Will a 4.7K potentiometer dim an LED or other bulb in a 12V DC circuit?

I want to put an LED in an automotive circuit to "read" the position of a potentiometer that is internal to a 12v motor. The motor, which opens a valve in the climate control heater, has an internal 4.7 kOhm potentiometer which delivers a signal to the climate control ecu. I will no longer be sending that signal to the ecu but rather would like to use it to dim an led or other bulb. I would like it to run on 12v nominal, I would prefer not to have to reduce it to 5v. If I just hook the power supply up to 12v and run the ground through the motor's potentiometer will the led/bulb dim as resistance is increased? A relatively gradual dimming from open to close would be nice but even if it is much more abrupt that will still accomplish what I want.
Brian

You could probably get it to do what you want, but it will be rather limited.

I don't know the size or power rating of the potentiometer, but you should never overload any part of a potentiometer. Many potentiometers are rated as low as 50 mW, some as low as 10 mW. Have a look at these:-

https://uk.farnell.com/webapp/wcs/s...ze=25&showResults=true&pf=110156870,110156873

The physical size will give you a reasonable idea about the rating.

If you have a 125 mW, 4.7 kΩ potentiometer, the current at maximum power is just over 5 mA, so you should never put more than 5 mA though any part of that potentiometer.

To get the LED to turn off at one end rather than dimming slightly, you need to connect 12 V across the potentiometer, and connect the wiper to the LED. With 12 V across a 4.7 kΩ potentiometer, you get around 2.5 mA with no load.

If you have a red LED and a 12 V supply, the LED will drop about 2 V, and so the remaining 10 V will need a current limiting resistance of 4 kΩ to keep the current down to 2.5 mA so that the total current is never more than 5 mA in any part of the potentiometer.

That will give you a quite linear change in brightness, but you will only get 2.5 mA in the LED so it will never be very bright.

If your potentiometer is smaller and the power rating is less than 125 mW, you will need a different circuit.

carwhisperer said:

I want to put an LED in an automotive circuit to "read" the position of a potentiometer that is internal to a 12v motor.

Click to expand...

The reason most automotive sensors work on 5 volts is because the 5 volts is a nice regulated DC voltage to work with. Here is a potential problem you face in an automotive circuit.

carwhisperer said:

The motor, which opens a valve in the climate control heater, has an internal 4.7 kOhm potentiometer which delivers a signal to the climate control ecu.

Click to expand...

If I apply 12 volts to a pot, any pot and the other end of the pot to ground and look at the pots wiper (output) I can figure that at 50% of pot travel I will get 6 volts. Heck at 10% I will get 1.2 volts and at 90% I will get 10.8 volts. That assumes a true 12 volts across the pot. Should that 12 volts vary at all, as automotive systems do, then all bets are off. So is the 12 volts a fixed regulated 12 volts or can it vary between say 12 and roughly 14 volts? Because if the supply varies so goes the output. That said there are ways to do what you seem to want to do. You can take a LED from off or dim to full intensity with some external circuitry but it won't happen easily with just the output of a pot.

**broken link removed** are pretty common. They work on a wide input voltage of 9 to 17 volts (automotive) and have a 0 to 5 volt input = 4 to 20 mA output. While not a fancy PWM type LED intensity control they will drive a LED having a forward current of 20 mA and show you what you want to see.

Just My Take....
Ron

Have you actually measured the output voltage range from the pot? My guess is that, since it's intended to be fed to an ECU, it could well be in the 0-5V range.

Diver300 said:

To get the LED to turn off at one end rather than dimming slightly, you need to connect 12 V across the potentiometer, and connect the wiper to the LED. With 12 V across a 4.7 kΩ potentiometer, you get around 2.5 mA with no load.

If you have a red LED and a 12 V supply, the LED will drop about 2 V, and so the remaining 10 V will need a current limiting resistance of 4 kΩ to keep the current down to 2.5 mA so that the total current is never more than 5 mA in any part of the potentiometer.

That will give you a quite linear change in brightness, but you will only get 2.5 mA in the LED so it will never be very bright.

Click to expand...

YOu guys are awesome. Thank you so much for taking an interest. I have read through your responses and I see that I need to be cautious. Forgive me if this is a dumb question. If I connect 12V across the pot then will the wiper wire put out positive voltage? So If I put in a fixed 4kohm resistor in series along with an LED and then connect to ground it should turn off at one end and work the way I want? This would be assuming that the pot can handle 5mA, correct?

What if I don't need it to turn off at one end? Do I still need to add a resistor to the circuit?

I plan to supply my own voltage to the pot. If it is intended for 5v and I send it 12v is that a problem in and of itself? I would think not as it would be too much amperage that would damage it, not voltage, correct?

