One for our electronics experts.

Where's the input to the regulator? Is that the vertical line at the top right? Is it battery 12V, or does it go through other components on the way? What is the component on the left side attached to the collector of Q3?

My guess is it's something like a 3.6v zener with another diode in series to make 4.3v, and then a base-collector drop in Q1 to make 5v. But I'm not sure of the feedback arrangement, nor whether it's relevant given the Q3 tap. It's very hard to tell anything without context.

Theo

Shown as being direct from the 12v input.

Goes to pin 11 on the processor IC. An LFC1041AE

Thanks, Theo. If it helps, voltage with the suspect zener in place reads supply volts negative side, 0.012v the other.

Would a bit of help trying to determine the kind of zener (ZD2) you need help?

I?ll talk you through some basic analysis of part of the circuit:

Start with Q3: The base / emitter voltage will be between 0.6 and 0.7 V.

Therefore the voltage across R9 is between 0.6 and 0.7 V.

Therefore current in R9 between 0.6 and 0.7 mA

We can ignore the base current, so this flows through R6, R8, R9, and ZD2.

We can now find the voltage at the junction of R6/ZD2

R6+R8+R9 = 15300 ohms.

At R6/ZD2 between Vmin=15300x0.0006 and Vmax=15300x0.0007

So between: 9.18V and 10.71 V

The other end of the zener is at nominal battery / alternator voltage.

The question is, what have they assumed that is.

Well, zeners come is standard values and you have a rough idea what the battery voltage is.

A 5.1 V one on the 10.71V would suggest a battery / alternator of 15.81V - bit high.

A 3.3 V one on the 10.71 would be 14.01 V & 12.48 on the 9.18 limit.

Promising but possibly a bit low.

A 3.9V zener on 10.71 gives 14.61V & 13.09 on 9.81 - more promising.

A 4.3 zener take you over 15V, which I think is high but it could be?

I think it is probably a 3.9V zener. As for rating, it doesn?t seem to be passing much current but look for one of similar size.

As for the existing zener, it should ?look? like a diode if tested on a meter with a diode test function BUT it will appear to be miss-labelled (ie the wrong way around). That is normal - seriously.

Brian

Can you explain that, Brian? All the zeners I have here are marked with a band at one end - same as diodes. And on my DVM diode test function band to negative works with both.

Ok, first why, then a suggested demo.

Zeners are used as references, they ?break down? (conduct) in the REVERSE direction when the marked voltage is reached. In the other direction, they will show a 0.6 to 0.7 volt drop, like a normal diode.

Now, take a diode and a zener.

Using a multimeter with a diode test range, apply the probes to each, one way and the other.

You should see the forward voltage, typically, 0.6 to 0.7 V.

Note which end the probes are applied to in each case.

That's what I did. I use the diode test function often.

And on both an ordinary diode and a zener, you connect the meter neg lead to the band end of the diode. And get a reading of the forward voltage drop - about 0.6v Reverse the connections and you may get a rubbish reading or none at all.

Just to summarise.

The supposed faulty zener removed from circuit tests out OK using the DVM diode test function. Reads 0.624v. Reverse the connections, Fluke reads OL.

Wired it up to a 12v supply using a series 15K0 resistor from non band end to ground. Positive to banded end. Fairly similar as the circuit

Volts across the diode 0.

Do the same with a 3.9 new zener.

Volts across that zener 2.9v.

I'm wondering if it is actually a zener diode? The circuit obviously isn't the original Lucas one - it is much more recently drawn than 1980. So could be simply a mistake?

By rubbish you mine it indicates open circuit or high resistance? If so, it is probably ok.

I think I may have confused you.

Let me try again.

On semiconductors, the ?arrow? indicates the direction of flow of conventional current.

In a normal diode, generally, you have conventional current flowing in the direction of the arrow most of the time. In never follows the other way (ignoring leakage current).

A zener is normally used so no conventional current flows in the direction of the arrow. (Some circuits may differ, most don?t.) Instead, the zener is used in reverse, so it ?breaks down? and maintains a constant voltage.

You will note the zener you asked about APPEARS the wrong way up in terms of the arrow and the flow of conventional current UNTIL you accept it operates as above.

(Emphasis not for you, it is for the idiots who stalk me from the Radio group. )

There are actually two modes of break down, one below about 6.7 V one above it. One has a positive temp coef, one a negative. Around 6.7v you see both effects and the temp coefficients cancel out. They are still all called zeners.

