hello! trying to increase the range of my Digital volt meter. looking for some help with the math. if you look at the attached pic i have a voltage divider. the impedance of the meter itself is 9.1Mohms. i just need some help calculating the values i need. i would like to load the circuit down as little as possible. could someone show me the math to find both a 100:1 and a 1000:1 ratio? thank you in advance and let me know if you have any questions

 

Are you trying to measure voltages HIGHER than your meter input range?
When you put a resistor (R2) across the meter your lowering its input resistance.

So a 10 meg resistor in parallel with your meter means r2 * meter/R2 + meter

so your meter will now have an input resistance of 10x9.1/19.1 = 4.764Meg


100:1 would be R1= 471.68Meg ohms , R2=10 Meg ohms
1000:1 would be R1 = 4759.6 Meg ohms , R2 = 10 Meg ohms

 

Are you trying to measure voltages HIGHER than your meter input range?
When you put a resistor (R2) across the meter your lowering its input resistance.

So a 10 meg resistor in parallel with your meter means r2 * meter/R2 + meter

so your meter will now have an input resistance of 10x9.1/19.1 = 4.764Meg


100:1 would be R1= 471.68Meg ohms , R2=10 Meg ohms
1000:1 would be R1 = 4759.6 Meg ohms , R2 = 10 Meg ohms

yes, increasing its max. could you please elaborate about the R2? how does the circuit function with and without it? is it necesarry? do commercial probes use an r2?

since this is a resistive divider, will it work equally as well with AC?

 

You need to create a voltage divider, but by putting your meter across R2 effectively lowers its value, so you need to calculate using parallel resistors formula.

so find out what the combined resistance will be, the multiply by your ratio and work out for R1, which i did for you already.

 

could i use a pot for R2 as a calibration/accuracy adjustment?

since this divider is resistive, will it work equally well for AC?

 

Put a 90MΩ resistor in front of it. A DVM is typically 10MΩ input impedance. It would probably suffer on accuracy (though attention to detail could make a difference) but it would be a X10 reduction in scale and a X10 increase in impedance.

For a X100 reduction in scale and increase in resistance use 990MΩ.

R2 is built into the meter itself, it is the 10 MΩ I mentioned.

You can use 5% resistors and tweak them using parallel and series combinations to higher accuracy. If you are thinking of using this for high voltage you had better have some very good potting compound, and they tend to be slightly conductive when wet (a bad thing).

 

Put a 90MΩ resistor in front of it. A DVM is typically 10MΩ input impedance. It would probably suffer on accuracy (though attention to detail could make a difference) but it would be a X10 reduction in scale and a X10 increase in impedance.

For a X100 reduction in scale and increase in resistance use 990MΩ.

R2 is built into the meter itself, it is the 10 MΩ I mentioned.

You can use 5% resistors and tweak them using parallel and series combinations to higher accuracy. If you are thinking of using this for high voltage you had better have some very good potting compound, and they tend to be slightly conductive when wet (a bad thing).

thanks for the reply bill. could you dumb it down a little more for me? how did you get 990Mohms where dodgydave got 471.68Mohms

 

You don't need the resistor R2, just use R1 and the values will be

100:1 = 900.9 Meg
1000:1 =9090.9 Meg


your Dvm is 9.1meg resistance, so
multiply the dvm resistance by 99 for the 100:1 scale
and multiply by 999 for the 1000:1 scale.

( my previous calculations were based on using both R1 and R2)

 

I think you need to redo your math. It is a really simple voltage divider equation. Going with the original drawing, and figuring R2 = 10MΩ, the meter has no impedance (R2 is the meters impedance). I'm not sure where you got the 9.1MΩ, 10MΩ is a pretty standard value for most meters.

Then:

(1/10) Vcc = (R2 * Vcc) / (R1 + R2)

Vcc/10 = (R2 * Vcc) / (R1 + R2)

(Vcc R1 + Vcc R2) / 10 = Vcc R2

Vcc R1 + Vcc R2 = 10 Vcc R2

Vcc R1 = 9 Vcc R2

R1 = 9 R2

If R2 = 10MΩ then R1 = 90MΩ

The other variation:

Vcc R1 = 99 Vcc R2 or R1 = 990MΩ

I did this in my head first time around, voltage dividers are pretty bread and butter. It doesn't really matter in the long run, it will be hard to see 1% error on this.

 

Assuming that your DVM will also measure current and resistance, an alternate approach is to get some high resistance resistors and put them in series with the DVM, which is switched to measure amps. You can use the DVM to measure their resistance. Use resistors that are as high as possible while still maintaining the full resolution of the DVM. If necessary put several in series after measuring each separately. In this way you can maintain most of the meter's accuracy without having to buy a precision high resistance resistor.

The resistance of the DVM when measuring current should be inconsequential relative to the big resistors. With the resistance and the current you can calculate the measured voltage.

I've used this technique to measure voltages in the 15KV range with a meter that would only go to 1KV. It's been a while but I think that I had a string of 10MΩ resistors to minimize the loading on the high voltage supply and any resistive heating due to the current. The meter was set at the 2 milliamp or 200 microamp range.

However high voltages can be lethal. The resistors needs to be on the high voltage side and the meter on the ground side of whatever you're measuring. You should connect it to the circuit before powering it up. This is not something to be disconnected while the high voltage is present.

 

I like that.
Take a 1GΩ resistor in series with the ammeter set to 200μA.
10kV into 1GΩ will give a reading of 10μA.

 

yes, increasing its max. could you please elaborate about the R2? how does the circuit function with and without it? is it necesarry? do commercial probes use an r2?

since this is a resistive divider, will it work equally as well with AC?

What do you mean by AC? As in, what frequencies are you talking about? If sufficiently low, then it should work okay. At higher frequencies, the stray capacitance will mean that you need to compensate it. But if you are using this with a DVM, I suspect you are at low enough frequencies for it not to matter too much.