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?

--
*I want it all and I want it delivered

Dave Plowman dave@davenoise.co.uk London SW
To e-mail, change noise into sound.

In article <59da028d2edave@davenoise.co.uk>,
Dave Plowman (News) <dave@davenoise.co.uk> wrote:
> 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.

My typo. It read 12v.

--
*Many hamsters only blink one eye at a time *

Dave Plowman dave@davenoise.co.uk London SW
To e-mail, change noise into sound.

Dave Plowman (News) <dave@davenoise.co.uk> wrote:
> In article <59da028d2edave@davenoise.co.uk>,
> Dave Plowman (News) <dave@davenoise.co.uk> wrote:
>> 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.
>
> 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.

3.5 is the min I’d expect.

On Sat, 16 Apr 2022 14:14:35 +0100, "Dave Plowman (News)"
<dave@davenoise.co.uk> wrote:
>In article <59da028d2edave@davenoise.co.uk>,
> Dave Plowman (News) <dave@davenoise.co.uk> wrote:
>> 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.
>
>My typo. It read 12v.
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.
--
Dave W

In article <5f5p5hh3tatld1oss722b8lm6hhe6fuvho@4ax.com>,
Dave W <davewi11@yahoo.co.uk> wrote:
> On Sat, 16 Apr 2022 14:14:35 +0100, "Dave Plowman (News)"
> <dave@davenoise.co.uk> wrote:

> >In article <59da028d2edave@davenoise.co.uk>,
> > Dave Plowman (News) <dave@davenoise.co.uk> wrote:
> >> 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.
> >
> >My typo. It read 12v.

> 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.

--
*You can't have everything, where would you put it?*

Dave Plowman dave@davenoise.co.uk London SW
To e-mail, change noise into sound.

On 4/18/2022 6:51 AM, Dave Plowman (News) wrote:
> In article <5f5p5hh3tatld1oss722b8lm6hhe6fuvho@4ax.com>,
> Dave W <davewi11@yahoo.co.uk> wrote:
>> On Sat, 16 Apr 2022 14:14:35 +0100, "Dave Plowman (News)"
>> <dave@davenoise.co.uk> wrote:
>
>>> In article <59da028d2edave@davenoise.co.uk>,
>>> Dave Plowman (News) <dave@davenoise.co.uk> wrote:
>>>> 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.
>>>
>>> My typo. It read 12v.
>
>> 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.
>

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

https://www.mupdf.com/releases/index.html

Create a folder to work in. Put the mutool.exe and the .pdf file,
and then when it dumps the contents via "Extract", just look
through the bits and pieces, for the stuff you want.

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

"Dave Plowman (News)" <dave@davenoise.co.uk> wrote:
>
> 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.

I think this is the circuit - page 6 or page 10:
https://www.triumphwedgeowners.org/uploads/3/0/4/2/3042952/_lucas_4cu_ecu.pdf

I started doing an analysis and then ran into the weeds. One thing that
bothered me though: there's no feedback after Q1. Supposing the load
current changes, you'd expect the voltage drop across emitter/collector of Q1 to go
up. But I don't see how the circuit compensates for that. There is that
tap from pin 11 of the IC, but I'm doubtful it's a feedback tap.

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

Theo

In article <t3ldmc$1vnl$1@gioia.aioe.org>,
Paul <nospam@needed.invalid> wrote:
> > 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.
> >

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

> https://www.mupdf.com/releases/index.html

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.

--
*Eagles may soar, but weasels don't get sucked into jet engines *

Dave Plowman dave@davenoise.co.uk London SW
To e-mail, change noise into sound.

In article <OJg*9Q6Ly@news.chiark.greenend.org.uk>,
Theo <theom+news@chiark.greenend.org.uk> wrote:
> "Dave Plowman (News)" <dave@davenoise.co.uk> wrote:
> >
> > 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.

> I think this is the circuit - page 6 or page 10:
> https://www.triumphwedgeowners.org/uploads/3/0/4/2/3042952/_lucas_4cu_ecu.pdf

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.