Sorry I confused you. Not my best explanation. A busy day.

My typo. It read 12v.

That suggests it either isn?t a zener, a bit unlikely unless the circuit is wrong.

It has decided to stop being a zener. Even less likely! Given it seems not to have died going by the diode check.

It is possible you need to increase the 12V to push it into break down.

Also, the 3.9v zener should be closer to 3.9 than 2.9.

well done brian you are always helpful

Couldn't you post the whole circuit not just a part? Your zener is being supplied by 5V, so it's probably a crowbar to limit any nasty voltage spikes. I've just worked out part of a switching PSU in order to change the stabilised output from 12V to 9.5V. There is a zener diode across the output rated at 16V.

I'm sort of thinking that too. It's to cope with spikes.

I don't think you could read the entire circuit as a JPEG that I could provide a link to. The master copy I have is a PDF - and that has to be zoomed in to read individual parts, even on a 24" monitor.

It is a really odd circuit as drawn. You might want to try downloading LT spice (free) and run it in the simulator. My best guess from looking at it is that the whole thing looks a little implausible.

My best guess of intended operation is that Q2/Q3 is intended to be a soft start so that there is a short delay before it powers up pads E,K. The chain R6,R8,R9 & C5 forming the shorter time constant and R3/C2 the longer one. The circuit seems to be very clunky to me.

I can put a ballpark estimate on the maximum voltage that ZD2 can be before it becomes impossible for Q3 to switch on as 14v - 0.6v*15.3 ~

For the purposes of a quick suck it and see test replacing ZD2 with 6.8k ought to get something like the right voltages at pads E & K. Regulation will be terrible though it really wants a zener diode in there.

Check the current draw looks acceptable!

BTW If C2 should ever fail open circuit it will fry ZD1 by putting Q2 & Q1 into hard conduction.

The more normal classic old school series regulator would have the zener diode and capacitor in the leg of the base to ground and the beefy transistor configured as an emitter follower. Or alternatively as a shunt regulator which is less efficient. Your circuit is neither.

This isn't a bad intro to the two sorts of transistor based regulator:

It was designed in the 70s by, I'm told, Lucas Aerospace.

The mupdf includes command-line mutool.exe . The tool is cross-platform, and if you're on Linux, might be in your package manager.

mutool extract fileofimages.pdf

The idea is, it extracts the image as stored inside the file, at "captured" resolution. Whatever someone did to create the PDF, that's the result that will be extracted.

This avoids the need to Print Screen and other such things.

You could also do it with GIMP, but that isn't always set up properly for this in all cases. The unfortunate part, is they could just stick an engine right in the tool, but instead, it's done via "add-on" fiddly bit. Whereas writing up the mutool one, is more straight forward.

LibreOffice and some part of the Microsoft suite, might also work, but only worth a try if they're installed and you're fluent. I think LO Draw might be able to import a PDF, but when output later, it'll be bitmaps rather than vector based.

I use mutool for jobs like this. If that doesn't work, my GIMP is second in line. LO is third. Mutool should really give a superior result.

Paul

I think this is the circuit - page 6 or page 10:

Or would the drop across the series resistor R1 be enough to lower the emitter voltage and turn it on to compensate?

Theo

Not a problem here as my CAD prog loads PDFs.

I was more meaning how to make it available on here as the only hosting service I use is imgBB.

That's the one. Don't think Lucas ever made details available. Much thanks to the chap who must have spent ages drawing it out.

All a bit over my head, Theo. I'm more repair than design. ;-)

I've been checking a different version which I think is good - just for volts and so on. That Vcc is spot on 5v. The snag being the broken one is to Oz spec - engine in a lower state of tune to UK cars - and no lambda sensors like that TR8 NAS spec RV8. So the thoughts are the mapping will be different.

My guess is the main processor may be faulty, and pulling Vcc down (although total current drain is similar between them). Or, as you say, there is feedback from it to the regulator via Pin 11. And a search for LCF1041AE gets me nowhere. I do have used spares but all from UK spec ECUs.

I'm really just looking at this as a sort of x-word puzzle rather than commercially, (as it were). I concluded years ago it's much easier to just convert to MegaSquirt if having problems. My own car ran weak at low throttle openings and defeated attempts to fix it. With the MS, it now pulls like a V8 should from idle onwards.

FWIW, several aftermarket companies did so called adjustable versions. But not entirely successful. And before it was so easy to get an aftermarket user programmable replacement. Other thing is MS gets rid of the very expensive AFM which also wears out.