Electronics > Beginners
Capacitors don't work like I thought, please help me understand.
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FriedMule:
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/yx9opv9f

Extra 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.
Wimberleytech:

--- Quote ---And for some strange reason, the simulator shows AC animation, while it is DC.


--- End quote ---

Putting 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/y2w6g6vk

It is a cute simulator for sure, but not foolproof.
TimFox:
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), and
B). You can’t change the current through an inductor instantaneously (results in infinite voltage).
soldar:

--- Quote from: TimFox on April 15, 2019, 06:52:52 pm ---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), and
B). You can’t change the current through an inductor instantaneously (results in infinite voltage).

--- End quote ---

It's called duality.
gilligan:

--- Quote from: KL27x on April 15, 2019, 05:15:03 am ---
--- Quote ---I'm only working with DC, so I assume that it will just not let current pass.  No?
--- End quote ---
In 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.

--- End quote ---

Yes, I was floating the gate... I knew better than this.

I swear, floating grounds will be the death of me (hopefully not literally ;) ).
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