> I started doing an analysis and then ran into the weeds. One thing that
> bothered me though: there's no feedback after Q1. Supposing the load
> current changes, you'd expect the voltage drop across emitter/collector
> of Q1 to go up. But I don't see how the circuit compensates for that.
> There is that tap from pin 11 of the IC, but I'm doubtful it's a
> feedback tap.

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

> Theo

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.

--
*What do you call a dinosaur with an extensive vocabulary? A thesaurus.*

Dave Plowman dave@davenoise.co.uk London SW
To e-mail, change noise into sound.

On 19/04/2022 15:38, Dave Plowman (News) wrote:
> In article <OJg*9Q6Ly@news.chiark.greenend.org.uk>,
> Theo <theom+news@chiark.greenend.org.uk> wrote:
>> "Dave Plowman (News)" <dave@davenoise.co.uk> wrote:
>>>
>>> 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.
>
>> I think this is the circuit - page 6 or page 10:
>> https://www.triumphwedgeowners.org/uploads/3/0/4/2/3042952/_lucas_4cu_ecu.pdf
>
> 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.
>
>> I started doing an analysis and then ran into the weeds. One thing that
>> bothered me though: there's no feedback after Q1. Supposing the load
>> current changes, you'd expect the voltage drop across emitter/collector
>> of Q1 to go up. But I don't see how the circuit compensates for that.
>> There is that tap from pin 11 of the IC, but I'm doubtful it's a
>> feedback tap.
>
>> Or would the drop across the series resistor R1 be enough to lower the
>> emitter voltage and turn it on to compensate?
>
>> Theo
>
> 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.

D3 is a red herring, presumably trying to stop load dumps getting
through Q1.

It's a very strange way to simply get +5V. I'm wondering if ZD1 is being
used as a shunt regulator, with the arrangement connected to the base,
and the 1.2R resistor Q1 providing a constant current. Are Q1 and ZD1
fairly beefy?

First thing to check is Q3 collector / Q2 base. Is this at 0V, or closer
to 3.3V. I assume this is some sort of power enable (on/off) from the MCU.

On 18/04/2022 11:51, Dave Plowman (News) wrote:
> In article <5f5p5hh3tatld1oss722b8lm6hhe6fuvho@4ax.com>,
> Dave W <davewi11@yahoo.co.uk> wrote:
>> On Sat, 16 Apr 2022 14:14:35 +0100, "Dave Plowman (News)"
>> <dave@davenoise.co.uk> wrote:
>
>>> In article <59da028d2edave@davenoise.co.uk>,
>>> Dave Plowman (News) <dave@davenoise.co.uk> wrote:
>>>> 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.
>>>
>>> My typo. It read 12v.
>
>> 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.

The wiki allows PDF uploads...

--
Cheers,

John.

/=================================================================\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\=================================================================/

In article <t3mkkf$usm$1@dont-email.me>,
Fredxx <fredxx@spam.uk> wrote:

> D3 is a red herring, presumably trying to stop load dumps getting
> through Q1.

> It's a very strange way to simply get +5V.

It was designed some 50 years ago. ;-)

> I'm wondering if ZD1 is being
> used as a shunt regulator, with the arrangement connected to the base,
> and the 1.2R resistor Q1 providing a constant current. Are Q1 and ZD1
> fairly beefy?

> First thing to check is Q3 collector / Q2 base. Is this at 0V, or closer
> to 3.3V. I assume this is some sort of power enable (on/off) from the MCU.

Reads 0.58v (Pin11) on the faulty one. 0.006v on the spare unit which
shows 5v Vcc.

--
*IF A TURTLE DOESN'T HAVE A SHELL, IS HE HOMELESS OR NAKED?

Dave Plowman dave@davenoise.co.uk London SW
To e-mail, change noise into sound.

