Basic question about volts

First of all, I am completlely new to electronics and have just begun trying to learn, so I apoligize if this is a super simple question.

I have a power source that is 5 volts. Is this 5 volts per circuit? or total?

i.e. if I have an LED that uses 2 volts, does that only leave 3 volts for other items to use? i.e. could I only connect 2 2 volt LEDs?

mike c

Reply to
mike c
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You can use any number of LEDs. Put a 150 ohm resistor in series with each one, and you can put any number of LED-resistor combos in parallel with each other. When you put items (such as LED-resistor combos) in parallel with each other, all receive the full supply voltage.

Just don't draw more current than is available from the power supply. Each LED-resistor combo should draw about .02 amp.

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You can put two 2V LEDs and a resistor in series with each other if the resistor is 47 ohms. You can put this load in parallel with other loads on the same power supply.

For more on resistors for LEDs, check out

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- Don Klipstein ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

The voltage of the supply is the force that drives current through any components you connect across the supply. As long as you do not exceed the current rating of the supply, it will maintain nearly 5 volts, regardless of how many current paths you connect across it. Within each of those paths, the total of all the voltage drops in series will add up to the total voltage of the supply.

So if you want to connect an LED that requires about 2 volts before it passes the required current (LEDs produce light roughly in proportion to the current passing through them), you will have to put something in series with it to consume the rest of the supply voltage while passing the needed current. Typically, this is a resistor. Ohm's law relates resistance, voltage drop and current. If you wanted .01 ampere (10 milliamps) to pass through both the resistor and LED, while the LED dropped 2 volts, but your power supply supplied 5 volts, you would need a 300 ohm resistor in series with the LED so that it would drop 3 volts while passing .01 ampere (because 3 volts divided by .01 ampere = 300 ohms).

For devices like resistors, that pass current in proportion to the voltage across them, you can think of ohms a word that means 'volts per ampere'.

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John Popelish
Reply to
John Popelish

Thanks. That helps clarify things.

One more questions. If I have multiple LEDs, can they all use the same resistor?

i.e. the 5 volt wire comes in, and has a resistor connected to it. Then multiple LEDs are connected to the resistor, and the each LED is connected to the ground.

Is that possible? Is this running the LEDs in Parellel?

mike c

John P> passing the needed current. Typically, this is a resistor. Ohm's law

Reply to
mike c

Thank you very much for explaining that.

mike c

Byr>

Reply to
mike c

Yes. Under certain circumstances.

That's not one of the circumstances.

It is running the LED in parallel. It's a dangerous game though because one LED will take almost all of the current leaving the others with none. And since you'll choose the resistor to spread the current evenly among the LEDs, that current hogger will get way too much current and burn up.

The circumstance that you can run all the LEDs off a single resistor is when you put them in series. When a circuit is run in series, then each item in the string will receive equal current. This means that the LEDs will have even brightness.

But in order to accmplish this you must have sufficient voltage for each LED. Blue and White LEDs can require as much as 5V each whereas red and greens will use between 2V and 3V. So if you have 3 red LEDS with a voltage of 2.1V then you'd need 6.3V minimum available to drive the string. You would then set the resistor to drop the remaining voltage at the current you wish the drive the string. So for example of you have a 9V source and you wanted to drive the string at 10ma then you would set the resistor value at

R=V/I R=(9-6.3)/.01 = 270 ohms.

If you are running in parallel, then each LED should get its own resistor.

BAJ

Reply to
Byron A Jeff

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