LED clock too dimm

[prenavin]

More current please

Well, i pulled out one leg of the resistor and connected my ammeter in series. I measured the volt drop Of each LED, it 1.71V. Its not the high bright type, normal GaAs- model BL-B5134 bright red. I want to get 60mA through them when connected directly. So that when i multiplex them by 6, the average current will be 10mA.

 

[audioguru]

Hi Prenavin,
You didn't see the spec's for yor LEDs did you?
They do have a fairly wide angle (35 degrees) but they aren't very bright (1.2mcd) and their absolute maximum current is only 7mA!

 

[bonxer]

Re: More current please

I want to get 60mA through them when connected directly. So that when i multiplex them by 6, the average current will be 10mA.

You are confusing series and parallel. If you chain them all together end to end, then they will all have the exact same current which the resistor does. If, on the other hand, you connect them in parallel with all anoces connected to each other, and all cathodes connected, then each one will use a fraction of the current coming through the resistor.

And as everyone else is saying, yes, you will see all the magic smoke coming out of your diodes if you put 60mA through them.

 

[prenavin]

I have made an amazing discovery thats gonna change the electronics industry forever. He he he ! I finally understand what you guys have been telling me over the past week.

Firstly, if i had 10 LEDs in series, that would create a volt drop of 17V. If i supplied 17V to these LEDs, there would be no need for a current limiting resistor, as these LEDs would bias themselves. If i wanted to increase the current through them, i would have to INCREASE THE VOLTAGE. Aha! At 25V, 10mA should flow through them, with no current limiting resistor. More voltage = more current. Of course using a resistor is a good design practice when high voltage allows more than 20mA through the LEDs.

Sometimes i get caught up in the complex stuff and tend to forget the basics.
Is there any formula to calculate voltage, for the required current.

PS: Audioguru, thanks alot. i definately need to change my LEDs.

 

[Nigel Goodwin]

Re: More current please

I want to get 60mA through them when connected directly. So that when i multiplex them by 6, the average current will be 10mA.

You are confusing series and parallel. If you chain them all together end to end, then they will all have the exact same current which the resistor does. If, on the other hand, you connect them in parallel with all anoces connected to each other, and all cathodes connected, then each one will use a fraction of the current coming through the resistor.

And as everyone else is saying, yes, you will see all the magic smoke coming out of your diodes if you put 60mA through them.

You appear to have missed the most relevent word "multiplex", this means rapidly switching between six chains of diodes, giving an average of 1/6th (or 10mA in this case) to each chain of diodes.

However, this 10mA still exceeds the maximum for these diodes.

Perhaps he ought to post a diagram to show us what he is doing?.

 

[audioguru]

Hi Prenavin,
You must use a current-limiting resistor with LEDs because their voltage changes with temperature and they won't all be the same.
Also, look at the voltage vs current curve of a low-voltage red LED. It is almost vertical: 1.6V at 1mA, 1.7V at 10mA, 1.8V at 40mA etc.
I have been using Fairchild's MV8141 and MV8191 red LEDs and they are much brighter than yours. https://www.fairchildsemi.com/ds/MV/MV8191.pdf
Here's their voltage curve: