Ohm's law

[popsik]

I've noticed that Ohm's law is used like everywhere!!! But it's hard for me to understand how to use it... for example:
if i have an LED which runs on 1.5 V but i don't have 1.5 V battery, i only have 12V battery, and if i attach that LED to 12V battery it would burn emediatly, so i want to know how strong the resistor must be to resist those 10.5 volts, and how to know how many Volts equals to 1 Ampere, or how many Amperes equals to 1 Volt.
I wish you could help me with this, because if i wont unerstand this i would be stuck int this level forever... Thank You

 

[ericgibbs]

hi,
Consider just one LED, that has a voltage drop of 2Volts and your supply is 12Vdc

You subtract the LED voltage from the supply voltage, thats 12V - 2V =10V, this means you have to drop 10V across the resistor.

Lets assume your LED requires 20mA [0.02A] to work, using Ohms Law equation [R=V/I] 10V/0.02 = 500 ohms.

So for any single LED,,, its Resistor Value = [Vsupply - Vled]/LED Current Where the LED Vforward and LED current are read from the datasheet for that LED.

 

[Nigel Goodwin]

Ohms law is simple - but you need to know how to use it. An LED ISN'T a resistor, so it doesn't apply directly.

You calculate the resistor based on the voltage you want to drop (which is the supply voltage minus the LED forward voltage drop), and the current you want to pass.

 

[popsik]

thank you very much it helped me a lot!!!

 

[Jugurtha]

I've noticed that Ohm's law is used like everywhere!!! But it's hard for me to understand how to use it...

General writing: V=Z*I I being the current.
Z being the Impedance (R, 1/JwC, jwL "Resistive, Capacitive, Inductive").

if i have an LED which runs on 1.5 V but i don't have 1.5 V battery, i only have 12V battery, and if i attach that LED to 12V battery it would burn emediatly, so i want to know how strong the resistor must be to resist those 10.5 volts

There is some disambiguation needed here:

*The PN junction in the LED needs a certain direct potential (anode cathode) VAK called threshold tension Vs or V0 in order to be conductive (ID<>0)

*The LED will cause a tension drop of that amount.

*The LED has a certain resistance r, that is small.

The reason the LED would burn isn't "because" it is 12 volts, the reason it would burn is because the CURRENT circulating through the LED (ID) is so huge, it will cause an avalanche, a snow-ball effect. Back to Ohms law, once you pass the threshold tension, V=r*I so I=V/r .. This is big, as r is small.

If r=10ohms and V=12 Volts, you get I=1.2 Ampères. That's a lot for our poor LED.

The point is, it's not the tension that hurts, it's the current that hurts.. So, you want to put a Protection Resistor not to "resist a tension" because the tension will still be 12 volts, but to "limit" the current circulating in the circuit.

Without the Protection Resistor (Rp):

I=(V-Vs)/r (Vs the tension drop in the LED, r or the diode is too small, I will be huge )

With the Protection Resistor:

I=(V-Vs)/(r+Rp) (I diminishes).

and how to know how many Volts equals to 1 Ampere, or how many Amperes equals to 1 Volt.
I wish you could help me with this, because if i wont unerstand this i would be stuck int this level forever... Thank You

Here again, let's get back to Ohm's Law:

V=Z*I , Z in Ohms, V in Volts, I in Ampères. So {Volt}=={Ohm}*{Ampère}

So, {Ohm}=={Volt/Ampère}.


It is an equation with three parameters, which means that you ABSOLUTELY NEED TWO KNOWNS to get the THIRD.

Let's say you want a Current I of 1 Ampère, you can get it with

10Volts/10Ohms=1 Ampère. (The 10 Ohms can be Two 5 Ohms Resistors in Series, or Two 20 Ohms Resistors in Parallel, whatever combination you wish)

2KVolts/2KOhms=1 Ampère. Whatever you want and is smart according to your context.


So if you want a current of 20 mA (milli-Ampère = (1/1000) Ampère) You'll have:

V=Vs+(Rp+r)*I ... Let's calculate the value of the Protection Resistance to have that current...

(Rp+r)*I= V-Vs

Rp+r=(V-Vs)/I

Rp=(V-Vs)/I - r

Suppose Vs=1.5 Volts
V= 12 Volts
r= 10 Ohms

Rp= (12-1.5) Volts/(20*1/1000)Ampères - 10 Ohms

Remember, Volt/Ampère==Ohm

Rp= 525 Ohms - 10 Ohms = 515 Ohms .. (Example for educational purposes only ..)

If there is anything you didn't get, feel free. (If so, would you please tell which level you're in to get more pertinent explanations)

 

[BrownOut]

'Course, r in an operating LED is very small, and almost always ignored in the calculation of Rp.

 

[DerStrom8]

I suggest you print this out and keep it in your wallet for future reference:

**broken link removed**

It's a diagram of all the parts of ohm's law, and it's helped me a lot in the past.
Good luck!
Der Strom

 

[MrAl]

Hi,


Ohm's Law is probably the best place to start as others have clearly pointed out. Using the 'internal resistance' of the LED isnt that good of an idea though and doesnt seem to help, but it does help to think of the internal resistance as being zero and the dynamic performance of the LED described in another way. This really narrows down the variations that could come up.

The basic idea is to use the characteristic voltage alone and knowing the source voltage come up with a resistor value that provides the correct current to the LED. This gets us started. There are some problems that come up however that could be very serious. If the power supply voltage can change even by a small amount when the ratio of supply voltage to total LED voltage is close to 1 we could see a sharp rise in current when the power supply voltage rises even a little bit. This means that it's always a good idea to leave a little 'head room' so the supply can vary and not bother the LED too much. Also note that this problem can be even worse than the problems that come up with a change in characteristic voltage with temperature.

To study these effects we can look at one particular white Nichia LED. We can first come up with an approximate equation that allows us to change the characteristic voltage easily and then make any changes to supply voltage or LED voltage that we want to and then see how much this affects the current.

To start with, we can use this equation:
vLED(i)=a3*i^3+a2*i^2+a1*i+a0

This is simply a third degree equation in 'i' which is current through the LED. Note that the left side is in units of volts so that makes a0 in units of volts too.
For the LED we will use here we define the constants as:
a0=2.943133813722055
a1=38.2849159319738
a2=-247.5791299320233
a3=-1225.06004859014

and the range of validity for the current 'i' in the equation is 2ma to 40ma which covers many purposes.

Now in the series circuit with R1 being the LED series resistor, we have:
i=(Vs-vLed)/R1

where Vs is the source voltage and vLED is the LED voltage. Note that the LED voltage is also on the left side and since vLED is dependent on 'i' also we can write:
i=(Vs-vLed(i))/R1

Now we substitute the equation for vLED(i) above and after simplifying we end up with:
Vs-a3*i^3-a2*i^2-a1*i-R1*i-a0=0

This is still a third degree equation in 'i', and it can be solved analytically but doing this numerically with a numerical solver works good too. With this equation we can vary Vs or R1 and see the effects on the current i (when we solve for i). As noted above, a0 is in units of volts so to get an idea what happens when the temperature changes we can vary a0 by the amount we would find on the data sheet for the particular LED. For example, if we find that the LED voltage varies by -10mv per deg C then we can subtract 0.010 from a0 for each degree C we expect to find in the real world application.