# How many LED's can a 9v circuit power?

Discussion in 'The Projects Forum' started by tarquinkrikery, Apr 15, 2009.

1. ### tarquinkrikery Thread Starter New Member

Apr 15, 2009
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Hi there,

Sorry, I'm a newb, I've not made any kinda circuits since school and that was some time ago!

I'm building myself a pedal board for my guitar pedals- which all run off a 9v power supply (1500ma supply). I have 10 pedals using a total of around 700ma between them.

What I thought would be nice, was to have the pedals mounted on thich perspex with LED's underneath, giving a cool glow to the board. I wanted to use the 9v supply to power the LED's as I assumed this was probably be the best and most sensible route (since I already have it available!). I was thinking of running them off a switch and also running 6 x 3 LED's in series.

My question is if 9v is enough to power these - I really don't quite understand he principles of current! I know I need around a 220Ohm resistor if I hope to have 3 x 2.2v Green LED's in series, but that's about as far as my knowledge goes!

Sorry if this is a really dumb question!

2. ### alainp New Member

Apr 15, 2009
3
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Hi,

the answer is yes. With 3 LED's in series the total voltage drop will be 6.6V; thus you will have 9V-6.6V=2.4V across the resistor. The resistor will limit the current in the LED's according to Ohm's law: i = V / R = 2.4V / 220 ohm = 10.9mA. Depending of the type of LED you are using you may need to adjust the value of the resistor. E.g. if you need 20mA the resistor value will be 2.4V / 0.020 A = 120 ohm.

3. ### Darren Holdstock Active Member

Feb 10, 2009
262
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Not a dumb question at all, and the arrangement you suggest is just fine. You'll get about 11 mA of bias current, which should be nice and bright with a modern green LED. I'd say 3 LEDs in series is about the limit for a 9 V supply - don't be tempted to slip in a 4th as the resultant small (200 mV) voltage drop across the resistor will make the current difficult to control.

Beware of earth loops from running several pedals off the same supply - you'll notice it as increased mains hum. The best way around this is to have an individual, isolated supply for each pedal rather than one big supply for all of them. This, of course, is a massive pain, and explains the persistence of the PP3 battery in effects pedal land.

4. ### tarquinkrikery Thread Starter New Member

Apr 15, 2009
8
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Thanks Alain,

But really I waned to know if the 9v DC supply could cope with 6 "rows" of 3 led's in series? I.e can I have 18 Leds in total (6 x 3-in series) running off the same power source?

I'm not sure how the whole current/voltage thing works works!

Thanks!

5. ### tarquinkrikery Thread Starter New Member

Apr 15, 2009
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Sorry Darren you posted whilst I was replying!

So 3 in series in fine. Cool, can I have 6 sets of 3 running from the same power supply? i'm sure I read somewhere that voltage doesn't drop across different routes/channels from the circuit board. Sorry I'm really crap at explaining stuff as I don't know what I'm talking about! lol

6. ### alainp New Member

Apr 15, 2009
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No problem, you can add as many sets of 3 LED's in parallel, as long as the total current does not exceed what your power supply can give. 6 sets times 11mA gives 66mA, you have much more than that available.

7. ### italo New Member

Nov 20, 2005
205
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the maximun leds that you can you use is 1.5 a/.02 or75 parallel leds strings or 225 total LEDS. then again if the 9v is regulated and stay 9v at this load of 1.5a then you may sring 4 LEDS in series with a 10 Ω limiting then the leds increases to 300 LEDS that is asuming a 20ma per string. if you use 15ma per string total LEDS are 400. that is a lot of LEDS

8. ### tarquinkrikery Thread Starter New Member

Apr 15, 2009
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Brilliant!

Thanks everyone for your help - I wasn't quite sure about this to start off with, but now I understand a lot more fully!

9. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Figure a 9V can handle around 250ma, it can do more, but not reasonably. Even at that current it will have a very short life.

I accidently shorted an off brand alkaline once, it provided 1A continous at no voltage.

Last edited: Apr 15, 2009
10. ### Audioguru New Member

Dec 20, 2007
9,411
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The OP has a 9V/1.5A power supply, not a little 9V battery.

11. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
If you have plenty of current, LED series parallel array wizard will give you all sorts of configurations. The only limitation is a single Vf (LED Voltage, changes a bit with color) per calculation.

12. ### tarquinkrikery Thread Starter New Member

Apr 15, 2009
8
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Thanks Thatoneguy,

That's a really helpful little app! I'm sure I'll be using that in the future!

13. ### Darren Holdstock Active Member

Feb 10, 2009
262
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Quoting one's self is bad form, but what's a chap to do? I guess I'm going to have to offer the proof of why 4 LEDs won't fit nicely into a 9 V supply, and it's all to do with tolerancing. The figures I've quoted below are all from datasheets, and so are slightly different from the OP spec, but the principles remain.

The forward voltage drop across the LED will vary with batch, age, temperature and current - a green LED chosen at random has a Vf spec of 1.8 V (min), 2.0 V (typ) and 2.4 V (max), all at If = 20 mA, so we have to assume Vf can be anything from 1.8 to 2.4 V. The 9 V supply won't be exact either, and just for an example if a popular 7809 regulator is used, it can be anywhere from 8.64 to 9.36 V.

Let's take 3 LEDs in series with a 220 Ω resistor and assume the voltage supply is 9.0 V. This will give an LED current of 13.6 mA.
The highest LED current will occur when Vf is at a minimum (1.8 V) and Vsupply is at a maximum (9.36 V), giving If = 18 mA.
The lowest LED current will occur when Vf is at a maximum (2.4 V) and Vsupply is at a mininum (8.64 V), giving If = 8.2 mA.

Now let's try 4 LEDs in series, and keeping 13.6 mA current for LEDs with Vf = 2.0 V and Vsupply = 9V gives us a resistor value of about 73 Ω.
Using the same approach, the highest toleranced current draw is now 29.6 mA.
You can see a problem with the lowest current draw, in that 4 X 2.4 V = 9.6 V; there isn't enough supply headroom to bias the LEDs properly. They will glow dimly for sure, but the exact current draw has to be worked out iteratively from the LED I-V graphs.

So in summary, with 3 LEDs the forward current can vary between 8.2 mA and 18 mA, which is plenty spread for anyone, whereas with 4 LEDs the current can vary from way too small to a potential LED-popping 29.6 mA.

14. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
The 9V 1.5A Supply is in place of the batteries, and assuming it is a regulated supply, since it is for musical instrument replacement of 9V batteries, the current swing shouldn't be an issue.

If the supply is Not Regulated, and high brightness is required, then one of the constant current circuits by Bill Mardsen or sgtwookie would be a better choice.

Does the OP know if the supply is a regulated 9V, such as a switching supply, or has 'regulated' marked on it?