Why burn out a pot by trying to use it to make an LED glow dimly?
Use the pot to feed a variable voltage at a very low current to a transistor emitter-follower. Then the output current from the transistor can be high enough to make the LED very bright and the pot can turn it down to zero.

carwhisperer said:

YOu guys are awesome. Thank you so much for taking an interest. I have read through your responses and I see that I need to be cautious. Forgive me if this is a dumb question. If I connect 12V across the pot then will the wiper wire put out positive voltage? So If I put in a fixed 4kohm resistor in series along with an LED and then connect to ground it should turn off at one end and work the way I want? This would be assuming that the pot can handle 5mA, correct?

What if I don't need it to turn off at one end? Do I still need to add a resistor to the circuit?

I plan to supply my own voltage to the pot. If it is intended for 5v and I send it 12v is that a problem in and of itself? I would think not as it would be too much amperage that would damage it, not voltage, correct?

Click to expand...

Before I forget you may want to give some thought to what AudioGuru mentioned above. Let a transistor do the work.

Now if I read you correctly what you mention won't really work well if at all. Keep in mind a LED is a current device. When we look at a LED data sheet they publish an If or Forward Current rating and normally a Vf Forward Voltage. Let's say the If is 20mA which is a typical generic LED current. They are saying that when 20mA flows through the LED the forward voltage drop will be about 1.2 volts, assuming 1.2 volts is the Vf for that LED. When we use a LED we place a current limiting resistor in series with the LED. OK, so lets say I have a If of 20 mA and a Vf of 1.2 volts for my LED. I want to run this LED at normal brightness from a 12 volt supply so it works out like this: 12 Volts (Vsupply) - 1.2 Volts (LED Vf) = 10.8 Volts / .020 Amp (LED If) = 540 Ohms so my series resistor would be 540 Ohms for 12 Volt operation. Can you imagine what would happen placing a 4 K resistor in series with my LED?

Again, I would think about a current source like I linked to earlier or consider using AudioGurus suggestion of letting a transistor do the work. Your pot wiper drives the transistor and the transistor in turn does the work driving the LED.

Bedtime calls and I have a drive ahead of me in the morning.

Ron

carwhisperer said:

If I connect 12V across the pot then will the wiper wire put out positive voltage? So If I put in a fixed 4kohm resistor in series along with an LED and then connect to ground it should turn off at one end and work the way I want? This would be assuming that the pot can handle 5mA, correct?

What if I don't need it to turn off at one end? Do I still need to add a resistor to the circuit?

I plan to supply my own voltage to the pot. If it is intended for 5v and I send it 12v is that a problem in and of itself? I would think not as it would be too much amperage that would damage it, not voltage, correct?

Click to expand...

The pot will give a proportion of the voltage on its ends. So if you put one end at 0 V and one at 12 V, the wiper voltage with no load can be anywhere between 0 and 12 V, as long as the full mechanical range can be achieved.
An LED with a suitable resistor connected between the wiper and ground will see that 0 - 12 V. The wiper voltage will be affected by the LED current a bit. The pot has to be able to handle the full LED current plus the current through the pot.

You need the extra resistor whatever. That controls the current. Without it you will burn out the pot when the wiper goes to the 12 V end.

If you don't need it to turn off at one end, you could disconnect the end of the pot from 0 V. That would make the potentiometer into a variable resistor. The problem there is that with a 4.7 kΩ pot and a 4 kΩ series resistor, the LED current would only about halve when the pot was moved over its whole travel, which is may not give enough change in brightness.

It is the power that damages potentiometers, by heating up the track. The heating power is the current squared times the resistance, or the voltage squared divided by the resistance. On a potentiometer, you should not overheat any part of the track, so it is easiest to use the power rating of the pot to work out the maximum current that the whole track can take, and make sure no part of the track takes more than that maximum.

If it is intended for 5v and I send it 12v is that a problem in and of itself?

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It could be if the motor is itself applying a voltage to the pot. You haven't answered my question in post #4.

No I have not yet measured the voltage. My daughter has been visiting for Christmas and I haven't taken any time to look at it. I don't know how to get specifications for this pot since it is internal to the motor. If I disassemble it so that I can see it is there a way I can tell?

Please forgive me for not understanding everything being said. But would using an incandescent bulb perhaps be a better choice for this circuit than an LED?

An incandescent bulb would be a far worse choice than an LED. The bulb takes more current, and would not light at all with the current that you could get through the pot.

Alec's point is well made. If you can get 0 - 5 (or 0 - 12) V from the pot, then you can use that to control the LED.

Back to the beginning:

carwhisperer said:

I want to put an LED in an automotive circuit to "read" the position of a potentiometer that is internal to a 12v motor. The motor, which opens a valve in the climate control heater, has an internal 4.7 kOhm potentiometer which delivers a signal to the climate control ecu. I will no longer be sending that signal to the ecu but rather would like to use it to dim an led or other bulb. I would like it to run on 12v nominal, I would prefer not to have to reduce it to 5v. If I just hook the power supply up to 12v and run the ground through the motor's potentiometer will the led/bulb dim as resistance is increased? A relatively gradual dimming from open to close would be nice but even if it is much more abrupt that will still accomplish what I want.
Brian

Click to expand...

