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?
I have seen a good calculator, but do you think I can find it now? (Errr, no.)
Is this any good to you?:
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.
Hey CtC. The ADC inputs have a very high impedance, so almost no current can flow into those pins from the middle of the resistor bridge. That means that you can treat the whole system as a single resistor, with a total resistance equal to the pot’s maximum (10kΩ).
In this case the current draw will be: 12V supply - Ir = 12V/10.3kΩ = 1.17mA 7.2V supply - Ir = 7.2V/10.3kΩ = 0.70mA
I am glad to hear my little trimmers will work with no smoke and very little current draw. This seems to be a lot simpler and tidy way of doing this. Oh, and this was the last step… I have coded and tested the following (just need to put them together)
