Hello All

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
 
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

**broken link removed**
 
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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.
 
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.
 
I found this:
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"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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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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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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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.
 
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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one question: How are we multiplexing the led's. what we are using to multiplex them?
 
you would need something like a M4_1E:
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As a matter of fact look at MAX7219, MAX7221:
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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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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?
 
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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