Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Leads Help

Status
Not open for further replies.
measure the voltage of the battery in open circuit and then connect a very high variable resistance across the battery and put an ammeter in series and turn down the resistance while reading the ammeter watch the voltmeter until the current is so high that the voltage is going down strongly. Just before that is the maximum current the battery can supply.
 
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.

almost.

as blueroom said,
"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."


your battery, assuming its at a set voltage (it declines over time as the battery wears out, but ignore this for now) will provide whatever current is needed through the load to satisfy ohms law. of course this current has an upper limit. so if you put a 1 ohm resistor on there, with your LED (2v forward), then

9V - 2V = 7V
7V / 1Ohm = 7 Amps
which is, of course, way too much current, your battery wont do this much, itll just heat up and bad things will happen.

but say you had 100ohms

then 7V / 100Ohm = .07A = 70mA, which is more reasonable. whatever current is needed will be supplied by ohms law.

now lets say you've been running this LED for a while, and your battery is run down a bit, say its down to 7.7V out of the original 9V (this is with that 175Ohm resistor blueroom suggested):

7.7V - 2V (LED forward voltage) = 5.7V

5.7V / 175Ohm = .032A = 32mA

your LED will only be getting 32mA instead of the maximum of 40mA, so it will be dimmer.


by the way, its probably a good idea to run your LED at less than 40mA (the maximum) so as to make it last longer; if you need the extra brightness get one thats rated for, say, 50mA and run it at 40mA, or something like that.


ive had 9Vs running at maximum before, and the best they can do is about 500mA, or .5A, and this isnt for very long and they get fairly warm. possibly up to 1A in spikes, i dont remember too well.
 
Last edited:
You have a battery that is 9V when it is new.
You have one 2V LED in series with a current-limiting resistor.
Then the voltage across the resistor is 7V so its value is (9V - 2V)/30mA= 233 ohms. 240 ohms is the nearest standard value.

If you use two 2V LEDs in series then the current-limiting resistor value is (9V - 4V)/30mA= 167 ohms. 180 ohms is the nearest standard value.

If you have three 2V LEDs in series then the value of the current-limiting resistor is (9V - 6V)/30mA= 100 ohms.
 
OK you told me the formula is.
9V - 2V Divided by 30mA = 233 ohms. 240 ohms

I thought my Load the 30 mA LED is what they call Current and the Battery 9. Volt is the Voltage.

So it was E Divided by I = R

Why is my way off I thought this was the Ohms Law for finding Resistance??

Can you tell me why I am thinking this
 
OK you told me the formula is.
9V - 2V Divided by 30mA = 233 ohms. 240 ohms

I thought my Load the 30 mA LED is what they call Current and the Battery 9. Volt is the Voltage.

So it was E Divided by I = R

Why is my way off I thought this was the Ohms Law for finding Resistance??

Can you tell me why I am thinking this

i think i can...

an LED is a diode, and there is a voltage drop associated with diodes. your diode has a 2V drop.

Now, if your series circuit is (think clockwise) 9V, connected to resistor, connected to 2V LED, connected back to the 9V (- terminal), then before you figure out the voltage across your resistor you have to add ALL the series voltages. you have +9V, and then your diode "drops" 2V (so thats +9V - 2V). this means that, all total, there is 7V across your resistor.

Current is the same for all parts of a series circuit, and you need 30mA. So, NOW you can use the voltage at the RESISTOR (called a current-limiting resistor) to apply ohms law for the resistor, with E = 7V and I = 30mA, giving 233Ω.


if you're confused, look up Kirchoff's Voltage Law (KVL) . additionally, there is Kirchoff's Current Law (KCL) which is useful to know, but we arent using it here. KVL is the law applied in summing the voltages that I did above.
 
I have ask 2. things.

If you add the 9 Volt Battery and the 2 Volt LED you get 11 Volts how do you get 7. Volts??


Why are you connecting the parts going back to the - and not the +??

Electrons flow from - to +??
 
I have ask 2. things.

If you add the 9 Volt Battery and the 2 Volt LED you get 11 Volts how do you get 7. Volts??


Why are you connecting the parts going back to the - and not the +??

Electrons flow from - to +??

diodes do not add their voltage to the circuit; they are not sources.

they take AWAY their forward voltage. imagine it sort of like a negative battery.

thus, 9V - 2V = 7V



as for the connections, its just a general circuits convention. yes, electrons flow from - to +, but back when circuits was invented, they didnt know and had to guess. there was a 50% chance, and they got it wrong. so the conventional current is DEFINED as from + to -. doesnt really matter which way you say it is, as long as your consistent with your plusses and minuses it will all fall out the same in the end.

so when drawing a circuit diagram, its always habit for me to start at the + terminal and end at the - terminal of the voltage source or battery.
 
The battery is 9V.
The LED has a voltage drop of 2V.
Therefore the voltage across the series resistor is 9V - 2V= 7V.

Ohms Law says that the value of the resistor in a series circuit is the voltage across a resistor divided by the current.
 
I would like to Teach myself Electronics so I got a Book on just that.

Can I use this Group for help or is this Group just for Pros and things like that
 
This thread is about some of us trying to explain about a battery, an LED and a current-limiting resistor. It took a lot of time and 3 pages in this thread. It should have taken only one sentence in a grade 7 electricity class. Kids in grade 7 are 12 or 13 years old.

If you learn things from the book then we will answer a complicated question. Not a simple one. What is the name of the book? If you asked then most of us would have recommended The Art of Electronics which can be downloaded from the web.
 
I'd say im with audioguru, I'm willing to help but this was ridiculous. if you learn the basics first from the book, perhaps talk to a teacher or some other knowledgeable person about it and do some research on the internet, you'll find it to be a lot of fun I hope and people everywhere are always more willing to answer an informed question.

I don't want to turn anyone off to electronics, as its an excellent and intriguing field, and theres always a demand for more engineers. Have fun.
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top