I have a working '94 940 speedometer which I need to put into an '87 740. I know they are not plug compatible. I still need to make it work. I need your advice.

Problem: the '87 turbo sedan's speedo connects to the speedo sensor on the differential through the ABS system, but the '94 turbo wagon's speedo connects directly. Another problem: the '87 speedo sensor shares a ground with the fuel level sensor, but the '94 has separate grounds.

So far I have encountered three separate results.
1. If speedo and fuel level sensor grounds are electrically tied together, then the speedo reads 0 MPH and fuel guage is below empty.
2. If speedo and fuel level sensor grounds are separated per '94 wiring diagram and fuel sensor wired directly to combined instrument's fuel sensor inputs but speedo sensor is routed through '87's ABS system, then the speedometer reads about 0.2 of road speed. (When the car is going 40 MPH the speedo reads about 8 MPH.)
3. Same situation as 2 but speedo sensor is wired directly to speedo inputs (+ and -), the speedometer reads about 2x road speed. (When the car is going 40 MPH the speedo reads about 90. Whee.)

Does anybody have any sage advice for this? I have Volvo's wiring diagrams for both cars but neither one has a detailed schematic for the speedo's inner workings.

Thanks in advance for the help.

This issue is kind of special because of your early ABS eqiuppped car. The 87 should have a 96 teeth sender wheel in the differential. This was replaced by a 48 teeth wheel on 1989- ABS models. All 900 cars used the 48 teeth sender. BUT non-ABS 700 cars (until 1990) had a 12 teeth wheel, which is what the 700 speedometer requires. That's why the signal passes through a part of the ABS system on your car - this simply is a divider which gives the speedometer its required 12 pulse/axle turn signal from the 96 pulse input.

As you might already have figured, the 940 speedo uses the 48 pulse/axle turn signal as is (so does the 940 ABS system). This is why it reads a quarter of the correct speed (12/48=0.25) or double (96/48=2) speed depending on the way you connect it.

This also gives a not-so-simple solution to the problem, since you can't tap a correct, 48p signal for the 940 speedo anywhere. Solve the problem with a separate divide-by-two cirquit and use the original speedo signal (not the ABS system output).

/Martin
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65 121, 73 165

Thanks very much for the info. I had hoped for something simpler, but this is essentially what I expected.

I no longer have access to an oscilloscope to measure voltages on the signal and I'm not confident that my DVM can do the trick. Do you know the specs?

I assume that input and output are essentially square waves, measured with reference to the combined instrument's reference ground, and that it doesn't much matter which level transition (up or down) I use as long as I am consistent.

Hmm. I haven't designed or built any digital circuits since college 25 years ago. This ought to be fun.

The speedometer sender produces a regular sine wave with pretty high amplitude. Original speedometer inputs start with an ordinary RC filter, not sure if I remember correct but 10kohm/47nF feels familiar. Cut-off frequenzies will look weird if you calculate but it works in reality. Use something like it in your own cirquit or you'll risk loading the signal to much, which may cause ABS warnings/shutdown.

940 speedo's are pretty sensitive and will accept a square-shaped positive signal. Try to get the zero part as close to ground as possible, though, since the original design detects sine wave zero crossings. The speedo input is very tolerant, btw - you won't risk damaging the internal cirquit if you shorten or give it a high voltage.

/Martin
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65 121, 73 165

Benglar,
Are you an engineer?
-Gil

I am an engineer after a fashion: BS and MS in computer science, UC Berkeley. That means I've had training in digital hardware but not analog electronics. I can (and have) design a computer from scratch but I can't design an amplifier. However, I do know somebody who does know analog electronics who thinks he can make me a digital divide by two circuit. I'll post the results.

...And if my friend's circuit doesn't work I'm tempted to try the following rather messy idea.
1. Use the existing speedo sensor's sine wave signal as input to a digital circuit. Use the rising edge as a clock to trigger a D flip flop with Q-bar output tied to D input. The Q output toggles at 1/2 the input frequency in the digital domain.
2. Use the Q output to gate the speedo sensor's sine wave signal via a transistor to the speedo. When Q is low the signal would be zero. When Q is high it would pass 1 sine wave cycle -- in effect, odd-numbered sine wave cycles go through and the output is at the 0v reference level during the even-numbered sine wave cycles. This produces analog output without requiring things like op amps and phase-locked loops which I don't understand. Using low-resolution ASCII graphics, the signal would look something like this:

_ _ _ _ _
input: ( ) ( ) ( ) ( ) ( ) [imagine this is a sine wave]
0v ref: _) (_) (_) (_) (_) (_
_ _ _
output: ( ) ( ) ( )
0v ref: _) (_____) (_____) (_

If the speedo just senses the rising edge of the wave, this should work. But all bets are off if the speedo does a fast fourier transform or something else looking for the first harmonic...

Well, the ASCII graphics didn't work because the message board removes extra whitespace characters. Sorry, folks....