# Thread: How to build a circuit with decaying or increasing resistance?

1. ## How to build a circuit with decaying or increasing resistance?

I use a servo tester to operate a folding screen mechanism in my car.

On the outside is a potentiometer that feeds a signal to a chip inside that generates the signal to the servo that tells it to move in one direction or the other.

Problem: Currently, I have the potentiometer set to 'full open'. When I start the computer, the screen opens and everything is fine. When I shut the car off, I have to manually close the screen, which backdrives the servo.

What I'd like: I'd like to measure the resistance in the full open and full closed position and build a circuit that will either increase or decrease resistance slowly, perhaps over a period of 3 seconds or so.

My question: This seems like something simple to do but I don't quite have the electronics knowledge to figure it out. Can anyone either A) tell me the answer; B) give me the name of the type of circuit to research?

I know how to solder, use a multimeter and can create boards and follow diagrams. I also know the difference between resistors and capacitors and so forth. I just need a push in the right direction.

I'd appreciate any help.

2. Perhaps a digital potentiometer, eg : http://datasheets.maxim-ic.com/en/ds/MAX5402.pdf

There are many more out there.

3. You might want to figure out more about the servo tester and what it's doing first. Like say the pot is connected to 5V and ground, with the sweep (center pin) connected to an input that converts the value, with a low current draw. If you connected an electrolitic capacitor between the sweep and ground it would create a delay. Wait, it needs a resistor between the sweep and cap. Now turn it either way and it will take more time to reach the correct voltage, slowing the servo.

4. Originally Posted by Rob Withey
Perhaps a digital potentiometer, eg : http://datasheets.maxim-ic.com/en/ds/MAX5402.pdf

There are many more out there.
I can't tell, but it looks like this is just a digital version of the analog pot. Is there a way to control the speed that it changes resistance?

5. Originally Posted by Curiosity
You might want to figure out more about the servo tester and what it's doing first. Like say the pot is connected to 5V and ground, with the sweep (center pin) connected to an input that converts the value, with a low current draw. If you connected an electrolitic capacitor between the sweep and ground it would create a delay. Wait, it needs a resistor between the sweep and cap. Now turn it either way and it will take more time to reach the correct voltage, slowing the servo.
I'll take a look at that. I suspect this is how it works. I've looked inside it before and it is dead simple.

6. Originally Posted by Bugbyte
I can't tell, but it looks like this is just a digital version of the analog pot. Is there a way to control the speed that it changes resistance?
Yes, a PIC or similar. I thought that bit was obvious so I didn't mention it.

7. Here's the idea. The top is probably close to how it works now. The bottom is the idea. The relay would connect to ACC or you could replace it with a switch. When connection is made it will slowly increase to full. Switching off will decrease. I'm not sure of the values though. For the same speed on/off, the top would be 1/4 the resistance of the bottom one. Is that right? Well, maybe the bottom resistor should connect to NC. That would be easier. They could both be the same.

8. Awesome! This looks like it might work. I'll post a couple of photos of the servo controller innards later today and if it looks right, I'll try this circuit out.

9. Okay, it looks like the circuit Curiosity shows will work if the right values are put on it.

Here's a photo of the controller innards:

A closer image of just the circuitry:
do=displayimg&imgid=7071[/IMG]
I can only faintly make out the number on the PIC but it *looks* like 12C6752 where Curiosity shows a 12F675. That sounds like the same or something similar, does that sound right?

The black button is a battery eliminator circuit test button and is not necessary for operation.

And the underside of the board (the pot is at the bottom of the photo):

The capacitor on the board is a 10mf 35v

Could I somehow insert the circuit Curiosity suggested in place of the potentiometer and switch it with a relay? How would I figure out the values for the cap and the resistor?

10. All the PIC12 chips have the same pinout. I just guessed that's what it was.

Ok, this is better. Both resistors would be the same.

It's going to have a curve though. It slows down as it reaches full charge. Then reverse draining. Which might look kinda cool.
Here's a calculator. I punched in 5V, 8K, 100uF, 4000ms and got close to full.

http://www.csgnetwork.com/rctimecalc2.html
or this
http://ourworld.compuserve.com/homep..._Bowden/rc.htm

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