# Hello All

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#### Brian Hoskins

##### New Member
The main application for a conventional diode is not it's forward voltage drop but the fact that current is only able to flow in one direction through them (apart from a small leakage current when reverse biased). As Hero said, the forward voltage drop is dependant on current, and to a smaller amount other factors such as temperature. As an example, if you read the datasheet for a general purpose diode like the 1n4001 you'll see a graph which plots forward voltage against current. The device is rated up to 1Amp, and from 0 - 1Amp the forward voltage varies by 0.4V - this is why they're not very good voltage droppers, unless the circuit current is known and remains fairly constant.

Brian

#### AtomSoft

##### Well-Known Member
ok im trying this and the LED is getting 1.75v . i am having trouble calculating the I = E / R. I get some wierd number ."0.0002459419" Can someone explain. or would this just be 245mA

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#### Pommie

##### Well-Known Member
If you remove the 10k resistors and just keep the 330Ω you will then get a voltage of approximately 2V across your LED. The other 3V will be across the resistor and therefore the current will 3/330 = 9mA. To make your LED bright you want around 20mA through it and so the resistor should be 3/0.02 = 150Ω.

Mike.

#### AtomSoft

##### Well-Known Member
How can i calculate like you just did? Like how did you calc that when a 330Ω is placed there it would generate 2v on the led? I see how to get info from ohms law but like to create a outcome based on input is not yet understood lol.

#### AtomSoft

##### Well-Known Member
I found this:
http://www.doctronics.co.uk/resistor.htm

"A typical LED requires a current of 10 mA and has a voltage of 2 V across it when it is working. The power supply for the circuit is 9 V. What is the voltage across resistor R1? The answer is 9-2=7 V. (The voltages across components in series must add up to the power supply voltage.)

You now have two bits of information about R1: the current flowing is 10 mA, and the voltage across R1 is 7 V. To calculate the resistance value, use the formula:
R = V / I

Substitute values for V and I:"

So in mines it would be like

5V in and LED eats 2v so 3 left over
R = 3 / .010 = 300ohms

right ?

Now is it possible to use 3 100ohm in series to make a 300ohm resistor ?

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#### Pommie

##### Well-Known Member
Correct, LEDs will conduct more and more current until the voltage across them is 2V. As more current flows a bigger voltage is dropped across the resistor. When the current is high enough the voltage across the resistor is 3V and the current stops increasing.

Mike.

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#### AtomSoft

##### Well-Known Member
ok with this same concept i can manually control a 7 segment led right?

Also the forward voltage/current is the input the component actually needs to work right.

I tried this with a 7seg display works 100%. Give me a min to up a pic of it

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#### AtomSoft

##### Well-Known Member

i like it lol

of course a microcontroller or pic w/e its called will replace the wiring in future. lol

#### Pommie

##### Well-Known Member
It's looking good. Just be aware that the segments are sharing the resistor and so the more segments lit the duller they will be. This can be rectified by having a separate resistor for each segment.

Mike.

#### AtomSoft

##### Well-Known Member
ok cool. Will note. Its just that its like 2am here and everyhere is closed so i didnt want to blow it lol. Um also i took a Dual 7 SEG LED off something i had around but its funny it only has 9 pins

I got it to light up tho just going to make a pinout now

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#### AtomSoft

##### Well-Known Member
Look but how would i control each seperately ?

#### Pommie

##### Well-Known Member
You have to multiplex them. See link.

Mike.

#### AtomSoft

##### Well-Known Member
Cool just looked at it makes sense. I guess timing is everything lol im gonna take a small sleep now . Ill be on in like 3-4 hr

##### Member
very informative!!

##### Member
one question: How are we multiplexing the led's. what we are using to multiplex them?

#### AtomSoft

##### Well-Known Member
Ok im having a issue sort of. I connected to resisters in a series on a 5v supply
R1 = 100 ohms
R2 = 150 ohms

My voltage drop is ok i end up with 3 volts in the middle
but i end up with only 5mA instead of 20mA how come?

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#### kchriste

##### New Member
Forum Supporter
You should have 20ma.... Some meters will drop a little voltage in current mode, but it shouldn't be enough to make THAT much of a difference. I assume the meter is in series with both the resistors. Maybe it is defective?

#### AtomSoft

##### Well-Known Member
oh didnt notice something lol meter was grounded . I get about .016 now is that normal?

#### kchriste

##### New Member
Forum Supporter
It's a little low. You might want to check your meters accuracy against another one; Just connect both meters in series with a resistor and a voltage source, forming a loop, to see if they read the same.

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