why is their a resistor wire going to voltage regulator in cluster?

so im working on rewiring a maverick and have found that their is a resistor wire going to the voltage regulator in the instrument cluster.
does anybody know why it has this wire, what the resistance of the wire needs to be and what would the effect be if the regulator was to get full battery voltage?

 

The instrument voltage regulator usually supplies gas and temp gauges in most vehicles so normally there is the load of two gauges vs one(three if oil pressure is included).... Sooo I'll guess it's to limit current because load of one gauge isn't enough to drop voltage to the proper working level, meaning the gauge will likely read high without it... Of course if Ford used a specific regulator for one gauge I may be all wet, but I'm thinking one part number fits all...

What color wire & year is this???

 

From what my uncles taught me many years ago.. that resistor just cuts juice levels so the idiot light becomes more sensitive to charge output related issues.

That's why we've all seen those dimmly lit idiot lights when the system just starts acting up until they get progressively brighter as output dies off and it's finally totally pooched.

 

this wire only feeds the gas gauge circuit.
the idiot lights have their own power supply circuit.

 

I don't know if this is it but the logic seems to "fit". The resistor may be related to reading ohm's from the sending unit. Ford likely did something similar as Chevy to achieve the same results i.e. more accurate fuel gage reading and smoother needle sweep.

http://www.chevyclassicsclub.com/the-gm-fuel-gauge-mystery-1965-newer/

 

My understanding is that it prevents "sloshing" of the needle. Modern cars use circuitry for this effect.

I installed VDO gauges and the fuel level is all over the place when the fuel is moving around in the tank.

 

It is an inrush current limiter. Because the gauge circuit is so dynamic with the ICVR (CVU) heating coils and 10-70ohm sender, the current limiter keeps the circuit safe (better than a slow fuse can). Without the resistor, if all the coils are cold, the contacts closed, and the tank is full (or shorted), initially the circuit would experience an inrush current of atleast 1.2A (V=I*R; I = V/R = 12/10 = 1.2A). Using the resistor, lowest (full tank) resistance is 8.5ohms+10ohms=18.5ohms, so the max inrush current is about 12/18.5 = 0.65A. If the sender has failed (to ground), then the current cannot exceed about 1.5A. Now, why Ford chose to operate within those current windows is beyond me. It has to do with the ICVR, gauge and circuit design.

EDIT: It also makes sense, as noted by Kevin and injectedmav, that this can have the effect of "slowing" the gauge down by limiting the amount of current the circuit can use. I always thought of it as the "pulsing" of the circuit was what kept the gauge from swinging. This resistor definitely influences the pulsing.

 

So do I need to make sure that I have a certain resistance on that wire and if so how much

 

Not sure... the schematic I found shows a resistance of about 8.5 ohms. See if you can measure the resistance of the entire ICVR feed wire.

A new style, solid state ICVR may not even need the resistive wire.

 

The wire is part of the signal conditioning and filtering portions of the circuit. It's a pretty basic clipping filter. The idea is that if you clip the DC voltage to 8.5-9V you wont boil and explode the ignition coil slamming the gas pedal and getting a voltage spike out of the alternator. Old coils couldn't handle prolonged exposure to voltages over 13 volts before letting the smoke out.

Also, I guess you guys didn't realize this wire was the resistive wire that limits the voltage sent to the ignition coil.

Although, I should probably double check the expanded circuit to make sure. I am pretty certain this wire limits the whole circuit. Either that or it is further clipping what was clipped by that wire. Trying to work off memory from last year diagnosing my car.

 

I think we are talking about two different resistive circuits. Bryant is talking about the resistive circuit feeding the ICVR, not the ignition coil. Are these same circuit?

Also, I think you are referring to a voltage divider, not a clipper. A clipping circuit requires a diode and are used in AC or PWM circuits. But yes, voltage division is indeed used in the RUN position for the coil. A bypass of that circuit fried my dads coil on his IH tractor. That coil got smoking hot!

 

They are two different circuits...

Look at the posted diagram, gas gauge is fed from Accessory vs Ign(coil on that diagram)... It's no doubt for current limiting, more so for the regulator vs gauge...

BTW the anti slosh damping that was mentioned really isn't necessary on theses old bi-metal types, their response is is snail pace slow vs a modern coil type gauge...

 

You are probably correct. I should really not post **** from work because I can't look at what I have written down on it at home.

I know for a fact though that the circuit from the ignition switch to the fuel gauge is a near perfect linear proportional controller such that the fuel gauge can be used as a real time ammeter/voltage meter. So I am still fairly certain that the 8.5 ohm resistance wire is there specifically to clip the voltage so that voltage spikes don't disrupt the simulated 5V output.

 

That voltage regulator is nothing more than a glorified signal flasher(The posted diagram depicts exactly what it is)... Only difference is load of the bulbs provide current for flasher operation, while the internal heater does same in regulator...

The regulator is continually making and breaking the circuit(chopping) to provide a average 5v as far as what the gauge sees... The reason there is no
"twitching" on the gauge, is the bi-metal types are so numb takes two three seconds to respond and by that time the regulator contacts are closed again... A coil type gauge would flutter in a circuit such as this...

 

What are you explaining exactly? The voltage regulator may be nothing more than a glorified flasher but that doesn't change what the actual circuit does. It mixes a reference voltage with a signal voltage to produce a graduated result. If you reduce or increase either voltage it makes the fuel gauge read incorrectly.

The picket fence style (chopping) regulator produces a 5-volt clipped waveform that mixes with the signal from the fuel sender. The fuel sender has a much more controlled signal which is why affecting the reference voltage has an extreme variance.

Use an externally powered starter to fire an engine on a dead battery and as the battery charges the fuel gauge will rise with the battery level until it balances with the signal from the fuel sender and shows an accurate fuel reading. Or, put in a dead battery and put a charger on it while the accessory circuit is engaged, doesn't matter.

It is basically the same as a simple DC motor controller.