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Forum Systems : Homebuilt Breadboard forum

Hello,
i am using a potentiometer but the led doesn't seem to react to it. I still have no idea about breadboarding, have just bought my first one. can you tell me if i am building this circuit right? http://www.foroselectronica.es/f43 [...] -7185.html

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Message edited by beginner0 on 01-27-2012 at 02:24:50 PM

Reply to beginner0

It doesn't look right. Please post the maker and model number of the breadboard. I can only guess without that which holes connect with which. Also, your picture does not show the whole thing. Where does the green wire go? Anyway, normally an LED will not have a resistor in parallel with it but a series current-limiter. I would expect resistor-pot-LED across the supply.

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Reply to Petrofsky

Petrofsky wrote :

Hi, ok:

breadboard - MB3. it says nothing else on it.

the green wire goes to an interruptor which is also connected to the battery.

it seems that the led must have a resistor, according to our teacher, but i dunno how to put it on the breadboard.

Reply to beginner0

beginner0 wrote :

I guess that depends on what you're trying to achieve.

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Reply to FinneousPJ

FinneousPJ wrote :

I guess that depends on what you're trying to achieve.

the light (LED) should be brighter/dimmer when i move the central part of the potentiometer, that's all. I know I haven't connected it properly, but-

Reply to beginner0

An LED is a semiconductor junction, pure and simple. A silicon one drops 0.6 volts. That's the magic of a junction---it always drops the same voltage when it is forward biased. If you connect the battery across one as you have done, it will pop, because you are asking it to drop the entire battery voltage, and it will try to draw enough current from the battery to bring the voltage down to 0.6. There needs to be a resistance in series with a LED to drop the voltage difference---a 12-volt battery will drop 11.4 volts across the resistor. By Ohm's law, the resistor will essentially dictate the current in the circuit, and therefore the brightness of the LED. A potentiometer might present zero ohms when turned all the way down, so you can't use it as the current-limiting resistor. There has to be a fixed resistor in series with it. Like I said, connect it -supply/cathode/anode/1k/pot/switch/+supply.

Is this a school project or something?

What I think you should do is plan what you need and want, draw a circuit diagram and try to recreate it. That's a learning experience.

Reply to FinneousPJ

OK. Let's assume you have a supply voltage of 12 volts. 20 ma through an LED will cause it to illuminate about medium brightness. I wouldn't put more than 30 ma through the LED. So the main current limiting resistor should be (11.4 volts/30 ma) or 380 ohms. Next larger 5% value is 390 ohms. 1/2 watt resistor is large enough. 1 Kohm is a good size for the variable resistor.

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Reply to jsc

jsc wrote :

Hmm... by the moment, i am an absolute beginner and don't have to calculate anything. Besides, i am sure i got all the elements needed for this job, i just took the list to a shop and bought them.
All i'd like to understand is how to connect a potentiometer and those two resistors. I mean, how should i put them so that they become part of the circuit in which there is an led? i don't want to short or spoil it

Reply to beginner0

well, finally i got it, but still have a question: power reaches the leds through the potentiometer, but when i adjust this one, i don't see any changes in the lights. More precisely, i have seen them go brighter/dimmer only at the beginning, but now they don't seem to vary.

what's wrong?

[url=http://www.foroselectronica.es/f43/interruptor-3-patas-ponerlo-la-protoboard-7193.html#post16691][/url]

Reply to beginner0