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Help with voltage divider for boat trim sensor

Gort

New Member
Hello all,
I am trying to get a boat trim sensor and meter configured, this is a 12 V DC system. The sender that attaches to the trim unit is a potientometer, and shows 145 ohms at “full up travel”, and 62 ohms at full down. The meter requires 0v at full up and 8.5 volts at full down.
I created a voltage divider with a 500 ohm pot for adjusting the range, and I get the best results at about 122 ohms, but I am only getting about 60% of the indication on the trim meter (basically a voltmeter).
Somehow I need to increase the scaling, as I am stuck with the sender on the pot. Not sure how to proceed. TIA.

Papabravo

Well-Known Member
I'm not sure why you think a voltage divider is the right way to do this.

rjenkinsgb

Well-Known Member
The meter requires 0v at full up and 8.5 volts at full down.
What is the resistance of the meter itself?

crutschow

Well-Known Member
If you want 0V to 8.5V from that sender resistance range with a 12V supply, you could use an opamp circuit to generate that.
Is that something you could build from a schematic?

Gort

New Member
What is the resistance of the meter itself?

Gort

New Member
I'm not sure why you think a voltage divider is the right way to do this.
How else? I am stuck with the “variable resistor” sender at the ohm range quoted, and with my limited knowledge of electronics, a voltage divider seemed like the easiest choice if it could be made to work.
Do you have something helpful to suggest?

Gort

New Member
If you want 0V to 8.5V from that sender resistance range with a 12V supply, you could use an opamp circuit to generate that.
Is that something you could build from a schematic?
Yes, have used I can probably find an op amp circuit online, will it use another potentiometer to adjust the scale?
If you want 0V to 8.5V from that sender resistance range with a 12V supply, you could use an opamp circuit to generate that.
Is that something you could build from a schematic?
I have used 741s along time ago in some hobby experiments, is that what I need? I could then use a potentiometer to adjust the scale?

crutschow

Well-Known Member
That's way too high for a mechanical meter.
Or is the meter powered from 12V?
I have used 741s along time ago in some hobby experiments, is that what I need? I could then use a potentiometer to adjust the scale?
741's won't work well with a single supply in this application.

I'll work on a circuit using a single-supply type op amp, but may not finish it until tonight or tomorrow.

Papabravo

Well-Known Member
How else? I am stuck with the “variable resistor” sender at the ohm range quoted, and with my limited knowledge of electronics, a voltage divider seemed like the easiest choice if it could be made to work.
Do you have something helpful to suggest?
Well excuse me, but your reasoning was not clear from your original post. Based on your answer and other information I'll second the suggestion from crutschow to use an opamp circuit. Voltage dividers are usually appropriate with fixed resistors. When one leg of the divider is variable, it is like grabbing a tiger by the tail because the behavior is counter intuitive. Even with fixed resistors, when the load impedance changes you also get counter intuitive behavior. An opamp will effectively decouple the sensor from the indicator.

If I was going to do it, I would use two equal valued resistors on either end of your sensor powered by the +12V supply. The current would be nearly constant as long as the power supply remains at +12V. This will create a differential voltage based on the wiper position. This volage will be limited to some range inside the power supply rails, eg. 5-7 Volts DC. Using an operational amplifier with an appropriate gain and offset setting you can measure the voltage and use that to drive the meter.

The impedance of the meter seems unusually high, I guess more details would be appropriate.

EDIT: Here is a starting point. This is not a unique design; other choices of resistor values are possible. I picked 10 mA as a reasonable current for purposes of understanding. I've modeled the trim sensor as two resistors in series; the minimum value of 62Ω the 83Ω potentiometer give the range up to 145Ω. The result is a swing of jus under a volt over the range of the potentiometer. If your supply voltage drops a bit this will stll work but the range will be a bit smaller. This not a problem for the opamp to handle.

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Gort

New Member
That's way too high for a mechanical meter.
Or is the meter powered from 12V?

741's won't work well with a single supply in this application.

I'll work on a circuit using a single-supply type op amp, but may not finish it until tonight or tomorrow.
Thank you, I appreciate you taking the time to help. It is not urgent.

