I'd like to have an effect of an LED slowly turning on and then off. Can I do this with a capacitor? How do capacitors work? Do they hold a certain max voltage, they store current? help For instance, if a capacitor says: Leakage current: after 1 min:0.03CV or 4uA whichever is great; after 2 min.: 0.01CV or 3uA whichever is great what does that mean? If the capasitor is described as 180uF, 16V, 105C what's that mean??
180uF is the capacitor's capacity (180 microfarads) 16V is the maximum voltage it can handle i think 105C is its maximum temperature
The capacitor itself won't give you the throbbing effect, all it will do is gradually bring the LED to full brightness, but then it will stay at full brightness until you shut off, then it will throb down. This article is for CC inverters, but I'm sure macro or linear could tell you how to change it for LEDs.
That's really the effect I want! I'm going to be making an array of these so I need each LED setup to be as simple as possible. This will save me money. I'm also looking at using small bulbs rather than LEDs. I'll be experimenting a bit. I'd like the light, when the power is turned on, to slowly glow up to full bright, and then when the power is taken off, to slowly dim to out.
So, if I have a small 5 volt bulb, in series with a capacitor. Would I put 5 volts to the cicuit to light the bulb? Will the capacitor ever store more voltage or ampers than the power supplying it? Would the aformentioned cap work with a 5 volt bulb @ .25A
No. In parallel perhaps. A cap looks like an open circuit to DC once it's charged--putting it in series will give you a dark bulb. In parallel, once you apply the 5V, the cap will begin to charge, so the bulb will ramp up to full brightness which will be reached when the cap has reached 5V. At that point, the cap still doesn't conduct, but no further energy is beingg stored in it. Now when you shut off the voltage supply, the bulb will stay lit, because it will draw energy from the capacitor until the voltage across the cap is zero. The time this takes is calculable, we'll get there in a minute. The circuit should be parallel, and in that case the voltage across the cap will never exceed the supply voltage. Looks okay to me. You'd want to know a little bit about the bulb too, to do a full analysis. Mainly it's value in ohms. (Or a decent approximation.) Then you can go after the time constant, or a decent guess at it. Tau represents the time constant of an RC circuit, an it's calcualted easily enough: tau = R*C The significance is that after time = 5*tau, the cap will have either fully (more than 99%) charged from zero or discharged from full. Regardless of the supply voltage. Neat, eh? So assume your bulb looks like a 5 kohm resistor. tau = 5 000 * 0.000 180 = 0.9 seconds your bulb will take 4.5 seconds to make the on-off transition (or off-on).
The dim-out is no problem, a high-value capacitor in parallel with the led/resistor will discharge fairly slowly (if you keep the maximum current to 15mA or less). You may need a diode in the supply line so the capacitor can only discharge through the led. Bringing the light up slowly is harder. A capacitor charging curve is very steep at first, then gradually flattens off. To get it more linear you need a constant-current gizmo in the charging circuit. Here's an idea to play with (I've not built this, so all values are guesses): The 317 is used as a current regulator and passes 15mA max. Some volts are lost across it, R1 and the diode, so the capacitor only gets to about 9v.
That's the good news. The less good is that 63.2% of the action happens in the first tau, you're 86.5% there after 2t and 95% after 3t, so it ain't very linear. I've a led indicator on a bench psu, when I switch off nothing seems to happen for a second or so then it dims fairly smoothly to black over a few seconds. But that's starting at about 18v with IIRC a 10,000uF cap so YMMV.
Thanks for the help everyone, I'm going to be experimenting soon. I'm making a whole board of these so I'd like to keep each light, photosensor, cap, etc as low as possible. Under a dollar each if I can manage it. It's ultimately a final project for a class in technology and art i have.
So if I've got a 2.25V @ .25A bulb can I use a 180uF, 16V cap? Also, could you explain that equation a little more for me, I didn't quite understand. Lastly, can I put enough voltage through a 40mW, 20V photocell with out burning it out with the power required to run the light?
I get the feeling that a capacitor that small would not really give an effect as it would probably discharge within a second. Maybe a 2200uF one might be better.
Especially with a bulb drawing that current. Using the formula, R is 2.25/0.25 = 9, so with 180uF, Time Constant (tau) = RC = 9 x 180/1,000,000 = 1.62 milli-seconds. A 0.25A bulb isn't really suitable, a 12v 60mA bulb (R=200) and a 10,000uF capacitor gives you tau=2 seconds, which should give about a 5 second fade to black.
I'm under some budget constraints, if I wanted to keep the bulb I've got and still get a 1+ second fade, how many uF are we talking about? 2200 like stevey mentioned?
Well.... 5*tau => 1s tau => .2 .2 => 9*C .2/9 => .0222... F Your capacitor would have to be at least ~22k mF
Of course a much better idea is to use a transistor or mosfet to drive the bulb, and a small capacitor on the base/gate with resistors positioned in positions to enable a slow charge and discharge.
