Help me understand  how capacitor works. - Page 1
Forum section: electronics / beginners

author: kafor1
author_rank: Contributor
timestamp: January 04, 2019, 02:46:49 pm
content: So I've been trying to understand capacitor for a day now, i watched videos on yt, read wikipedia and other sites that would be useful to help me learn capacitors and in the end I say frick it, I'll just ask for help here.. Note: Im a beginner, incredibly stupid and my english is bad so plz bear with me.1. Refer to the picture 1A (number is in the upper left), This is a simple circuit I found on the book, in this circuit there are 2 (NO) switches. Switch 1 is for allowing the positive voltage to pass through the circuit and switch 2 is for shorting the capacitor. Pressing the switch 1 charges the capacitor as shown in the meter, pressing SW2 discharges the charge by shorting it. So in my head it will let current flow for a period of time2.(refer to 1B), this time the capacitor is before the resistor, same concept pressing SW1 charges, SW2 discharges, in my head oh so capacitor and resistor can be interchanged, the only diffirence is the discharge time is slower than the first, since the first discharges instantly, its because its not directly shorted and the resistor is in it.3. (refer to 2A), this time ive added LED, pressing switch will charge capacitor and will let current flow for a period of time. Switch2 discharges the charge, Ive directly shorted the Capacitor this time and it discharges instantly. the LED glows and then is fading, understandable outcome.since it will only let current flow for a period of time4. (refer to 2B) this time capacitor is before the resistor, same output led glows then starts to fade. ive concluded that resistor and capacitor can be interchanged.5. (refer to 3A) look at the circuit, as u can see ive paralled the resistor between the resistor and capacitor, Pressing the SW1 will charge the capacitor, the led movement is from no light to fully bright. then when i release the SW1 it starts to fade. I really don't know why. i havent even pressed the SW2 yet to short capacitor. i thought the charge would remain even if i stopped pressing SW1 unless i discharge it. plz explain that to me6. (refer to 3B) look at the circuit, ive reversed it again, this time the capacitor is before the resistor, ive paralled it to LED Same concept as 3A. BUT when i press the SW1 it glows immediately and fades. then the charge still remains, i still have to press the SW2 to discharge it UNLIKE 3A where i dont have to press SW2 to discharge. oh man i really don't know why someone explain this to me. i think resistor and capacitor can't be interchanged7 (refer to 4) then here comes a simple oscillator circuit, ill explain this from what i understood correct me plz if im wrong, this circuit has 4 resistors, 2 capacitors and 2 transistors. 2 lower value resistor and 2 higher value resistors. current will flow to the lower value of resistor through the parallel capacitor this will charge to the pin 1 capacitor fast since the value of the resistor is low and then the current will stay on the collector of transistor, from past experiment current passing from capacitor decreases in a period of time, does that also work in this circuit?. if so then the voltage in the collector will keep decreasing up to 0. Current will pass through the higher value of resistor then to the pin 2 of the capacitor. uuhh it is already being charged at pin 1 does that mean it can be charged both pins at the same time? the book says it will slowly charge the base of the 2nd transistor since the resistor value is high. is the base of the transistor really charging from low to high or high to low, if its high to low it doesnt make anysense. and if its low to high thats not how capacitor works. since capacitor only let a current flow a period of time. i don't know. AND how the heck are the capacitors able to discharge in this Circuit without shorting its terminals? plz explain this to me..8. (refer to 5) then here comes another circuit this is basically chaining 555 timer and the output connecting the 555 timers is a 0.01uF ceramic capacitor. it says that the input of the timer will receive only a very brief pulse . is it possible to use only capacitor without resistor in series in this circuit. and how the heck are these capacitors able to discharge? I mean if it is charged then it shouldn't deliver anymore output and this circuit should only run for 1 cycle only? unless the capacitor is discharged9. (refer to 6) aaah yes this is the final circuit regarding why i don't understand capacitors. this is an alarm circuit, I won't further explain how this circuit works, just the part with capacitor. So basically the capacitor resistor is paralled to the input of the 555 timer, meaning the voltage in the input of 555 will go from 9v until 0. when it reaches 1/3 of the input voltage the timer will trigger. what i don't understand in this circuit is. if this circuit is turned off the charge of the capacitor is gone, based from my past knowledge it should not charge if its already charged before. unless its discharged but i dont know how. or what if turning off the power discharges the capacitor?circuit from 1-6 is based off from a book make electronics, all rights deserve to the owner of the book.

