Resistor/Voltage calculation Puzzle/woes.

[Brian Guralnick]

Ok, here is a simple 2 resistor series circuit.

(1024v) -> R1 <-(Vprobe)-> R2 <- (0V).

Now, to calculate voltage at (Vprobe), I use the formula:

1024 * ( R2 / ( R1 + R2) ).

Now, if I know thw voltage at (Vprobe), & I know the value of R1, however, R2 is now the unknown value. How do I calculate the unknown R2?

Example value 1:
R1 = 47000 Ohm
Vprobe = 200v
R2 = ???? Ohm

Example value 2:
R1 = 47000 Ohm
Vprobe = 900v
R2 = ???? Ohm

 

[richie00boy]

As you know the supply voltage and Vprobe you can work out the voltage across R1, and as you know the resistance of R1 you can then work out the current.

I = (1024 - Vprobe) / R1

Now this same current flows down the whole circuit. As Vprobe is the potential across R2 and you know the current, the resistance of R2 is simply:

R2 = Vprobe / I

 

[Brian Guralnick]

Thanks richie00boy. When I think about it in terms of current consumption, things get too simple. I don't know why I didn't think of it that way the first time around. It was like I was treating the whole thing as a math problem to be solved & not a simple circuit.

 

[Enzo]

Or calculate the thing:

200v is about .195 of 1024v, so that is the voltage divider ratio.

for clarity I leave out the "k" but assume the 47 means 47k and the X is in k ohms.

X/(X + 47) = .195

solving

X = .195X + 9.17

or

.805X = 9.17

which makes

X = 11.4k within the accuracy of my numbers.

Using the current method suggested above I got about 11.1k. Pretty close. They should yield the same result, but the differences are from rounding in the process. With your numbers we really only have a two digit resolution anyway.

Basic algebra.

 

[HFGuy]

use KCL !!!!!!!!!!!!



(Vprobe - 1024) (Vprobe - 0)
------------------- + ----------------- = 0
R1 R2

 

[HFGuy]

Or with the right formatting


(Vprobe - 1024)/ R1 + (Vprobe - 0)/R2 = 0

 

[AndrewT]

Hi Brian,
Enzo got it right.
R2= Vprobe*R1/(V-Vprobe)=200*47/(1024-200)=11k408.
You need a small correction for the input impedance of the probe. Note that 11k << 1M0/10
For a 1M0 probe the corrected value of R2=11k54.

The answer is so close, the correction does not matter and with a 10M probe even less so.

For your 900V reading the probe correction becomes very important.
uncorrected R2=341K13 Note 341K>1M0/10
corrected R2=517k75 for 1M0 probe, corrected R2=353k18 for 10M probe.

how to do the correction for 1M0 probe:-

corrected R2= uncR2*probeR/probeR-uncR2 (this is parallel resistors)
eg. corrR2=341.13*1000/1000-341.13=517.75 all values in kohms.

If the R2 value is less than 10% of probe value do not apply correction, if R2 value is more than 10% of probe impedance apply correction formula.

 

[jackinnj]

one thing about resistors and high voltages (over about 400V) some (carbon comps) become non-linear.

I don't know your setup, but assuming you are using an opamp to measure a voltage -- it is helpful to buffer before you amplify -- AND protect the input with clamp diodes.

 

[Enzo]

And for that matter film resistors have voltage ratings too. They are made as a spiral conductive traces wrapped around an insulating core. If two adjacent turns have enough voltage difference between them, they can arc. The space is tiny. I suppose it is a helical rather than spiral trace.

Yes, my calculations assumed an ideal infinite-impedance measuring instrument. Just earlier today I was helping someone fathom out a problem and his inconsisyeny voltage readings were due to his meter impedance loading the circuit.

ANd a caution about accuracy: We cannot extend calculations to results like 11.23456k. even if the numbers come to that. The initial values were 1024 volts, 200 volts and 47k. 47k is only 2 significant digits, so any results will have only an accuracy of 2 significant digits. That means all we can state with confidence is that it will be 11k, or round up to 12k if warranted.

 

[AndrewT]

Hi,
I used 11k408 so that the enquirer could reproduce the calculations himself, NOT to imply super accuracy. Similarly the correction, also quoted to 5 sig fig was given to show the futility of correcting for probe impedance if the probe is well above the impedance being measured.

This was confirmed by my last advice to ignore the correction if .....

If the R2 value is less than 10% of probe value do not apply correction, if R2 value is more than 10% of probe impedance apply correction formula.

This situation applies to the 900V test reading if using a one meg probe.