# need help with my circuit im a beginner

Discussion in 'General Electronics Chat' started by metalca, Sep 11, 2014.

1. ### metalca Thread Starter New Member

Sep 11, 2014
16
0
im working on a infrared illuminator, im powering it using a 9v battery, im using 20 leds, in 5 rows of 4 leds,
the battery drains after 10 mins, im not sure if I have the right resistors or the right amount of resistors on the circuit,
resistors I have
0.25w
270ohm
carbon film
ammo packed
1/4 watt- tolerance 1%

the leds I have
10mm
forward voltage; 1.5--2V
backward voltage; 5V
forward current; 20mA
wavelength; 850nm
angle; 30 degree
luminous power; 300mw
low frequency
infrared ray

that's the specs of the two parts im using, I eed help with amount of resistors or correct resistors, or if it cant be done what and how many resistors would I need for a 12v circuit, im new to this a ay help would be great.

Apr 5, 2008
17,103
3,001
3. ### metalca Thread Starter New Member

Sep 11, 2014
16
0
im a amateur only started few weeks back all the electronic talk an symbols are all chiese to me I haven't a clue as im only new, I thought I had the light made perfect then battery died after 10 mins an I tried another light same thing

Apr 5, 2008
17,103
3,001
Hello,

You have 5 strings of 4 leds with an 270 Ohms resistor.
The battery voltage is about 9 Volts.
The 4 leds take 4 X 1.5 Volts = 6 Volts.
The voltage over the resistor is the difference = 9 - 6 = 3 Volts.
The current through the resistor ( = the current through the leds ) = 3 Volts / 270 Ohms = 0.0111A = 11.1 mA.
5 strings will take 5 X 11.1 mA = 55.5 mA.
This could be quite a strain for the small battery.

Bertus

5. ### wayneh Expert

Sep 9, 2010
13,594
4,394
I agree, it's that simple; too many lights and not enough battery.

20 LEDs at 300mW each is 6W. A typical 9V battery might provide 400mAh to your lights until the voltage has sagged so far it will no longer work. If it really maintained, say, 8V (it won't), that would be 3.2W•hrs or about one half hour at 6W. Ten minutes is a little short, but not at all surprising.

The batteries may have some useful life left in them for other uses, though.

Last edited: Sep 11, 2014
6. ### metalca Thread Starter New Member

Sep 11, 2014
16
0
What about a 12v circuit how many resistors would i need on it.

Apr 5, 2008
17,103
3,001
Hello,

What will be the 12 volts powersource?

Bertus

8. ### metalca Thread Starter New Member

Sep 11, 2014
16
0
4 AA batteries

9. ### metalca Thread Starter New Member

Sep 11, 2014
16
0
Sorry think its 8 AA batterys

Apr 5, 2008
17,103
3,001
Hello,

4 AA batteries will be 4 X 1.5 Volts = 6 Volts for alkaline batteries and 4 X 1.2 Volts = 4.8 Volts for the rechargable NiMH batteries.

This will be much lower as the wanted 12 Volts.

Bertus

11. ### wayneh Expert

Sep 9, 2010
13,594
4,394
Well, you can place 5-7 LEDs in series. If you want to use identical strings, 4 strings of 5 LEDs is the choice. Then you calculate the resistor as Bertus has shown, as would also be detailed in the link he provided.

[didn't see the AA comment in time]

12. ### metalca Thread Starter New Member

Sep 11, 2014
16
0
Ill work something out thats for the help

13. ### metalca Thread Starter New Member

Sep 11, 2014
16
0
If i have a 12v powersupply (8x AA batteries) what resistors would i need an how many

14. ### metalca Thread Starter New Member

Sep 11, 2014
16
0
Im sorry to go on its just its new to me an i wouldnt be the smartest about lol. A 12v holds 8AA batteries if each batterys 1.5v thats 12v altogether then how do i work out what resistors i need. The link doesnt work for me

15. ### wayneh Expert

Sep 9, 2010
13,594
4,394
First off, your cells may not be 1.5V. That depends on the battery chemistry and state of charge.

But it's not hard to make the calculation: In any electric circuit, the total voltage drops across all the devices in series. So if you have 5 LEDs in series with a 12V battery, those LEDs will drop, say, 5 x 1.5Vf = 7.5V across themselves. The resistor must drop the rest, or 12V - 7.5V = 4.5V.

Ohms' law, V = I•R, is used to calculate the proper R. In this case 4.5V = 0.02A • R and thus R=225Ω. That's to achieve 20mA. If that's the maximum spec for your LED, then you should aim below that, meaning you should use a higher value resistor to lower the current to 10-15mA. Your 270Ω resistors would be fine.

Apr 5, 2008
17,103
3,001
Hello,

The leds are 1.5 to 2 volts.
For the calculation of the resistor, I will use the lowest value as that will give the highest current.
The leds are rated for 20 mA, lets take 10 % lower value to be safe, so 20 - 10 % = 18 mA.
The supply voltage is 12 Volts, this woulld give a max of 12 / 2 = 6, but then there is no voltage left for regulation, so lets take 5 leds.
5 leds will take 5 X 1.5 Volts = 7.5 Volts.
The voltage across the resistor will be the difference betwen the supply voltage and leds voltage => 12 - 7.5 = 4.5 Volts.
The value of the needed resistor will be 4.5 Volts / 18 mA = 250 Ohms.
The closest E-series value is 270 Ohms, this will give you a current of 4.5 Volts / 270 Ohm = 16.66 mA.

Bertus

17. ### metalca Thread Starter New Member

Sep 11, 2014
16
0
How many resistors would i need on my circuit then. With 20 leds. 5 rows of 4 leds

Apr 5, 2008
17,103
3,001
Hello,

If you want to use 4 leds in a string, the regulation will be better, but you will loose more power.
For 4 leds you will need the following resistor:
(supply voltage - n X ledvoltage) / wanted current =>
(12 - 4 X 1.5) Volts / 18 mA =>
6 Volts / 18 mA = 333.333 Ohms.
The closest E-series resistors are 330 Ohms and 390 Ohms.
The 330 Ohms resistor will give you a led current of 6 Volts / 330 Ohms = 18.1818 mA.
The 390 Ohms resistor will give you a led current of 6 Volts / 390 Ohms = 15.38 mA.
You will need a resistor for each string of leds, so 5 resistors.

Bertus

19. ### metalca Thread Starter New Member

Sep 11, 2014
16
0
I will start learning my electonic components an calculations use guys are great ill order some resistors an give it a shot. If i order 330ohm resistors an use 5 with the.circuit i have