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windozeuser

Member
Hi As Some Of You Know Im Building a small 12 volt power supply i figured it well pump out 12 to 13.6 volt's at 1.3 amp's My Question is How do i lower the current with out lowering the voltage (I Think it is impossible cause of ohm's law but i dont know for sure)

I want to be able adjust the current output from 500 Ma to 1 amp

Or well it matter? Well the Device Just Use The Required Current it need's
 

Sebi

Active Member
I mean for this purpose simply solution the L200 chip.
 

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windozeuser

Member
Well it aint a real good or big supply

If Just Uses a Step Down Transformer To Turn 120 VAC to 12 VAC Then Goes Through A Full Wave Rectifier And A 1000uF Cap Then A 7812 Voltage Regulator And Then The Output Goes into a 12 volt Power Accessary Port (Like On a Car) And I Was Wondering What I Can Do To Limit The Current Output

Like Say A Device I Connect Need's Only 1 amp of current and the supply pump's out 1.8 amps well the device get damaged?

I Was Thinking Of Using A Pot. I Know that well limit the current But well it lower the voltage by much

Plus Can you Give Me A site with a free and good circuit Design Software

Thank's
 

Roff

Well-Known Member
I guess I should have written the answer in a larger font. You apparently missed it (see my previous post). The load will only draw as much current as it needs. Think about this: Your car battery is capable of supplying as much as 600 amps, and yet it doesn't send your radio up in a ball of flames.
If you want to use the full current capability of your 7812 (only if the load needs it, of course), you need a bigger cap than 1000uF. With a 1 amp load, the ripple on the input of your 7812 will be about 8 volts, and since the peak voltage from your rectifier is probably only about 15 or 16 volts, the output will also have ripple. You need at least 4000uF. 10,000uF would be better.
 

windozeuser

Member
THANK YOU VERY MUCH RON :D

I still Dont get the current thing lol :x Why do u need a resistor on a led to limit the current if it well only draw 20ma from a say 1 amp supply?

Or am i missing something :cry:

Doesnt the current stay constant in a series circuit Through out every componet and only the voltage drop's.......

If u can explain it better it Would be very helpful

Thank's a bunch
 

Roff

Well-Known Member
If you connect your 12 volt supply, which can supply over an amp, directly to an LED, the LED will die immediately. You need to understand Ohm's law before you can hope to get very far with electronics. Do a Google search on "ohm's law". Make sure you get the spelling right. :roll:
The short answer is, you have to put a resistor in series to limit the current to your LED.
The voltage drop across an LED is relatively independent of the current flowing through it. Different types of LEDs have different forward voltage drops, but let's say you have an LED that has a forward voltage drop of 2 volts, and you want to run 20ma through it from your 12 volt supply. The resistor will have 10 volts across it (12v -2v = 10v). SInce R=V/I (one form of Ohm's law), where V is the voltage across the resistor and I is the current through it, the resistor needs to be

R=10V/0.02A (remember 20ma=0.02A),
R=500 ohms.

OK, let's say you changed your mind - you only want 10ma through your LED. The voltage across the LED will be a little less, but for practical purposes we can say it is still 2 volts. Now,

R=10V/10ma
R=1000 ohms (1kohm).

Like I said, do some studying. You will get a lot more enjoyment out of electronics.
 

windozeuser

Member
Yes i Know Ohm's Law

And The Math For Current, Power, Votage, and resistance

You Confused me about the Device drawing only the current it need's thats why i said that about the led i know u need a resistor to limit the current you kinda just confused me

If i wanted to change the current i would need a different resistor, thats why im gonna use a pot on the output

So if i had a 1 amp power supply that means the supply can have a max current of 1 ampere

and i connected a device that drawed 600 Ma there would be 400 Ma Left Over Would i need a resistor for the extra ampage ???

R = 12 volts / 0.4 amp = i got 30 ohm resistor correct me if im wrong
 

fat-tony

Member
I'll try to explain this a little better to you.

We're going to ignore LED's for the first while, since they are diodes, and diodes don't exactly behave the same way that most passive devices do.

