Voltage Divider Mismatch

[bryan1]

G'day Guy's,
I'm on designing up a circuit for that pic32 module I have here and decided on using a voltage divider to measure the battery voltage. I put the AAA rechargeable pack on my smart charger and found the max charging voltage was 8 volts.

So went up to my shed and played around with some resistors to form the voltage divider. Measured the resistance with my fluke 865 DMM 6.806K - 14.873K works out to 2.185:1 using maths right.......

First I used a breadboard and got totally different results

14.8K - 6.8K is a 2.176 ratio used a single digit for resistance measurement

5.7 to 1.815 = 3.14:1
9 to 2.840 = 3.17:1
11.2 to 3.523 = 3.18:1
14.3 to 4.521 = 3.16:1

though how in the hell can the ratio be one full digit out so decided to solder up the resistors and do some measurements freestyle.

Resistors measured again with the fluke to 3 decimal places

6.806K and 14.873K works out to 2.185:1 on paper

14.40 input to 4.470 output = 3.220:1
13.33 input to 4.141 output = 3.219:1
10.96 input to 3.406 output = 3.217:1
9.64 input to 2.994 output = 3.219:1
7.74 input to 2.404 output = 3.219:1
4.14 input to 1.286 output = 3.219:1
2.80 input to 0.870 output = 3.219:1

So the breadboard wasn't the culprit so what doing maths on paper and real life tests show something totally different......

Got me totally confused now.....

Regards Bryan

 

[DerStrom8]

I may be misreading this, but what do you mean by "the ratio"? Voltage dividers don't work by the ratio of one resistor to the other. Let's take this scenario: 5v -- 10Ω -- 5Ω -- 0v, where the 10Ω resistor is R1 and the 5Ω resistor is R2.

The formula used to find the voltage at the node between the resistors is this:

Vo = Vin * R2/(R1+R2)

Therefore, using the previous values, you get:

Vo = 5v * 5Ω/(10Ω+5Ω)
Vo = 5v * 1/3
Vo = 1.66667v

The ratio of the two resistor values gives you 2:1, which is nowhere near the correct answer.

Matt
V

 

[dougy83]

The ratio of input voltage to output voltage is [LATEX]\frac{V_{out}}{V_{in}}=\frac{R_2}{R_1+R_2}[/LATEX]

 

[Mr RB]

I think Bryan did a real late night...

...
14.40 input to 4.470 output = 3.220:1

Another point is to allow more overhead if you are measuring a 12v system, they can easily go over 15v. As a general figure if the make the ADC max out at 20v DC input it gives you a lot of safety overhead to still get a reliable reading in fault conditions, and still allows enough ADC resolution for decent accuracy.