Gort

New Member
Well excuse me, but your reasoning was not clear from your original post. Based on your answer and other information I'll second the suggestion from crutschow to use an opamp circuit. Voltage dividers are usually appropriate with fixed resistors. When one leg of the divider is variable, it is like grabbing a tiger by the tail because the behavior is counter intuitive. Even with fixed resistors, when the load impedance changes you also get counter intuitive behavior. An opamp will effectively decouple the sensor from the indicator.

If I was going to do it, I would use two equal valued resistors on either end of your sensor powered by the +12V supply. The current would be nearly constant as long as the power supply remains at +12V. This will create a differential voltage based on the wiper position. This volage will be limited to some range inside the power supply rails, eg. 5-7 Volts DC. Using an operational amplifier with an appropriate gain and offset setting you can measure the voltage and use that to drive the meter.

The impedance of the meter seems unusually high, I guess more details would be appropriate.

EDIT: Here is a starting point. This is not a unique design; other choices of resistor values are possible. I picked 10 mA as a reasonable current for purposes of understanding. I've modeled the trim sensor as two resistors in series; the minimum value of 62Ω the 83Ω potentiometer give the range up to 145Ω. The result is a swing of jus under a volt over the range of the potentiometer. If your supply voltage drops a bit this will stll work but the range will be a bit smaller. This not a problem for the opamp to handle.
Now that is helpful, I will order what I need. I thank you for taking the time to work this up. Will follow up when I have results.

Papabravo

Well-Known Member
Hold on there a minute. Will a continuous current draw of 10 mA be a problem for the +12V power source. If it is, then we can adjust that current consumption down to a more convenient level. I believe it is a mistake to order parts before the design is complete.

Gort

New Member
Hold on there a minute. Will a continuous current draw of 10 mA be a problem for the +12V power source. If it is, then we can adjust that current consumption down to a more convenient level. I believe it is a mistake to order parts before the design is complete.
Thank you. This is being run off of a 50 ah battery being recharged by the alternator on the outboard, so I think 10ma would be no issue, but if there is no reason to draw that much I would like to be as efficient as possible. I appreciate your input and will not order stuff yet.

Gort

New Member
Hold on there a minute. Will a continuous current draw of 10 mA be a problem for the +12V power source. If it is, then we can adjust that current consumption down to a more convenient level. I believe it is a mistake to order parts before the design is complete.
Ok, the meter measures 146 ohms, I went to the wrong terminal before by mistake. Also, see attached picture, it is 5v full travel, not 8.5. I apologize for the error.
Thanks for the help.

Gort

New Member
Ok, the meter measures 146 ohms, I went to the wrong terminal before by mistake. Also, see attached picture, it is 5v full travel, not 8.5. I apologize for the error.
Thanks for the help.

Papabravo

Well-Known Member
So, if I understand correctly:
1. The DC resistance of the meter is 146Ω
2. You want the output of the amplifier to output 0V for full UP and 5V for full down.
Is that correct?

Gort

New Member
So, if I understand correctly:
1. The DC resistance of the meter is 146Ω
2. You want the output of the amplifier to output 0V for full UP and 5V for full down.
Is that correct?
Exactly, with my 62 to 145 ohm sender, thanks.

Gort

New Member
So, if I understand correctly:
1. The DC resistance of the meter is 146Ω
2. You want the output of the amplifier to output 0V for full UP and 5V for full down.
Is that correct?
yes, and 62 ohms from the sender at full down, 145 ohms for full up.

Papabravo

Well-Known Member
I'll look at the rest of this tomorrow. In the meantime, crutschow may have an opamp circuit suggestion. I have not purchased any for a while and with global supply issues I don't have a good candidate off the top of my head. One more question: are you going to fabricate one on a breadboard or are you going to layout a printed circuit board. I ask because SMT parts can be hard to work with on a breadboard and most available parts come in those packages.

crutschow

Well-Known Member
Below is the LTspice simulation of a circuit that should do what you want.
It uses a common, single-supply op amp with a transistor output buffer to generate the 5V to 0V for a sensor resistance variation (Rsender) of 62Ω to 145Ω (yellow trace), to drive the meter (simulated by R_Meter).

(Interestingly, all the gain and offset resistor values can be calculated directly without having to solve any simultaneous equations).

I added the common LM317 voltage regulator to generate a constant 9V for the circuit which minimizes any change in the meter output voltage due to variation in the battery voltage.

If you have any problem buying the parts, let us know, as substitutes should be available.

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