Extracting the Signal

[pebe]

I have no idea how a Penske meter (dwell / tach) works, I assume it picks up the 12V from the coil to power the tach. But whenever the contact breaker goes o/c then the voltage across the coil will go highly negative.

So it likely power is fed through a diode to charge a capacitor for the supply rail. When the contact breaker goes o/c, the diode cuts off and the negative voltage on the coil is fed as a trigger to the counter.

But I can't be sure how to connect it up or what the component values would be.

[dw85745]

If you don't mind spending your time, and you think it would help looking at the Meter Schematic, I'll be glad to post it.

[pebe]

Ah yes. If you have the schematic then it should be possible to design something suitable.

[dw85745]

pebe:
Great. Will post by this weekend.

[dw85745]

Tach_p5.jpeg (733.43 KiB) Viewed 53451 times

Tach_p3.jpeg (600.37 KiB) Viewed 53451 times

Tach_p1.jpeg (685.57 KiB) Viewed 53451 times

Tach_p2.jpeg (786.55 KiB) Viewed 53451 times

Tach_p4.jpeg (522.6 KiB) Viewed 53451 times

Page 5 is schematic, Page 3 is Parts.
Rest pages howto of tach portion of unit.
If you need dwell portion will upload.
Didn't know how may pictures forum would allow me so put in order of importance.

Pebe: Thanks again for taking a look see.
FWIW: These are in jpeg format. Hope OK, if not please advise.

[dw85745]

Always nice to see a picture of the unit.
Didn't scan as well as I thought.
The toggle switch in upper right area under meter is setting for 8 (up) - 4(middle) - 6 (bottom).
The rotating switch from left to right is:
point-resistance
dwell
Low RPM
High RPM.

Tach_p6.jpeg (1.31 MiB) Viewed 53450 times

[pebe]

OK. The schematic is an awful thing to follow, with the control switches shown as wafers rather than as usually shown in a circuit. I got lost each time I tried to follow their functions in each position. The top switch is badly drawn and it is uncertain whether the slider to the RH top switch connects to pin2 at the 2 o'clock position or to pin4 at the 5 o'clock position.

It appears that the meter is powered only by the negative pulse at the contact breaker when it opens. A pickup on the HT lead would not have enough energy, but it could be applied to a 555 timer to generate a suitable pulse. I could make a circuit that may be OK, but I cannot guarantee it. Are you Interested?

[dw85745]

pebe:

Somehow missed your post so late in responding.
I can physically open the unit and see where things connect that are in question if that will help?
If so, give me a list of all things in question, and I will do my best to physically trace them.
From your post I need to do the top switch and the "turn" NOT "toggle" switch?

[pebe]

The schematic/wiring diagram layout is too complex to understand without redrawing it as a logically laid out circuit, and that would be too time consuming.

From the schematic, it looks like negative pulses from the red lead are integrated by R2/R3/R6 and C6 to give a mean DC current through the meter proportional to RPM. So I suggest not modifying the meter, but generate pulses that simulate the action of the contact breaker, instead.

I can give you a simple 555 circuit if you want to try it.

[dw85745]

pebe:

I you don't mind putting in the effort I would appreciate it.
Would like something as part of the circuit though to protect the meter from any high voltages if possible.

[pebe]

David:
I found I had to unravel the switch positions in order to find out how the circuit worked, but it didn’t make sense. Usually in tachometer circuits, the incoming pulse triggers a monostable to give a fixed length pulse (the ON condition). That is followed by a variable time (the OFF condition) until the next pulse arrives. The ratio of ON:OFF times is a measure of the speed that the tach is measuring.

I expected to find Q1 and Q2 as the monostable, but cannot see how it works – unless I have misread the switch connections. Perhaps Ken Moffett or one of the other experienced regulars can shed some light on it.

However, it appears that the tach is powered by positive pulses from the coil when the contact breaker opens (not negative as I had assumed). Assuming a max speed of 10000RPM and 6 cylinders, the max pulse width must be less than 1ms so in my circuit I have made the 555 pulses 600us wide. The pulses are clamped at 6V in the tach by D4 so to protect it I have added R5 in my circuit. I forgot to label the transistor. Any small signal NPN will do.

I think there is a fair chance it will work.

Ignition Pulse generator.GIF (4.86 KiB) Viewed 53160 times

[dw85745]

I found I had to unravel the switch positions in order to find out how the circuit worked, but it didn’t make sense.
............... I expected to find Q1 and Q2 as the monostable, but cannot see how it works – unless I have misread the switch connections. Perhaps Ken Moffett or one of the other experienced regulars can shed some light on it.

pebe:
Thanks for all your efforts on my behalf.
As previously indicated, I can "physically" open the tach unit and trace any connections.
Don't know if it will help at this point to answer any questions still left open, but will be more than happy to do it.
If not, will wait a bit to see if any comments then go with your BEST effort.

[dw85745]

No additional input.
Hopefully can get some parts I need this weekend, and get bread boarding.
Thank you again pebe for your help on my behalf.

David

[dw85745]

pebe:

Got the final parts I needed today.
I'm going with a 2N3904 transistor if that is acceptable?

Also, if you have the time, would like to really understand how the circuit functions
rather than just taking benefit of your expertise. It may be way beyond my capabilities at this point,
but this is what "I think" I understand so far:

-------------------------
Signal Part of Circuit.
-------------------------
1) R1 reduces the voltage (signal) from the induction pickup.
2) This reduced voltage (signal) drives the -- Common Emitter configuration -- transistor "base".
3) The driving of the base causes an electron flow from the Emitter to Collector.
In addition, by design the transistor causes the output signal to be 180 degrees out of phase
4) This out of phase signal (extra electrons), coupled with the 12Vdc (reduced by R3) causes the 555 to periodically trigger (pulsate).

>> What I don't understand is why D1 and R2 are needed instead of the the transistor emitter just connecing
directly to the ground (0v)?

-----------------------------
Power Side of the Circuit
-----------------------------
The 12Vdc and associated reduced values are need for the 555 to function.

>>I don't understand why C1 is needed rather than the voltage through R4 just terminating at Pins 6 and 7 on the 555?

-------------------------------
Still to Do
------------------------------
1) Determine what's going on inside the 555 regarding why the signal phase again changes (rotates 180 degrees).
2) How all the values were calculated.
.... a) How do you calculate the current for a 12Vdc battery in order to use Ohms to calculate resistor values?
'''''''''''''''' My understanding battery current can vary based on the battery.
.... b) If the voltage on the spark plug wire is say 30000 volts, what kind of voltage will result from using an inductor pickup -- and why is using a 1K resistor rather than something larger OK for this voltage?

[pebe]

dw85745,
I'll do a write up for you at the weekend.