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Trying to wire leds to a 9v battery. I am confused by different instructions. Answered

Some say the resistor gets attached from the positive battery lead to the first led. Others say that the resistor is attached to the neg pole of last led and the ground wire of the battery. What the Hey?


So just think of series leds as subtracting from the voltage, and its the remainder that damages your leds. diodes are semiconductors that can handle a voltage up to a point, then they just become regular old conductors. It's the current that damages them, and the current flowing through them that damages them, and that current can only flow if the voltage is too high.

so 9v in
and 2v-2v-2v-2v out
leaves 1 volt left. all the diodes are saturated at 2 volts each, so 2.25 volts will damage them. That 56 ohms is to reduce the last volt to a manageable 0 extra volts.

Yes, I get that much, and Thank you for taking the time to type that up.I am confused as to Why some diagrams show resistor on positive lead,and some have it on the negative lead.

because unlike diodes, resistors resist wherever they are, in any polarity. You could add it between the leds in the strand if you wanted to be really convoluted.

There are only a few times you need to pay attention to where the resistor actually is, and that's because things like transistors need specific voltages and currents at certain places in the circuit.

Electricity doesn't care (*usually) if the potential difference (voltage) is a push or a pull, positive or negative, with regard to passive components like resistors, they just resist that voltage wherever they are, and the circuit responds to that effect.

You didn't tell us enough about your project. But, if you're trying to wire up led lights to a 9 volt battery it does not matter which pole you put the resistor on. Since it's a circuit the resistor will do it's resisting. If the circuit is something else you must give us more info since in other circuits it MAY matter which pole connects.

My apologies!
My plan is to connect a 9v battery to 4 2v yellow leds in series. The diagram calls for a 56 ohm, 1/8 Watt resistor. Have batteries, leds, and (ta-da) resistors. I don't know enough about electricity. I thought the resistor would need to be between the positive wire and the first led, to keep the full 9v from running head first into the poor little leds. O.o

It doesn't matter if the resistor is at one end or the other. You could also have it in the middle of the string of LEDs. Just as long as you have the resistor in there. As others have said the resistor is there to slow the current flow through the LEDs.

Think of the circuit like a loop of marbles. Your power source is a power wheel that take one marble and pushes it into the rest. While that wheel may be able to push 1 marble a second your resistor or a choke point slows it down so that only 1 marble passes through the wheel every 5 seconds. No matter where that choke point is in the loop the marbles are only able to move at 1 marble per 5 seconds cause the wheel is only getting the next marble every 5 seconds. And the marbles are all grouped up just outside the output of the wheel so when it shoots the next marble out it doesn't go very far very fast. The same goes for the electrons in the wires and LEDs. They are all grouped up waiting for the battery to push out an electron. When the circuit is closed the battery is able to push out an electron causing all the electrons in the chain to be pushed. The resistor slows the speed at which the electrons are being push and it takes slightly longer for an electron to return to the battery and the next one to be pushed out.

Believe it or not that is the very basic idea of what is going on there. Of course there is a lot more to it.

Huh. OK.
For some reason I was seeing the electrons like a train that would plow through anything until it was slowed by the resistor. Thanks!