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Author Topic:   Trim Gauge Resistor

hank119 posted 10-21-2009 03:34 PM ET (US)
I'm re-powering my 1984 Montauk with a rebuilt 1995 Evinrude 90-HP engine. The trim sending unit on the engine checks out with my ohmeter test. However I think I have a defective trim gauge on my boat. The gauge appears to be original equipment. I'd like to replace it, but I notice that there is an external 45-ohm resistor placed between the [I terminal] on the gauge and the [S terminal on the] gauge for the tan wire to the sending unit. My measurement of the sending unit resistance is 0 to about 84 ohms.
Why this external resistance? Of course, if I bought a new gauge I could always hang the external resister on the new gauge. I would like to why.
Hank
fishgutz posted 10-21-2009 04:54 PM ET (US)
Hmm, perhaps electrical noise suppression?
jimh posted 10-21-2009 06:54 PM ET (US)
The resistor value is probably 47-ohm. A 45-ohm resistor would be an odd value. The purpose of the resistor is to limit current flow in the trim gauge circuit. It is an integral part of the circuit. If a replacement gauge does not have an external resistor, it probably has one internal to the gauge enclosure.
hank119 posted 10-22-2009 11:58 PM ET (US)
Jimh
Thanks for your reply. I also thought at first that protecting the gauge might be the reason. However,the voltage drop across the external resistor and the gauge are the same.The worst case is when trim is UP, ie. resistance between point S and G is close to or at zero That is, full battery voltage is still felt across the gauge. The external resistor would make no difference within the gauge.
I can't figure this one out. I'll just get a new gauge.
jimh posted 10-23-2009 12:28 PM ET (US)
I do not think the purpose of the resistor is to protect the gauge.
jimh posted 10-23-2009 12:48 PM ET (US)
Here is what we know about the trim gauge circuit from its externally visible components:
The trim gauge is a three-terminal device, with connections I, S, and G. We know that the voltage at I is nominally 12-volts, and the voltage at G is nominally 0-volts. There are two known resistors; a fixed resistor R1 connected between I and S, and a variable resistor R2 connected across S and G. The variable resistor is the trim sender unit, and its resistance varies with the position of the motor.
If we ignore the influence of the gauge itself, we can compute the voltage at S for various settings of the variable resistor (in the trim sender unit). For the case where R2 = 85-ohms, the voltage at S will be:
12 x 85/(85+47) = 7.7-volts
For the case where R2 = 0-ohms, the voltage at S will be
12 x 0/(85+47) = 0-volts
We now see that the voltage at the three terminals is as follows:
I = 12-volts
G = 0-volts
S = 7.7 to 0-volts
Since the gauge indication varies, we have to conclude that inside the gauge the meter movement of the gauge is arranged in such a way as to be influenced by the voltage that appears at S. This could be done by connecting the meter as a voltmeter between S and either of the other terminals with a fixed voltage.
When R2 is at 0-ohms, the current flow in the circuit is limited by R1 to
12/47 = 0.25-ampere
When R2 is at 85 ohms, the current flow in the circuit is
12/132 = 0.09-ampere
When the current flow is maximum, the voltage drop across R1 will be 12-volts. This means the power dissipated in in R1 will be
12 x 0.25 = 3-watts
The size of the external resistor on the back of the gauge corresponds well with a resistor of 3-watts of power dissipation. To give some margin, the OEM manufacturer may have used a 5-watt resistor.
jimh posted 10-23-2009 12:53 PM ET (US)
Placing the resistor external to the case was probably done in order to prevent excessive build up of heat inside the gauge. Although 3-watts does not sound like a lot of heat, if contained in a small space, it could raise the ambient temperature. Also, the gauge likely has an incandescent lamp inside for illumination. When the lamp is illuminated it will also create heat. This may have proven to require the R1 resistor be moved to an external location.
Chuck Tribolet posted 10-23-2009 07:43 PM ET (US)
A three-watt resistor would be a big one.

