All About Circuits Forum  

Go Back   All About Circuits Forum > Electronics Forums > Homework Help


Homework Help Stuck on a textbook question or coursework? Cramming for a test and need help understanding something? Post your questions and attempts here and let others help.

Reply   Post New Thread
Thread Tools Display Modes
Old 12-09-2009, 04:59 AM
Ooomph Ooomph is offline
New Member
Join Date: Dec 2009
Posts: 2
Question 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. Itís 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.
Attached Images
File Type: jpg rcLED circuit.JPG (7.1 KB, 51 views)
Reply With Quote
Old 12-09-2009, 05:34 AM
Paulo540's Avatar
Paulo540 Paulo540 is offline
Senior Member
Join Date: Nov 2009
Location: Central Coast, CA -8
Posts: 188

I like your idea but I reckon its gonna take about 6-7 farads at 5v to supply 30mins @ 10mA, and white (and blue) LEDs are generally around 3.6v operating.

Last edited by Paulo540; 12-09-2009 at 05:57 AM.
Reply With Quote
Old 12-09-2009, 07:31 AM
av.bhargav av.bhargav is offline
New Member
Join Date: Oct 2009
Location: Chennai
Posts: 7
Send a message via Skype™ to av.bhargav

Hi paulo540 can you please explain me how you have calculated the capacitance as 7 farads

Last edited by av.bhargav; 12-09-2009 at 07:36 AM.
Reply With Quote
Old 12-09-2009, 07:50 AM
Paulo540's Avatar
Paulo540 Paulo540 is offline
Senior Member
Join Date: Nov 2009
Location: Central Coast, CA -8
Posts: 188

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.
Reply With Quote
Old 12-09-2009, 08:22 AM
ping ping is offline
Junior Member
Join Date: Sep 2009
Posts: 13

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


R2 = (5-2.4)/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


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; 12-09-2009 at 08:58 AM.
Reply With Quote
Old 12-10-2009, 04:01 AM
Ooomph Ooomph is offline
New Member
Join Date: Dec 2009
Posts: 2

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
Reply With Quote
Old 12-12-2009, 07:10 AM
ping ping is offline
Junior Member
Join Date: Sep 2009
Posts: 13

No problem.
Reply With Quote
Reply   Post New Thread

, , , ,

Related Site Pages
Section Title
Worksheet Thyristor application circuits
Worksheet Class A BJT amplifiers
Worksheet Bipolar junction transistors as switches
Worksheet Time constant calculations
Worksheet Series-parallel DC circuits
Worksheet Kirchhoff's Laws
Textbook The Unijunction Transistor (UJT) : Thyristors
Textbook Biasing calculations : Bipolar Junction Transistors
Textbook Special-purpose diodes : Diodes And Rectifiers
Textbook Voltage multipliers : Diodes And Rectifiers

Similar Threads
Thread Thread Starter Forum Replies Last Post
Help! EMG Circuit Ocelot The Projects Forum 4 07-16-2011 03:18 PM
Power-On delay circuit tzitzikas General Electronics Chat 19 08-03-2009 05:49 AM
My first project -- how to use a circuit board? koggit General Electronics Chat 18 06-16-2009 07:21 PM
How to design an egg timer circuit? JLam Homework Help 2 03-21-2008 04:36 PM
Please help designing ekg circuit igloo The Projects Forum 3 04-05-2007 07:58 PM

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump

All times are GMT. The time now is 10:08 AM.

User-posted content, unless source quoted, is licensed under a Creative Commons Public Domain License.
Powered by vBulletin
Copyright ©2000 - 2014, vBulletin Solutions, Inc.