I'm making some simulations in Tinkercad and trying different circuits to turn on a light bulb.
Why does an LED need some resistance and a lightbulb does not?
I'm only using a breadboard and a 9V battery.
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\$\begingroup\$ lookup what is a diode? and compare with tungsten bulb \$\endgroup\$– Tony Stewart EE75May 28, 2020 at 13:20
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7\$\begingroup\$ Because LEDs and lightbulbs operate on fundamentally different principles and have different drive requirements. \$\endgroup\$– HearthMay 28, 2020 at 13:26
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28\$\begingroup\$ You DO need a resistance ... but the lightbulb is it. \$\endgroup\$– user16324May 28, 2020 at 13:42
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1\$\begingroup\$ I can't understand why there were 6 answers to a question so simple. \$\endgroup\$– mguimaMay 29, 2020 at 23:10
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1\$\begingroup\$ Because resistance would be futile \$\endgroup\$– user541686May 31, 2020 at 11:16
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6 Answers
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With LED's, a small increase in voltage will result in a large increase in current. So it is really hard to get just the right voltage to keep an LED at the right brightness. If you let the voltage just get a tiny bit too high it may destroy the LED.
What makes it even harder is that as the LED gets hot, the current will also increase. Naturally when you power it up it will tend to get hot. As a result it is just too much trouble to drive an LED with a voltage. Some form of current limiting usually has to be put in place. It doesn't have to be a resistor, but that is probably the most simple way to do it.
LED light bulbs have circuitry integrated into them that overcomes all these problems.
Old-fashioned incandescent light bulbs (including halogen bulbs) are different. The part that lights up is made from a thin tungsten wire that glows when it gets hot. The wire has resistance which limits the current automatically. This resistance is also what causes it to heat up. And, icing on the cake, the resistance goes up with temperature, so incandescent light bulbs are really stable when powered from a voltage source.
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2\$\begingroup\$ But there are limits(?). Using double the nominal voltage is probably not a good idea. \$\endgroup\$ May 29, 2020 at 17:52
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\$\begingroup\$ @PeterMortensen for LED's if you control the current the voltage will take care of itself. For incandescent bulbs, they need to be used at their rated voltage. At 2x rated voltage they will glow very bright and then fail. I didn't mean to imply in any way that it is safe to over-volt incandescent bulbs. \$\endgroup\$ May 29, 2020 at 21:05
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1\$\begingroup\$ In the old days (maybe even now) workers sometimes used to get injured if they put 220v (or worse, 110v) inspection lamps across 415v in a three-phase setup. The bulbs went BANG. \$\endgroup\$ May 30, 2020 at 10:20
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2\$\begingroup\$ It took me a long time to figure out that your first sentence is the key to understanding LEDs. We start off in electronics with fixed resistances and a nice linear Voltage vs Current graph. Not so with LEDs. Good explanation \$\endgroup\$ May 30, 2020 at 17:08
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The "problem" of an LED (and each diode) is that it has not constant (static, ohmic) but "dynamic" resistance that decreases when the voltage applied to the diode increases. Thus, in Ohm's law, both voltage and resistance change but in opposite directions - I = Vinc/Rdec. As a result, the current through the diode vigorously increases... and if the applied voltage is quite higher (9 V here) than the diode threshold voltage (typically 2 V for an LED), the current and accordingly the power, will become unacceptably high. To solve the problem, we connect a resistor in series. The operation of this network is visualized in Fig. 1.
Fig. 1. LED presented as a voltage-stabilizing dynamic resistor
When we turn on the power supply, the (input) voltage VIN increases from zero to maximum. In the graphical representation, its IV curve (including the resistance R) moves to the right (translates). At the same time, the diode begins decreasing its static resistane RST so its IV curve rotates counterclockwise. As a result, the operating point A slides up along the vertical part of the diode IV curve. The current variations are significant while the voltage drop VD (VF) across diode does not change - the diode differential resistance is zero.
The incandescent lamp also does not possess constant resistance. However, in contrast to an LED, its dynamic resistance increases when the voltage applied across the lamp increases. Now, in Ohm's law, both voltage and resistance change but in the same direction - I = Vinc/Rinc. As a result, the current and accordingly the power, more slowly increase... and will not become unacceptably high.
