Dodge RamCharger Central banner
Not open for further replies.
1 - 1 of 1 Posts

25,730 Posts
Discussion Starter · #1 ·
I am going to explain the basics of how an Ammeter works. An Ammeter measures Current, the unit of measure is Amperes (Amps).

First, some Basic Electronics. The first thing I learned was Ohm's law, which simply means: Voltage=Current X Resistance. Using some math, this also means that Current=Voltage / Resistance.

Now, if you take a fixed resistance, the Voltage measured across it will vary with the current that flows through it. For example, in the BASICS diagram below, we will give the resistor a value of 1 Ohm. If the battery is 12 Volts, we can figure the current:
Current= 12 (Volts)/1 (Ohm) ; So Current=12 Amps

One more example--Resistor=3 Ohms and Battery=6 Volts:
Current= 6/3 ; So Current=2 Amps.

Also, using the first equation--Voltage=Current X Resistance. If Resistance is 2 Ohms and Current is 3 Amps:
Voltage= 3 (Amps) X 2 (Ohms) ; Voltage=6 Volts.

So, once again, with a fixed resistance, the voltage across it will vary with the current.

This is the principle an Ammeter works on. The current is passed through a fixed resistance (called a "Shunt" or "Shunt Resistor") and the resulting voltage is measured and displayed on a meter that is calibrated (marked) accordingly.

Now refer to the Series Circuit diagram below. R1 is short for Resistor1 and TP1 is short for TestPoint1. The resistors are in "Series" configuration. In a series circuit, Resistance is additive-which just means you add the value of R1 to R2 to get the Total resistance. Give R1 a value of 40 Ohms and R2 a value of 80 Ohms. R1 + R2 = 120 Ohms.

Also, in a series circuit, Current is constant-which just means that R1 and R2 will have the same amount of current flowing through them. So if we take R1=4 Ohms, R2=8 Ohms, and give the battery 12 Volts, we can solve for Current.
Current= 12 (Volts)/ 12 Ohms ; Current=1 Amp.

Since current is constant in a series circuit, this means R1 and R2 each have 1 Amp, so we can solve for the voltage across each resistor:
R1: Voltage= 1 (Amp) X 4 (Ohms) ; Voltage= 4 Volts (measured between TP1 and TP2)
R2: Voltage= 1 (Amp) X 8 (Ohms) ; Voltage= 8 Volts (measured between TP2 and TP3)

This next diagram combines what I have covered so far. From here on out, these things will remain constant: Battery is 12.5 Volts and the Shunt Resistor will be .01 Ohm. If we measure the voltage between TP1 and TP3 in this diagram, it is 12 Volts.

One more formula now--for Power (measured in Watts)--Power=Current X Voltage ; which also means Current=Power / Voltage. This formula is useful when adding accessories. For example, a pair of 150 Watt ([email protected] Volts) offroad lights. 2 X 150 = 300 Watts.
Current= 300 (Watts) / 12 (Volts) ; Current=25 Amps.
If we use these offroad lights as a load, we can calculate the resistance of the load.
Resistance=Voltage/Current ; Resistance= 12 (Volts)/ 25 (Amps) ; Resistance= .48 Ohms.
(Note that I used the [email protected] volts for the calculations, not the actual battery voltage)

Now to insert the values we have into the Diagram above:
Overall Resistance = Shunt Resister + Load = .01 + .48 = .49 Ohms.
Current = Voltage / Resistance = 12.5 / .49 = 25.5 Amps
Shunt Voltage = Current X Resistance = 25.5 X .01 = .255 Volts
Load Voltage = 25.5 X .48 = 12.245 Volts

If the math lost you, don't worry about it. I mainly did it to show that the Ammeter actually drops a small amount of voltage (about 1/4 Volt in this example). Measuring Voltage from TP1 to TP2 would show .255 Volts, with TP1 being more positive than TP2. The needle on the Ammeter (in this case one of Dodges FINE stock ammeters) would go a few "ticks" toward "D" (Discharge) to show that the battery was providing current to the load.

The next diagram illustrates this-it also shows where the Alternator ties into the circuit.

Conditions: Engine OFF, Alternator NOT charging, Offroad Lights ON.

Now we add the Alternator into the circuit. Two more constants--Alternator Output

Voltage=13.5 Volts (Measured from TP2 to Ground); Alternator Max Current=60Amps.

Please "bear" with me on this next one. A battery, once charged, will have the same voltage as the Alternator is putting out. So for this next example, assume the battery is in a slightly discharged state (like right after starting the engine) and being recharged. So the battery Voltage is STILL 12.5 Volts (Measured TP1 to Ground).

The math:
Voltage Shunt=Voltage TP2 - Voltage TP1 = 13.5 - 12.5 = 1 Volt
Current Shunt=Voltage Shunt / Resistance Shunt = 1 / .01 = 100 Amps

Hmmm, our 60 Amp Alternator is putting out 100 Amps? Wouldn't it be nice if it was that easy? Actually, what would happen here is the alternator would put out it's max capacity (60 Amps) and the Ammeter would most likely be "Pegged" at "C" until the battery recharged, then they (both the Alternator and Ammeter) would taper off until the battery was fully charged. This is providing nothing burned up in the meantime...

One thing to note here--the main power for the vehicle comes off the Alternator side of the Ammeter. Because of this, once the battery is recharged, the Ammeter will be centerlined,

showing neither a Charge nor Discharge as long as the Alternator is working correctly and providing sufficient electrical power to run the vehicle and all accessories. I based these diagrams from 1981 and later wiring diagrams. Unfortunately, I am not sure where 1980 and earlier models placed the vehicle load in relation to the Ammeter-perhaps someone can clue me in.

Next let's look at what happens if too much of a load is put on the alternator. Remember the Offroad lights from before? Suppose we added six of them instead of just two. Since two was 300 Watts, six will be 900 Watts.
The math: Current= 900 (Watts) / 12 (Volts) ; Current=75 Amps.
Resistance=Voltage/Current ; Resistance= 12 (Volts)/ 75 (Amps) ; Resistance= .16 Ohms.
(Note again that I used the [email protected] volts for the calculations, not the actual battery voltage)

I started to do the math for the diagram above, but it was deeper than I wanted to get into here. In short, the Alternator is going to provide it's full 60 Amps. The remaining Current (Amps) will come from the Battery, so the Ammeter will show a Discharge. If you run the lights too long, the battery will run out of "juice", the overall voltage will drop, the lights will begin to dim...if the lights aren't turned off, either something in the charging system will "fry" or the lights will continue to pull the voltage down until the ignition cannot function or both...

From reading posts on the forum, then looking at my wiring diagrams, I noticed there were two different Ammeter types used on our Dodge RCs and Trucks--Internal Shunt and External Shunt. The diagrams from my Haynes show that 1980 and earlier were Internal and 1981 and later were External. Thanks to boba for bringing this to our attention!

Here are the differences.

1980 and prior years.
The main difference is where the Shunt is. In earlier models, they ran the charge circuit through the Ammeter-this literally put the entire charge current through the instrument cluster, which turned out to be a fire hazard as reported by some.

1981 and later years.
In later models, they used a section of the Charge wire as the Shunt and just ran the sense wires to the Instrument cluster.

Since a 101 Course wouldn't be complete without homework...figure this out. How would moving power connection for the offroad lights, both the 300 Watt and 900 Watt Versions, so they are connected directly to the battery affect the reading of the Ammeter?

I hope this makes sense and can even help someone understand how an ammeter works in conjuction with their charging system.

Hopefully I'll get some feedback that will make this a better answer to a FAQ.

1 - 1 of 1 Posts
Not open for further replies.