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Electronics => Electronics => Topic started by: garriwilson on March 12, 2008, 08:31:39 PM

Title: LED circuit
Post by: garriwilson on March 12, 2008, 08:31:39 PM

Hey everybody. In my engineering club, we're making this electric circuit with LEDs. We're using breadboards for the whole thing and we're supposed to make letters with them.

The battery goes first, then the LED?? Isn't the resistor (which is after the LED) supposed to be first to limit the current going to the LED? Or does the placing in the circuit not matter?

Here is the link, go to page 5, you'll see the schematic:

http://www.fsea.org/pdf/EC2%20Electric%20Circuit%20Breadboardpdf.PDF (http://www.fsea.org/pdf/EC2%20Electric%20Circuit%20Breadboardpdf.PDF)

Title: Re: LED circuit
Post by: Soeren on March 13, 2008, 07:25:27 AM

Hi,

Quote from: garriwilson on March 12, 2008, 08:31:39 PM

The battery goes first, then the LED?? Isn't the resistor (which is after the LED) supposed to be first to limit the current going to the LED? Or does the placing in the circuit not matter?

<it doesn't matter the least where the resistor is placed, as long as it is in the circuit, it will limit current.

Title: Re: LED circuit
Post by: ed1380 on March 13, 2008, 09:14:42 AM

normaly I'd go with what soeren said, but this site has it on the positive side. http://led.linear1.org/1led.wiz
maybe they chose it for simplicity and not a real reason. but I cant be the judge of that.

Title: Re: LED circuit
Post by: paulstreats on March 13, 2008, 10:30:50 AM

i think if its on the negative side, it still has to pass through the led in order to sink through the led. I imagine that the initial startup would have a high current like this before it levels out accordingly, so putting it on the positive side will stop the initial spike

Title: Re: LED circuit
Post by: bukowski on March 13, 2008, 10:46:49 AM

unless an LED has wierd properties, there wouldn't even be an initial spike, current throughout the circuit remains the same. you can put the resistor where ever you want.

Title: Re: LED circuit
Post by: Trumpkin on March 13, 2008, 10:57:59 AM

IT DOESN'T MATTER WHAT SIDE THE REISISTOR IS ON!!!  ;D What Bukowski said

Title: Re: LED circuit
Post by: garriwilson on March 13, 2008, 06:33:15 PM

OH, I thought the current goes from - to +. So, the current is at all places at the exact instant when the battery is  plugged in? Doesn't it take time for the electrons to go from - to + ?

Title: Re: LED circuit
Post by: paulstreats on March 13, 2008, 10:20:49 PM

thats what i thought.

Normally elctrons move -to +. But conventiannal current moves + to -

Surely there is a time when the current flows through the led before it hits the resistor and reduces, There are a lot of physics theories pointing to this but there is nothing factual(other than all the world placing the resistor between the + and the anode)

Title: Re: LED circuit
Post by: garriwilson on March 13, 2008, 10:33:41 PM

If I were to attach the resistor to one of the battery terminals, then what you are saying that I could put it on + or -, my choice? The LED won't burn out?

Title: Re: LED circuit
Post by: Trumpkin on March 14, 2008, 08:28:49 AM

yes, if you want to you might as well put it on the + side. I'm almost positive it doesn't make a difference.

Title: Re: LED circuit
Post by: bukowski on March 14, 2008, 01:24:58 PM

Current in a series circuit is the same throught the circuit no matter where the components are placed. No, the LED wouldnt burn out. It does take time for electrons to travel the circuit (speed of light), but the potential difference and resistance influencing current flow is instantaneous. 

Title: Re: LED circuit
Post by: garriwilson on March 14, 2008, 05:15:56 PM

Thanks guys! It's very clear now.  ;D