Wiring pot as voltage divider across polarities

[BobK]

A voltage control circuit does not "output a current". How are you measuring this? Are you putting an ammeter between the signal and ground? If so, do not do this, you are likely to destroy something. An ammeter is essentially a short circuit.

The input using the control circuit will draw only as much current as it needs, independent of how current the voltage source can supply. It is typically in the microamp region.

Bob

 

[guskenny83]

A voltage control circuit does not "output a current". How are you measuring this? Are you putting an ammeter between the signal and ground? If so, do not do this, you are likely to destroy something. An ammeter is essentially a short circuit.

The input using the control circuit will draw only as much current as it needs, independent of how current the voltage source can supply. It is typically in the microamp region.

Bob

Thanks for your reply.

So if i have a 7805 circuit running to a pot which acts as a voltage divider, that will be fine to use as a control voltage signal then?

Thanks for your advice, my background is in mechanical engineering, so my knowledge of electonics isnt very great. I sometimes get stuck on the whole "electrons flowing through a pipe like water" (ie. voltage/pressure, current/flow-rate) analogy, which i know isnt very accurate, but fluid mechanics makes a lot more sense to me. Hehe.

Thanks again. This site has been invaluable to me!

 

[guskenny83]

Thanks for your reply.

So if i have a 7805 circuit running to a pot which acts as a voltage divider, that will be fine to use as a control voltage signal then?

Thanks for your advice, my background is in mechanical engineering, so my knowledge of electonics isnt very great. I sometimes get stuck on the whole "electrons flowing through a pipe like water" (ie. voltage/pressure, current/flow-rate) analogy, which i know isnt very accurate, but fluid mechanics makes a lot more sense to me. Hehe.

Thanks again. This site has been invaluable to me!

One final question.

If i was to want to add two buttons, that when pressed, one of which gives a connection to ground, and the other gives a connection to the 5vDC. Can i just wire them up directly to a jack with a 10k resistor connected from the jack side of the switch to 5vDC (for the ground button) and to ground (for the 5vDC button) to act as pull-up/pull-down resistors respectively?

Or is there something else that i should do as well?

 

[BobK]

What you describe is two buttons, which of which will normally output 5V and can be pulled to ground when the button is pushed. The other would normally output ground, and would be pulled to 5V when pushed. These are suitable to connect to an input that does not draw much current. Is that what you want?

Bob

 

[duke37]

If you reduce the current, the voltage will reduce.
The gubbins will take the current it requires. Make sure the voltage does not exceed what it can cope with.

 

[Martaine2005]

Can I ask a stupid question here?
I cannot get my head around a VR having 0V in the centre.
If it's 520k VR , wouldn't the centre point for the wiper be 260k?
Sorry, that really confuses me. An explanation for an idiot would be good.

Martin

 

[guskenny83]

Hi martin, the way i understand it is that by connecting the pot as a voltage divider (not variable resistor) across -5/+5vDC instead of GND/5vDC, you are able to get 0vdc in the centre because that is the midpoint between the two voltages. All the way to one side of the pot you will get pure -5vDC and all the way to the other side you will get +5vDC and anywhere between those will give you a difference in potential.

For example, if it was a linear pot like this:

-5vDC [--------|---------------------------] +5VDC

Connecting something to the wiper terminal would give you around -2.5vDC

Whereas the same configuration connected across GND/5vDC:

GND [--------|---------------------------] +5VDC

Would give you around +1.25vDC

That is how i understand it to work anyway (please correct me if im wrong, anyone!)

I hope that was helpful to you

 

[guskenny83]

What you describe is two buttons, which of which will normally output 5V and can be pulled to ground when the button is pushed. The other would normally output ground, and would be pulled to 5V when pushed. These are suitable to connect to an input that does not draw much current. Is that what you want?

Bob

Thats exactly what I wanted to know. Thank you.

 

[Martaine2005]

Hi martin, the way i understand it is that by connecting the pot as a voltage divider (not variable resistor) across -5/+5vDC instead of GND/5vDC, you are able to get 0vdc in the centre because that is the midpoint between the two voltages. All the way to one side of the pot you will get pure -5vDC and all the way to the other side you will get +5vDC and anywhere between those will give you a difference in potential.

For example, if it was a linear pot like this:

-5vDC [--------|---------------------------] +5VDC

Connecting something to the wiper terminal would give you around -2.5vDC

Whereas the same configuration connected across GND/5vDC:

GND [--------|---------------------------] +5VDC

Would give you around +1.25vDC

That is how i understand it to work anyway (please correct me if im wrong, anyone!)

I hope that was helpful to you

Thanks, that makes perfect sense.
I notice that Bob explained the voltage divider earlier. I missed that bit.

Martin