# resisters for LED's

Discussion in 'The Projects Forum' started by lotusmoon, Jul 24, 2013.

1. ### lotusmoon Thread Starter Member

Jun 14, 2013
203
4
Hi I am a beginner
I am making an LED flasher of 55 LED's.
there are 6 lines of 8 LED's and 1 line of 7 LED's.
The start voltage will be 15 volts prior to the 555 chip and drop to about 13 volts by the time it leaves the 555 chip.
I will be having a resistor before each line of LED's. I would like each LED to run at about 20MA, but am unsure what to put in the ohm's law equation as the voltage. Is it 13 volts, or the voltage drop of an LED, or the full voltage drop of the series of LED's, or the voltage leaving the LED's.?

2. ### mcgyvr AAC Fanatic!

Oct 15, 2009
4,770
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You need to know the forward voltage rating of the LED's also.. (see their datasheet)

3. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
You aren't trying to power all these LEDs from the output if the 555, right? Perhaps you could post a schematic if what you are describing...

Your LEDs will be destroyed with 20 mega-amps through them, better stick to 20mA...

If you don't have the datasheet, you could forward bias the LEDs, and measure the voltage across them...

4. ### LDC3 Active Member

Apr 27, 2013
920
160
The resistance needed would be (roughly)
R = (13V - 8 * LED voltage drop) / 20mA
or
R = (13V - 7 * LED voltage drop) / 20mA

With 140mA of current being supplies through the 555, it may get hot.

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5. ### lotusmoon Thread Starter Member

Jun 14, 2013
203
4
Sorry there have been a few more replies since i started writing this -

There are different LEDS. the values I have been using to see how many i can have in series is - IF = 100 mA, tp = 20 ms VF typ some are 1.3v others 1.5 or 1.8. Should i take an average value of the line of LED's?
below is a copy from the data sheet -
BASIC CHARACTERISTICS -
Forward voltage
IF = 100 mA, tp = 20 ms VF typ - 1.5 max - 1.8 V
IF = 1 A, tp = 100 µs VF 2.3

Forward voltage
IF = 100 mA, tp = 20 ms VF typ -1.35 max 1.6 V
IF = 1 A, tp = 100 µs VF typ- 2.6 max - 3 V

Last edited: Jul 24, 2013
6. ### LDC3 Active Member

Apr 27, 2013
920
160
To get the most accurate resistance, you would subtract the forward voltage of each LED in the string from the supply voltage and then divide by the current. Since the forward voltage is variable, you would need to measure each LED to increase the accuracy.

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7. ### lotusmoon Thread Starter Member

Jun 14, 2013
203
4
thank you all

8. ### lotusmoon Thread Starter Member

Jun 14, 2013
203
4
I was going to use a transistor to deal with this heat but I had stepped back from it as I did not understand the different calculations concerned. But that was from the internet and from this forum people level the info at a level that I can understand. so if possible could some let me know how to do this.
I was looking at npn transistors.
the things I think i need to understand are. -
I was going to have the npn on the negative wire from the LED's
I think there needs to be a ratio in volts between the collector and base I am unsure what this should be.
I was also told I needed a capacitor after the emitter, I don't know what this is based on or how to work out what value it is.
I do not have a schematic but the basics of what I am doing are covered earlier in this thread, thank you

9. ### wayneh Expert

Sep 9, 2010
12,130
3,052
You're getting close. The emitter goes to ground. No capacitor needed if you're using the 555 to switch the transistor. You might use a capacitor if you were driving a speaker, but you are not.

You WILL need a resistor from the 555 to the base of the transistor. Whatever current you need for your LEDs, aim to put 10% of that onto the base of the transistor. For instance 10mA on the base to switch 100mA LED current.

The 555 output will be close but not quite equal to the power supply voltage. Look it up. Then use ohm's law to calculate the right resistor: R = ∆V/I where I is your base current, for example 10mA, and ∆V is the output voltage of the 555.