author: tpowell1830
author_rank: Frequent Contributor
timestamp: January 04, 2019, 03:15:15 pm
content: QuoteI will take question 1. You are mostly correct, however, you are incorrect about SW1, as it is a normally closed switch, so pressing SW2 will discharge the capacitor and the voltage across the capacitor will drop to zero until you release SW2, at which time it will start charging again. If you press SW1, it opens that part of the circuit and the capacitor will not charge until you release SW1. Pressing SW1 and SW2 simultaneously will discharge the capacitor and there will be no voltage across R1 and the capacitor will not charge.Here is your challenge: Assuming that you are pressing both SW1 and SW2 for a long enough period to completely discharge the capacitor, how long will it take to charge up the capacitor to the voltage shown on the meter when you release both SW1 and SW2?HINT:

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 04, 2019, 03:25:49 pm
content: Q2: Both circuits have the same charge and discharge time assuming, of course, that only one switch can be closed at a time. In both circuits, when the capacitor is charging, there is a series circuit consisting of C in series with R. The only difference is where the resistor is placed in the series path. For 1A, the resistor is on the 9V side of the series connection. For 1B, the resistor is on the 0V side of the series circuit.

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 04, 2019, 03:30:39 pm
content: Q3/4: You are correct. BTW, the 470 resistor is not required because current is already limited by the 1K resistor.Q5: When you release Sw1, the charge on the capacitor flows out through the LED, R path. The LED in series with R provides a current path to discharge the capacitor.

author: bsfeechannel
author_rank: Super Contributor
timestamp: January 04, 2019, 04:44:40 pm
content: 1A. Correct, provided S1is NO (the schematic shows it NC).1B. The capacitor will discharge as instantly as 1A.2A. Correct.2B. Yes.3A. The capacitor is discharging through the LED.3B. This circuit is like 1A, 1B, 2A and 2B. The capacitor has no path to discharge, except for the switch S2.4. When the left transistor is off, the left cap charges through the 4k7 resistor and the base-emitter junction of the right transistor (which is nothing more than a diode). The right transistor then turns on. Meanwhile, the right cap, that was already charged is suddenly connected to ground via the collector of the right transistor, making the base of the left transistor reversed biased (that's why the transistor is off in the first place). The only path for the current to flow now is through the right 470k resistor. The right capacitor "slowly" discharges through this resistor until the voltage at the base of the left transistor is positive, the base-emitter junction is forward biased again and the left transistor turns on. This will make the left capacitor, which charged up long ago, to apply the same negative voltage to the base of the right transistor, which will turn off, and now it's time for the left capacitor to slowly discharge through the left 470k resistor and the rest of the story you know.5. The 0.01µF capacitors will discharge through the 10k resistors when the input returns to its original voltage.6. The capacitor will discharge through the +Vcc supply pin (pin 8 ) of the 555 itself.Hope this helps.

author: bson
author_rank: Unknown
timestamp: January 05, 2019, 12:10:23 am
content: When analyzing capacitors in circuits it's worth remembering thatits voltage will settle at the point where no current flows. Its voltage changes with current, and when there is no current it means its voltage has stopped changing. This means you can apply KVL, KCL and circuit analysis - simply set all capacitor currents to zero and go around the loop(s) to determine their voltages for some particular inputs or condition. This is exceedingly handy in many, many circuits.

author: rstofer
author_rank: Super Contributor
timestamp: January 05, 2019, 04:09:50 pm
content: Everything there is to know about the charge and discharge characteristics of that circuit is here:https://www.electronics-tutorials.ws/rc/rc_1.htmlThe link covers the 'charge' case and there's a link at the bottom of the page to a discussion about the 'discharge' case.I take every opportunity to play with MATLAB (and by extension Octave) so here is a plot of the charge and discharge voltages as a function of time.

author: ocset
author_rank: Super Contributor
timestamp: January 05, 2019, 07:05:13 pm
content: Also here is your circuit in LTspice simulator .LTspice is a free download from linear.com

author: spec
author_rank: Frequent Contributor
timestamp: January 06, 2019, 12:57:10 pm
content: Hi kafor1You say that you would like to understand how a capacitor works. May I offer a bit of advice- you are going about it the hard way, a bit like trying to understand how an engine works by taking a car for a drive.Instead, I would suggest focusing just on the capacitor itself and get its characteristics fixed firmly in your mind, before thinking about any applications. By the way, you can do the same with all other electronic components: resistors, inductors, transformers, transistors.And within a few months you could be posting replies on EEV.