Device to act as 0 - 9V source

[Twafs]

Hi
I want to have a portable device that will output 0-9V.
Source is a 9V battery.
I am trying to get this to work with a 10k Ohm potentiometer as a voltage divider.

I currently have it wired as this:
**broken link removed**

But only with the potmeter turned 20% the voltage drops to 0, and i want to have the full range on the potmeter to get it accurate.
Also the output voltage doesnt care if the potmeter ground is connected to the battery, so im guessing its used as a regular resistor and not voltage divider?

Do i need a smaller value potmeter or is there something else wrong?
Formula for calculating the right potmeter?

The voltmeter is wired into the ciruit as i want to be able to read what im putting out, could this cheap 3-digit voltmeter be the problem?(too low internal resistance?)

Thanks

 

[pinout]

Can you confirm what the voltmeter is reading in comparison with the DVM read out, say at maximum, 50% and when the DVM reads zero.

Assuming they read similar then the DVM must have a low input impedance as you suggest, but I would not expect it to be so low or actually read zero.

What is the DVM voltage range and the resolution, i.e. what does a "1" in the least significant position represent?

 

[chuckey]

If the pot is working OK, the voltage would only drop to zero when the pot is at one end. it sounds if your pot is damaged. A lower value pot would be better, also make sure its a linear one (not log or anto log).
Frank

 

[barry]

This is a terrible idea(sorry). This will only work if your load is very high impedance, otherwise, any current draw will cause a voltage drop across the pot.

 

[shatruddha]

I agree with barry!! If not a good idea to use potential divider. But I'll correct him here, this potential divider will work if your load is having low impedance than the potentiometer, otherwise most of the energy will flow through the potentiometer resistor and that would be waste of energy.
Since here you are asking about something portable, that surely means that it should have optimum usage of energy, here in your case battery. So rather than going with potentiometer, I would suggest you to design a small buck converter. Its pretty easy, will certainly use a bit more number of components than your resister divider, but that will be more efficient.

 

[barry]

But I'll correct him here, this potential divider will work if your load is having low impedance than the potentiometer, otherwise most of the energy will flow through the potentiometer resistor and that would be waste of energy.

At the risk of sounding like a flamer, NO! NO!! NO!!!.

If the load is low impedance you will drop nearly ALL your voltage across the pot. This is fundamental electronics!! Let's say the pot is set at 50%. With no load, the output will be 4.5 volts. Great. Now say you've got a 100 ohm load. That means you've got a voltage divider with 5K on the top, and approximately 100 ohms on the bottom. Your output voltage will be 0.17 volts.

 

[Twafs]

Can you confirm what the voltmeter is reading in comparison with the DVM read out, say at maximum, 50% and when the DVM reads zero.

Assuming they read similar then the DVM must have a low input impedance as you suggest, but I would not expect it to be so low or actually read zero.

What is the DVM voltage range and the resolution, i.e. what does a "1" in the least significant position represent?

The voltmeter wired into the circuit is a little 3 digit 2 wire voltmeter, range from 2,5V-30V. Under 10V it shows 2 decimal points and over 10V it shows 1 decimal point.
And it seems pretty accurate, measuring the new 9V battery gives me a reading of 9,14V.
So when the pot is turned to one side it measures 9,14V but it hits 2,5V when just turning it somewhere around 20% of the way, and i get this result no matter if the ground on the potmeter is attached or not, like i was only using just the wiper and one side of the pot.

If the pot is working OK, the voltage would only drop to zero when the pot is at one end. it sounds if your pot is damaged. A lower value pot would be better, also make sure its a linear one (not log or anto log).
Frank

The pot is new so it should be ok

This is a terrible idea(sorry). This will only work if your load is very high impedance, otherwise, any current draw will cause a voltage drop across the pot.

It is for testing equippment designed to operate on a 0-10V signal, so i assume the load would be a high impedance.

I agree with barry!! If not a good idea to use potential divider. But I'll correct him here, this potential divider will work if your load is having low impedance than the potentiometer, otherwise most of the energy will flow through the potentiometer resistor and that would be waste of energy.
Since here you are asking about something portable, that surely means that it should have optimum usage of energy, here in your case battery. So rather than going with potentiometer, I would suggest you to design a small buck converter. Its pretty easy, will certainly use a bit more number of components than your resister divider, but that will be more efficient.

