Electronics > Beginners
Capacitors don't work like I thought, please help me understand.
gilligan:
--- Quote from: FriedMule on April 14, 2019, 09:13:37 pm ---
--- Quote from: gilligan on April 14, 2019, 08:50:01 pm ---
--- Quote from: alsetalokin4017 on April 14, 2019, 08:30:05 pm ---If you want the mosfet to turn off you have to provide a path from Gate to ground somehow, usually through a pull down resistor.
--- End quote ---
But now I can't get the gate to open at all. :(
--- End quote ---
An 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?
--- End quote ---
Yes, that was the experiment... just trying to wrap my head around components with these little circuits.
FriedMule:
--- Quote from: FriedMule on April 14, 2019, 09:13:37 pm ---
--- Quote from: gilligan on April 14, 2019, 08:50:01 pm ---
An 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?
--- End quote ---
Yes, that was the experiment... just trying to wrap my head around components with these little circuits.
--- End quote ---
Okay 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:-)
Wimberleytech:
I suggest you get your head around a circuit that does not include a mosfet. Like this.
http://tinyurl.com/y4zbmqbl
KL27x:
--- 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.
KL27x:
--- Quote ---The current through the large capacitor is going to be very small but enough to charge the gate up to nearly 5v.
--- End quote ---
In 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.
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