RC circuit basics - Page 1
Forum section: electronics / beginners

author: tester43
author_rank: Regular Contributor
timestamp: January 21, 2020, 12:51:27 pm
content: So.. I was preparing debouce basic circuit for the button based on single ceramic capacitor and resistor.I found some old 1nF capacitors and 1M ohm resistor to see how it will be charging with 3,3V rectified power supply.It is charging as expected: voltage on the capacitor 'leg' is increasing to flat out in several milliseconds.The part that was a suprise was that it topped out to ~2,8V. Far away from 3,3V of the supply.So I started to test the influence of resistor on the voltage and when changed to 86KOhm it topped out to 3,1V. Of course in shorter time.That would mean that there must be a current flow, which is not really expected as the voltage is really rectified and capacitor should load and block any current. There is nothing that would be the load on it.How should I understand it? What did I miss?

author: RoGeorge
author_rank: Super Contributor
timestamp: January 21, 2020, 01:16:16 pm
content: Good observation!The instrument has internal resistance. In order to measure, some small current will go through the instrument, and together with the 1Mohm resistor, will make a voltage divider.(Offtopic: 'So' is for drawing a conclusion, not good to use as a first word.)

author: thinkfat
author_rank: Unknown
timestamp: January 21, 2020, 01:45:04 pm
content: Quote from: RoGeorge on January 21, 2020, 01:16:16 pmAlso, the capacitor will have a leakage current.

author: tester43
author_rank: Regular Contributor
timestamp: January 21, 2020, 03:09:18 pm
content: Declared oscilloscope input impedance is 1 MΩ+-1%.... I am guessing that in terms of DC it will be simply 1 MΩ resistance (?) and then it will allow some small current to go through the probe as we have voltage divider there leading to ... 1/2 voltage (because 1M resistor + 1M probe?)?

author: mikerj
author_rank: Super Contributor
timestamp: January 22, 2020, 11:14:06 am
content: Are you sure you don't have a 10x probe? That would provide a 10M impedance which would put the final voltage closer to your observation.

author: Vovk_Z
author_rank: Super Contributor
timestamp: January 22, 2020, 11:56:59 am
content: Quote from: tester43 on January 21, 2020, 03:09:18 pmYes, but we usually use oscilloscope with 10/1 divider (so it'll have 10 MOhm input impedance). There are very few reasons to use it with 1/1 mode.

author: thinkfat
author_rank: Unknown
timestamp: January 22, 2020, 12:05:27 pm
content: I guess what you've learned from this is, that RC elements with large resistors are somewhat unpractical

author: tester43
author_rank: Regular Contributor
timestamp: January 22, 2020, 12:48:17 pm
content: what I suspect right now is that I might have an issue with breadboard. I noticed using the voltmeter that withopencircuit I have a voltage drop over the resistor. Without any load... as there should be no magic it's not normal...edit: it's not breadboard - it's magic or too high current going through the measuring device.I have 3,3V rail. I plugged ONE end of resistor (1M) in the rail - other end is hanging.Measured rail voltage 3,3Measured "hanging" resistor voltage 2,9V. With oscilloscope and multimeter the same resultedit2:after short calculation using ohm's suggestion it appears that 0,4V over 1MOhm resistor is very low current...

author: mikerj
author_rank: Super Contributor
timestamp: January 29, 2020, 05:25:15 pm
content: Quote from: tester43 on January 22, 2020, 12:48:17 pmA DMM can present a significant load when measuring high impedance circuits, i.e. you don't have an open circuit like you think you have. Most DMMs will have an input impedance of ~10Meg, the same as a 10x probe on a scope.

author: rstofer
author_rank: Super Contributor
timestamp: January 29, 2020, 06:11:01 pm
content: It gets worse depending on the device technology. For example, such a circuit driving the reset input of a TTL circuit, say a 74193, would be expected to flow 1 mA into the pin while holding 2V minimum. So, given a 5V source, the resistor can only drop 3V at 1 mA so a maximum value of 3k Ohms. I would probably use 2.2k or 2.7k because there will be some leakage through the capacitor and I would always want some margin.Reset circuits don't stand in isolation. You need to include the output requirements. I picked TTL because it is illustrative of the process. CMOS will be a lot more accommodatingPage 7, VIHand IIspecs for 74193https://www.ti.com/lit/ds/symlink/sn54ls193-sp.pdf

author: TheDood
author_rank: Regular Contributor
timestamp: January 29, 2020, 10:39:36 pm
content: Quote from: tester43 on January 21, 2020, 12:51:27 pmJust an idea and sounds like others have said similar (I'm just a beginner so not sure I'm comprehending all the comments), but using the same R value, try a different size cap, something bigger in the μF range, see if the voltage drop is less or reduced compared to the V drop of the 1nF. Maybe your multi is flowing cap current through itself during measuring, and enough to reduce cap V on a 1nF cap. When you use a smaller R (86kΩ vs 1MΩ), the possible current in the RC cct increases, so it's able to replenish the charge (that's being lost across your multi or leaking out the cap) at a higher rate than if a large R were having to be flowed through to refill your cap?EDIT-Idk if I'm close at all so just keep that in mind... Maybe leakage is adding to multimeter current theft, perhaps even more of a factor than multimeter current theft..0.5V ÷ 1,000,000 Ω = 500nA0.2V ÷ 86,000 Ω = 2,325nARate in = rate out, but capacitor V levels off, so like a bucket with a hole in the bottom the more coulombs filling the capacitor the more pressure of water leaking out the bottom until there's enough of the bucket filled up to generate enough pressure at the hole in which the rate in = the rate out.

author: tester43
author_rank: Regular Contributor
timestamp: January 30, 2020, 10:16:41 am
content: my capacitors were "new old stock" - it's a bag of ceramic caps from closure of company that was building 8 bit computers for automation ... in the early 1990s... in east Europe. Maybe some of them were not so great. Anyway new box of tantal caps solved the issue.Side note: electrolytic caps from 90s still work ok

author: thinkfat
author_rank: Unknown
timestamp: January 30, 2020, 11:05:45 am
content: See, told ya.