nelsontj

I have an old Sun Super Tach II that I want to connect to the battery charger to see if the needle will go to the 1800 RPM mark like it is supposed to. How do I connect it to the battery charger? The tachometer has four wires coming out of it. I have two black wires, one green wire and one yellow wire. One black wire goes to the (battery) terminal and other balck wire connected to the ignition coil. The green wire connected to the ground. The yellow wire is insignificant (for the test) for it powered the internal lamp. Do I connect the two black leads to the positive battery charger clamp and the ground to the negative battery charger clamp? ...Or???
Any other ideas for testing it? I don't have a car available that has the old style ignition coil. The only car I have available, at the moment, is a Mercury with Coil On Plugs. I don't have access to a signal generator either.

MDchanic

I am assuming that the logic behind this is that a battery charger provides a non-smoothed pulsed DC waveform, straight off the bridge rectifier, and that since the AC input is a 60hz sine wave, then after passing through the rectifier it should be pulsed DC at 120hz, which would be the correct number of pulses per second to run an 8 cylinder 4 cycle engine at 1,800 RPM or 30 revolutions per second.

IF your battery charger does, in fact provide a pulse DC waveform (I know older chargers do, but I'm not sure about newer ones with lots of electronics), then, according to the online instructions, you would connect the black wire to the (-), the red wire to the (+), leave the white wire dsconnected, and touch the green wire to the (+) in order to get the gauge to register 1,800 RPM.

In your case, you say you have two black wires, so I have no idea what to tell you.



- Eric

nelsontj

You've got the theory down. By the diagram you sent I can figure out the wire scheme. Those are newer instructions. My tachometer goes back to when it was the Sun Electronics Company in Illinois. I think Actron makes them now under the Sunpro name (or, some little chinaman is making them for Actron ) I'll give it a try. Thanks!

MDchanic

I'm pretty sure the wiring colors are the same, because they sounded familiar to me from when I installed one many years ago.

Good luck!

- Eric

RetroRanger

Relevant thread so im adding to it.

From Erics post above how did he arrive at “120hz, which would be the correct number of pulses per second to run an 8 cylinder 4 cycle engine at 1,800 RPM or 30 revolutions per second” ?

Im going to test a tach on the bench. I figure w the above info 15hz = 1 rpm, so to test at say 750, 2500 and 5000 rpms ill use 50hz, 166.66hz, and 333.33hz.

can anyone mansplain to me how Eric got the original 120hz = 1800 rpm.?

Fun71

This site shows how a full wave rectifier works as Eric described above.

https://www.electronics-tutorials.ws/diode/diode_6.html

RandyS

In theory, with a full bridge rectifier, there would be virtually pure DC at the output terminals. Where is all of this hertz theory coming from?

old greybeard

Just the bridge rectifier will leave you with 120hz modulated DC. Capacitors after the bridge charge and discharge smoothing out the ripple. The more you spend the better the filtering. Most inexpensive battery chargers don’t go to that much trouble as it’s not nessecary. Leaving a slight 120hz ripple.
I’m not sure about the math. Wouldn’t 120hz per second equal 7200rpm, divide by 8, shouldn’t it indicate 900 rpms? A 6 cylinder would be 1200 rpm. 1800 for a 4 cyl.

RetroRanger

LOL i should have been clearer. I get the dc ripple its the

"which would be the correct number of pulses per second to run an 8 cylinder 4 cycle engine at 1,800 RPM or 30 revolutions per second” ?

how did he arrive at 15hz = 1 rpm ??

MDchanic

Okay, WARNING: if you say my name three times, I appear.

• A two-stroke ("Otto-cycle") internal combustion engine fires its spark plugs once every two revolutions.
• Thus, a V-8 (or a straight-eight for that matter) will fire four times per revolution, or every 90°.
• 60 RPM = 1 rev. per second, so 120 x 60 =
• 7,200 RPM = 120 revs per second.
• So, if you have a pulsed-DC signal at twice the US household power frequency of 60hz, that will be at 120hz, which is 120 times per second.
• But, as above, the V-8 has four pulses for each revolution, so 7,200 / 4 = 1,800 RPM at 120 ignition pulses (or coil pulses) per second.

Make sense?

- Eric

edit: and, to be clear, 15hz is one quarter of 60hz, so at four pulses per revolution, and 60 revolutions per minute, 15hz must be 1 RPM.

Fun71

Welcome back!

old greybeard

Thanks. That makes sense, forgot to factor being a 4 stroke.

rocketraider

😳😬🤯

And here we all thought the loup-garou had got you!

RetroRanger

Hi Eric

thanks for posting. That makes it clearer.

i tested my tach on the bench last night using a function generator. I found the tach read closer to what i expected when i used an ~80% duty cycle square wave input VS 50 or 20% duty cycle. Ie 333.33hz read 5k w 80% and just a little above or below 5k w 50 or 20% duty cycle.

I think this because the input signal w 80% is closer to what a car may produce, can anyone confirm that ? I couldnt easily find anything on what the actual tach signal on a car is.

on another note 1 tach (mallory)worked fine as above. 3 other tachs (all sun) did not show any output w the same setup. Im wondering if the input signal needs to be refined more. The 3 that didnt work all drew about 15mA but showed no output.

i used a 50-333.33hz, 0-10v square wave, 50-80% duty, pulse width 1.5mS @50%…maybe needs more voltage or different pulse width? Or maybe all 3 suns dont work?

Fun71

Hmm. I don't know the answer, but remember that the dwell angle is 30* so think of what that waveform would look like.

oddball

I would want to generate at least 12V, keep duty cycle under 50%. The "duty cycle" is when the coil is charging. Typical dwell is 30deg, firing separation is 90deg, which means a typical ignition ""duty cycle"" is about 33%.
The tach will be latching onto the rising or falling edge, but will have some aggressive noise filtering which is probably squashing out your signal. The firing event is on the falling edge, but counting either edge is fine.