Can Resistance Increase Brightness in LED Diodes and Light Bulbs?

In summary, the increased resistance of the LED caused the voltage to increase faster than the current and the power consumed by the LED increased.
  • #1
Bassalisk
947
2
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
 
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  • #2
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.
 
  • #3
Averagesupernova said:
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.
 
  • #4
Bassalisk said:
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 Rp, both in series with a resistor Rs and an ideal voltage source U, where Rs can be thought of to also model internal resistance in the battery, then I get the power consumed by R as

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

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

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

when

[tex]R = \frac{R_s R_p}{R_s+R_p}[/tex]

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.
 
  • #5
Ok i think i understand. I will have those equations checked. Thank you very much.
 

What are LED diodes and how do they work?

LED stands for light-emitting diode, which is a semiconductor device that emits light when an electric current passes through it. They work by allowing current to flow in one direction, resulting in the release of energy in the form of photons (light).

How do LED diodes differ from traditional incandescent light bulbs?

LED diodes use significantly less energy than traditional incandescent light bulbs, making them more energy-efficient. They also have a longer lifespan, with an average of 25,000-50,000 hours compared to incandescent bulbs' 1,000 hours. Additionally, LED diodes do not emit as much heat, making them safer to use.

What are the benefits of using LED light bulbs?

LED light bulbs have several benefits, including energy efficiency, longer lifespan, and lower heat emission. They also do not contain harmful substances like mercury, making them more environmentally friendly. LED bulbs also have a wide range of color options and can be dimmed, providing more control over the lighting atmosphere.

Can LED light bulbs be used in any type of lighting fixture?

Yes, LED light bulbs come in various shapes and sizes, making them suitable for use in most lighting fixtures. They are commonly used in residential, commercial, and industrial settings for both indoor and outdoor lighting.

Are LED light bulbs more expensive than traditional incandescent bulbs?

Initially, LED light bulbs may have a higher upfront cost than traditional incandescent bulbs. However, considering their longer lifespan and energy efficiency, they can save money in the long run by reducing energy bills and the need for frequent replacements.

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