What about a transistor circuit? As its for testing 0-10V equipment id be happy if it didnt lower the voltage much more than it already is.
Long batterylife isnt that necessary, but helps...

Any other method of making this device is appreciated

Thanks for quick and helpful answers !

 

[barry]

Okay, I'll just ask the question: What is your load? This will determine the answer.

 

[pinout]

Try and repeat without the load. This will show the potentiometer characteristics.

So when the pot is turned to one side it measures 9,14V but it hits 2,5V when just turning it somewhere around 20% of the way, and i get this result no matter if the ground on the potmeter is attached or not, like i was only using just the wiper and one side of the pot.

Good reading at 9V i.e. potentiometer is effectively out of circuit.
At 20% jumping to 2.5V indicates something in your load is changing at that point.
So the top of the potentiometer is acting like a series resistor as confirmed by the ground removal observation.

So as per previous poster, what is your actual load?

 

[Twafs]

.

.

I have never connected it to any load, only the voltmeter to test if the potmeter was working as a voltage divider, which it dont...

 

[barry]

I have never connected it to any load, only the voltmeter to test if the potmeter was working as a voltage divider, which it dont...

Then it's not connected as shown, or the pot is bad.

BTW, you're not going to be able to supply 0-10V out of a 9V battery....

And just because your pot is new, doesn't mean it can't be damaged.

 

[pinout]

Yes, I agree with Barry the pot is bad.

Is it possible you confused the wiper with a track end?
Applying volts to the wiper and turning the pot down will head towards zero ohms and high(ish) currents burning out the end of the track.
Then it will behave just like a series resistance as if the ground connection wasn't there!
It would have also provided the 9V at the max rotation the other way......

Looking at your drawing could the battery go from gnd to the wiper and
The output to the meter go from gnd to the other end of the pot?

 

[Twafs]

Yes, I agree with Barry the pot is bad.

Is it possible you confused the wiper with a track end?
Applying volts to the wiper and turning the pot down will head towards zero ohms and high(ish) currents burning out the end of the track.
Then it will behave just like a series resistance as if the ground connection wasn't there!
It would have also provided the 9V at the max rotation the other way......

Looking at your drawing could the battery go from gnd to the wiper and
The output to the meter go from gnd to the other end of the pot?

0-10v or 0-9v, whatever, good enough.

Physical wiring atm:
Switch hooked up between battery + and potmeter +, temporary connections made with those keyring thingies.
**broken link removed**

Note the little difference of the potmeter and the reading of the voltmeter
**broken link removed**

Sorry for upsidedown photo, stupid phone-.-

 

[barry]

That's not a great picture, but it sure looks like you've got the battery connected to the wiper.

See post #11.

 

[ads-ee]

The labels on the "pins" that are on the red circuit board are labeled S + - so if those are correct it's hooked up to the battery correctly.

I'd disconnect the potentiometer and check using the "tabs" at the top of the potentiometer and measure the resistance across the outer pair with a multimeter.

 

[barry]

What did you mean when you said the voltmeter range is 2.5-30V? Can it measure below 2.5V? And have you tried using a DVM to verify the readings from your little meter?

 

[Audioguru]

It is obvious that the pot is a logarithmic volume control, not a linear pot.
Without a load when it is turned to half then its output voltage is 0.45V to 0.9V. If it is turned down 20% then its output voltage might be 2.5V.

 

[barry]

It is obvious that the pot is a logarithmic volume control, not a linear pot.
Without a load when it is turned to half then its output voltage is 0.45V to 0.9V. If it is turned down 20% then its output voltage might be 2.5V.

You're probably right.

 

[Venkadesh_M]

not only a logarithmic pot but also in linear pot if the potentiometer resistance is high and if the load resistance is low the same non linearity will arise ..... try a pot with lower resistance if your source doesn't support try a voltage regulator (output voltage controllable), if you still thought it was inefficient you can use a switched regulator (If you want linearity)..

In all the images the pot is in center position according to the load the output voltage varies.....
For better operation the Potentiometer resistance shd be lower than load resistance.....

 

[shatruddha]

Hi barry!!
sorry to misread you, I was thinking something else.

BTW for Pot based circuit, I think your device will turn up more into a device dependent voltage as well, since your device impedance will matter a lot when you are setting a voltage.
I would still go on with a buck converter, making one is pretty easy, you can just google it and you'll find some really easy solutions. You just need a capacitor, inductor, a mosfet and a comparator. And of-course a pwm signal