On 20/04/2022 14:44, Dave Plowman (News) wrote:
> In article <t3mkkf$usm$1@dont-email.me>,
> Fredxx <fredxx@spam.uk> wrote:
>
>> D3 is a red herring, presumably trying to stop load dumps getting
>> through Q1.
>
>> It's a very strange way to simply get +5V.
>
> It was designed some 50 years ago. ;-)
>
>> I'm wondering if ZD1 is being
>> used as a shunt regulator, with the arrangement connected to the base,
>> and the 1.2R resistor Q1 providing a constant current. Are Q1 and ZD1
>> fairly beefy?
>
>> First thing to check is Q3 collector / Q2 base. Is this at 0V, or closer
>> to 3.3V. I assume this is some sort of power enable (on/off) from the MCU.
>
> Reads 0.58v (Pin11) on the faulty one. 0.006v on the spare unit which
> shows 5v Vcc.

First of all that doesn't make sense. Assuming your measuring wrt GND.

In essence the current through Q1 is determined by a diode drop across
R1, so for 1.2R that's going to be 0.5A and for 1.8R 0.33A

But for that I would expect 0.6V on the base of Q2 to supply, through
R4, the bias current to flow in D1 and D2 and lower Q1 base wrt 12V by 2
diode drops. It's a PNP transistor. Are you sure you haven't got the
faulty/good measurements reversed?

Next I would look at the chain ZD2 / R8 / R9 and compare voltages. They
look as if they are designed to shutdown the 5V through an over-voltage
on the 12V input.

Volts dropped across R1 should be ~0.6V

Can you check again at pin11, or just in case the diagram is wrong, at
the base of Q2?

After some research Q1 (17344) is a big beast as I expected, a TO66
https://en.wikipedia.org/wiki/TO-66
Also ZD1 is a 1N5341B which is a 6.2V 5W zener, again quite beast.

On 4/20/2022 7:49 PM, Fredxx wrote:
> On 20/04/2022 14:44, Dave Plowman (News) wrote:
>> In article <t3mkkf$usm$1@dont-email.me>,
>>     Fredxx <fredxx@spam.uk> wrote:
>>
>>> D3 is a red herring, presumably trying to stop load dumps getting
>>> through Q1.
>>
>>> It's a very strange way to simply get +5V.
>>
>> It was designed some 50 years ago. ;-)
>>
>>> I'm wondering if ZD1 is being
>>> used as a shunt regulator, with the arrangement connected to the base,
>>> and the 1.2R resistor Q1 providing a constant current. Are Q1 and ZD1
>>> fairly beefy?
>>
>>> First thing to check is Q3 collector / Q2 base. Is this at 0V, or closer
>>> to 3.3V. I assume this is some sort of power enable (on/off) from the MCU.
>>
>> Reads 0.58v (Pin11) on the faulty one. 0.006v on the spare unit which
>> shows 5v Vcc.
>
> First of all that doesn't make sense. Assuming your measuring wrt GND.
>
> In essence the current through Q1 is determined by a diode drop across R1, so for 1.2R that's going to be 0.5A and for 1.8R 0.33A
>
> But for that I would expect 0.6V on the base of Q2 to supply, through R4, the bias current to flow in D1 and D2 and lower Q1 base wrt 12V by 2 diode drops. It's a PNP transistor. Are you sure you haven't got the faulty/good measurements reversed?
>
> Next I would look at the chain ZD2 / R8 / R9 and compare voltages. They look as if they are designed to shutdown the 5V through an over-voltage on the 12V input.
>
> Volts dropped across R1 should be ~0.6V
>
> Can you check again at pin11, or just in case the diagram is wrong, at the base of Q2?
>
> After some research Q1 (17344) is a big beast as I expected, a TO66
>  https://en.wikipedia.org/wiki/TO-66
> Also ZD1 is a 1N5341B which is a 6.2V 5W zener, again quite beast.
>

Do you have a datasheet for the RCA 17344 ?

It might be from, like, 1978.

Now, a theory would be, the 17344 is a PNP Darlington pair.

The other thing I was thinking, is Pin11 could be a disable
input, causing the 1041AE to stop driving the injectors,
so the injectors don't get currents from a load dump. But then
the readings don't make sense.

