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ContinuousWave: Small Boat Electrical
Mystery of the Trim Gauge
|Author||Topic: Mystery of the Trim Gauge|
posted 03-03-2013 09:21 AM ET (US)
I have a conventional TRIM gauge that I have been investigating to see if I could figure out just how the circuit works. The device has been a bit of a mystery to me for several years. I decided to investigate it more thoroughly.
On the older TRIM gauge there are three terminals for the gauge: I, S, and G. (This notation is fairly conventional. The I is for the ignition circuit that supplies the 12-Volt power. The S is for the sender connection that supplies the signal. The G is for ground.) There is an external resistor connected between I and S. I removed this resistor and measured it; the resistance was 83.5-Ohm. With the external resistor removed, I measured the resistance between I and S as 95.8-Ohm. When in parallel, the combination would have a resistance of about 45-Ohm. This is apparently why there is a 47-Ohm resistor used in the engine wiring harness if electronic gauges are used and there is no conventional trim gauge; it duplicates the resistance that would be in the circuit if a conventional trim gauge were used. A value of 47-Ohm is a standard value, and close enough to 45-Ohms.
I also measured the resistance of the gauge between S and G: it was 166.2-ohms. The external resistance of the TRIM SENDER is connected across these same terminals. The TRIM SENDER is typically a 0 to 100 ohm rheostat. When the sender was at 100-ohms the combination with the internal resistor would be about 62-Ohms.
I also measured the resistance from I to G (without the external TRIM SENDER); it was 262-Ohms, or the sum of the resistances form I to S and S to G. This suggests that the resistances are in series. With the TRIM SENDER at 100-ohms, the resistance from S to G would be about 62-Ohms. This implies the total resistance I to G with the TRIM SENDER at 100-Ohms would be about 45 + 62 = 107-Ohms.
Next, I applied 13.2-Volts to the gauge at I and connected a 100-Ohm resistor between S and G to simulate the TRIM SENDER. I measured the current flow: 0.12-Ampere flowed into I and returned at G. According to Ohm's Law, the resistance of the circuit was therefore 13.2/0.12 = 110-Ohms. This is a good agreement with my measured resistances.
I did not have a TRIM SENDER on hand, but I used a couple of fixed resistors to simulate the TRIM SENDER values. With the TRIM SENDER at 100-Ohms, the TRIM gauge read near the "DN" end of the scale.
Next I put a 1.5-Ohm resistor between S and G, to simulate the other end of the TRIM SENDER range, near 0-Ohm. With 13.2-Volts applied the TRIM gauge read "UP" and I measured 0.27-Ampere of current. This is in good agreement with the expected current, which would be 13.2/46.5 =0.284-Ampere. (My meter has only a 10-Ampere scale, so there can be some error in these low current readings.)
This narrative is a hard way to describe things. Let me try to draw with ASCII. I will show the external connections to the TRIM gauge:
13.2 Volts------>(I)---83.5-Ohm---(S)----100-Ohm---(G)-------Power Common
This produces 0.12-A flow and the meter reads "DN."
13.2 Volts------>(I)---83.5-Ohm---(S)----1.5-Ohm---(G)-------Power Common
This produces 0.27-A flow and the meter reads "UP."
It is easy to see that the TRIM SENDER is close to 100-Ohms when in the gauges points to DN and close to 0-Ohms when in the gauge points to UP.
What is not apparent to me is which terminals of the gauge actually have the meter connected to them. To test this, I modified the circuit. I did not connect anything to (G). Instead, I just took the 100-Ohm resistor lead from (S) and connected it to Power Common:
To my surprise, the TRIM gauge pegged hard to the UP position.
I had to scratch my head on that one. I had been expecting the meter to be connected between (S) and (G), and when I removed the connection from (G) there would be no current in that part of the circuit, which would imply that the meter should not move at all.
Where is the meter movement connected? What is going on inside the TRIM gauge?
Pondering what might be inside the TRIM gauge, I came up with the idea of the meter being arranged in a bridge circuit. The circuit seems to be like this:
The use of an external resistor across the I and S terminals always had me puzzled. But this is really a very good idea. It allows the manufacturer of the gauge--which is very likely not the outboard engine maker--to make a universal gauge. The value of the resistor installed externally at I and S is related to the range of the TRIM SENSOR that is connected between S and G. The internal resistors values are selected to work well with a range of external resistors values. This allows the gauge to be used with various ranges of TRIM SENSOR resistances. For example, suppose another outboard engine manufacturer used a TRIM SENSOR that varied from 25 to 200-Ohms. By changing the value of the external resistor connected between I and S, the same trim gauge could be used with that new range of values for TRIM SENSOR.
The bridge arrangement also explains very nicely why the meter reacted so unexpectedly when the ground connection to G was removed. The bridge balance was completely upset and much more current flowed through the meter. That is why the meter pegged so hard.
MORE ABOUT RESISTANCE MEASUREMENTS
The resistance of the coil of an ammeter is typically very low, perhaps only a fraction of an Ohm. When I measured the resistance from the three terminals (with the external 83.5-Ohm resistor disconnected), I was probably measuring the coil resistance of the meter in those measurements.
Solving those relationships leaves RM = 0, which may be close to the truth. The resistance of the meter coil is very low, perhaps only 0.1-Ohm. It may be too low to see in the measured values unless very accurate resistance measurements are made.
Another clue that perhaps the meter is wired into a bridge circuit is the meter itself. It appears to be a Zero-center meter, that is, a meter that shows current flow in two directions, with the rest position in the center of its scale. When there is no current flow in the meter the dial pointer drifts toward the center.
With this deduction, I think I have solved the mystery of the trim gauge. I would love to disassemble an old TRIM gauge to see if the internal circuitry is the same as my hypothesis. If anyone has an old TRIM gauge that no longer works, I would love to take it apart to see exactly what is in there. I do think the bridge circuit I describe (above) is very likely what will be found.
posted 12-07-2013 01:46 PM ET (US)
I also shown how a bad ground in the TRIM gauge circuit could be replaced by the lighting circuit lamp filaments when they were cold. This is another common and often mysterious problem with TRIM circuits. See
for the analysis with detailed schematic diagrams.
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