Here are a few things worth considering. The pot which is internal to the motor creates a signal based upon motor shaft position. That position is a function of the valve position driven by the motor. That pot output signal was considered at design time for the motor to drive the ECU be it a temperature ECU or whatever the signal drives a computer input. Automotive sensors generally run on 5.0 DC Regulated which is provided by the ECU. Things are done this way because automotive power (voltage) varies, heck. it could be between around 12.5 volts to 14.5 volts. Therefore the pot's output would vary with the pot's excitation voltage. Again, this is why the ECU or computer provides a regulated voltage to the sensors it reads. The analog inputs to the ECU are normally high impedance so the sensors driving them do not provide and current capability to speak of. All of that is considered at design time.

Here is the problem. At 12 VDC the current through a 4.7 K pot will be about 2.55mA with the pot's wiper open circuit, at 5 VDC that current is about 1.06mA again, wiper open circuit. These sensor pots are not designed to source current to drive a load. Trying to even drive a LED directly simply isn't going to happen without likely destroying the pot and an incandescent bulb is out of the question. You need to signal condition that pot's output. You can either use a voltage to current converter as I suggested earlier to drive a LED or use a transistor as AudioGuru suggested but you will, in my humble opinion need something for signal conditioning.

While I heven't a clue what your final objective is I do know this much. The pots output voltage will be a direct function of a valves position (closed to open). You want an indicator be it LED or Incandescent lamp to vary proportional to the pot's output. Rather than vary the intensity of a single indicator you may want to consider the use of a simple
LM3914 Dot/Bar Display Driver. Rather than a single LED you can have 10 LEDs that afford a good indication of valve position from 0 to 100% open or closed. The 3914 takes care of signal conditioning and has what it takes to directly drive either 10 individual LEDs or a small LED bar graph.

Ron

If the little pot tries to drive an incandescent light bulb then instead of the filament in the light bulb glowing at 2000 degrees C then maybe the resistive track of the pot will glow at 2000 degrees C?

audioguru said:

If the little pot tries to drive an incandescent light bulb then instead of the filament in the light bulb glowing at 2000 degrees C then maybe the resistive track of the pot will glow at 2000 degrees C?

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Yes, but only for a short time.

Ron

What we don't know yet is whether (a) the motor contains a purely passive pot intended to be interrogated by an ECU actively applying an external voltage to determine its setting, or (b) the motor itself applies a (possibly regulated) voltage to the pot so that the ECU can passively sense a voltage signal from the pot wiper.

alec_t said:

What we don't know yet is whether (a) the motor contains a purely passive pot intended to be interrogated by an ECU actively applying an external voltage to determine its setting, or (b) the motor itself applies a (possibly regulated) voltage to the pot so that the ECU can passively sense a voltage signal from the pot wiper.

Click to expand...

I agree and this would go much better if the original poster could better explain this:

I want to put an LED in an automotive circuit to "read" the position of a potentiometer that is internal to a 12v motor. The motor, which opens a valve in the climate control heater, has an internal 4.7 kOhm potentiometer which delivers a signal to the climate control ecu.

Click to expand...

Really don't know what the pot's actual excitation is or exactly what the signal to the climate control ECU is. Most of what I have posted is based purely on speculation. If the original poster returns with more information, including an answer to your question this would go much, much better.

Ron

Once again thank you for the replies. Here is a picture of the wiring diagram schematic of the motor. I don't know if that sheds any more light on the subject.

The array of LEDs is a great idea. However, space limitations likely will not permit that. I actually need to control 3 of these motors. One positions a drum to direct air at face/defrost/fan, one opens/closes the fresh air door and one opens the heater valve. So 3 reversible momentary switches and 3 indicator lights is what I'm shooting for. Then I need a rotary multi position switch for the fan and also a thermostat switch of some sort to cycle the AC compressor. If it helps any, this is for a 1995 Ferrari 348. I wanted to avoid mentioning that but WTH at this point. The climate control on these cars is problematic and one guy even resorted to adapting manual controls from a Miata. I think since the motors are already there and they have built in sensors I can use them as intended. However, I think it is prudent to have some sort of indicator so I don't sit there with my finger on the switch while the motor has hit it's stop.

The schematic is indeed helpful. Now we know that the pot is volts-free if disconnected from the ECU. We can work from there and come up with a solution for indicating wiper position. But apart from dimming a LED you will still need some way of switching the motor current off when either limit of travel is reached.

Here's one possible LED brightness control circuit, but it may not be suitable if you also intend using the same pot in a servo loop to control drum/valve position proportionally.
The three curves show current variation with pot wiper position and supply voltages of 12V, 13V, 14V.