Paul

On 21/04/2022 08:37, Paul wrote:
> On 4/20/2022 7:49 PM, Fredxx wrote:
>> On 20/04/2022 14:44, Dave Plowman (News) wrote:
>>> In article <t3mkkf$usm$1@dont-email.me>,
>>>     Fredxx <fredxx@spam.uk> wrote:
>>>
>>>> D3 is a red herring, presumably trying to stop load dumps getting
>>>> through Q1.
>>>
>>>> It's a very strange way to simply get +5V.
>>>
>>> It was designed some 50 years ago. ;-)
>>>
>>>> I'm wondering if ZD1 is being
>>>> used as a shunt regulator, with the arrangement connected to the base,
>>>> and the 1.2R resistor Q1 providing a constant current. Are Q1 and ZD1
>>>> fairly beefy?
>>>
>>>> First thing to check is Q3 collector / Q2 base. Is this at 0V, or
>>>> closer
>>>> to 3.3V. I assume this is some sort of power enable (on/off) from
>>>> the MCU.
>>>
>>> Reads 0.58v (Pin11) on the faulty one. 0.006v on the spare unit which
>>> shows 5v Vcc.
>>
>> First of all that doesn't make sense. Assuming your measuring wrt GND.
>>
>> In essence the current through Q1 is determined by a diode drop across
>> R1, so for 1.2R that's going to be 0.5A and for 1.8R 0.33A
>>
>> But for that I would expect 0.6V on the base of Q2 to supply, through
>> R4, the bias current to flow in D1 and D2 and lower Q1 base wrt 12V by
>> 2 diode drops. It's a PNP transistor. Are you sure you haven't got the
>> faulty/good measurements reversed?
>>
>> Next I would look at the chain ZD2 / R8 / R9 and compare voltages.
>> They look as if they are designed to shutdown the 5V through an
>> over-voltage on the 12V input.
>>
>> Volts dropped across R1 should be ~0.6V
>>
>> Can you check again at pin11, or just in case the diagram is wrong, at
>> the base of Q2?
>>
>> After some research Q1 (17344) is a big beast as I expected, a TO66
>>   https://en.wikipedia.org/wiki/TO-66
>> Also ZD1 is a 1N5341B which is a 6.2V 5W zener, again quite beast.
>>
>
> Do you have a datasheet for the RCA 17344 ?

No, I've only gone along with forums regarding this ECU.

> It might be from, like, 1978.

Yes, likely of this vintage.

> Now, a theory would be, the 17344 is a PNP Darlington pair.

If it was a Darlington then it would have an indeterminate base current
from being biased with two diodes so I feel it is unlikely. BICBW

> The other thing I was thinking, is Pin11 could be a disable
> input, causing the 1041AE to stop driving the injectors,
> so the injectors don't get currents from a load dump. But then
> the readings don't make sense.

Yes, those were my thoughts. But I felt that was too much information
and was thinking of a more piecemeal approach especially with the
discrepancy with readings.

Either way, as you agree, the readings don't make any sense.

Fredxx <fredxx@spam.uk> wrote:
> Yes, those were my thoughts. But I felt that was too much information
> and was thinking of a more piecemeal approach especially with the
> discrepancy with readings.
>
> Either way, as you agree, the readings don't make any sense.

Here is someone's beermat version of the circuit, which is at least drawn in
a more sensible organisation:
http://www.jpurnell.com/rr/LucasEFI/lucas4cu.htm

and some pictures:
https://i.ebayimg.com/d/w1600/pict/264894798928_/Lucas-4CU-Rover-SD1-TVR-Range-Rover-fuel.jpg
https://i.ebayimg.com/images/g/-ewAAOSwAyNfgjbJ/s-l1600.jpg
(calls transistors Tn rather than Qn so might be a different version)

That picture has ZD2 marked '24'. According to:
https://www.logwell.com/tech/components/zener.html
diodes with that suffix might be:
1N724: 27.0V 0.4W (unlikely)
1N5224: 2.8V 0.5W
1N4624: 4.7V 0.5W

I'm guessing it's not in the BZX... series since a BZX..24 would be a 24V
zener, which doesn't sound right. The existing zener being a 1N5341,
suggests it's either the 2.8 or the 4.7V. Would either of those make sense?

Now you mention it, Dave's readings do slightly make sense. Because pin 11
is at the base of Q2, the base-emitter junction means it won't rise above
0.7v - that might be the voltage Dave is measuring. For some reason on the
other one the base of Q2 is at zero volts. Either that's pin11 holding it
down, or Q3 is. That could be because Q3 is being jammed on, eg if ZD2 was
short circuit.

