Electronics > Beginners
Voltage divider question, what R to use
lpaseen:
I'm need to measure 0-14V with a MCP3424 ADC with as much precision as the wallet allows. At first I was thinking that I don't want to load what I measure to much so I put in 5.8M and 1M and started testing. Soon I realized the obvious, the ADC has an input impedance that causes a big error in the basic math. 2.5M parallel with 1M = 714285 ohm and it gets way worse if I use the PGA, 2.5M/8 makes my lower R closer to 238K. Putting a DVM or a scope on the lower R also impacts the over all current and with that the voltage over the lower resistor.
Ok, now when I know that I can adjust the math in the app but I still have the issue of whatever this is the right path. Maybe I should just go with some lower values, maybe 220K and 33K. I still need to compensate for the ADC load and so on but maybe it's some other reason to do one way or the other, reasons like that it's easier/cheaper to get 0.1% resistors at 220K than 5.8M.
Sidenote, I found out that what this contraption is going to measure will be the voltage on the different power rails on a PC motherboard so the load it creates should not be a problem.
Nerull:
The MCP3424 wants a low impedence input. It measures voltage by charging a capacitor and timing how long it takes to discharge, the source impedance needs to allow this in a reasonable time.
beanflying:
Resistive divider unless you use very low value resistors are going to load the ADC. The Datasheet on a quick glance shows this. they actually mentioned ideally Zero.
Consider an op-amp buffered input instead of just a resistor divider. Depending on the 'accuracy' you need look seriously at the Temp Co of the resistors instead of their % accuracy and trim appropriately with a trimpot this will get you a more stable input again over a range of environmental conditions.
Lots of reading here http://www.ti.com/lit/an/sloa098/sloa098.pdf
Zero999:
The problem is the voltage on the potential divider changes, when a sample is taken, as the built-in sample and hold capacitor takes a charge. If the bandwidth is low, try a capacitor across the lower resistor in the potential divider, to provide a charge to the sample and hold capacitor. Start with 1nF.
Here's a handy tool for calculating the resistor values in a potential divider.
https://www.random-science-tools.com/electronics/divider.htm
magic:
LTC2400 datasheet treats this topic in a lot of detail.
A sufficiently large filter capacitor is an improvement, but it only smoothens the ripple and turns it into a constant DC current which still flows through the resistors and causes some error, by Ohm's law.
The magnitude of that current may depend on temperature to some extent.
IIRC, for the LTC2400, a few komhs of input resistance was supposed to be enough to get decent accuracy without any opamps and filter capacitors. With ATX voltages, such dividers would only draw a few mA of waste current on each rail.
Remember that the output resistance of a divider is equal to the two resistances in parallel.
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