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so I've recently begun my venture into electrical engineering, just got myself a breadboard with some jumper cables, a pack of assorted LED's and a few 200 ohm resistors.

I'm trying to understand how to further limit voltage to my LED using a 9V battery and multiple 200 ohm resistors but I've come up with the following question / problem.

enter image description here

Judging from this image: can you please tell me if this is wired properly? I have 7 200 ohm resistors connected in this circuit and it doesn't appear to be dimming the LED at all. When I take away 3-4 resistors, it's still just as bright. @ 1400 ohms, I figured the LED's would be significantly dimmer but alas they're not.

The specification of the LED i'm trying to power with my 9V battery is, Green, 2.9-3.1V with a forward voltage of 20mA

Anyhow, I'm having a heck of a time tying to wrap my head around this and figure it's something small I'm missing. When I touch all 7 resistors connected to this circuit, I can definitely feel a significant amount of heat coming off of them.

Would anyone be able to assist me with this issue I'm having? any advise or criticism is appreciated.

Thanks all.

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    \$\begingroup\$ You are connecting the resistors in parallel, you don't want that because it will kill your LED. Put them in series, make one single long chain of the resistors and the LED. At 9V you want at least 2 or 3 resistors in series \$\endgroup\$ – jippie Oct 27 '12 at 22:34
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First of all, lovely picture. This makes it easy for me to see what's going on.

Looks like you are adding the resistors in parallel and so each additional one is reducing the total resistance.

Try hooking them in series instead and you should see the LED get dimmer.

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  • \$\begingroup\$ Thank you for the information, my experiment appears to be functioning properly now that I've connected the resistors in series. \$\endgroup\$ – Clu Oct 28 '12 at 2:39
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    \$\begingroup\$ Great answer, further reading for the OP: en.wikipedia.org/wiki/Series_and_parallel_circuits \$\endgroup\$ – Robert Atkins Oct 28 '12 at 3:55
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Using 9V source and 3V drop on LED choose the series R to control the current with 6V across R. Use higher R not lower. e.g.

  • 6V / 1mA = 6K ohm
  • 6V /10mA = 600 ohm
  • 6V/ 20mA = 300 ohm Pr= 120mW in R

whereas you are doing is burning out the LED..

  • 6V/ 200 ohm = 30 mA Pr= 180mW in R
  • 6V /100 ohm = 60 mA Pr= 360mW in two R ... etc oh oh

With 9V battery and 2 similar 3V LED's in series. 9V - 3 - 3 = 3V across R 3V / 20mA max = 150 ohm min. then 1500 ohm for 2 mA

This approximation works well for low power LED's, but when you get into high power LED's the voltage tolerance on the supply and variation of LED voltage whenn running at say 300 or 1000 mA, R gets much smaller and more critical of your assumed values and also factor the internal resistance of the power source and LED as well becomes inmportant in the 1ohm range. Here the LED has an ESR of ~10~15 ohm, so you can neglect.

Think of LED's as Zeners with a tolerance.

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