Capacitors don't work like I thought, please help me understand. - Page 1
Forum section: electronics / beginners

author: gilligan
author_rank: Contributor
timestamp: April 09, 2019, 09:10:55 pm
content: Why does this work?http://tinyurl.com/y679uqr8

author: jpanhalt
author_rank: Super Contributor
timestamp: April 09, 2019, 09:16:38 pm
content: What did/didn't you expect it to do?

author: PartialDischarge
author_rank: Super Contributor
timestamp: April 09, 2019, 09:17:51 pm
content: Not the kind of circuit you’d want to build but, the gate to source connection is also a capacitor, a small one.In a resistor divider, the large resistor takes most of the voltage, in a capacitive divider, the small cap takes most of it.The current through the large capacitor is going to be very small but enough to charge the gate up to nearly 5v.Id recommend you to watch this video which is somewhat related

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 10, 2019, 02:18:59 am
content: The voltage across a capacitor isV= 1/c (integral) i dtSo you reset the capacitor so that it has zero volts across it (zero charge). In order for the voltage on the capacitor to change there must be a current to charge it according to the equation above.There IS NO CURRENT PATH. The input resistance of the mosfet is infinity. Since no current flows, the voltage across the cap never changes--remains zero volts. So the Vgs of the mosfet does not change either.Surprisingly, this is a useful concept--with a few additional components.

author: gilligan
author_rank: Contributor
timestamp: April 14, 2019, 08:22:54 pm
content: I always assumed that Caps blocked voltage.So when the switch was opened, after the cap drained then there would not be anything on the gate of the MOSFET.But it seems like this will just latch the MOSFET open and just run continuously even when removing supply to the Gate.

author: james_s
author_rank: Super Contributor
timestamp: April 14, 2019, 08:29:08 pm
content: Why would you assume that capacitors block voltage? Capacitors store charge, to store a charge you have to have a complete circuit to put energy into the capacitor and if you want to use that charge you have to have a way of completing a circuit to get the charge out of the capacitor.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: April 14, 2019, 08:30:05 pm
content: If you want the mosfet to turn off you have to provide a path from Gate to ground somehow, usually through a pulldown resistor.

author: gilligan
author_rank: Contributor
timestamp: April 14, 2019, 08:49:17 pm
content: Quote from: james_s on April 14, 2019, 08:29:08 pm"Capacitors are widely used in electronic circuits for blocking direct current while allowing alternating current to pass."I'm only working with DC, so I assume that it will just not let current pass. No?

author: gilligan
author_rank: Contributor
timestamp: April 14, 2019, 08:50:01 pm
content: Quote from: alsetalokin4017 on April 14, 2019, 08:30:05 pmBut now I can't get the gate to open at all.

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 14, 2019, 09:06:59 pm
content: Quote from: gilligan on April 14, 2019, 08:49:17 pmIf you think of a cap as a rechargeable battery with a capacity.It will charge until it is filled and then it stop charging and thereby stop more power going trough, thereby blocking DC.AC on the other hand keeps charging while positive and discharging while negative, so the cap will never get filled and therefore do AC keep "working".The cap will try to release it's charge and as soon you turn off the power, the cap will deliver its charge until depleted.

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 14, 2019, 09:08:06 pm
content: QuoteWhat do you mean?Try this:http://tinyurl.com/y5w5ybl6

author: james_s
author_rank: Super Contributor
timestamp: April 14, 2019, 09:13:02 pm
content: Quote from: gilligan on April 14, 2019, 08:49:17 pmIt's a little more complex than that, any time you have a changing voltage you have something like AC. In a steady state, a capacitor will charge up to whatever potential is across it and then it will block the flow of DC once it reaches equilibrium, but then what? In order to do something useful you have to have a way to charge and discharge the capacitor. What are you trying to accomplish? What do you want the mosfet to do?

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 14, 2019, 09:13:37 pm
content: Quote from: gilligan on April 14, 2019, 08:50:01 pmAn easy way to understand why your circuit is not working, is thinking of power as super lazy. If you follow your path from the battery and to ground / negative, you'll see that the power can run trough your switch via the lower 10K and trough to ground, totally avoiding the obstacle your mosfet and upper 10K is making.EDIT: Also, if you look at your circuit. You have positive on booth sides of your cap, so nothing is going on there. You only get a flow if you go from positive to negative.Am I right in guessing that you are trying to make a delay, so when you switch off, the mosfet stays on for a short while longer?

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 14, 2019, 09:18:28 pm
content: Quote from: james_s on April 14, 2019, 09:13:02 pmYou are perfectly right but I thought that a simply but not totally correct explanation was better, for later to learn more about what is in reallity is going on. Like the flow of electrons, resistance, conductance, heat and much more do in his circuit. :-)

author: gilligan
author_rank: Contributor
timestamp: April 14, 2019, 09:43:09 pm
content: Quote from: Wimberleytech on April 14, 2019, 09:08:06 pmAh, I blew my capacitor and didn't notice... you'd think I'd learn as i've done that a hundred times now.

author: gilligan
author_rank: Contributor
timestamp: April 14, 2019, 09:50:52 pm
content: Quote from: FriedMule on April 14, 2019, 09:13:37 pmYes, that was the experiment... just trying to wrap my head around components with these little circuits.

