Weird resistor voltage divider problem

RDL2004

Hi,

I am not an electronic genius, its just a hobby for me and I have run into a bit of strangeness that I don't understand.

I am planning to add a voltage display to a variable voltage power supply using an ADC0831 and a BS1. The power supply can go to 19.99 volts, so I knew I was going to need to re-scale the volts using a resistor divider.

I breadboarded up 4 equal ohms resistors (looking to get 25% of 20 volts), but the voltage I read was not correct between the 1 and other 3 (it was 1.66 volts with the power supply set at 5 volts), I tried just using 2 to get 1/2 the voltage, but that didn't read right either. I checked and verified that the resistors were within 5% (I plan to use 1% and trimmers in the final implementation). I checked the divided voltage with 3 different meters and basically got 3 different readings. Across the entire string of resistors (either 4 or 2) the voltage read correctly, it was only when I tried to pick off a divided voltage that it wouldn't read right.

Then I began to wonder if the problem was because I was using 10 meg resistors, so I tried again with 15K resistors and this time all the voltage readings were correct, both with 2 resistors to get 1/2 the total voltage and with 4 to get 1/4 of the total volts.

Is there a simple explanation why the 10 meg resistors didn't read right, or was I doing something wrong?

Thanks for any insight.

- Rick

steve_b

If you were measuring the same way in both instances then I'd guess there wasn't enough current to give you a reading across the resistors....????

figure that you're using 4x10megohm resistors....that's 40megs (hurray for math)...and at 5Volts that's only 0.125uA or there abouts.

When using voltage dividers, you need to know how much current you want to use as well and work backwards from there.

So, say you have a 10volt supply and only need 5volts...but you also need that 5volts to supply a chip that needs 100mA peak current. Uh oh....already having troubles with math....someone will have to jump in and bail me out I think!!

Well...start with this....if you are 'halfing' the supply voltage, we know that that requires 2 equal value resistors in series so that the sum of the voltage drops across the resistors should equal the supply.
So...lets plan for 200mA in this chain to be sure we didn't skimp on the math...
so 10Volts / 200mA is 50ohms. So your 2 voltage divider resistors would be 25ohms and 25ohms.
NOW....something else to be wary of is the power you're drawing....
Power(watts)=Voltage * Current = 10V * 200mA = 2Watts. So be sure you have the appropriate rated resistors!

Anyhow....most Ics don't require that much current....but this was just an example!

Hopefully Paul or some of the others can smack me if I'm wrong!! haha Been years since I've had to do the calculations! To be honest....I put the stuff in and if it gets hot...I replace it with a higher wattage one!! lol

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Steve

"Inside each and every one of us is our one, true authentic swing. Something we was born with. Something that's ours and ours alone. Something that can't be learned... something that's got to be remembered."

KenM

RDL2004 said...

Then I began to wonder if the problem was because I was using 10 meg resistors, so I tried again with 15K resistors and this time all the voltage readings were correct, both with 2 resistors to get 1/2 the total voltage and with 4 to get 1/4 of the total volts.

Is there a simple explanation why the 10 meg resistors didn't read right, or was I doing something wrong?

Thanks for any insight.

- Rick

If your are using your typical digital volt meter, I believe the reason for your results are·from the fact that when·you connect your volt meter to the circuit, the circuit "sees" your meter as about a·10 Meg resistor.

Your volt meter (like a 10 Meg resistor)·in parallel·with 15k ohms has little effect on the 15k. ( I assume you know about resistors in parallel).

Now your meter (~10 Meg)·in parallel with 10 Meg will have a large effect on the 10 Meg resistor.

Does this make sense to you?

Steve also brings up an excellent point about the very low current in the divider circuit you created with 10 Meg resistors.· Without any calculations if I wanted to make a divider quick and off the cuff I would certainly start with values much lower than 10 Meg, likely anything close 10K to 20K area.


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Ken

Post Edited (KenM) : 10/14/2005 5:42:28 AM GMT

KenM

To take this a step further, you can prove to yourself the digital voltmeter loads the circuit.

If you build the simple circuit the voltage at A will be about 5 volts, and the voltage at B will be about 2.5 volts (assuming your meter is similar to the dozens of others out there).

Ohms law does not lie.


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Ken

Post Edited (KenM) : 10/14/2005 6:06:00 AM GMT

RDL2004

Thanks guys. I see I was just forgetting to consider the effect of the meter on the circuit. Ken's explanation makes perfect sense. It was late at night and I think I was going brain dead. Steve has a good point about the low current also. I wasn't really going to use 10 megs in the final project, those were just the first ones I grabbed out of my tub of resistors for the initial prototype on the breadboard.

Thanks again

- Rick

steve_b

Depending on what your use is for the 'dropped down' voltage, I'd suggest getting the appropriate regulator to do the job. That way you don't have to worry about the voltage divider moving up and down with the power supply where the regulator will just keep outputting the wanted voltage so long as it sees an input slightly larger than what it's supposed to output.

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Steve

"Inside each and every one of us is our one, true authentic swing. Something we was born with. Something that's ours and ours alone. Something that can't be learned... something that's got to be remembered."