Hmm, the network at the base of Q3 divides the voltage at the anode of ZD2
by 15.3. Vbe(on) for the BC184 is 0.55-0.7v, which multiplied by 15.3
gives us 8.4 to 10.7. I wonder if this is an overvoltage detection network.

Let's say the zener is 4.7v. If the battery voltage were to go above
8.4+4.7 = 13.1, the system shuts down by turning on Q3 and yanking down the
base of Q2. If the transistor turned on at 0.7, that would be 10.7+4.7 =
15.4v.

There will likely be sufficient tolerance of the resistors and transistor
that might mean this would need to be more accurately tuned via component
selection (eg transistor grading).

But, if this is so, I don't see how the 5V regulation works...

Theo

On 21/04/2022 15:23, Theo wrote:
> Fredxx <fredxx@spam.uk> wrote:
>> Yes, those were my thoughts. But I felt that was too much information
>> and was thinking of a more piecemeal approach especially with the
>> discrepancy with readings.
>>
>> Either way, as you agree, the readings don't make any sense.
>
> Here is someone's beermat version of the circuit, which is at least drawn in
> a more sensible organisation:
> http://www.jpurnell.com/rr/LucasEFI/lucas4cu.htm
>
> and some pictures:
> https://i.ebayimg.com/d/w1600/pict/264894798928_/Lucas-4CU-Rover-SD1-TVR-Range-Rover-fuel.jpg
> https://i.ebayimg.com/images/g/-ewAAOSwAyNfgjbJ/s-l1600.jpg
> (calls transistors Tn rather than Qn so might be a different version)
>
> That picture has ZD2 marked '24'. According to:
> https://www.logwell.com/tech/components/zener.html
> diodes with that suffix might be:
> 1N724: 27.0V 0.4W (unlikely)
> 1N5224: 2.8V 0.5W
> 1N4624: 4.7V 0.5W
>
> I'm guessing it's not in the BZX... series since a BZX..24 would be a 24V
> zener, which doesn't sound right. The existing zener being a 1N5341,
> suggests it's either the 2.8 or the 4.7V. Would either of those make sense?
>
> Now you mention it, Dave's readings do slightly make sense. Because pin 11
> is at the base of Q2, the base-emitter junction means it won't rise above
> 0.7v - that might be the voltage Dave is measuring. For some reason on the
> other one the base of Q2 is at zero volts. Either that's pin11 holding it
> down, or Q3 is. That could be because Q3 is being jammed on, eg if ZD2 was
> short circuit.
>
> Hmm, the network at the base of Q3 divides the voltage at the anode of ZD2
> by 15.3. Vbe(on) for the BC184 is 0.55-0.7v, which multiplied by 15.3
> gives us 8.4 to 10.7. I wonder if this is an overvoltage detection network.
>
> Let's say the zener is 4.7v. If the battery voltage were to go above
> 8.4+4.7 = 13.1, the system shuts down by turning on Q3 and yanking down the
> base of Q2. If the transistor turned on at 0.7, that would be 10.7+4.7 =
> 15.4v.
>
> There will likely be sufficient tolerance of the resistors and transistor
> that might mean this would need to be more accurately tuned via component
> selection (eg transistor grading).
>
> But, if this is so, I don't see how the 5V regulation works...

The +5V regulation is with ZD1. It's used as an awful shunt regulator.
Q1 is a constant current source from 12V supplying power to the 5V rail
with ZD1 dissipating the extra current.

The above hand drawn schematic link is confusing for some as it shows
Q1, a PNP, essentially upside down from a current flow perspective.

