1. I am modifying a cooling system in a machine I have. The fan is controlled using a voltage from a temp sensor circuit between 0v and 3v for full power. I have found a 3v power source on the board, so I wish to add a pot that will vary the voltage from 3v to a minimum value, of around half that, or less.

My thought was to connect both ends of the pot to the 3v rail, but put two (maybe three) diodes in series with one end. Then the middle connector to the fan circuit.
The voltage drop across the diodes would mean that one end of the pot would only be getting 1.6 (or less) volts. So I would be varying the voltage between the two values.

There is no drive current being taken to drive the fan, it is simply a reference voltage.

Would this work? or am i missing something? I am basically trying to get the pot to control the fan speed without it being able to be switched off (0v), or going from off to full speed in a fraction of a turn. But I am worried the diodes might affect the other circuitry that uses the 3v rail.

2.

3. I can't quite picture what you mean when you say you will connect both ends of the pot to the 3 volt rail. There wouldn't be any current flowing in a circuit with both ends at 3 volts.

Why not just put a fixed resistor in series with the potentiometer, equal in value to the potentiometer's resistance, and connect the wiper of the potentiometer to the fan control circuit? The voltage on the wiper of the pot would then vary between 3 and 1.5 v.

4. Yeah that is a much better way to do it. Doh!

ill show you what I meant though:

I thought the .7 voltage drop across the resistors would do the trick.

5. Your idea could work, but only if the resistance of the pot is very high compared to the input resistance of the fan controller, which is already high, as you state.

If the resistance of the pot is low relative to the controller, then the voltage divider equation will show that the voltage is always about 3 volts on the wiper of the pot, no matter where the potentiometer is set. Then your diodes would never be forward biased enough to conduct.

6. I did think it was a bit of an odd way to do it, which is why I was looking for advice. I think it was too early in the morning for electronics.
Thanks for the explanation Harold!

7. I think a more conventional circuit would look like this:

8. Leszek, yes that looks better.
I decided to just go with the resistors, but I'm still a bit stuck. Turns out that the controller actually is only sensitive to changes between 1v and 2v, so I need the pot brush to sway between these from the 3.28v supply.

I was going to use some 10K pots I have, so I was thinking, i do the first voltage divider to get 2v, then make another with the pot to sway between 1v and 2v:

will this work? but this is verging on too many components as I am doing this without using a circuit board, the circuit will be sitting in a plastic case with the pot sticking through it.

Any thoughts? Am I over complicating things? It has been a while since I have done voltage divider circuits.

9. Leszek, I read about the zener diodes. will this work, if i have understood correctly?

EDIT: I dont see any zeners with reverse breakdown voltages below a few volts, so i think im on a dead end with that.

10. Originally Posted by harvestein
Leszek, yes that looks better.
I decided to just go with the resistors, but I'm still a bit stuck. Turns out that the controller actually is only sensitive to changes between 1v and 2v, so I need the pot brush to sway between these from the 3.28v supply.

I was going to use some 10K pots I have, so I was thinking, i do the first voltage divider to get 2v, then make another with the pot to sway between 1v and 2v:

will this work? but this is verging on too many components as I am doing this without using a circuit board, the circuit will be sitting in a plastic case with the pot sticking through it.

Any thoughts? Am I over complicating things? It has been a while since I have done voltage divider circuits.
I don't know how you are doing your voltage divider calculations. With just the 6.1K and the 10K, the voltage between them would be 3.28*6.1/(10+6.1)=1.23. When you add in the 10K pot and the 10K fixed resistor, you have to calculate the parallel resistance of those two in series, and the 6.1K in parallel.
The way I would do it would be to have three 10K resistors in series, with the middle one being a pot. The wiper of the pot goes to the fan control input. That way, if you had 3 volts, it would be divided into thirds, so at each node you have 1 and 2 volts. The wiper of the pot would then swing between 1 and 2 volts.

11. I am using the wrong formula for the calculation. I was just using the mutiplier between the two resistances for some reason I'll never know. So I was doing 10/6.1 then dividing the voltage by the answer.
Haha, what a fool. Been that long I forgot the voltage divider equation.

If I remember right, and from the maths you did, it should be:

Vref= Vcc * (R1/(R1+R2))

Is that correct now?

Also regarding the way I did my circuit, I was trying to make one voltage divider feed the Vcc for the second one if you see what I mean (despite the values being horribly wrong). In any case I am just going to throw in the two 10k resistors and the pot, should be accurate enough.

Thanks again Harold, I knew I should have got my textbook out!

12. Originally Posted by harvestein
I am using the wrong formula for the calculation. I was just using the mutiplier between the two resistances for some reason I'll never know. So I was doing 10/6.1 then dividing the voltage by the answer.
Haha, what a fool. Been that long I forgot the voltage divider equation.

If I remember right, and from the maths you did, it should be:

Vref= Vcc * (R1/(R1+R2))

Is that correct now?
Yes, that's the voltage divider equation.
Also regarding the way I did my circuit, I was trying to make one voltage divider feed the Vcc for the second one if you see what I mean (despite the values being horribly wrong).
I see what you mean, but you forgot that the second voltage divider will affect the voltage in the first, because you have put in a resistance in parallel with your 6.1K resistor. The formula for parallel resistance is R(parallel) =R1R2/(R1+R2). So instead of having a voltage at the first node of Vcc*(6.1/16.1) you would have Vcc*Rparallel/(Rparallel+10).

13. Now I remember why I liked digital electronics so much.

 Bookmarks
##### Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement