# Please look if the resistor I chose is the right one.

Discussion in 'The Projects Forum' started by Lightfire, Aug 5, 2011.

1. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Hello,

Catapult is in need of your help. Please look at the image I attached.

I use 51 Ω resistor and not 47 Ω because I guess 51 is much better.

Please help and any help would be? Greatly appreciated.

Catapult

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2. ### bertus Administrator

Apr 5, 2008
18,492
3,561
Hello,

There is a mistake in your drawing.
The resistor should be in series with the leds.

Bertus

3. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Thank you for notifying me. Here is the modified and corrected one. Please take a look. Thanks.

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4. ### bertus Administrator

Apr 5, 2008
18,492
3,561
Hello,

As for the value of the resistor.
You state the leds are 2 Volts (3 in series), the powersupply is 12 Volts.
The voltage accross the resistor will be 12 - (3 X 2) = 12 - 6 = 6 Volts.
You state you are using a resistor of 51 Ohms.
The current will be 6 / 51 = 0.118 A = 118 mA, wich will probably be to much for the leds.
If you want 20 mA in this case the resistor must be 6 / 0.02 = 300 Ohms.

Bertus

5. ### kubeek Expert

Sep 20, 2005
5,266
986
Still wrong, currentfrom + to - through a diode goes only in the direcrion of the "arrow", so you have all backwards.

Are you sure the leds have only 2V of Vf? Even then the resistor is wrong. It drops 6V at 20mA, so that is 300 ohms.

6. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Oh. I calculated the whole schematic with 12 volts. I forgot, it 9 volts. Sorry. And sorry for the backwards. So if it is 9 volt, it is now okay?

7. ### SgtWookie Expert

Jul 17, 2007
22,202
1,792
If you are going to be using the LEDs while your 12v SLA battery is charging, then the battery may be as high as 13.8V.

In that case, you need to calculate for the maximum battery voltage.
Rlimit >= (Vsupply - (Vf_LED * Number_of_LEDs_in_series) ) / Desired_Current
(13.8v - (2.0v * 3) ) / 20mA
(13.8v - 6) / 0.02
7.8 / 0.02
390 Ohms, which is a standard value of resistance.
For the wattage rating, you will need 7.8v * 0.02a * 1.6 = 249.6mW; 1/4 Watt is perfect.

If you operate the LEDs until the battery is discharged down to 12v or 50%, the LEDs will have about 15.4mA current flowing through them.

8. ### SgtWookie Expert

Jul 17, 2007
22,202
1,792
No, you still did not calculate the resistors correctly.
Start with this formula:
Rlimit >= (Vsupply - (Vf_LED * Number_of_LEDs_in_series) ) / Desired_Current
And then substitute your voltages, LED count, and current in the formula, and then tell us what you calculated.

9. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Okay.

My resistor should be 150 ohms. Isn't it?

Thanks

10. ### SgtWookie Expert

Jul 17, 2007
22,202
1,792
Yes, 150 Ohms is correct.

Now, please re-draw your schematic with the resistor values and the LEDs with their Vf and current ratings, and your battery voltage range, from full to discharged.

11. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
I don't know the discharge and full battery rate. But here is the schematic I redraw. I hope it's correct.

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12. ### bertus Administrator

Apr 5, 2008
18,492
3,561
Hello,

The leds are the wrong way around.
They are back-biased now and will not give any light.

Bertus

13. ### SgtWookie Expert

Jul 17, 2007
22,202
1,792
You have the battery shorted out, and the LEDs backwards.

Re-draw the schematic with the battery on the left side, positive terminal up.

Then add the two resistors to the right of the battery, oriented vertically.
Then connect the top of the two resistors to the battery positive terminal.
Then place three LEDs below each resistor, anode up, cathode down, in two vertical columns.
Then wire the LEDs in series, and connect the bottom cathodes to the battery negative terminal.

I thought I'd shown you that kind of schematic layout before, but perhaps you forgot.

Anyway, some general rules for schematic drawings are:
1) More positive voltages towards the top, more negative towards the bottom.
2) Inputs come from the left, outputs flow towards the right.

Since the only "input" is your battery; the power supply, it goes on the left, with the positive terminal up.
Since the only "output" is the light from your LEDs, they go towards the right.
The LEDs should be oriented so that the cathodes are pointing down, since that is the more negative part of the schematic.

You need to document the part number and manufacturer of the battery. It might be a "PP3" type, or other type. Once we know what the manufacturer and part number is, we can find the documentation for the battery.

14. ### iONic AAC Fanatic!

Nov 16, 2007
1,426
70
There are chargers out there that apply voltages as high as 14.4V, which would require a
larger resister. I have more than one charger that can reach this voltage level.

15. ### Audioguru Expert

Dec 20, 2007
10,495
1,165
A little 9V battery is 9V when new. Its voltage quickly drops to 7.2V then drops slower to 6V when it is said to be dead.
So your LEDs will dim as the battery voltage drops.

16. ### debjit625 Well-Known Member

Apr 17, 2010
790
186

When we apply positive to anode and negative to cathode we say the diode is forward biased and when we do the opposite we say the diode is reverse biased .When a diode is forward biased it conducts but when its reverse biased it does not (in some special diodes in reverse bias it also conducts).In your circuit you have reverse biased all your diodes, so it will not conduct i.e.. it will not illuminate.

Here is how it should be...

Good Luck

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