I have a 22uf capacitors hooked up to a 5 volt power source to power a 3.2 LED I am using a 220 ohm resistor to bring down the voltage, why isn't the LED slowly dimming and then turning off? I had it running for a while. I also have it connected in parallel. I am pretty new to circuit boards and all. Also when would it be appropriate to add a capacitor? like in what situation on a small circuit board should I use one
Why isn’t the LED slowly dimming and eventually turning off
capacitorcircuit-design
Related Solutions
I must have missed this when it was asked in January.
This is a well described question and Al's answer to part of his own question was very good. He subsequently deleted it, but hopefully it will get undeleted sometime soon.
I'll address the core questions first and then come back and talk about some clever circuit aspects.
Q: So now I have one old 15uF, and one new 22uF [in series]. ...Will there be problems?
A: Probably not.
When you charge two capacitors in series so that the same current fklows through both capacitors, as happens here, the larger capacitor will experience a smaller voltage rise. This will be very approximately in inverse proportion to their capacitance. The two capacitors are close in nominal value (15/22 =~ 0.7) Electrolytic capacitor values may vary more widely than this (depends on specification). The older capacitor has probably lost some capacitance with age. So, the older small one will probably have a higher voltage to start when charging finishes. This will offset the capacitor voltage midpoint.
However, as you rightly note in your deleted answer (please undelete), when the capacitors discharge they will be electrically in parallel bu=t behind diodes so that the somewhat higher voltage capacitor will start to discharge first and when the output voltage gets down to the voltage of the lower voltage cap the second cap will "join in" seamlessly.This will have some effect on capacitor ripple currents and the higher voltage MAY stress the old cap more, but overall it should work OK. Arguably, a new cap that is not the same as the old one should be at a somewhat LOWER capacitance so that it takes more of the stress. BUT should be OK.
This is Al's picture of the discharge process. Whichever capacitor is at higer voltage will discharge first.
Q: Those caps are surrounded by a lot of diodes. I expect that normally the potentials around and between those caps are -162V, 0V, +162V. When I replaced one of them by a different one, I probably moved the center potential out of ideal zero. Does it matter here?
A: As above. This is the heart of the Valley Fill circuit. The caps charge to ABOUT Vinpeak/2. All should be well enough.
Q: Note that the reason why there are two strange capacitors instead of one 400V one is probably just the space.
A: No. As above. this provides passive power factor correction by very substantially spreading the conduction period of the input diodes. It also provides Vsupply at half Vin peak during the valley period.
Q: The 0R5 resistors on emitters of each transistor are now 0R56. I don't understand ... if it's dangerous change or not.
A: This is OK. The emitter resistors are current sense resistors which provide voltage drive via the diode D1 D2 to trigger SCR1, which terminates the current switching half cycle via D3. I'd have to spend more time on this circuit to get all the nuances and I'm pretty sure it's not 100% correct, but it gives a reasonably good idea of what happens. Increasing the resistors to 5R6 from 5R increases the voltage across them by a factor of 5.6/5 ~= 12% so they will cause the circuit to turn off at very slightly lower currents causing very slightly lower brightness. You would be very unlikely to see the difference visually.
Valley Fill Circuit:
A Valley Fill Circuit is a piece of brilliant black magic from the beginnings of time that allows surprisingly good power factor correction into a resistive load - which a constant brightness high frequency inverter tends to provide.
Rather than continue to sing their praises - here are some references to basic and more clever versions and some discussion. Well worth acquainting oneself with if you have not met them.
IR (amongst market leaders) AN1074 - New valley fill circuit - A new Circuit for Low-Cost Electronic Ballast Passive Valley Fill with additional Control Circuits for Low Total Harmonic Distortion and Low Crest Factor - passive magic refined.
+____________________________
A very clever circuit that appears to offer substantial gains over the traditional circuits Improved Valley-Fill Passive Current Shape - 1997
- The original valley-fill current shaper permits input current conduction from 30° to 150°, and then from 210° to 330°. Due to the discontinuities from 0° to 30° and from 150° to 210°, substantial amount of harmonics were introduced into the input current waveform. This article presents an improved version of the valley-fill circuit which extends the conduction angle to near 360°, thus lowering unwanted harmonics as well as improving power line current waveform. Improvements are made with passive components. SPICE simulations compare original circuit with different improved versions of the circuit. 98% power factor is achievable with this new circuit.
