0
\$\begingroup\$

In all the physics i learned and practiced many subjects But understanding Ohm's law used to be a big problem for me... Here is what i mean :

We have a 5 volt battery. And a 3 volt/ 20 mA led. So general approach to this schema is i learned: The led needs 3 volts. So we should control 5-3 = 2 volt with a resistor .

V = I*R & R = V/I Here as i assume : R = 2 V / 0.02 mA = 100 ohm resistor we need .. I hope this is correct..

But in this formula we use the voltage that resistor will control : 2 volts .. Since this voltage is resisted against; 3 volts will reach the Led. But we find the current in formula using the voltage resistor will control and what it has to do with the LED's required Current ? Cause only 2 volts reaching to LED.

I really do not understand the logic behind this ; connection this two unrelated values.. Usually tutorials uses Pipes analogy.. Even with pipe analogy i can not clarify this .. Example : In a pipe system , If a narrow pipe is stopping 2/3 of water ; the water tribune placed after this will receive 1/3 of the water. So no way to use 2/3 water to calculate anything about the water tribune and no effect on this tribune at all.

Is there any logical explanation of this ?

Thanks in advance from confused person..

\$\endgroup\$
  • 2
    \$\begingroup\$ "Cause only 2 volts reaching to LED." What? \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 9 '18 at 13:47
  • \$\begingroup\$ I can't completely follow your reasoning due to the unusual English but I have the distinct feeling your argument is equal to the 'turtle and achilles paradox'. And it is not 'stopping' the water it is limiting the total flow of water through the system. \$\endgroup\$ – Oldfart Feb 9 '18 at 13:51
  • \$\begingroup\$ LED's light intensity varies directly with current through it and there's maximum limit on current above which LED gets fried up. By thumb rule, I "restrict" current through LED to be 20mA and I calculate resistor such that R = (5-3)/(20mA), this is not strict limit and varies with person, LED, temperature it will be subjected to and perhaps other factors. The max rating will be available in data sheet (but shopkeeper in my country has never given me one, in my whole life of buying LEDs since I was in school, quite understandable with cheap Rs. 5 = (1/16 $) LEDs I buy) \$\endgroup\$ – Deep Feb 9 '18 at 13:57
  • \$\begingroup\$ The total resistance in the series circuit is the the sum of 100R (which is a fixed value) + the effective resistance of the LED (when 20mA flows). NB The effective resistance of the LED varies with current. \$\endgroup\$ – JIm Dearden Feb 9 '18 at 13:59
  • \$\begingroup\$ Sorry for language problem.. But if it is limiting the 2/3 of water what it has to do with the amount of water the tribune need to run efficiently ( 2 volts in the problem) ?.. I mean if we want to calculate something related with tribune it sound logical to use the amount of water tribune received.. Not the amount of water limited... Maybe i am missing something ..? \$\endgroup\$ – user7685914 Feb 9 '18 at 14:03
2
\$\begingroup\$

The calculation you're using is an approximation, but it's a good one. It assumes that the voltage across an LED is constant, no matter what the current is. This is not actually true, but it's close, because LEDs are diodes, and thus exhibit a characteristic diode IV curve (ignore reverse voltage for now):

enter image description here

Notice that as the voltage increases, the current increases very quickly. For reasonable currents, from the minimum operating current to the maximum current before the LED breaks, the voltage will remain pretty constant, +/-10%. If you have a constant voltage power supply, that means that the voltage across your resistor will also be constant, no matter the current, due to kirchoffs voltage law:

schematic

simulate this circuit – Schematic created using CircuitLab

You can see here that as we change the resistance, the voltage across the LED stays the same, so the voltage across the resistor stays the same. The current increases as we decrease the resistor.

\$\endgroup\$
1
\$\begingroup\$

Simple enough: LEDs are highly non-linear and DO NOT follow Ohm's law at all. The relationship between voltage and current of an LED is quite complicated. For most practical purposes it can be simplified to a "constant voltage drop" model. Therefore you can't define the "resistance" of an LED. Your math is actually correct, 100 Ohms will get you roughly 20 mA through the LED.

Ohm's law simply states that for a resistor voltage and current are proportional. However, an LED behaves very different from a conventional resistor so you can't apply Ohm's law to it.

\$\endgroup\$
1
\$\begingroup\$

The LED does not behave like a resistor. The simple version of Ohms Law only applies to linear devices that don't "react back". In the pipe analogy the LED behaves like pressure reducing valve - it subtracts pressure. Here volage is analogous to pressure.

\$\endgroup\$
0
\$\begingroup\$

We have a 5 volt battery. And a 3 volt/ 20 mA led. So general approach to this schema is I learned: The led needs 3 volts. So we should control 5-3 = 2 volt with a resistor.

OK.

V = I * R and R = V/I. Here I assume: R = 2 V / 0.02 mA = 100 ohm resistor we need .. I hope this is correct..

Correct.

But in this formula we use the voltage that resistor will control : 2 volts .. Since this voltage is resisted against; 3 volts will reach the Led. But we find the current in formula using the voltage resistor will control and what it has to do with the LED's required Current ? Cause only 2 volts reaching to LED.

... Usually tutorials uses Pipes analogy.. Even with pipe analogy i can not clarify this .. Example : In a pipe system , If a narrow pipe is stopping 2/3 of water ; the water tribune [turbine] placed after this will receive 1/3 of the water. So no way to use 2/3 water to calculate anything about the water tribune [turbine] and no effect on this tribune [turbine] at all.

Your turbine understanding is not quite correct. Let's try this:

  • You have a turbine. It will run at maximum speed at 10 L/s. For 10 L/s I need 0.4 bar pressure difference between the input and output. That's about 4 m "head" of water. (10 m = 1 bar = 1 atmosphere.)
  • You have a river and a dam. It is 8 m higher than the outlet. Now your turbine will have 0.8 bar pressure. The flow will be much faster (higher current) and your turbine will run too fast and break up. (The blades will fly off or the rotor imbalance will shake it to pieces.)
  • To solve this we add a valve on the feed to add some resistance to the water and limit the current to 10 L/s. Now the turbine can run at the rated current.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1.
(a) The correct battery (head of water) for the lamp (turbine).
(b) Double the head gives double the current. The bulb (turbine) will be destroyed.
(c) R1 limits the current (flow), introduces a voltage (pressure) drop and the lamp (turbine) is run with the correct current (flow).

... and no effect on this tribune [turbine] at all.

That is the idea. The turbine doesn't know anything about the resistance.

enter image description here

Figure 2. The water analogy for a diode is a one-way check valve. Source: What is an LED?

From the article:

If you look at the check-valve in the figure above, it should be clear that the spring normally keeps the ball in position and prevents back-flow. When “forward-biased” the ball shut-off can be moved against the spring but it will take some initial pressure to move the ball. This results in a pressure drop across the valve: the pressure downstream will be less than the inlet pressure.

In a similar manner the PN junction causes a voltage drop. For silicon it is about 0.7 V. Since there is a PN junction in the base-emitter of your transistor you can expect a 0.7 V drop across it when forward biased.

Pushing the analogy a little further, we can also see that further pressure drop will occur due to the constriction of the valve. The more water we push through the valve the more the pressure will drop. This will be added to the initial pressure drop required to open the valve in the first place. The resultant pressure drop graph will look remarkably like one of the I vs V curves below.

enter image description here

Figure 3. Typical IV curves for various colours of LEDs.

I hope that helps. (The LEDnique article is mine.)

\$\endgroup\$
0
\$\begingroup\$

Lets examine the combined I-V curve for LED to Ground, in series with resistor to +5v

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.