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#1




Resistor Capacitor LED circuit
i need help figuring our the values for the resistors and capacitor in this circuit.
It's easier if I just reproduce the question here: A white LED lamp (L) operates at 2.4 V, 10 mA. It’s energy is supplied from a capacitor, C. The lamp should operate for at least 30 minutes before needing to be recharged. To charge the lamp, it is connected to a 5 V supply through resistor R1 and terminals aa. When the lamp is docked to be charged, the switch is automatically flipped to the left position (disconnecting the lamp) and it is restored to the right side when the lamp is undocked. Charging should take less than 30 s (now you see why it uses a capacitor, not a battery). What values of R1, R2 and C would you recommend. I think I have figured out R2 by subtracting 2.4 V from 5 V to get 2.6 V and then using the LED current to figure out the resistance. I got 260 Ohms. Does that sound right? I'm not sure how to proceed on to the next step. Do I need to figure out a time step for the capacitor? Thank you for any help with this. 
#2




I like your idea but I reckon its gonna take about 67 farads at 5v to supply 30mins @ 10mA, and white (and blue) LEDs are generally around 3.6v operating.
Last edited by Paulo540; 12092009 at 05:57 AM. 
#3




Hi paulo540 can you please explain me how you have calculated the capacitance as 7 farads
Last edited by av.bhargav; 12092009 at 07:36 AM. 
#4




Well, I just did a really rough estimate based on the overall voltage and the time/watts.
the Circuit will use .05 (.01amps x 5v) w/sec or Joules to operate and 30 mins would require 90 Joules of energy. So, to find capacitance needed you go... C = 2j/V squared so 2*90 / 5*5 = 180/25 = 7.2 Farads Of course there are other variables but it does show that capacitors make poor energy storage at low voltages. 
#5




OK here goes. I am going to try to explain it in the most basic way to help anyone else.
Please feel free to correct any mistakes I may have made. I am making the assumption that the LED current should not exceed 10ma. So to calculate the value for R2 is actually rather simple if you have worked with basic LED circuits. R2 = (52.4)/10ma = 260 ohms Now the tricky part is calculating the capacitance and it requires a bit of insight and thinking. Remember that a capacitor might be handy for its ability to charge very quickly but its not ideal when being used in a situation like this, because as the capacitor discharges its voltages drops. Looking at the formula relating charge(Q) capacitance(C) and voltage we see that C=Q/V Rearranging this formula gives you V=Q/C which implies that the charge on a capacitor is directly proportiional to the voltage across it. Since the voltage is a result of the charge on the capacitor, the voltage across the capacitor is dependant on the charge. So as the capacitor discharges the voltage gets less across the capacitor. When you initially hookup the LED and resistor to the capacitor a current will flow out of the capacitor. That current will be 10ma if you are using the 260 ohm resistor. Now think about what current is for a second. Electric current is the flow of charge(Q) and we measure the rate of flow of charge when working with currents. The ampere is the rate of flow of charge when we have 1 coulomb of charge passing a point every second. As current flows out of a capacitor its loosing charge at rate equal to the current. So at 10ma its loosing charge at a rate of 10milliColoumbs per second.Now here comes the punch line... ....after 1 second the capacitor will have lost 10mC of charges making its stored charge less. This means the voltage will also decrease. This in turn will make the current flowing decrease slightly. the lower current means less coloumbs per second leaving the capacitor, so it will take longer for the voltage to decrease. But as the voltage decreases the current will also decrease and this in turn makes means the current even less which makes the voltage decrease at a lower rate and so on... This happens infinitestmly and the net effect is that the voltage doesn't decrease at a constant rate, but instead the rate at which the voltage decreases slows down. If you plot the voltage across the a capacitor as a function of time you will see that the voltage discharge plots as curve and is therefore an exponential function of time. ... Its much like when you release a balloon that is open the pressure decreases and means air flows out at a lower rate which in turn means the pressure decreases but it does so more slowly which means that the rate flow also decrease but more slowly etc etc. Its something you kind have to think about a few times to see what I mean. The net effect of this is that your current is going to decrease exponentially making the LED get dimmer until the voltage across the cap reaches 2.4 V at which point the LED will go off. This is because of the "conservation of energy rule" of the universe. There is a fancy exponential formula describing the relationship you can find it here on wikipedia http://en.wikipedia.org/wiki/RC_circuit Looking at that formula you can see its an exponential formula. V(t) is what the voltage across the capacitor will be after t seconds if we have a capacitence of C with an initial charge of Vo discharging through a resistance of R. e is eulers constant. Read that again. Now you have said that you want the LED to burn for 30min which is 1800 seconds. You will charge the cap to 5V so Vo = 5V and you are discharging it through an LED with a resistor of 260Ω in series. So you have Vo =5V , V(t) = 2.4, R=260 and t =1800 how do you work out C? With some mathematic jumbling using logarithms I get a formula for 1/C = (R/t)ln (V(t)/Vo). Working that out gives me a value for C = 9.43 Farads. These are all approximations and I think the LED might effect it a bit, to be honest I never thought about how an LED might effect a simple RC circuit. Also I am making assumptions about the voltage dropping to exactly 2.4V. Depending upon the LED's used and the tolerances and assumtions about how far you want the voltage to drop these values will vary a bit. But at the end of the day you will be looking at a capacitor of least 6 or 7 Farads. As you can see there are many ways to approach this problem. Since most capacitors are usually less than a Farad over 6 Farads is a large value. There are super capacitors on the market which have high values of capacitance. You can get little 1F ones running rated at 5V and put a few of them in parallel but they you would be running them close to their maximum voltage. There are bigger supercaps which are very expensive and are used for energy storage but it will be too bulky. To summarize: Just remember that the capacitor looses charge and therefor voltage and this will mean your LED will dim as it drains the cap and even though the rate at which the voltage drops decreases which means rate at which it will dim will get less over time its going to get dimmer which won't be good if you are using the LED as lamp. So what options do you have? Well you need to have a steady current for optimum brightness. You would need something like a buck convertor or some kind of DC to DC convertor that can run off a cap and power the LED. The problem is that the convertor itself will consume power and this will load the caps even further so it would have to be very efficient for this kind of application. If it were as simple as using a few super caps like this everyone would have been doing it a long time ago. Last edited by ping; 12092009 at 08:58 AM. 
#6




Thank you very much, this is an excellent response, it helped me out a lot. i was struggling with the formula part of it. Everything you explained made sense to me as it was presented in such a clear way, thanks for taking the time to explain it

#7




No problem.

Tags 
capacitor, circuit, led, led circuit capacitor, resistor 
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