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LED lowers Voltage

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audioguru said:
The resistor in series with the LED limits its current and adjusts its brightness.
120 ohms makes it very bright.
150 to 220 ohms makes it not so bright.
270 to 1000 ohms makes it dimmer.
1.5k to 10k ohms makes it very dim.
15k to 100k ohms makes it extremely dim.
Mhm..I just put 120 ohms on there, and I still got a steady 4.95 :)

I was also wondering, lets say you put 270 ohms on there, will it take less battery, like..would my voltage go up to 4.97 mabye?
 
The voltage regulator has an output of 4.95V if the battery is more than 7V. Then if the LED is 2.2V its current from the regulator is (4.95V - 2.2V)/120 ohms= 22.9mA. If you use 270 ohms then the regulator's voltage is still the same but the current is reduced to (4.95V - 2.2V)/270 ohms= 10.2mA.

The datasheet for the regulator shows that its output voltage drops only typically 0.01V if its load is increased from 0.01A to 1.5A.
 
Um...you just confused me so much haha. So right now..my voltage is about 4.95. I THINK your saying that...the voltage wont be reduced, but if I use 270ohms it will stay the same(but the LED will be dimmer) right? Will anything else happen though? Can you also explain why a 9v wont charge a ipod?
 
kaspar389 said:
I just put in a new duracell(I was using some cheap walgreens one) and I have 4.99 volts.
Because the loaded voltage of your cheap 9V battery dropped below 7V?
I use a low-dropout 5V regulator for my 9V battery projects because it still works perfectly when the battery voltage drops to 5.5V.

I don't know how many mAh the battery in your ipod is rated at. If it is less than the approximately 500mAh of an alkaline 9V battery then it won't be fully charged and the 9V battery will be dead.
 
You have 4.99V from a 5V regulator?

That's perfectly acceptable as the tollerence is normally about 5%.
 
audioguru said:
Because the loaded voltage of your cheap 9V battery dropped below 7V?
I use a low-dropout 5V regulator for my 9V battery projects because it still works perfectly when the battery voltage drops to 5.5V.

I don't know how many mAh the battery in your ipod is rated at. If it is less than the approximately 500mAh of an alkaline 9V battery then it won't be fully charged and the 9V battery will be dead.
YEah, the cheap battery quickly fell to 7v, so I just bought some duracells.

Ugh, this is all so confusing.


I believe the battery is 400 mah.
 
The Energizer datasheet for their alkaline 9V battery shows that its voltage drops to 6V in 20 hours when the current is 25mA which is a capacity of 500mAh. If the current is 200mA then it is at 6V in 1.5 hours and its capacity is only 300mAh.

Energizer has curves on their datasheet for their 9V alkaline battery that has the voltage quickly falling to 7.2V then slowly dropping lower. Your voltage regulator will not regulate for most of the life of the battery.
 

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Im so ****** confused its not even funny. Can you explain all this MAH and crap. What does this mean? I dont get this. All Im doing is hooking up a 9v battery, to a 5v resistor, then I have the 120ohms resistor to the LED.
 
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kaspar389 said:
Im so ****** confused its not even funny. Can you explain all this MAH and crap. What does this mean? I dont get this. All Im doing is hooking up a 9v battery, to a 5v resistor, then I have the 120ohms resistor to the LED.


MAH is the current capacity of a given battery, it tells one how long a given battery can supply a load before being discharged. You can wire 8 AA cells in series for 12volts total but it won't have the same MAH capacity as a large 12volt car battery. Make sense?

So a battery has two main specifications, it's nominal voltage rating in volts and it's current capacity rating in MAH (milliamp/hours) or AH (amp hours).

Batteries come in two main types, primary types (non-rechargable) and rechargable. There are many different types of batteries that have different benifits Vs costs, esp in the rechargable types.


Lefty
 
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kaspar389 said:
Im so ****** confused its not even funny. Can you explain all this MAH and crap. What does this mean? I dont get this. All Im doing is hooking up a 9v battery, to a 5v resistor, then I have the 120ohms resistor to the LED.


Yes it can be confusing, do you want to learn all this stuff or do you soally just want to make a charger???

mA(milli Amps) is what your application (LED,or any other load) will draw from the battery, the resistor, "resists" the amount of current(mA) passing though the LED.

if you use a resistor that does not "resist" inuf current(mA) then the LED will be really bright and burn out.

LED's have a milli Amp rating
and a voltage

eg: if your LED is rated at 3Volts and 25mA of current.
then you will need to find a resistor to suit 25 mA of current

and this is how you find the correct resistor

say your power supply is 5V

and your LED is 3V

and the mA (milli Amps) needed to light the LED is 25mA

so you do the followong:

5V - 3V (LED) = 2V

so now

2V / 0.025A(25millAmps) = 80

so you would need a 80ohms resistor to get 25mA supplied to the LED

i hope this makes sense, iv tried to explain it as noob friendly as possible
 
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Hi!

I’m glad, that you are interested in electronics. The LED must be fed with current generator. The LED drop voltage is 1,9…4V depending on it’s color. If you connect the LED directly to 5 Volts, the LED and the source supply may damage. Always use a serial resistor for LED’s. R=(US-UL)/I. US=supply voltage (5V from USB), UL=LED voltage (2,7V in your project), I=LED current (in general 10…20mA => 0,01..0,02A). In your project R=(5V-2,7V)/0,02A=115 OHM. The nearest standard value is 120 OHM. If you want to charge your portable device from battery, you must use a higher opacity off batteries then the portable device. Use 6pcs batteries of 1,5V connected serial. Use R14 or R6 batteries, or sealed lead-acid battery of 2…10Ah.
 
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