In article <t3q65k$7n8$1@dont-email.me>,
Fredxx <fredxx@spam.uk> wrote:
> On 20/04/2022 14:44, Dave Plowman (News) wrote:
> > In article <t3mkkf$usm$1@dont-email.me>,
> > Fredxx <fredxx@spam.uk> wrote:
> >
> >> D3 is a red herring, presumably trying to stop load dumps getting
> >> through Q1.
> >
> >> It's a very strange way to simply get +5V.
> >
> > It was designed some 50 years ago. ;-)
> >
> >> I'm wondering if ZD1 is being
> >> used as a shunt regulator, with the arrangement connected to the base,
> >> and the 1.2R resistor Q1 providing a constant current. Are Q1 and ZD1
> >> fairly beefy?
> >
> >> First thing to check is Q3 collector / Q2 base. Is this at 0V, or closer
> >> to 3.3V. I assume this is some sort of power enable (on/off) from the MCU.
> >
> > Reads 0.58v (Pin11) on the faulty one. 0.006v on the spare unit which
> > shows 5v Vcc.

> First of all that doesn't make sense. Assuming your measuring wrt GND.

Everything measured reference to ground unless otherwise stated.

> In essence the current through Q1 is determined by a diode drop across
> R1, so for 1.2R that's going to be 0.5A and for 1.8R 0.33A

> But for that I would expect 0.6V on the base of Q2 to supply, through
> R4, the bias current to flow in D1 and D2 and lower Q1 base wrt 12V by 2
> diode drops. It's a PNP transistor. Are you sure you haven't got the
> faulty/good measurements reversed?

0.85v between supply positive and Q1 base

> Next I would look at the chain ZD2 / R8 / R9 and compare voltages. They
> look as if they are designed to shutdown the 5V through an over-voltage
> on the 12V input.

> Volts dropped across R1 should be ~0.6V

Voltage across R1 is 0.15v However, on both boards it is 1R2 rather than
1R8 as shown in the schematic.

> Can you check again at pin11, or just in case the diagram is wrong, at
> the base of Q2?

> After some research Q1 (17344) is a big beast as I expected, a TO66
> https://en.wikipedia.org/wiki/TO-66
> Also ZD1 is a 1N5341B which is a 6.2V 5W zener, again quite beast.

--
*Marriage changes passion - suddenly you're in bed with a relative*

Dave Plowman dave@davenoise.co.uk London SW
To e-mail, change noise into sound.

On 21/04/2022 18:20, Dave Plowman (News) wrote:
> In article <t3q65k$7n8$1@dont-email.me>,
> Fredxx <fredxx@spam.uk> wrote:
>> On 20/04/2022 14:44, Dave Plowman (News) wrote:
>>> In article <t3mkkf$usm$1@dont-email.me>,
>>> Fredxx <fredxx@spam.uk> wrote:
>>>
>>>> D3 is a red herring, presumably trying to stop load dumps getting
>>>> through Q1.
>>>
>>>> It's a very strange way to simply get +5V.
>>>
>>> It was designed some 50 years ago. ;-)
>>>
>>>> I'm wondering if ZD1 is being
>>>> used as a shunt regulator, with the arrangement connected to the base,
>>>> and the 1.2R resistor Q1 providing a constant current. Are Q1 and ZD1
>>>> fairly beefy?
>>>
>>>> First thing to check is Q3 collector / Q2 base. Is this at 0V, or closer
>>>> to 3.3V. I assume this is some sort of power enable (on/off) from the MCU.
>>>
>>> Reads 0.58v (Pin11) on the faulty one. 0.006v on the spare unit which
>>> shows 5v Vcc.
>
>> First of all that doesn't make sense. Assuming your measuring wrt GND.
>
> Everything measured reference to ground unless otherwise stated.
>
>> In essence the current through Q1 is determined by a diode drop across
>> R1, so for 1.2R that's going to be 0.5A and for 1.8R 0.33A
>
>> But for that I would expect 0.6V on the base of Q2 to supply, through
>> R4, the bias current to flow in D1 and D2 and lower Q1 base wrt 12V by 2
>> diode drops. It's a PNP transistor. Are you sure you haven't got the
>> faulty/good measurements reversed?
>
> 0.85v between supply positive and Q1 base

I believe the design voltage would be be nearer 1.2V if the circuit was
working as intentioned.

>> Next I would look at the chain ZD2 / R8 / R9 and compare voltages. They
>> look as if they are designed to shutdown the 5V through an over-voltage
>> on the 12V input.
>
>> Volts dropped across R1 should be ~0.6V
>
> Voltage across R1 is 0.15v However, on both boards it is 1R2 rather than
> 1R8 as shown in the schematic.