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 14, 2019, 10:40:55 pm
content: Quote from: FriedMule on April 14, 2019, 09:13:37 pmOkay great.First do I think that if you have a horizontal positive rail at the top and a horizontal negative ral at the bottom in the start, that will make it a lot easier for you to keep track on what is going on. :-)If you place the cap across from positive to negative, you do charge the cap up and everything after that, will get the power from either the power supply or the cap.Try to make a new version thinking of that and show it here, then we'll take from there, until you know how it works:-)

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 15, 2019, 12:42:43 am
content: I suggest you get your head around a circuit that does not include a mosfet. Like this.http://tinyurl.com/y4zbmqbl

author: KL27x
author_rank: Super Contributor
timestamp: April 15, 2019, 05:15:03 am
content: QuoteIn the simulation, do you see DC current passing through the cap?If you remove the capacitor completely, it should do the same thing. There does not need to be any current between a MOSFET gate and source for the transistor to remain on, indefinitely. When the switch is open, the gate is floating. Whatever charge has last been left on the gate will theoretically remain there, forever, unless the simulation includes the gate leakage of the FET. Maybe I don't understand the simulation, though.This is one way to test if a signal or logic level FET is working, out of circuit. Take your multimeter and put it on diode test. Put one probe on gate and one on the source. Then lift the probes and change the meter to continuity test; put the probes on source and drain to test continuity. Do the diode test again, but reverse the probes on the gate and source. Then repeat the continuity test. One outcome will show continuity, because the FET gate is saturated. And the other will be negative, because the gate is below threshhold, and the source to drain will be high resistance (or will have a diode drop, if the probes are reversed... still negative continuity test).In the diode test, there is a voltage (typically about 3V) applied through the probes, but limited to only a small current. This will charge the gate. The charged gate will retain its state long enough for you to do a continuity test, if the FET is out of circuit.**I bet this is what MasterTech's video demonstrates.

author: KL27x
author_rank: Super Contributor
timestamp: April 15, 2019, 05:44:11 am
content: QuoteIn an otherwise great explanation, I would change "current" to "coulombs." The total amount of coulombs or charge that will flow will be small. The current will be relatively high, since there's no impedance or resistance in the circuit. It will flow for only a short duration, though, after the circuit is initially powered on. Think of it as the first quarter of an AC waveform that "passed" through the (initially uncharged, presumably) capacitor, then just got stuck there.

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 15, 2019, 07:02:12 am
content: I have tried to make the LED delay, so you can see how it can be done. :-)The simulated circuit have to charge for a while, because some strange behavior in the simulator.And for some strange reason, the simulator shows AC animation, while it is DC.http://tinyurl.com/yx9opv9fExtra fun: If you place the resistor on the let side of the cap and make it a 10uK, you will get a LED that slowly turns on after you turn the switch on and slowly off again when you switch off.

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 15, 2019, 12:19:27 pm
content: QuotePutting a voltage source directly across a capacitor challenges the simulator. Must be what is going on. Normally a switch is modeled with some resistance, but I bet this one is not. As you can see, putting a very small resistor fixes the animation.http://tinyurl.com/y2w6g6vkIt is a cute simulator for sure, but not foolproof.

author: TimFox
author_rank: Super Contributor
timestamp: April 15, 2019, 06:52:52 pm
content: The simple explanation taught me early in my first electronics course was:A). You can’t change the voltage across a capacitor instantaneously (requires infinite current), andB). You can’t change the current through an inductor instantaneously (results in infinite voltage).

author: soldar
author_rank: Super Contributor
timestamp: April 15, 2019, 07:37:17 pm
content: Quote from: TimFox on April 15, 2019, 06:52:52 pmIt's calledduality.

author: gilligan
author_rank: Contributor
timestamp: April 16, 2019, 12:46:49 am
content: Quote from: KL27x on April 15, 2019, 05:15:03 amYes, I was floating the gate... I knew better than this.I swear, floating grounds will be the death of me (hopefully not literally).

author: David Cutcher CEG
author_rank: Contributor
timestamp: April 17, 2019, 01:48:16 am
content: I don't want to prolong your agony, but what are you expecting to happen here?Nor do I want to sound presumptuous. But Don't depend on simulation programs to learn electronics.I suggest if you want to see the action of a Capacitor, set up the following circuit on a bread board. Then try different size caps.OH, and for R1, should be 220k.David Cutcher CEG "certified evil genius"

author: gilligan
author_rank: Contributor
timestamp: April 23, 2019, 02:50:57 am
content: Quote from: David Cutcher CEG on April 17, 2019, 01:48:16 amWithout a scope, this shows a lot of interesting data that would be much harder for me to witness.

author: David Cutcher CEG
author_rank: Contributor
timestamp: April 23, 2019, 12:36:22 pm
content: Glad that you found this useful.The text is set up for the complete beginner, and I assume that beginners don't have access to scopes. But scopes become necessary as a tool to visualize what we can't see. In the text, I introduce scopes using a freeware program. I asked the community here for a recommendation for a real intro level scope. Suggestions All over the map, but the winner was for the Owon vds 1022i.I have a class (for intro to electronics) starting on Udemy on July 1st. Get back to me if you are interested. Just reply to this post. I need some beta testers.David Cutcher "Certified Evil Genius"

author: MrAl
author_rank: Super Contributor
timestamp: April 23, 2019, 01:26:17 pm
content: Hi,What it looks like is that the large capacitor leakage would keep the MOSFET turned on part way. The feedback would cause some intermediate voltage level on the drain if there was a drain resistor. With no drain resistor there will just be some level of current flow in the drain and that seems to be what you are seeing.Keep in mind that the input resistance of the MOSFET is extremely high so it only takes a small amount of leakage current through the cap to keep the mosfet turned on somewhat. The more the cap leaks, the more the mosfet it turned on.