Can Resistance Increase Brightness in LED Diodes and Light Bulbs?

[Bassalisk]

Hello,

I did an experiment home, i took a power supply 9V and hooked it up in series with 1k Ohm resistor, then after that I hooked it up to a parallel connected 2 LED diodes. So both diodes were on the same voltage drop. I don't know exact calculations behind it but it doesn't matter. Here is the thing. I hooked a small metal wire to the one electrode of the LED diode, and I heated up that wire. That LED diode started glowing brighter and the other LED diode started dimming. Now this is against my intuition. I have 2 conflicts here.

I know that in semiconductor, when u give energy to it, u get more charge carriers ergo you can have more current. But today my professor told the same experiment, but with no heating, with light bulbs. He said that if one light bulb has more resistance, it will glow brighter. Of course, he was talking about AC current, but I don't think that matters.

Now how can this be? If resistance goes up, the current must go down! I=U/R! Can someone explain what is really going down with this phenomena? Are 2 cases(LED and bulbs) both explainable in the same fashion or is there a difference?

Thanks

 

[Averagesupernova]

2 LEDs in parallel is generally an engineering no-no. Slight variances in LEDs will cause one of them to hog all the current. This is what most likely happened in your LED experiment. By heating the wire you also heated the LED which changed its properties enough to lower the voltage drop required for it to turn on.
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I don't understand what your professor is saying. You cannot increase the resistance and maintain the same voltage and have a brighter incandescent light bulb. Maybe he is saying a higher resistance with the same current will increase the brightness.

 

[Bassalisk]

2 LEDs in parallel is generally an engineering no-no. Slight variances in LEDs will cause one of them to hog all the current. This is what most likely happened in your LED experiment. By heating the wire you also heated the LED which changed its properties enough to lower the voltage drop required for it to turn on.
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I don't understand what your professor is saying. You cannot increase the resistance and maintain the same voltage and have a brighter incandescent light bulb. Maybe he is saying a higher resistance with the same current will increase the brightness.

I understand, I will ask him about that problem with steady current. Thanks for the diodes part though.

 

[Filip Larsen]

Now how can this be? If resistance goes up, the current must go down! I=U/R!

I think you want to look at power through that resistance, not just current.

If we take a simple circuit with a resistor R in parallel with another resistor R_p, both in series with a resistor R_s and an ideal voltage source U, where R_s can be thought of to also model internal resistance in the battery, then I get the power consumed by R as

P = \frac{R}{((R_s/R_p+1)R+R_s)^2} U^2

which can be seen to increase from P = 0 at R = 0 to a maximum value

P_{max} = \frac{1}{4}\frac{R_p}{R_s(R_s+R_p)}U^2

when

R = \frac{R_s R_p}{R_s+R_p}

after which P will decrease back down to zero as R goes to infinity. Providing the model fits your situation, that is that the battery can be modeled as an ideal voltage source with internal resistance, the LED's can be modeled as simple resistors, and the emitted light intensity of each LED is proportional to the power through it, then it seems that there is a range of R for which the power through R in fact will increase when R increases and it may very well be what you observed. Or in other words, what seem to happen is that, as you increase R the voltage over R increases faster than the current drops so that power consumption of R effectively increases.

Just for the record you should know that I have not checked the above result with any reference, so please make your own calculations and conclusions. It should be a good exercise anyway to do the calculations yourself and make some plots of voltage, current and power over the various components as a function of R.

 

[Bassalisk]

Ok i think i understand. I will have those equations checked. Thank you very much.