Voltage dividers and A/D minimum impedances - Page 1
Forum section: electronics / beginners

author: jolshefsky
author_rank: Regular Contributor
timestamp: July 20, 2014, 12:49:55 pm
content: I'm curious about impedance matching with voltage dividers. The specific instance is the A/D input pin on a microcontroller. Since the µC's internal circuit works by sampling into a small capacitor, if the signal source has too high an impedance, the capacitor won't completely charge, and you'll get erratic readings.But let's say I'm sampling a battery and have a 10:1 divider like this:(V+)|---| | 10K--- µC A/D sampler|+---- V(out) ----/ *----| |--- ---| | 1K C--- ---| |(gnd) (gnd)The impedance at the V+ and gnd potential points is nearly zero ohms, so on the one hand, the impedance of V(out) seems like it should be 10K||1K = 909 ohms. But if the capacitor needs to increase charge, it's the 10K path that counts, so is the impedance weirdly unbalanced whether charging or discharging?

author: cyr
author_rank: Frequent Contributor
timestamp: July 20, 2014, 01:18:31 pm
content: It's 1k||10k for any input voltage. More specifically, your circuit is equivalent to a voltage source of (V+ / 9.09...) volts in series with a single 909... ohm resistor.http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theoremThink of it like this: If the cap starts out discharged it will indeed only be charged through 10k, but from a higher voltage than the output voltage of the divider which means the current is greater.In fact, any change in voltage over the cap will be driven by a larger change in voltage at the input and a larger current through the 10k resistor than if the voltage divider wasn't there.

author: jolshefsky
author_rank: Regular Contributor
timestamp: July 20, 2014, 01:24:55 pm
content: Thanks!Quote from: cyr on July 20, 2014, 01:18:31 pmThat looks like it would be wicked useful. I've always had trouble analyzing circuits that had a number of resistors in series and parallel ... I'll have to remember to use this when I get into that trouble. (P.S. I had a very-minor minor in electrical engineering as part of my computer science degree, so I know Ohm's law like mad and know enough of Kirchoff's rules to make some very messy algebra that I can never quite make useful, and that's about it.)

author: David Hess
author_rank: Super Contributor
timestamp: July 20, 2014, 02:19:40 pm
content: The resistance as far as the ADC input is the Thevenin equivalent resistance or 909 ohms.Some ADCs require a low impedance during their entire conversion cycle but with others only the settling time is affected. Usually if high bandwidth is not required, just adding a capacitor across the ADC input is sufficient to lower the AC input impedance enough for proper operation.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 20, 2014, 02:31:47 pm
content: Yes, in fact, the ADC should specify what it is as well: usually something like 10k ohms in series with whatever the capacitor looks like. So you can guess, if it's less than half, say 5k Thevenin -- it's not going to win you any more performance.And for something slow like a power line, the resistance can be even higher, as long as the sampling is okay. You might have to sample twice, so that the capacitor gets partially charged, does a conversion, then gets more fully charged and converts the correct number. The charge isn't usually destroyed by the conversion process (it's usually a charge approximation, so it's comparing the capacitor to smaller and smaller capacitors charged to VREF or 0V, and once the process is done, it's within an LSB of where it started), so you can repeat in this fashion.I haven't actually observed if there's such a thing as charge transfer between channels (say, from sampling CH0, then CH1, then..), but there should be, to some extent. So this would be one way to solve that.As long as the sample rate is low, you can use a quite large resistance divider, bypassed with a cap much larger than the ADC sampler capacitance, and get the right number on the first sample.For battery operated purposes, you might even use an extremely large resistance (megs) and buffer that with a micropower op-amp. Though I think the bypass capacitor will do just fine.Tim

author: Andreas
author_rank: Super Contributor
timestamp: July 20, 2014, 03:24:39 pm
content: Quote from: T3sl4co1l on July 20, 2014, 02:31:47 pmCaution: usually this works at room temperature.But leakage current of the pin may be as high as 1uA (see datasheet) at least at higher temperatures.In this case a 1Meg resistor will create a voltage deviation of 1V.So some datasheets define a 2.5K theventhin equivalent for newer devices or 12 bit ADCs.With best regardsAndreas

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 20, 2014, 03:36:47 pm
content: Quote from: Andreas on July 20, 2014, 03:24:39 pmOf the micropower op-amp???Or did you mean to quote the proceeding paragraph?But yes, guaranteed leakage of MCUs is generally poor (it's probably much better in practice, but they won't guarantee it). Something else to keep in mind.Tim

author: Andreas
author_rank: Super Contributor
timestamp: July 20, 2014, 05:02:23 pm
content: I have understood that you suggest to use alternatively to the micropower op-ampa bypass capacitor together with the high ohmic resistor(s).With best regardsAndreas