The voltage is consistent with base voltage.

>> Can you check again at pin11, or just in case the diagram is wrong, at
>> the base of Q2?

If this voltage is at zero then that can explain why Q1 is not
conducting sufficient current. The voltages around Q2 and R4 will
probably explain what's going on and may suggest Q1 is faulty.

>> After some research Q1 (17344) is a big beast as I expected, a TO66
>> https://en.wikipedia.org/wiki/TO-66
>> Also ZD1 is a 1N5341B which is a 6.2V 5W zener, again quite beast.

On 21/04/2022 18:15, Fredxx wrote:

<snip>
>
> The +5V regulation is with ZD1. It's used as an awful shunt regulator.
> Q1 is a constant current source from 12V supplying power to the 5V rail
> with ZD1 dissipating the extra current.

But ZD1 is a 6V2 zener, not 5V. The circuit is clearly wrong IMO.

--
Cheers
Clive

Clive Arthur <clive@nowaytoday.co.uk> wrote:
> On 21/04/2022 18:15, Fredxx wrote:
>
> <snip>
> >
> > The +5V regulation is with ZD1. It's used as an awful shunt regulator.
> > Q1 is a constant current source from 12V supplying power to the 5V rail
> > with ZD1 dissipating the extra current.
>
> But ZD1 is a 6V2 zener, not 5V. The circuit is clearly wrong IMO.

I think ZD1 is a clamp: if the rail goes above 6v2 then it will accept the
extra current. It'll also clamp any negative spikes on the rail to -0.7v.
If the rail is at 5V then I wouldn't expect much current to flow through
ZD1 and it would be essentially open circuit.

But then I'm not clear how the voltage *is* regulated. Unless it's some
strange current mirror from ZD2?

Theo

On 21/04/2022 19:51, Clive Arthur wrote:
> On 21/04/2022 18:15, Fredxx wrote:
>
> <snip>
>>
>> The +5V regulation is with ZD1. It's used as an awful shunt regulator.
>> Q1 is a constant current source from 12V supplying power to the 5V
>> rail with ZD1 dissipating the extra current.
>
> But ZD1 is a 6V2 zener, not 5V.  The circuit is clearly wrong IMO.

Most stuff can handle 5.5V without issue, absolute maximum rating of
logic is generally 7V. As you say it may not be correct.

From a design POV I don't see the issue with a proper regulator, ie in
discretes or something like a 7805 which I'm sure existed in the vintage
era of this circuit.

On 21/04/2022 20:35, Theo wrote:
> Clive Arthur <clive@nowaytoday.co.uk> wrote:
>> On 21/04/2022 18:15, Fredxx wrote:
>>
>> <snip>
>>>
>>> The +5V regulation is with ZD1. It's used as an awful shunt regulator.
>>> Q1 is a constant current source from 12V supplying power to the 5V rail
>>> with ZD1 dissipating the extra current.
>>
>> But ZD1 is a 6V2 zener, not 5V. The circuit is clearly wrong IMO.
>
> I think ZD1 is a clamp: if the rail goes above 6v2 then it will accept the
> extra current. It'll also clamp any negative spikes on the rail to -0.7v.
> If the rail is at 5V then I wouldn't expect much current to flow through
> ZD1 and it would be essentially open circuit.

ZD1 is a 5W device, consistent with being a shunt regulator. I can't see
any active control of the 5V, unless there is something strange
happening in U101. I can't find a datasheet of a LFC1041AE and assume
that Pin 11 is a logic port.

> But then I'm not clear how the voltage *is* regulated. Unless it's some
> strange current mirror from ZD2?

I'm pretty sure the sole purpose of ZD2 is protection.

At a raised voltage of the 12V rail Q3 will turn on, turning Q2 off and
with it any current through Q1 making the +5V fall.

