# LED and pull-up resistor

## Main Question or Discussion Point

I am getting ready for my lab tomorrow and one of the questions is this:
what happens if you connect LED to a pull-up resistor circuit/switch. First connect LED's anode to the switch and cathode to the ground. Next connect LED's anode to +5V and cathode to the switch. Describe the results.
ok, here's what I think: since current flows from anode to cathode and in first case LED's anode is connected to the pull-up resistor, so there's current through it and it's going to light. In second case, wouldn't it burn? since there's no resistance between the LED and the source and LED's have little resistance, so it's going to shine brighter?
If i am wrong could you please explain? i doubt i got this right.
ps: we have not put the circuit together yet, so it's hard for me to visualize.

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berkeman
Mentor
Yes, if you put more than a couple volts across an LED, it will burn up. The forward voltage drop across most LEDs is around 2V. So to put 10mA through the LED from a 5V supply, you would use a series 3V/10mA = 300 Ohm resistor.

Some LEDs come with built-in resistors, and those don't require an external current-limiting resistor. But jellybead LEDs are just diodes.

berkeman said:
Yes, if you put more than a couple volts across an LED, it will burn up
then either I am missing the question or why are they telling us in the lab to hook the anode of the LED to the +5V source and cathode to the switch? if it will burn up.
Oh, and we are using just jellybead LEDs.

Thanks again.

Averagesupernova
Gold Member
What is on the other side of the switch? It doesn't matter what part of the circuit the resistor is in as long as it is in series with the LED. I guess I would have to say there is not enough information for me to give a definite answer.

berkeman
Mentor
EvLer said:
then either I am missing the question or why are they telling us in the lab to hook the anode of the LED to the +5V source and cathode to the switch? if it will burn up.
Oh, and we are using just jellybead LEDs.

Thanks again.
Probably just a poorly worded assignment. Hopefully the lab went well for you.

Heck, I remember back a zillion years ago when I was just starting to learn electronics (end of high school I think), and I went to buy some LEDs from Radio Shack to play with. The recommended hookup circuit picture on the back of the package showed the LED backwards in the circuit! I was thinking to myself, "you have to zener the LED diode to get light out of it???" I asked the sales guy at Radio Shack about it, and he says yeah, that's how they work. :grumpy:

Thanks for the replies, yeah... the lab is coming up today. And unfortunatelly i do not know what would be connected to the switch. This is a "tough question" at the end of the lab that we are supposed to answer. So, i hope my partner and i will figure it out once we build the circuit.
thanks anyway!

Ok, so the lab is over! woo-hoo! :tongue2:
What the question meant was basically that in first case the LED would not be very bright because part of the voltage would be sucked up by the resistor (series connection) and there would not be enough voltage for the LED to shine, when in second case, the LED is directly connected to the voltage source, so it would light up brightly. Seeing stuff on the bread-board helps so much!
Oh, and it would not burn, since (like berkeman mentioned) some LEDs do have small resistors built in. Well, actually, the kits they made us buy had 3V LEDs, which would burn, but the instructor exchanged those for 5V with resistors built in (black dot?).
So...one lab down 15 more to go :surprised

Averagesupernova
Gold Member
Ahhhhhhhhhh. So your instructor is to blame for your confusion. As far as I am concerned an LED with a resistor built in is convenient and all but for students is not a wise choice. So what if a few students blow them up? That's how we learn. Just glad you got it figured out.