Chuck
jimh posted 10-23-2009 10:23 PM ET (US)
You can see the external resistor on an OEM OMC Trim Gauge in an illustration in my article
http://continuouswave.com/whaler/reference/conventionalGaugeRigging.html
Here is the image:
http://continuouswave.com/whaler/reference/images/revengeGaugeReRig/ 9NewDashRearWiring.jpg
The resistor is large enough to be a 3-watt resistor.
jimh posted 10-24-2009 10:50 AM ET (US)
It is reasonable to assume the trim gauge is wired as a voltmeter in this circuit, not as an ammeter. The basis for this is the variation in the voltage and current. Voltage at terminal S varies from 0 to 7.7 -volts. Current varies from 0.09-ampere to 0.25-ampere. From observation of the trim gauge, we can see that under certain conditions, the meter indication is zero. Since the current flow in the circuit is never zero, it seems unlikely the trim gauge could be wired as an ammeter in the circuit. Since the voltage at terminal S does go to zero, this corresponds better with the trim gauge indication. Therefore it seems reasonable to conclude the trim gauge acts as a voltmeter connected to terminal S.
hank119 posted 10-25-2009 10:42 PM ET (US)
Jim
Your analysis of the trim gauge circuit was very helpful.By your calculations, for the voltage read by the gauge to be between 0 and 7.7 volts it would be reading the voltage drop directly across the trim sending unit, ie. between points S and G. Alternatively, it would read between 4.3 and 12 volts if it were connected between S and I. This is easily verifiable with an external voltmeter.
Most helpful is your photo of the back of the trim gauge in your article showing a resistor across points I and S. The gauge looks exactly like mine. The resistor seems to be the same size as mine.
Thanks for your help, Jim. I'm going to get an exact replacement of this gauge. I like the arrangement of the scale. The needle goes up when the the engine is up, and it goes down when the engine is down. I haven't seen any new ones like that. That external resistor arrangement can't be too bad. The unit lasted 25 years.
Hank
swist posted 10-30-2009 11:26 AM ET (US)
Before you do anything, you might want to double check the sending unit one more time under operational conditions. Failures of trim readings are much more likely to be sensor-oriented due to the harsh conditions in the vicinity of the engine tilt pivot point.
For example, if you manually turn the pot with your fingers which watching the ohmmeter, you may be going much faster that it actually moves in real life, and dead spots will go by so fast that the meter will not waver.
You may very well have a bad gauge, but since marine trim sensors are not known for their longevity, it may be worth giving it a second look.
In any case you might want to replace both since even if it is not the sending unit, it probably soon will be, given how long it's been there.
jimh posted 12-07-2013 01:43 PM ET (US)
More investigation into the TRIM gauge circuit revealed how it actually works, shown in
http://continuouswave.com/ubb/Forum6/HTML/003348.html

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Author	Topic: Trim Gauge Resistor
hank119	posted 10-21-2009 03:34 PM ET (US) I'm re-powering my 1984 Montauk with a rebuilt 1995 Evinrude 90-HP engine. The trim sending unit on the engine checks out with my ohmeter test. However I think I have a defective trim gauge on my boat. The gauge appears to be original equipment. I'd like to replace it, but I notice that there is an external 45-ohm resistor placed between the [I terminal] on the gauge and the [S terminal on the] gauge for the tan wire to the sending unit. My measurement of the sending unit resistance is 0 to about 84 ohms. Why this external resistance? Of course, if I bought a new gauge I could always hang the external resister on the new gauge. I would like to why. Hank
fishgutz	posted 10-21-2009 04:54 PM ET (US) Hmm, perhaps electrical noise suppression?
jimh	posted 10-21-2009 06:54 PM ET (US) The resistor value is probably 47-ohm. A 45-ohm resistor would be an odd value. The purpose of the resistor is to limit current flow in the trim gauge circuit. It is an integral part of the circuit. If a replacement gauge does not have an external resistor, it probably has one internal to the gauge enclosure.
hank119	posted 10-22-2009 11:58 PM ET (US) Jimh Thanks for your reply. I also thought at first that protecting the gauge might be the reason. However,the voltage drop across the external resistor and the gauge are the same.The worst case is when trim is UP, ie. resistance between point S and G is close to or at zero That is, full battery voltage is still felt across the gauge. The external resistor would make no difference within the gauge. I can't figure this one out. I'll just get a new gauge.
jimh	posted 10-23-2009 12:28 PM ET (US) I do not think the purpose of the resistor is to protect the gauge.
