[thomas]

Quote:

I stand corrected.

Actually, i realized my reply was partially wrong, the chart AOS linked to was what i meant to say...

The resistance of a diode shoud be infinate at low voltages and very close to zero at high voltages like you said, (rather than reaching a constant as a said),but the reason that never happens is because you can't supply enough voltage to completely forward bias the diode. Instead the resistance of the LED is ALWAYS in the non-linear range when it is producing light, if you supply enough voltage to completely forward bias it (ie zero resistance), it would probably burn it out.

And the diode's resistance? it varies depending on voltage, as AOS said- there's no way of finding out without measuring the current. I usually find it to be in the range of 100-1000 ohms when it is producing some light, but only a small change in voltage will take it outside that range, and different LED types/colours will also change it.

Quote:

All the explanation does not suit my knowledge. Except stereth's. You are not measuring the LED forward bias voltage but measuring the battery suply voltage. The voltage drop is due to the internal resistance of the battery when high current passed through. Remember that voltmeter has high resistance so that when you measure at no load will give you higher voltage

Everyone posting so far has agreed on this point- the voltage measured is the source voltage, and its lower than the voltage measued on an open circuit because of the internal resistance of the battery...


Quote:

I square R is heat, not the energy to light up your LED. Power equation is IxIx(R_led+R_battery) + V(voltage for light)I

My calculation was about the total power dissipated by the LED, including heat and light. I wrote it in that form to clearly show that the resistance could not be zero (ie the diode is not fully biased) while the LED is emitting light.

I'm guessing that your equation is a convention of optoelectronics, which i have not studied, and that you've simply seperated the joule heating from the light production. I wrote the equation in that way to "prove" that the LED must have non-zero resistance when it is emitting light. Isquared* R may not have any practical applications (since you don't know R), but it is useful in explaining this subject.

 

[jasonhanjk]

I agreed with Thomas but maybe not with AOS.

Thomas said Quote:

The resistance of a diode shoud be infinite at low voltages and very close to zero at high voltages

Totally agree!

Quote:

I usually find it to be in the range of 100-1000 ohms when it is producing some light, but only a small change in voltage will take it outside that range, and different LED types/colours will also change it.

This is because you are measuring it at the curves of before forward bias.

Aos said Quote:

You can estimate resistance as for example (1.8V - 1.6V) / (30 - 7 mA) = 8.7 Ohm

Same like diode, resistance should be zero. Of course if you meant act like a resistor then it is acceptable.

The graph clearly shows the increase in voltage increases current but you cannot use it to calculate resistance. Remember LED have almost the same characteristic as diode.

In actual practice we find the voltage drop accross the LED and the current required. By then you calculate the series resistor required. Internal resistance of LED is a don't care.

Supply voltage = 10V DC
Voltage drop for LED = 1.7V
Supply current = 0.02A
Resistor = (10-1.7)/0.02 = 415 ohm
In the market there is no 415 ohm so choose 10k ohm resistor and follow the meta42 design.

To calculate required resistor, follow this link, this web designer forgot to add open and close bracket for the formulae.
http://home.cogeco.ca/~rpaisley4/LEDcalc.html

If somebody had read through the meta42 or the new PPA design, it use zener diode and CRD to stabilize voltage and current accurately.
That is why resistor is not favoured when supply voltage increase to 10.5V, your little LED has only 2.0V max and it is receiving 2.2V.
Using the same formulae and current remain the same.

Liteon produced very good DVD rom but mine broke when I bought it.

I write this lengthy article is not to prove someone is wrong, but to share the same mistake that I had encountered previously.
Really embarrased me last time...

 

[aos]

Quote:

Instead the resistance of the LED is ALWAYS in the non-linear range when it is producing light, if you supply enough voltage to completely forward bias it (ie zero resistance), it would probably burn it out.

Oh, this is probably the cause of the above misunderstandings. I don't know if this is the case or not, just that for this particular diode datasheet claims that 20mA is its normal operation, and that is well within llinear region. Also datasheet claims 40mA as max continuous current and 200mA as max peak current. So there is plenty of leeway to operate in the linear region.

Quote:

The resistance of a diode shoud be infinite at low voltages and very close to zero at high voltages

Which can also be seen from the slope of the curves. If the diode was ideal, you'd have horizontal line at 0 Amperes until the forward voltage is hit, then a vertical line. In practice, you have very high impedance and then small transition region, then low impedance. But several Ohms isn't zero just as several Mega or Giga Ohms aren't infinity either. You CAN ignore that in most cases and that's what engineering is all about - knowing when and what to ignore and when not to.

Quote:

The graph clearly shows the increase in voltage increases current but you cannot use it to calculate resistance. Remember LED have almost the same characteristic as diode.

Of course you can. We are talking "small signal" resistance, i.e. if the diode operates at a particular point of the curve and then we look at small variations (i.e. operating point moves very slightly up or down the curve). Usually the assumption is also that the curve is linear (as long as changes are small the approxmation holds), but in this case it really is linear. This is a common engineering practice.

Quote:

In actual practice we find the voltage drop accross the LED and the current required. By then you calculate the series resistor required. Internal resistance of LED is a don't care.

Yes, because diode resistance can be ignored when calculating the current setting resistor - it is much less than that of the series resistor. Normally you don't need to care about diode resistance - in my example you can just say that you need 500Ohm to get 20mA current at 10V. I did the calculation in order to show why LED can be used as voltage reference - and you need to take into account the resistance of the diode in that case. Because majority of the change in voltage drop will be taken by the current setting resistor, not by the diode, because resistor has so much larger resistance.