix the problem?

Update-1 The output voltage is set at 3.3V @ 25degC, but drops to 1.9V at -5 degC.

Disclaimer I haven’t designed the circuit, so I'm not aware why certain design decisions were made. I'm not an expert electronics engineer, typically I program the damn device ;). However I'd really like to fix that one with your help.

Solution As adviced I've replaced Q11 and Q12 with a pair of BC847 transistors and it works fine now.

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What is the output voltage and how much does it drop per degree K? What variation with temperature were you expecting, and why? –  Brian Drummond Feb 7 at 12:08
The output is set at 3.3V @ 25degC. It drops to 1.9V @ 25degC. I expect the device to operate from -20 degC to 50 degC. –  bvr Feb 7 at 12:23
So, that's about 0.05V/degK –  bvr Feb 7 at 12:34
1.9V @ 25degC?? How does it cope with small load changes and full load current? –  Andy aka Feb 7 at 12:39
Sorry, my mistake 1.9V @ -5 degC –  bvr Feb 7 at 12:41

3 Answers 3

up vote 2 down vote accepted

Why are you using Darlington transistors in a low-voltage design? To begin with, the VBE of 1.4V is greater than your reference voltage — I'm surprised this circuit works at all! As it is, R32 has just millivolts across it, and R27 can have at most 10× that value across it (assuming Q12 is cut off completely).

For another thing, the minimum VCE of 1.5V is going to limit the available voltage to drive the MOSFET, which is operating in a sub-threshold region.

At a minimum, you need to select non-Darlington transistors for your differential amplifier. You don't need a huge amount of current gain in this circuit configuration anyway, but single transistors with gains of 200 or more are available. You may be happier switching to a design that uses a low-voltage rail-to-rail opamp.

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Could unmatched transistors be an extra source of the problem too? Either thermally coupled or not. –  jippie Feb 7 at 13:36
Well, a monolithic matched pair (plenty of those available, too) would give better overall performance, but all bets are off until all devices are operating within a valid regime. –  Dave Tweed Feb 7 at 13:52
Possibly significant that Vbe vs temperature is -2mV/K (per transistor) or a 120mV increase for a Darlington cooling from 25C to -5C. –  Brian Drummond Feb 7 at 15:44
@DaveTweed I think its a valid point about Vbe. I understand why current solution might not work. I've replaced Q11 and Q12 with a pair of BC847C transistors. I'll report if that helps. –  bvr Feb 7 at 17:02

Not enough information - ie you have more that you could have given but didn't. MOSFET IRF9321 datasheet here

Reference diode LM285 datasheet here

Darlington PMBTA13 datasheet here

I'll assume that you are testing this with Vin = 4V2 (as per diagram). You don't say and you should as if I am correct about the problem then it may work OK on eg 5V.
What is Vin when testing?
What value is Vbat?

Your diagram has committed an Olin fail, with text over-writing important information.
Bigger and still munted version here

R30/R31 are 178k/100k to make a 2.78:1 divider.
If output is meant to be 3V3 that implies Q11 and Q12 bases should be 3.3/2.78 = 1.187V = 1.2V.
Closest regulator = 1.235V.
As you are using 1% resistors you could have got nominally if not actually closer than that. No problem though.

Fig 10 page 4 shows that gate threshold voltage increases with falling temperature.
The graph shows 2V at 0 C.
That's probably 'typical'.
Page 2 shows Vgsth as -1.3 / -1.8 / -2.4 min / typ / max, so you MIGHTY expect it to need as much as about -2.6V to get 50 uA.

The darlington transistors in the long tailed can saturate at 1 Vbe minimum.
The datasheet says Vsat = 1.5V but the current is "rather higher" than you are using.
A single transistor can saturate with Vce approaching zero, and this notionally applies to the upper transistor in the darlington pair. However, as it steals current from the lower transistor base as it saturate you won't get upper collector all the way down to lower collector. So Vsat will be 1 Vbe + some (probably :-) ).

You lose some headroom in the drop across R32 in the "tail" but as it's voltage is 1.235 - 2 x Vbe it is not much voltage - I'd have thought it was almost dangerously low as 1.235 - 2 x Vbe is about ill conditioned.

Overall you appear to have close to zero headroom at best, and a small increase in Vgsth could drive you against one limit. Whether you have any depends on the factors above - what is Vgsth actual?, what does the darlington saturate at actually?, does the long tailed pair behave if V_R32 is very low (and how low is it?).
What ... ?

Easy check - operate with Vin = say 6V.
If it works OK with temperature at 265 K, reduce Vin until it faults.

IF headroom is the problem, changing the darlingtons for single bipolars will give you at least a Vbe more at the loss of gain.

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Increasing Vin won't help with the Darlington pair; you'd have to pull the bottom of R32 below ground in order to get it operating with the left-hand base fixed at 1.2V. As it is, R32 has just millivolts across it, and R27 can have at most 10x that value across it. –  Dave Tweed Feb 7 at 16:29
I've replaced Q11 and Q12 with a pair of BC847C. Am I right that another approach would be to replace Q10 with a PMOS with lower Vgsth? –  bvr Feb 7 at 17:08
@DaveTweed - I doubt we disagree on any major aspect. More Vin will (potentially) help headroom for the MOSFET gate drive, but will not help the lack of voltage on the 'tail' for which as I noted "1.235 - 2 x Vbe is about ill conditioned." Increasing Vin is an easy check. If it fixes or somewhat fixes things it shows one is on the right track (and if it doesn't it proves nothing.). || Changing to a 2V5 version of the reference will help tail voltage - at the loss of gate drive voltage. A FET with much lower Vth should be easy enough to find. –  Russell McMahon Feb 8 at 4:43
I don't understand why you don't see it -- any additional Vin will simply appear across Q11. It won't help the differential pair, and it won't help with the MOSFET drive, either. Yes, increasing the reference voltage would help immensely. –  Dave Tweed Feb 8 at 5:13
@DaveTweed - I think we are talking at cross purposes. I'm pretty sure we both well enough understand the circuit. When you say 'it won't help the differential pair' I think you are (much) more succinctly putting what I said above - " ... will not help the lack of voltage on the tail ..." etc. || IF Q11 can be turned on as fully as it is able to go (which we both agree is marginally likely) the more Vin allos more Q1 collector swing = more FET gate swing . || IF the mahjor problem is in the LTP (as it may well be) then this will not help. IF the LTP is JUST working then more Q11 swing helps. –  Russell McMahon Feb 8 at 11:00

The LM285D voltage reference is probably not the best choice. It has a typical temperature coefficient of +80 ppm / degC. A 40 degC drop from ambient would be equivalent to a 0.32% drop in its terminal voltage. Is this the sort of voltage drop seen? This is only a typical figure and it could be a lot higher. See also figure 5 in the data sheet.

I guess you're using the 1.235V version of the device trying to achieve a 3v3 supply (maybe the 2.5V version for a 7V supply) - look at the voltage across the reference as temperature changes - this might indicate this part being the problem.

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I've checked that before. While its true that the reference voltage drops slightly I don't think its the reason for the huge voltage drop. –  bvr Feb 7 at 17:03

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