# Working out resistors? HELP!!

Discussion in 'General Electronics Chat' started by Dan1996, Jul 6, 2012.

1. ### Dan1996 Thread Starter New Member

Jul 6, 2012
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Hi, I am trying to make a mock F1 steering wheel with light up LED's, but I don't know how to find out the what strength resistor's to use or if it is even possible
Basically I would like to power 25 LED's in 7 different series (with a 9v battery), but they all have different drop rates. 4 have 4 LED's and 3 have 3LED's, but they are all different colours. The LED's only work at 20mA. I think the drop rate for 3 of the series have 8.8v, 2 have 6.6v drop, 1 has 5.1v drop and 1 has 6.8v drop (sorry if thats really confusing!).

Basically, can it be possible to power all 7 series even though they all have different drop rates and as each series will have it's own switch will the amount of voltage going through each series change when some of the switches of off?

Thanks
Dan1996

2. ### jurgensw New Member

Jan 6, 2012
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If u subtract the voltage drop from the battery volts you get the voltage across the resistor.
Using ohms law to determent the resistor value for 20 mA.
R = V/I.
R = 2.4/.02 = 120 ohm
W = 2.4*.02 = 0.048 W

Mar 24, 2008
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4. ### Dan1996 Thread Starter New Member

Jul 6, 2012
3
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Thanks guys, do you think it will be possible to power all 7 series with just a 9v battery?
thanks

5. ### Wendy Moderator

Mar 24, 2008
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Yep, but if you read the tutorial you will note that color denotes Vf, which tells you something about efficiencies. In other words, Red LEDs properly designed will last longer due to their lower dropping voltage.

Colors are important, define them in advance.

The length of time a 9V will power many LEDs will be short, but usable. Define a minimum time you would like.

6. ### WBahn Moderator

Mar 31, 2012
23,214
7,001
I don't think it's too practical since you say that three of the series have a voltage drop of 8.8V. Assuming the battery really is 9V and that the drop across the string at 20mA really is 8.8V, that means you are dropping only 200mV across the current limiting resistor at 20mA (making it a 10Ω resistor).

So what's bad about this? Well, it gives you no margin. If a string uses 8.9V (or the if the battery is only 8.9V), then your current is cut in half. If the string uses only 8.6V, then your are putting twice as much current as you want.

Worse, a 9V alkaline will drop below your 8.8V string's forward voltage almost immediately. Furthermore, 140mA (7 strings at 20mA each) is a pretty heavy load for one of these batteries. You can expect a typical Duracell (Coppertop or Procell) to be down to below 7V in only an hour.

And, yes, as you switch some of the series on and off the current in the others will change because the battery voltage will change as the total load changes. Normally this effect is minor, but you are operating with such low margins that you will probably see very noticeable effects even with a new battery and the effects will get progressively more pronounced quickly (as in within ten minutes or so).

7. ### Wendy Moderator

Mar 24, 2008
21,421
2,949
Can't argue with that, the way around is to design for 7V, and allow a little head room for the resistor. This is why colors matter.

If you want to go with something that regulates a bit better let us know, or at lease let us know the mix of diodes you are wanting.

There is nothing wrong with using more than one 9V battery.

8. ### Dan1996 Thread Starter New Member

Jul 6, 2012
3
0
Thanks guys, I hadn't even considered that the battery would lose strength over time, I'll be able to make adjustments to make it work, thanks again for all the advice