You've got a 12V supply (btw, I posted in that thread too for ya). Anything you hook up to it (minus LEDs) is going to have an internal resistance. This internal resistance is what determines the current. If R(load) = 30ohms, then you are going to get your 0.4A of current flowing through it.

If you were to put a 30ohm resistor in front of it though, then your total resistance would be 60ohms, and you'd get half the current.

V = I * R(total)

This is why most devices just draw as much current as they need.

Now, to try and explain diodes. Diodes have what's called a forward-voltage. This is the voltage that needs to be across them before they start to conduct. If we assume that we're talking about ideal diodes (which don't actually exist, but are a good approximation), then a diode has infinite resistance below it's forward-voltage, and 0 resistance after the forward voltage. This is why you need a resistor in series with an LED to keep it from burning out. Since it doesn't have any resistance (well, very little), piles and piles of current will flow through it, and heat it to the point that it melts. When you add a resistor in series, it limits the amount of current that can go through.
 

windozeuser

Member
Thanks :D

I Think i get it now, A device has a rated current it's internal resistance can handle and it well be fine but if the current is exceeded the device's internal resistance cant limit the current and u have a barn fire lol

And i would only need a resistor for devices with very low internal resistance and current rating's like the base of a transistor and a led

Do Led's act as regular diodes, If The Reverse Voltage is exceeded it well pass it and burn it's self up lol 8)

Thanks
 

Nigel Goodwin

Super Moderator
Most Helpful Member
windozeuser said:
Thanks :D

I Think i get it now, A device has a rated current it's internal resistance can handle and it well be fine but if the current is exceeded the device's internal resistance cant limit the current and u have a barn fire lol

Not quite got it!.

To put it very simply (ignoring lots of things) an LED will always have roughly 2 volts across it - in fact they are often used as voltage references. So if you put 12 volts straight across it the LED will try and pull this down to 2 volts - so if you apply ohms law V=IxR, rearranged as I=V/R you get I=10/0 - 10 volts for the difference between 12 and 2, and zero ohms for the wire between the battery and the LED. Now 10/0 is infinity - which is impossible, but this is a very rough illustration - and in any case it will be as much current as the battery can supply, and the LED will go seriously BOOOMMMMM!.

So for LED's don't think 'internal resistance' (it doesn't really apply) but think 'voltage drop'.
 

fat-tony

Member
windozeuser said:
Thanks :D

I Think i get it now, A device has a rated current it's internal resistance can handle and it well be fine but if the current is exceeded the device's internal resistance cant limit the current and u have a barn fire lol

Exactly. Normal, passive devices have no limits on the current, because they limit the current themselves. Any kind of semiconductor (transistor, diode, LED...) all will have limits on what they can dissipate.

The barnfire makes me smile :D

Do Led's act as regular diodes, If The Reverse Voltage is exceeded it well pass it and burn it's self up lol 8)

Thanks

Yes indeed they do. LEDs, however, have much lower tolerance to reverse voltage. https://www.electro-tech-online.com/custompdfs/2003/11/107393PDF.pdf is a datasheet for a typical 11 cent diode. If you look down near the bottom of Page 2, it states Reverse current @ 5V -> 10uA. This means that if you allow more than 10uA through it backwards, it will likely be destroyed.

If you know that your LED is going to be in a situation where it will be reverse biased, your best bet probably to have a different diode (such as a dirt cheap 1N4004) connected backwards to your LED. (See attached picture). This way, if the has a voltage greater than the output (and the difference is greater than Vforward), the LED will light up. If the output is higher than the input (and > Vforward for the diode), then all the current will be pulled away from the LED.
 

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windozeuser

Member
Thanks for all your help!

Lol i blew a few led's before i hooked about 20 led's in parellel and connected it 240 main's they smoked and got real bright than the top's blew off of some of them


Do you recomened any very good book's and sites for basic electronics
 

windozeuser

Member
Thanks alot for all your help

I already read the dc volume from that site and now in the process of reading the ac and semiconductor volumes
 

fat-tony

Member
windozeuser said:
I already read the dc volume from that site and now in the process of reading the ac and semiconductor volumes

The semiconductor part is incomplete, however, if you have any questions about it, I can try to help you out.
 
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