jimh	posted 10-23-2009 12:48 PM ET (US) Here is what we know about the trim gauge circuit from its externally visible components: The trim gauge is a three-terminal device, with connections I, S, and G. We know that the voltage at I is nominally 12-volts, and the voltage at G is nominally 0-volts. There are two known resistors; a fixed resistor R1 connected between I and S, and a variable resistor R2 connected across S and G. The variable resistor is the trim sender unit, and its resistance varies with the position of the motor. If we ignore the influence of the gauge itself, we can compute the voltage at S for various settings of the variable resistor (in the trim sender unit). For the case where R2 = 85-ohms, the voltage at S will be: 12 x 85/(85+47) = 7.7-volts For the case where R2 = 0-ohms, the voltage at S will be 12 x 0/(85+47) = 0-volts We now see that the voltage at the three terminals is as follows: I = 12-volts G = 0-volts S = 7.7 to 0-volts Since the gauge indication varies, we have to conclude that inside the gauge the meter movement of the gauge is arranged in such a way as to be influenced by the voltage that appears at S. This could be done by connecting the meter as a voltmeter between S and either of the other terminals with a fixed voltage. When R2 is at 0-ohms, the current flow in the circuit is limited by R1 to 12/47 = 0.25-ampere When R2 is at 85 ohms, the current flow in the circuit is 12/132 = 0.09-ampere When the current flow is maximum, the voltage drop across R1 will be 12-volts. This means the power dissipated in in R1 will be 12 x 0.25 = 3-watts The size of the external resistor on the back of the gauge corresponds well with a resistor of 3-watts of power dissipation. To give some margin, the OEM manufacturer may have used a 5-watt resistor.
jimh	posted 10-23-2009 12:53 PM ET (US) Placing the resistor external to the case was probably done in order to prevent excessive build up of heat inside the gauge. Although 3-watts does not sound like a lot of heat, if contained in a small space, it could raise the ambient temperature. Also, the gauge likely has an incandescent lamp inside for illumination. When the lamp is illuminated it will also create heat. This may have proven to require the R1 resistor be moved to an external location.
Chuck Tribolet	posted 10-23-2009 07:43 PM ET (US) A three-watt resistor would be a big one. Chuck
jimh	posted 10-23-2009 10:23 PM ET (US) You can see the external resistor on an OEM OMC Trim Gauge in an illustration in my article http://continuouswave.com/whaler/reference/conventionalGaugeRigging.html Here is the image: http://continuouswave.com/whaler/reference/images/revengeGaugeReRig/ 9NewDashRearWiring.jpg The resistor is large enough to be a 3-watt resistor.
jimh	posted 10-24-2009 10:50 AM ET (US) It is reasonable to assume the trim gauge is wired as a voltmeter in this circuit, not as an ammeter. The basis for this is the variation in the voltage and current. Voltage at terminal S varies from 0 to 7.7 -volts. Current varies from 0.09-ampere to 0.25-ampere. From observation of the trim gauge, we can see that under certain conditions, the meter indication is zero. Since the current flow in the circuit is never zero, it seems unlikely the trim gauge could be wired as an ammeter in the circuit. Since the voltage at terminal S does go to zero, this corresponds better with the trim gauge indication. Therefore it seems reasonable to conclude the trim gauge acts as a voltmeter connected to terminal S.
hank119	posted 10-25-2009 10:42 PM ET (US) Jim Your analysis of the trim gauge circuit was very helpful.By your calculations, for the voltage read by the gauge to be between 0 and 7.7 volts it would be reading the voltage drop directly across the trim sending unit, ie. between points S and G. Alternatively, it would read between 4.3 and 12 volts if it were connected between S and I. This is easily verifiable with an external voltmeter. Most helpful is your photo of the back of the trim gauge in your article showing a resistor across points I and S. The gauge looks exactly like mine. The resistor seems to be the same size as mine. Thanks for your help, Jim. I'm going to get an exact replacement of this gauge. I like the arrangement of the scale. The needle goes up when the the engine is up, and it goes down when the engine is down. I haven't seen any new ones like that. That external resistor arrangement can't be too bad. The unit lasted 25 years. Hank
swist	posted 10-30-2009 11:26 AM ET (US) Before you do anything, you might want to double check the sending unit one more time under operational conditions. Failures of trim readings are much more likely to be sensor-oriented due to the harsh conditions in the vicinity of the engine tilt pivot point. For example, if you manually turn the pot with your fingers which watching the ohmmeter, you may be going much faster that it actually moves in real life, and dead spots will go by so fast that the meter will not waver. You may very well have a bad gauge, but since marine trim sensors are not known for their longevity, it may be worth giving it a second look. In any case you might want to replace both since even if it is not the sending unit, it probably soon will be, given how long it's been there.
jimh	posted 12-07-2013 01:43 PM ET (US) More investigation into the TRIM gauge circuit revealed how it actually works, shown in http://continuouswave.com/ubb/Forum6/HTML/003348.html