Let's Make Robots!

Calculating Resistor Bridge

***Update***

Well, I messed around with a lot of resistor combinations, and got really frusterated with sorting through my resistor box looking for all these different values. Instead, I grabbed a pot. I got this one --its a 15 turn, 10k trimmer. I did some preliminary tests and I would love if someone could check my math and calculate the current it is going to draw. Following is the values read with a measure thingie on either "side" of the center pin of the pot.

At 12v:

Between + and center 6.35k

Between - and center 3.98k

At 7.2v:

Between + and center 4.25k

Between - and center 6.06k

Thanks in advance

 

I would like to monitor my onboard battery packs. 2 are 7.2v packs, each with a 5v regulator going to my processors. The third is my 12v SLA drive battery. I need to make a few resistor bridges to bring my juice (pre-regulator) down to 5v when they are fully charged. Obviously, the center of these bridges each going into an ADC. I just can't seem to find a good resistor bridge caculator online. Anyone want to do some math on what I need or have a link for a good fill-in-the-blanks resistor bridge calculator?

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Have some 12 number of 1KΩ resistors in series, so that you can get 1V across each resistor.  You need 5V, so you can pull it out across 5 resistors.

(I'm an amateur,  and only few basics like this.  If its wrong, please inform me about the correct one).

I have seen a good calculator, but do you think I can find it now? (Errr, no.)

Is this any good to you?:

voltageDivider.png

If you choose R2 to be <= 10k, which should suit any micro's adc input, then calculate R1 and choose the next largest standard resistor value (to make sure Vout is less than 5V), you should be right.

Anyone want to check my maths? Don't want to fry Chris's robot!

P.S. Alternatively, maybe use a 10k trimmer and adjust to get 5V off a fully charged battery.

Yep, everything is in order here. If you use 10kΩ for R2 then you'll need a resistor >=14kΩ for R1. The current draw of the bridge at 12V will be less than 0.5mA.

Likewise for the 7.2V batteries, if you choose R2 = 10kΩ again, you'll need a resistor >=4.4kΩ for R1. The current draw of the bridge at 7.2V will again be less than 0.5mA.

Thanks for checking that, TeleFox. I didn't think of the bridge current draw.
No problem. Most of my projects are small and run on low power, so current consumption is always on my mind =)

Ohm's law is your friend: http://en.wikipedia.org/wiki/Voltage_divider

 

Hooray! I pass my algebra exam!  :D

Ohm's law is your friend: http://en.wikipedia.org/wiki/Voltage_divider

 

Thanks folks,

I love this site!

Update