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help calculating voltage divider

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lokeycmos

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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
 
A bit more information would be useful.

What is the range of your DVM?

What voltage are you proposing to measure?
How accurate do you want the readings to be?

JimB
 
Don't forget to ensure your resistors are upto the voltage rating. The last thing you want is flash-over that will at least wreck your meter or worse cause serious injury to yourself.
 
I made a power supply for testing tank cpacitors for an am transmitter used for welding steel tube, the supply had a lopty, for the meter on the front panel I used a 'christmas tree', basically just a load of standard resistors in series, you divide the apllied volatge by the no. of resistors to get the volts per resistor, I think I used 20.
If you google for thevenin calculations you'll find online calculators for dividers, usually the divider impedance is 10 times less than the load, your meter is 10m so the divider ought to be 1m, 1m might be on the low side it'll pull the supply a little, if your sure of the meters impedance and that it wont change with volatge you might be able to just put one resistor in series with it.
 
The formula for voltage divider is Vout=VinR2/(R1+R2)...so for 100:1 ratio you need .01=R2/(R1+R2) and similarly for 1000:1 you need .001=R2/(R1+R2)
So solve for R1......R1=100R2-R2 for the 100:1 divider and R1=1000R2-R2 for the 1000:1 divider.
You can use the input impedance of the meter (9.1 M) as the R2 resistor (don't need to use a seperate resistor for R2). In this case the values required are as follows:
100:1 R1=900.9 M
1000:1 R1= 9090.9 M
This will give you the maximum input impedance (minimal loading on your circuit). These values may be obtained by wiring many resistors in series. This will also give you a higher breakdown voltage than using a single resistor.
To obtain more reasonal values of resistance, you can use a physical resistor for R2 but you must now use the parallel value of R2 and the input impedance of your meter in the above calculations. This will result in lower values for R1 but will also lower the input impedance (more loading on your circuit). If you make R2 << 9.1 M you can ignore the effects of the input impedance of the meter and just use the simple value of R2.
An example calculation using a value of 100K for R2:
since 100K << 9.1M we can use R2=100K so for 100:1 divider R1=9.9 M and for 1000:1 ration R1=99.9 M. Again you can use multiple lower value resistors to obtain the required value. For the above example you could use nine 1.0 M and one 900 K resistors in series. Be sure to enclose them in an insulative tube to prevent electrical shock.
I used this technique to measure the high voltage (35 kV) on a CRT tube from an old Color TV or Computer monitor.
 
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