I have been having trouble trying to solve this homework problem. I am to design (on paper) a simple circuit using these materials (don't necessarily need to use all of them): -as many as 6 one volt batteries -a 25 ohm resistor -a 100 ohm resistor -a 3300 ohm resistor -a LED The LED needs at least two volts and 10 mA to light up, and anything over 25mW of power will burn it out. The closest I got was just using two volts of power and 125 ohms of resistance, which game me 16mA of current but my power was 32 mW and everything else I tried either gave me too much power with the small resistors or not enough current with the 3300 ohm resistor. Any hints/tips/ideas are appreciated.
Hello, Did you read the forum rule? http://forum.allaboutcircuits.com/showthread.php?t=3002 What have you done upto now to get the solution? Where are you stuck? Bertus
I tried using two batteries (two volts) and the 100 ohm and 25 ohm resistor. This game me 16mA of current but 32mW of power which is slightly too high. I tried other combinations but found that if I used the 3300 ohm resistor I'd never have enough current and without it I always have too much power.
Hello, Can you show us how you calculated the 16 mA? Of wich component may the power not be larger as 25 mW? Bertus
The LED can't have more than 25 mW of power running through it. I did I=V/R v=2 R=125, 2/125=.016 A or 16 mA. then for power P=V^2/R so 4/125 which equals .032W or 32 mW
Hello, What is the voltage drop of the LED (Vled) ? Iled = (Vbatteries - Vled) / Resistor_in_series_with_led Bertus
I am not sure. If I could build the circuit I could measure the voltage across the LED but I don't have the gear to build it and am not sure how to calculate it.
Hello, From your assignment: Vled will be 2 Volts at 10 mA. You can also read this thread by Bill_Marsden: LEDs, 555s, Flashers, and Light Chasers Can you now complete the task? Bertus
Right, but doesn't that make the power going through the LED over 25mW? P=V^2/R P=2^2/125 p=4/125 P=.032 W or 32 mW
Hello, Take a look at this post: http://forum.allaboutcircuits.com/showpost.php?p=117637&postcount=2 As you see there is a voltage drop over the led. The battery voltage must be higher as this voltage drop. The resistor limits the current using the voltage difference between the battery and the led voltage. Bertus
I can't wait to see the answer! I think I have it, but my answer is an example of a basic electronics confusion that I have.
I'm still confused though. The problem said there needs to be 2V to light the LED as well. This means I need more voltage than 2V then because there won't be 2V going into the LED, correct?
Yes, your circuit will have a slight voltage drop, that will lower the battery voltage before it reaches the LED.
Okay so I guess my main question is how do I increase the voltage and current without increasing the power through the LED? When I calculate it using 3300 ohm resistor my current gets way too low (even if I use the 6 Volts i am allotted) so I am assuming the 3300 ohm resistor cannot be used. I=V/R I=6/3300 I= 1.8 mA
Did you read bertus's link? What did you understand or didn't understand from it. I has the form of the circuit you need to make right in the beginning,
I read the link and I got that the LED needs 2 volts to turn on and then runs on current. I'm just not understanding how to apply it I guess. I've been trying to hammer it out but still struggling. I've been using V= IR to calculate voltage across the resistor and thus the voltage drops but I haven't been able to get it to work.
You know that your LED needs 2V to run. You also know that a current I will go through it, which will be in the area of 10mA. You know that a resistor will be placed in series with it. The resistor will have the same current as the LED. Try combinations of source voltages and resistances to get the best result. Remember that the voltage that is applied on the resistor is Vsource-Vled=Vsource-2.