Discussion in 'Electronic Basics' started by mike c, Dec 30, 2004.

1. ### mike cGuest

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

2. ### Don KlipsteinGuest

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.

--------------------------------

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
http://www.misty.com/~don/ledd.html

- Don Klipstein ()

3. ### John PopelishGuest

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'.

4. ### mike cGuest

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

5. ### Byron A JeffGuest

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

6. ### mike cGuest

Thank you very much for explaining that.

mike c