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High level tach, low level drive modification


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Hi, this is a thread I did over on the Link forums and it went down ok.
I've been trying to get it up in tech articles but Gaz no longer runs them and I can't get another admin to let me post in there, it can be transferred if deemed worthy later.
So I'll put it here with a bit more info for the layman as opposed to those on the Link forums.
You can also drive it with a coil (like from a relay) to step up the voltage but I think that's hoary :P
If we peer review this we can have a pretty good "how to" with a bit of luck.

As below:


Seeing as there is weird info online and not much is clear in regards to driving a high level tacho with a low level signal I figure I'd take mine apart and apply Ohm's Law as I figure it's got to be simple enough surely!

97 Mazda B2200, link atom g4.

"High level" tachs are driven from the neg post of the regular coil which can see 200 - 400V which is where the "high level" comes from, it's back EMF from the collapsing primary coil.
What we see from a computer and sometimes from an output from an igniter is 0-12V (more or less) and is known as "low level".
Now something that takes high voltage to drive, will not drive with low voltage, thankfully they're really only dragging down the current so we can replace the resistor with a little bit of Ohms law to match the new feed voltage.

What you will need:
Soldering iron and solder.
New resistor.

Here's how I did mine.

So I dragged out my tacho


You need to see what is + 12V, what is Ground and what is signal from coil.
Do this by tracing the ribbons or traces or wires that connect to your tacho. You'll see that they use the traces for +12V and earth for things like lights so you'll be able to figure out which is which there which leaves the other one to be your signal.

Now that you've found your signal, check to see what size the resistor is (hopefully there is one and I've not wasted your time).
Mine is a through hole resistor which is nice and simple, they may exist in surface mount too but those are easy also, don't be discouraged. You'll want a fine tip on your soldering iron though.


So mine is a 42k resistor. That sausage with bands around it and wires out either end. (http://www.digikey.com/~/media/Images/Marketing/Resources/Calculators/resistor-color-chart.jpg?la=en-US if you don't have a resistance tool on your multimeter) 
Now using Ohms law 200V / 42000 =  ~0.005 amps. (guessed at 200V, might be 400 but with such a large denominator it doesn't really matter)
Because we have only 12V to play with we go backwards to find the resistance needed for that amperage
12 / 0.005 = 2400, so we need a 2k4 resistor.

So I took out the 42k resistor, replaced with 2k4 resistor and put it back in.


Video of tacho working:


RPM sweep works nicely too, thanks link!

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Nit-picking, but


Mine is a through hole resistor which is nice and simple, they may exist in thin film



I think you mean surface mount here. Thin-film specifies the internal construction of the resistor and has nothing to do with the packaging type.

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To all engineers; sorry for the over simplification


The way the tachos work that are driven from the negative of the coil is as follows;


the coil connects to the 12V supply, and to the points in your dizzy. When the points close, it connects the neg of the coil to ground which gets current flowing. When the points open again a split second later, the coil will do its thing where it tries to keep the same power flowing through it, and to do that it manages to create a massive high voltage as the current backs off (google "how inductors work" if you wanna know how/why, but for now just accept that it does) When the points open, the coil negative is essentially not connected to anything, so its free to generate whatever voltage it likes.


so with the voltage of the primary coil rising, it also induces a current flow in the secondary winding in the coil, which gets even higher voltage, so high that it can bridge the spark gap of the spark plug.


back to the coil


so the voltage rises on the primary winding, and the tacho input uses that high voltage to trigger the tacho. If you just connect a 5V square wave into the input (like from an ECU) then it will just ignore that as its too low for the tacho to consider a valid trigger. To create a valid trigger, we need a higher voltage, so basically what we are going to do, is put a little tiny coil in the car that will also generate a high voltage to trigger the tacho.