IEEE abstract - of interest]The circuit with valley switching technique
And again High power factor correction circuit using valley charge-pumping for low cost electronic ballasts
That article is crap, and you should probably forget you ever saw it. Besides the fact that it's totally wrong, as you've discovered, it's full of half-truths that cultivate an incorrect understanding of how electrical phenomena actually work.
The plate on the capacitor that attaches to the negative terminal of the battery accepts electrons that the battery is producing.
wrong. Batteries don't produce electrons. In fact, nothing in your circuit produces electrons. As far as physicists have been able to demonstrate, charge is never created nor destroyed. A physicist can tell you how to make an electron by assembling more fundamental particles, but unless your circuit includes a particle accelerator or operates inside a star, there won't be any relevant electron creation or destruction in your circuit. Batteries pump electrons. They don't produce them.
In the first few paragraphs, that article uses the word charge in several senses. An experienced engineer can distinguish the senses by context, but the novice is more likely to confuse them. Go read Bill Beaty to immunize yourself against this misconception.
The bulb will get progressively dimmer and finally go out once the capacitor reaches its capacity.
Not exactly true. The bulb will go out once the voltage across the capacitor is equal to the battery voltage. When this happens, there can be no voltage across the bulb, and thus no current, thus no light. I don't know what they mean by capacity in this sense, but it seems to me that they are phrasing it this way to avoid explaining how capacitors actually work. You might define a capacitor's "capacity" any number of ways, but they haven't defined it at all. This kind of half-thinking isn't going to help you understand once you figure out that the explanation is incomplete.
A capacitor's storage potential, or capacitance, is measured in units called farads.
If you take "capacity" and "storage potential" as synonymous, as anyone would naturally do, this is wrong and self-contradictory. If this were true, then we could re-write the previous statement as
The bulb will get progressively dimmer and finally go out once the capacitor reaches its capacitance
...which is totally bogus. Ordinary capacitors don't change capacitance under normal operation conditions. Again, the problem here is they haven't fully defined the underlying concepts. If capacitance is a capacitor's storage potential, then what is it storing?.
A 1-farad capacitor can store one coulomb (coo-lomb) of charge at 1 volt.
Here, they try to define the stuff that the capacitor is storing, but it hardly makes sense. The problem is if you use words like "capacity", this implies there is some sort of "full" or "at capacity". But, there is no concept for an ideal capacitor. If you push 1C of charge through a 1F capacitor, the capacitor will have a voltage of 1V. For 2C, you get 2V, and 1000C gets you 1000V. You can push 1000C through a 0.5F capacitor also, but it will then be at 2000V. An ideal capacitor is never full, and you can push charge through it forever and it will never be "full". That's not a capacity, it's the ratio of charge to voltage, which is evident in the definition of farad (a farad is a coulomb per volt):
$$ F = \frac{C}{V} $$
Real capacitors, incidentally, have a maximum voltage, which if exceeded will damage or destroy the capacitor. So in this sense, a capacitor can be "full" and can have a "capacity". But, this is not how this article is using the word.
I could probably write an entire article on the problems with that article, but hopefully you get the idea. Please wipe your memory of what that article has said, and seek a more sound explanation.
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Best Answer
When the capacitor is powering the LED, the circuit is an RC circuit with an exponentially decaying voltage. The LED is only going to stay lit for as long as the voltage remains above the forward drop voltage. Since capacitor values are not very precise, and the LED doesn't produce much light at low currents, we can estimate the amount of time the capacitor can power the LED as the time constant of the RC circuit. It's not the exact time, but it will give you a ballpark estimate of the decay time for the fade out. The time constant for an RC circuit is the product of capacitance and resistance:
$$ \tau=RC=(220\Omega)(22 \mu \text{F}) \approx 4\text{ms} $$
The time to fade is only about 4 milliseconds! That's why you aren't seeing the LED fade off, because it happens too quickly. There are two things you can do. The first thing you might try is increasing the size of the resistor. This will make the time constant longer, but it will also limit the current, making the LED dimmer. The other thing you might try is using a larger capacitor. This will increase the time constant without effecting the brightness of the LED. You can figure out the value for yourself, but I'm guessing you want a fade out time of around 1s, which means you need a capacitor roughly 1000 times larger than what you have currently.