Hi, this question is not strictly Arduino related, but a general understanding question.
I've tried to imitate the capacitor circuit shown here (the one with the light bulb), only I used a tiny motor instead of the light bulb.
When the circuit is built with the capacitor (picture 1) it won't work at all - the explanation claims the bulb (motor) should be lit and then fade out - no such thing happens. When the capacitor is removed (picture 2) the motor is running as expected.
I've replaced the motor with LED and buzzer, but the results are the same.
I've read the Wikipedia entry for Capacitor, but the explanation was somewhat theoretical, and I couldn't find an example of a working circuit.
You'll need a lamp to see the momentary inrush current. Your motor isn't seeing enough current to run it. A really low current/voltage one (maybe a pager motor) might.
A capacitor is two plates separated by an insulator. It functions as a voltage storage device. BEAM robots like those over on SolarBotics actually use a solar cell to charge a capacitor which then fires a voltage trigger to dump the charge through the motor. Once the capacitor is charged up, the current stops flowing, at which point it effectively is an insulator.
The effect would be even better with an led and a resistor, the inrush would last longer due to the lesser current flowing. An ordinary 3mm indicator LED and a 1K resistor works nicely with a capacitor that size. Your motor would work with a bigger capacitor and a good battery
Thanks for the replies. I tried a LED with a resistor in a number of configurations, but got no results from any? Here are the specs of my motor.
Maybe there's something wrong with my capacitors? I took em out from an old walkman ;D (the same place I got the motor)
Edit: I got it sorta working with the LED! The capacitor should've been on the positive side of the led apparently. I got the LED to fade out, but when I replace the battery with a wire as suggested on the tutorial I don't get the blink they're talking about?
I got the LED to fade out, but when I replace the battery with a wire as suggested on the tutorial I don't get the blink they're talking about?
The inrush current (when battery gets connected) and discharge current (when battery is replaced by a wire) flows in opposite directions. So to get the LED to glow on discharge, you must first disconnect the battery, turn the LED around and then connect the wire.
Keep in mind that once the capacitor is charged, there will no longer be an inrush current when you connect the battery (the capacitor will remain charged even when you disconnect the battery). So you need to alternate between charge/discharge to see the effect you're after.
Capacitors "in" series with the current path (As opposed to being spanned across the power bus) act like an open circuit in Direct Current circuits but they do not act like an open circuit in an Alternating Current circuit. The circuit shown would commonly be called a DC Blocking circuit. If you are using 5VDC... then I would expect little or no current reaches the motor, LED, since the capacitor charges up so fast that the results are not what you expected!
There is probably a much better way to show the capacitor storage effect by creating an RC (resistor/capacitor) timing circuit. By controlling the current (with a resistor) that reaches capacitor, you have control over the rate of charge on the capacitor and you are more likely to create a result you could "measure" or see.
Why don't you see it? Your capacitor values and load are (as stated in the very theoretical reference) not sufficiently big enough to create the effect they are talking about. It's not that it doesn't happen... Essentially, you are getting an "instant charge" on the capacitor and you never see what they explain.
To get your desired results, you might need a capacitor that comes close to 1 farad in value. Like maybe an ULTRACAPACITOR.
Essentially, I think that you just need to take their simple abstract explanation and build your concepts and understandings from it.
That kinda sucks I must admit. Anywayz, I've seen the effect one way, I'll have to assume it works the other way around also. Thanks for all the replies!
If you use a 1K resistor in series with the 220uF capacitor and a LED, the LED should light up for about 0.22 seconds while the capacitor is charging. Likewise you will see the same bright-to-dim pattern if you reverse the LED and wire a short in place of the battery.
Your circuit may not be of much use as a flash-light, but 0.22s is still plenty to observe the effect of charging/discharging the capacitor. If you give it another go, you need to make sure however that the capacitor is fully discharged before attaching the battery and likewise fully charged before connecting the short wire.
BenF:
I've actually seen the led fade out on charge, but not on discharge when I reversed it, and replaced the battery with a wire.
Is the order of the reverse wiring important?
Is the discharge process faster then the charge?
replace the battery with a wire.
You will see it immediately you do the final step, so just touch the wire while looking at the LED.
Is the discharge process faster then the charge?
No, but while you are doing all that fiddling about the capacitor is discharging itself through natural leakage.
For a really good demonstration of charged capacitors, unplug your TV set, leave it for an hour. Then remove the back and have a feel around the scan coils. You will discharge the capacitor and you will never forget it.
Yeah, the joys of CRT TVs... You can die doing that though Be sure to stand barefoot in a pool of salt water when you are feeling around the scan coils. It really helps.
My dad tells a story of a kid he went to school with whose father was a TV repair man. The kid got a HV cap from his dad's stuff and would charge it up before school. Then he'd place the fully charged cap in his pocket all wrapped up. At school he'd volunteer to go up to the board and when another student would be up there too he'd sneak the thing out of his pocket, lean in a bit and touch a leg of the cap to the metal shelf at the bottom of the chalk board. 'twould zap the snot out of the other kid at the board. He eventually got caught when the capacitor melted to the metal shelf. Sounds like a great (if somewhat dangerous) prank.
For a really good demonstration of charged capacitors ...
Some would argue that "a burnt child dreads fire", but the true pragmatic engineer would take on your advice, and simply conclude that "burnt flesh smells awful". ;D
The short answer is that the capacitor you are using (220uF) is much too small for you to notice the effect if you are using the typical DC motor. As the LED example with 1k resistor mentioned, you'd get about 0.2s of dimming LED (short but noticable.) But trying to run your motor through a 1k resistor probably won't work at all, and the resistance of the motor itself is much smaller than 1k, so that the time that the motor would be "powered" through the cap is much shorter than 0.2s (and probably too short for the motor to turn at all.) If you can get one of those 10000uF caps pwillard was talking about, the experiment would be more likely to work.
Be sure to stand barefoot in a pool of salt water when you are feeling around the scan coils. It really helps.