A relay has a coil inside it. It works by energising the coil, which creates a magnet and pulls the switch inside (as a simple way of explaining it)


So to get a relay coil to generate a high voltage, we need to energise it. An output from an ECU will NOT be able to drive a relay directly, as a small 12V relay will draw ~200mA and the pin from the ECU is probably only good for 5-20mA or so, plus we need to recreate the points opening as we need the coil to be not connected to anything after energising it so it will generate a high voltage. To do this, we need something called an open drain driver, meaning it is either open (not connected) or it 'drains' current (making a connection between a device and a ground)


A transistor is basically an amplifier. you put some current into the input, and it will drive the output with 100x as much current AND it can be turned off, and when its off, it is completely open, like the points in a dizzy! so not only can we use it to make the open drain driver, but it also amplifies and drives the coil with lots of current even though we only draw a little bit from the ECU.


So we now have a coil and a transistor and are nearly done! a transistor is a funny beast, and the input of it wont actually go above 0.6V. This means that we cant just connect a 12V or 5V signal directly to the input because you cant connect 12V to 0.6V. That would be like connecting your battery terminals together, and you would let the magic smoke out. to connect the 5V or 12V signal to the transistor input, you need to use a resistor. A resistor is happy for one side to be 12V and the other side to be 0.6V. When that happens, current flows through the resistor. The more current, or the higher the voltage, the hotter it gets. To make sure thats all ok, you get to use Ohms Law.


V = I x R and P = V x I are the two you want to remember. Voltage (V) = Current (I) x Resistance ® and Power (P) = Voltage (V) x Current (I)


remember how the coil was going to draw 200mA and the transistor will amplify the input current by 100x? those were example values. The actual value could be different, especially for the coil of the relay, but when you go to jaycar, the relay will say if its 200mA or 400mA or whatever, so just re-do the calcs. The transistor will normally be 100-300x amplification, so just stick to the 'normal' ones like a BC337 or 2n2222 or something. If you're not sure if what you have is legit, just google it and see what the datasheet says for DC Current Gain or Hfe.

so to drive the transistor, we need to drive it with 200mA (0.2A) / 100x amplifcation = 2mA (0.002A) and the ECU pin would be good for 5-20mA, so 2 is def under that.


To get 2mA of current to flow, we need to flow 2mA of current through the resistor. So using ohms law to figure out the value resistor, we get the voltage across the resistor, which we know as its 12V-0.6V=11.4V or 5V-0.6V=4.4V and we know the current (0.002A). so if V=IxR, then R=V/I. Plugging in the stuff we know, 11.4/0.002=5700 and 4.4/0.002=2200 so the resistors we need are 2200Ohm for 5V and 5700Ohm for 12V


They dont actually make a 5700, but they do make a 2200. If you get a value thats not quite right, go for one thats lower by up to 1/2. Halving the resistor value, will double the current you draw from the ECU and will double the heat the resistor generates though, but 4mA is still safe. So a good value for 12V would be 5600, or even 4700. For 5V 2200 is pretty good, but you could go to 1800 or 1500 if you want. Resistors that size are defined as K though, so 1K5 for a 1500, or 2K2 for a 2200, so get a 1K8 for 5V and a 4K7 for 12V


Now to see what physical size you need to get. Use the chosen resistor value to work out what the new current will be. 4.4V/1K8=2.4mA and 11.4V/4K7=2.4mA as well (accident, i promise). With those new values, we can use P=VxI to get power dissipation. 4.4V*0.0024A=0.010Watts and 11.4*0.0024A=0.027W

Most normal through hole resistors you're all used to are 1/4W, so all those values are WELL below that, so a 1/4W resistor is fine. If your values dont work out, you will have to go for a physically bigger resistor that can handle more heat before it blows up.


So this is the way you wanna hook it all up, and use the math above to calculate the correct values of resistor etc so have fun :)



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Ned, dude, that was awesome.

Though for a how to I'm sure we could cut that down somewhat again :)

If you need any help 77 let us know, or better yet do a "so Bangbug and Ned said this works...." thread and we'll help you out (Ned, I've volunteered you).

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