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Leads Help

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biferi

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I am going to make a simple light that I can turn off and on by a Switch and need some help.

I am going to use a Lead as my Light Sorce and it needs 40 mA at 2. Volts to work.

And I will be useing a 9 Volt Battery for my Power Sorce.

Ok here is the thing I need help with?

I know how to use Ohms Law to findout the Resistance I will need but I don't know how many mA a 9 Volt Battery has???

I meen befor I hock it up to the Lead?????
 
9V
So 9V - 2V (LED forward voltage) = 7V
7V / 0.04 A (or 40ma) = 175 ohms

You need a 175 or 180 (a more common value) resistor.
 
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As I said in your other post, go to a battery manufacturer's website and look at the detailed datasheets.

A battery voltage drops as it runs down. Therefore the LED (not lead) will get dimmer and dimmer. Transistors or an IC can be added to keep the brightness constant.
A 9V alkaline battery voltage drops to 6V in about 16 hours when it starts brightly with a 40mA load. The voltage drops quickly when it is less than 6V.

My red LEDs have an absolute max continuous current of 40mA. I never operate them at more than 30mA so they can last longer.
 
Ware can I go on the Web to get all the Symboles for the Diferant componets?

Like Resisters Capaciters and things lik that??
 
OK I just found a Web Page with a lot of Electonic Symboles but one???

I would like the Symbole for a Push Button Switch the kind that you Push to turn on and you Push to turn back off.

Thanks
 
If I want to findout the Resistance I will need I know the formula is E Divided by I = R

But like you told me Batteries drop Voltage and some Current as they work so how do I know how mutch Current is in a 9 Volt Battery??

The 9 Volt Battery I am going to use in all my things will bve a reguler one you find in Smook Detecters.
 
Like I said, look at the datasheet for a 9V battery on the website of a battery manufacturer like Energizer.

The datasheet show curves of the battery voltage dropping at a certain current.
The voltage drops quicker at higher current.
Here is a 9V alkaline battery with a 53mA load and with a 400mA load:
 

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I whent to the Energizer Battery site and the Specs Sheet tol me nothing about mA or Voltage of the Battery.

Please tell me how to find out all of these things. Maybe not all stores give this information.
Thanks
 
I went to the Energizer Battery site and the Specs Sheet told me nothing about mA or Voltage of the Battery.
On the datasheet for Energizer's 9V alkaline battery there is a graph showing mAh from 625mAh when the current is 25mA to 300mAh when the current is 500mA.
They have a graph of how long it takes for the discharging battery voltage to drop to 4.8V and 6.0V when the load resistance is as you select on the graph.
They have a graph of how long it takes for the discharging battery voltage to drop to 4.8V and 6.0V when a constant load current is as you select on the graph.
They have graphs showing the voltage dropping with time when the battery powers a 620 ohm radio, a 270 ohm toy and a low current smoke detector.
 
I think I know ware I am getting lost.

I thought Batterys had a set mA just as they have a set Voltage.
How does this work then if the Current is allways diferant and the Voltage is say 1.5 Volt??
 
How does this work then if the Current is allways diferant and the Voltage is say 1.5 Volt??
Now you need to look at the datasheet for a 1.5V battery.
Energizer's AA alkaline cell is 1.5V only when it is brand new. Its voltage drops as it runs down because its internal resistance increases.
Its datasheet shows that its capacity is 2750mAh when its load current is only 25mA and drops to 1500mAh when its load current is 500mA.
The datasheet has graphs for constant power and constant current.
It has graphs showing the voltage dropping with time when it powers a remote, a radio, a light, a toy, a CD player, a tape player, a digital camera and a photoflash.

Energizer have some fairly new and fairly expensive AA and AAA lithium cells. Their voltage stays near 1.5V for most of their life and they have a higher capacity than alkaline cells.
 

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Ok I think the best thing for me to do is to start off new here and forget everything I asked.

You know I want to have a LED light up and I will be useing a 9. Volt Battery.
The LED needs 2 Volts at 40 mA to light up. And I know the 9. Volt Battery will be to mutch for it.

So to findout how mutch Resistance I will need I will use Ohms Law.

E over I = R

E = the Voltage 9 Volts and the I = Current that will be the Load 40 mA

So 9 Divided by .040 will be my Resistance YES or NO??
Just tell me if I have this mutch right if I don't just tell me NO and I think I will get it.
 
You would be better off using 2 AAA cells and a 27Ω resistor than a 9V battery and a 180Ω resistor. The LED will stay lit longer with the 2 AAAs.
 
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Ok I think the best thing for me to do is to start off new here and forget everything I asked.

You know I want to have a LED light up and I will be useing a 9. Volt Battery.
The LED needs 2 Volts at 40 mA to light up. And I know the 9. Volt Battery will be to mutch for it.

So to findout how mutch Resistance I will need I will use Ohms Law.

E over I = R

E = the Voltage 9 Volts and the I = Current that will be the Load 40 mA

So 9 Divided by .040 will be my Resistance YES or NO??
Just tell me if I have this mutch right if I don't just tell me NO and I think I will get it.
No.
The resistor will not have 9V across it. It will have only 7V across it when the 9V battery is brand new.
I think 40mA is the absolute max for the LED. Use 30mA and the LED will last much longer.
The battery also will last longer.
 
The second post in this thread, by blueroom, showed you how to do it.
 
Can someone help me with my formulas?

I think because I thought I was doing the formulas right that is why I got lost.

If I want to findout the Resistance I need I will need to know the Current the Load needs and the Voltage of the Battery.

So E over I = R this is how you findout resistance right????
 
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