10. ### Dodgydave Distinguished Member

Jun 22, 2012
4,997
745
Use a circuit like this for driving the leds with an NPN or better still a Mosfet n-channel

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11. ### wayneh Expert

Sep 9, 2010
12,130
3,052
Since you have plenty of voltage to drive the MOSFET, I agree it is a better choice than a BJT. No worries about gate current - you can connect it directly to the 555.

12. ### lotusmoon Thread Starter Member

Jun 14, 2013
203
4
Thank you for all of this feed back.
A few people have said there is plenty of voltage to drive a mosfet.
I was going to have the transistor on the return wire where there would not be much voltage after it had gone through all of the LED's. should I have the transistor on the positive feed wire where there will be 15 volts?

13. ### LDC3 Active Member

Apr 27, 2013
920
160
A n-channel MOSFET does not need a high voltage across the source and drain since there is very little resistance when it is fully on. It needs more than 5V (depends on the MOSFET) across the gate and source to fully turn on the MOSFET. Having the MOSFET in the low position of the string allows you to have the voltage needed to turn on the MOSFET. If you have an n-channel MOSFET at the high end of the string, you would need an additional 5V above the voltage at the drain (which you wouldn't have with a single voltage supply).

14. ### lotusmoon Thread Starter Member

Jun 14, 2013
203
4
thank you for this.
excuse my newness but can I check some things in my vocabulary.

"source and drain" is this the collector and emitter
"gate and source" is this the base and collector
" low position of the string" is this before the LED's so the voltage is still 15 v

15. ### LDC3 Active Member

Apr 27, 2013
920
160
The connections on a MOSFET are given different names then what is used on a BJT. The drain of a MOSFET is typically connected to where the collector would be connected, the source is connected to where the emitter would be connected. See here.

In the driver for a motor called an H-bridge, there are 2 high side transistors and 2 low side transistors. In this diagram, Q2 and Q4 are the high side transistors and Q1 and Q3 are the low side transistors. So, no the transistor is not at 15V.

16. ### wayneh Expert

Sep 9, 2010
12,130
3,052
No, "low" refers to the voltage, and a normal N-type MOSFET is used to switch the low side, the path to ground. This allows the source to be at a low voltage, which makes it easier to get an adequate relative voltage on the gate. A MOSFET wouldn't work well in an emitter-follower configuration, where the emitter voltage can rise up. A BJT works as an emitter-follower because you need only about 1V more on the base above the emitter voltage. A regular MOSFET needs 10V.

17. ### lotusmoon Thread Starter Member

Jun 14, 2013
203
4
thank you thats really useful.
ok so the gate will be about 13 volts from the 555. the voltage at the drain will be less than 3 volts. I don't know what the voltage at the source is I guess it is an average of the base and drain.Does this work or do I need to add or change any thing?

18. ### LDC3 Active Member

Apr 27, 2013
920
160
Actually, the drain is connected directed to ground, so it would be 0V. When the MOSFET is fully open, the voltage drop from source to drain is low (due to the low resistance), less than 2V (it may even be less than 0.5V).

Correction: Drain should be Source.

Last edited: Jul 26, 2013
19. ### lotusmoon Thread Starter Member

Jun 14, 2013
203
4
now I am a bit confused I may have this all wrong, I had thought that the drain was connected from the LED'S and source connected to earth. http://diyaudioprojects.com/Solid/ZCA/ZCA.htm
"The connections on a MOSFET are given different names then what is used on a BJT. The drain of a MOSFET is typically connected to where the collector would be connected, the source is connected to where the emitter would be connected. See here."
and with npn transistors i had thought that the collector connected from the LED's and the emitter connected to earth.

20. ### lotusmoon Thread Starter Member

Jun 14, 2013
203
4
I may be getting this the wrong way round.
from this quote does that mean that drain is connected directly to the earth.
And that the source is connected directly to the load in this case the LED's.?