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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

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    \$\begingroup\$ It is slowly dimming, for a certain definition of "slowly". \$\endgroup\$ Dec 25 '15 at 1:02
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    \$\begingroup\$ Could you add a schematic diagram to your question, please. Just hit "control-M" while editing your question to bring up the schematic editor. \$\endgroup\$ Dec 25 '15 at 1:06
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    \$\begingroup\$ Did you disconnect the power supply? \$\endgroup\$
    – horta
    Dec 25 '15 at 1:11
  • \$\begingroup\$ Why should it slowly dim - why are your expectations that it should dim slowly? Justify your expectations please. If you can't do that then provide a circuit diagram. \$\endgroup\$
    – Andy aka
    Dec 25 '15 at 1:43
  • \$\begingroup\$ If you are somehow expecting the LED to go off slowly because the capacitor charges up and 'takes away' the voltage from the LED, having the C parallel to the LED won't work - you'd have to have them in series. You'll find though the LED won't light up at all, or only blink very briefly as explained below. You could try a supercap (> 1F). \$\endgroup\$
    – RJR
    Dec 25 '15 at 2:38
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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.

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With the circuit you have shown above, the capacitor will delay the LED from lighting for a very short time after you apply power, an may keep it lit briefly after you remove the power, but as shown, with no switching, the capacitor has no effect.

Once the capacitor is charged to the LED forward voltage almost immediately after power is applied, the capacitor has no effect on the circuit until power is removed.

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  • \$\begingroup\$ ok, i added a push button but still the capacitor doesn't seem to be doing anything \$\endgroup\$
    – Tai
    Dec 25 '15 at 3:17
  • \$\begingroup\$ As the other answers state, the time constant of 220 ohms and 22 uF is only about 4 mSec - much too short for you to notice any effect. \$\endgroup\$ Dec 25 '15 at 4:23
  • \$\begingroup\$ so what could i do to see a difference? \$\endgroup\$
    – Tai
    Dec 25 '15 at 5:10
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If your circuit is something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

then you have 2 problems:

1) 5*(22µF)*(220Ω) = about 24mSec charge time, where your discharge time depends upon R2 (I recommend R2 = R1 so discharge and charge time are the same).

2) Power supplied to the load (collector side of the transistor) & to the control circuit (R and C at the base of the transistor) is the same. So, when your disconnect the power, the collector loses its positive voltage and can't pull current through the LED, regardless of the base voltage.

If this is the case, try something like this instead:

schematic

simulate this circuit

This way, when you disconnect the power, C1 (make this as large a cap as you have, probably) will continue to source current & discharge through the LED over a time of approximately:

5*(1000µF)*(1kΩ) = 5sec...

Hope this helps.

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Well basically what is occurring assuming a simple RC circuit is your tau is too small meaning your circuit's capacitor is discharging relatively quickly. Can you provide a diagram and maybe we can see whats occuring more clearly?

Not sure how this will show up but the basic equations for charging and discharging RC circuits are this. tau is defined as $$tau = Resistance * Capacitance$$

Charging $$V(t)=V_{0}(1-e^{-t/\tau })$$

Discharging $$V(t)=V_{0}(e^{-t/\tau })$$

Your current discharging EQ is \$5e^-((t)/(0.00484))\$ essentially this means your LED will only remain on for approximately 0.0095 seconds. Which as you can see is not a long time.

If you increase tau you slow the discharge rate. If tau is small. Then the discharge rate is close to e^(-x) which is a rather fast discharge. This is a lot of basic circuit stuff but let's get a diagram and we can see whats causing it. The simplest solution would be to use a larger capacitor just so you know

If you calculate it and graph a curve in your favorite software. i.e wolframalpha or MATLAB, and then see how long in seconds the voltage remains above the point where the LED stays on you will know what to expect. I can go more indepth but at this point I need to know more about the circuit.

Also you connected what exactly in parallel?

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