Fredxx <fredxx@spam.uk> wrote:
> On 21/04/2022 20:35, Theo wrote:
>
> ZD1 is a 5W device, consistent with being a shunt regulator. I can't see
> any active control of the 5V, unless there is something strange
> happening in U101. I can't find a datasheet of a LFC1041AE and assume
> that Pin 11 is a logic port.

The LFC1041 is a Ferranti part, so I suspect it's a custom logic chip made
for Lucas. The PDF I linked upthread has a block diagram. It looks a bit
like the Ferranti ULAs in 8-bit computers like the BBC Micro, but that might
only extend as far as the packaging. Ferranti did make transistors and
analogue chips too. So I suppose we can't rule out some analogue stuff in
there.

If Dave is measuring 5V on the rail, I don't see how ZD1 is regulating it.
If the rail was 6.2v, fine. But at 5v it's well into the non-conductive
region, eg on the OnSemi datasheet for the 1N5341:
https://www.onsemi.com/pdf/datasheet/1n5333b-d.pdf
the lowest conductive voltage is 5.89v (current = 175mA) while at 3v it has a
current of 1uA. If it's part of a shunt regulator there will be very little
current flowing through it with 5v across it.

I agree that tolerances can allow 5v logic to run at 6v, but that should
show when Dave measures it?

Theo

In article <OJg*wmhMy@news.chiark.greenend.org.uk>,
Theo <theom+news@chiark.greenend.org.uk> wrote:
> Fredxx <fredxx@spam.uk> wrote:
> > On 21/04/2022 20:35, Theo wrote:
> >
> > ZD1 is a 5W device, consistent with being a shunt regulator. I can't
> > see any active control of the 5V, unless there is something strange
> > happening in U101. I can't find a datasheet of a LFC1041AE and assume
> > that Pin 11 is a logic port.

> The LFC1041 is a Ferranti part, so I suspect it's a custom logic chip
> made for Lucas. The PDF I linked upthread has a block diagram. It
> looks a bit like the Ferranti ULAs in 8-bit computers like the BBC
> Micro, but that might only extend as far as the packaging. Ferranti did
> make transistors and analogue chips too. So I suppose we can't rule out
> some analogue stuff in there.

Connectional logic with the SD1 lot is that the 4CU is entirely analogue.
It can't be remapped in a conventional way. Those 4CUs used for racing etc
were modified mainly by frigging the signal from the AFM. And of course
larger injectors. The next Lucas generation - the 14CU - is digital and
can be re-mapped using software.

> If Dave is measuring 5V on the rail, I don't see how ZD1 is regulating
> it. If the rail was 6.2v, fine. But at 5v it's well into the
> non-conductive region, eg on the OnSemi datasheet for the 1N5341:
> https://www.onsemi.com/pdf/datasheet/1n5333b-d.pdf the lowest conductive
> voltage is 5.89v (current = 175mA) while at 3v it has a current of 1uA.
> If it's part of a shunt regulator there will be very little current
> flowing through it with 5v across it.

> I agree that tolerances can allow 5v logic to run at 6v, but that should
> show when Dave measures it?

UK (mapped) versions of the 4CU are plentiful used.

All I've basically done was to make a test rig based on a MegaSquirt Stim.
This, with an external PS and via an adaptor lead, provides power, tach
pulses (variable from 0 - 6500 RPM in practice) and pots giving 0-5v to
each of the ECU inputs - TPS, AFM, CTS, IAT, etc. LEDs across the injector
outputs. By looking at the injector pulse width on a scope, you can check
that reacts to a sensor voltage change. Not an exhaustive check, but does
usually show if there is an actual fault. Actually quite rare. Most of the
problems are with peripherals, rather than the ECU itself.

The one in question isn't UK spec. And was locking on one bank of
injectors at random - intermittently. Not good for the engine.

But the first thing I noticed was the low Vcc. Every other one I've looked
at has been pretty well spot on 5v.

I did manage to see the locked on injector output once at 0 RPM (no tach
signal) when they should of course be off. The injector drivers alternate.
Pin 23 on U102 was showing Vcc - Pin 24 0V. But, of course, when
attempting to look into it further, that fault went away.

--
*How can I miss you if you won't go away?

Dave Plowman dave@davenoise.co.uk London SW
To e-mail, change noise into sound.