A Bit Confused

Discussion in 'General Electronics Chat' started by xxpbdudexx, Jan 16, 2010.

  1. xxpbdudexx

    Thread Starter New Member

    Jan 16, 2010
    I bought some electrical components today. I bought a printed circuit board, a 9v battery connector, a five-pack of 100, 220, and 330 ohm resistors, and a red LED rated at 5v and .08a.

    I learned about stuff like ohm's law in physics class so I figured I would apply it. I hooked up the LED and one 100 ohm resistor and tried to calculate things like current through a 9 v battery and then tried to calculate voltage drop across the 100 ohm resistor.

    As you may be thinking to yourself right now, that was pointless. When I figured out the current and then plugged it back in to V=IR, it said the voltage drop across the 100 ohm resistor was all 9 volts. The LED still lit so I clearly did something wrong. But what?

    Do LED's carry a resistance? Is they do, I think that's where I went wrong. I would need to use the 100 ohm resistor + the resistance of the LED to get the correct current, and thus the correct voltage drop across the 100 ohm resistor. Only problem is, I can't find any resistance reading on any of the packaging for the LED. So I don't know.

    One more thing. The relationship between current and resistance confuses me. The voltage is a constant of the battery you put in, but adding more or less resistors changes the current? Let's say I have 1 100ohm resistor hooked up to a 9v. the current is 90 mA. But if I add another 100 ohm resistor, it changes to 45 mA. Why does this change the current, aside from the obvious mathematical answer? Why would adding resistors change the flow of electrons?

    I have a 5V LED rated at 80 mA. Will it be brighter at 80 mA, than say, 50 mA? So then what effect would adding or subtracting voltage do?

    I realize this is already a sprawling post, so I'll stop it there. Thanks for the help guys!
  2. SgtWookie


    Jul 17, 2007
    LEDs are rated for their Vf (forward voltage) at a given current. 80mA sounds rather high for a typical LED; that might be the peak current - which would be for a very low duty cycle. It is usually safer to use the typical Vf @ current rating.

    A 3mm or 5mm red LED of the newer super-bright type might be rated for a typical Vf=2.0v @ 20mA.

    So, to calculate the current limiting resistor, you subtract the Vf from Vsupply, and then calculate what value resistor you would need to get a 20mA current flowing through the series circuit.

    The generic formula for calculating LED current limiting resistors is:
    Rlimit >= (Vsupply - Vf_LED) / Desired_Current

    If you want to play it safe, you'll calculate:
    Rlimit >= (9v - 2v) / 20mA
    Rlimit >= 7 / .02
    Rlimit >= 350 Ohms.

    Then we look at our table of standard resistor values: http://www.logwell.com/tech/components/resistor_values.html
    and we scan the E6 thru E24 columns, and find that 350 Ohms is not a standard value, but 360 Ohms is.

    We can use the 360 Ohms with the 7v to see what the circuit current will be:
    7/360 = 19.444...mA - close enough.
    But, you don't have any 360 Ohm resistors - you do have some 330's though.
    7/330 = 21.212...mA - well, that's a bit much. The LED will be brighter, but you may risk shortening it's life.

    They don't have a "real" resistance per se; it's conductance. They have a nearly logarithmic voltage drop across them according to how much current is flowing through them.

    It's like putting a kink in a garden hose, or closing down the valve on your bathroom sink. The more resistance there is, the harder it is for current to flow.

    You control the brightness of an LED by current flow, not by voltage.

    By the way, 9v batteries are generally rated for around 150mAh - and that's at a 20-hour discharge rate, or 7.5mA. If you try to operate your LED at 80mA, you'll drain the battery flat in less than an hour.

    If you're sure you want to operate it at 80mA, then:
    Rlimit >= (9v-5v)/80mA
    Rlimit >= 4/.08
    Rlimit >=50 Ohms
    You could use two 100 Ohm resistors in parallel to get 50 Ohms.
    Your battery will be dead very, very quickly.

    With the 100 Ohm resistor, the current was roughly 4/100=40mA.
  3. xxpbdudexx

    Thread Starter New Member

    Jan 16, 2010
    That's what it says on the package. Maybe it doesn't matter, but I forgot to mention that it is a blinking one.
    Does the voltage drop actually have anything to do with the current?

    Alright. maybe a more direct way to put it is, how would you find the current of the circuit and how would you find the voltage drop across the 100 ohm resistor?
    That makes sense.

    I'm actually not even sure what I'm doing. I don't know how to find the current or anything. I did learn all this stuff, but it makes no sense with only one resistor.

    What is the 5 V for? Maximum voltage level?
  4. SgtWookie


    Jul 17, 2007
    OK, so it has some electronics along with the LED then.

    Yes. At lower current, the Vf will be somewhat less. If you run much more current through it though, it will have a short life.

    We already know what the current will be.
    The LED is rated for a Vf of 5v @ 80mA.
    So, you subtract that 5v from the supply voltage, and you have 4v remaining.
    Then you divide the remaining voltage by the desired current of 80mA, and wind up with 50 Ohms required. If you then wire a 50 Ohm resistor with the 5v @ 80mA rated LED across a 9v supply, you should wind up having 80mA total circuit current, 5v across the LED, and 4V across the 50 Ohm resistor.

    Poke around in our E-books; there are links to it at the top of every page on the website. ;)

    No, that's the voltage drop across the LED when it's rated current is flowing through it.
  5. xxpbdudexx

    Thread Starter New Member

    Jan 16, 2010
    Ok, alright. Thanks then.
    Last edited: Jan 16, 2010