answered May 28, 2020 at 20:20
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1\$\begingroup\$ This should be the top answer. The LED will drop only so much voltage, leaving the wiring to drop the rest. Since the resistance of the wiring is well under an ohm, to drop the voltage requires a current that is much more than enough to destroy the LED. (I once made an LED explode with a loud BANG by overloading it...) \$\endgroup\$ May 29, 2020 at 7:07
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\$\begingroup\$ Thanks... Interesting thoughts showing a tendency to think deeply... The conclusion is that we have to absorb the redundant voltage by some "elastic" element (resistor)... or by a combination of elements where some of them are "stiff (diodes)... but at least one of them is "elastic". \$\endgroup\$ May 29, 2020 at 9:17
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\$\begingroup\$ You lost me in the last sentence. "in Ohm's law, both voltage and resistance change but in the same direction". 1) In Ohm's law the resistance does not change; it is a constant (independent of the voltage) 2) If you meant current as a function of voltage, that is also the case for both the LED and bulb - both have a positive, but non-linear, resistance (otherwise it would a negative resistance (possible with active circuits)). The second derivative (derivative of resistance) is zero for Ohm's law, negative for the LED, and positive for the bulb. \$\endgroup\$ May 29, 2020 at 18:23
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\$\begingroup\$ @Peter Mortensen, interesting thoughts... I just wanted to explain the "dynamizing trick" where we change the voltage but, in the same time, someone changes the resistance. As a result, depending on the change direction, the current magnitude is increased (if the resistance was decreased) or decreased (if the resistance was increased). If we go too far with this trick, we will observe negative resistance - S-shaped in the first case (neon lamp) and N-shaped in the second case (tunnel diode). \$\endgroup\$ May 30, 2020 at 8:11
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\$\begingroup\$ @Peter Mortensen, I have illustrated the idea behind the voltage-stabilizing dynamic resistance in my answer above. It is based on the believe that basically, the diode is a kind of "resistor" but non-linear... and this non-linear resistor can be thought as of a dynamic (self varying) static resistor... \$\endgroup\$ May 31, 2020 at 18:31
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When you say lightbulb I take it you mean an incandescent light source? If so, the 9V battery has nowhere near enough juice to do any damage to its filament, which is usually made out of carbon, tungsten, or titanium. Two of the filament's requirements are high resistivity and high melting point, which is one of the needed features to make it glow and produce light. You might even say, it is its own "resistor"...
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\$\begingroup\$ Yes i'm refering to a incandescent light source. Thank you for the answer i understand now. \$\endgroup\$– VillanoMay 28, 2020 at 13:39
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3\$\begingroup\$ A 9 volt battery isn't necessarily just the little battery that you think it is. It can be one of these PP6 types so be careful about not pre-empting things. And of course any 9 volt battery or power source should not damage a 9 volt rated light-bulb. The filaments don't need high resistivity (I believe you might have meant resistance) either - their resistance is tailored to the voltage source and can hardly ever be regarded as high in value. \$\endgroup\$– Andy akaMay 28, 2020 at 13:54
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\$\begingroup\$ Note that the asker is using a simulation which may not model damage behavior \$\endgroup\$ May 28, 2020 at 14:00
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\$\begingroup\$ I doubt you would get one of these batteries to power up your breadboard experiments ;-) In any case, I do mean resistivity and not resistance (as in material coefficient of resistance). High resistivity does not equate high resistance either... Here is a link about incandescent filament materials and their requirements (from 10k feet view point) electrical4u.com/materials-for-lamp-filaments \$\endgroup\$ May 28, 2020 at 14:05
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1\$\begingroup\$ If you mean resistivity you are wrong: Tungsten, the most common filament material, is a pretty good conductor (has a relatively low resistivity of 5.60×10−8), twice as good as iron and magnitudes better than carbon. In order to achieve the required resistance the filament must be accordingly thin and long: Take a close look at the typical recursively spiraling tungsten filament. \$\endgroup\$ May 29, 2020 at 10:22
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Why don't I need a resistance when testing a light bulb circuit in a breadboard?
A light-bulb is pure resistance - it passes a current and it gets hot and it gets so hot that it glows and produces light. It is inherently an inefficient method of converting electricity into light. It has a voltage and power rating and therefore you apply the correct voltage and you get a power consumption of 10 watts, 20 watts, 40 watts etc.. and a lot of heat.
An LED also produces light and it produces light with maybe up to five or ten times more power efficiency than a regular light-bulb - that is one of its biggest attractions.
Graph from here.
Why does an LED need some resistance and a light bulb does not?
The downside is that an LED comes with restrictions of supply voltage and that means being careful when you apply voltage to it. Of course an LED could come with a built in resistor that allowed it to be used on the same voltage supply as a light-bulb but that misses the point of using an LED to produce light power efficiently.
So, if you don't need a highly efficient light source use an LED with a series resistor. If you are relying on the power efficiency of an LED (because you need to) then drive it carefully, effectively and efficiently.
answered May 28, 2020 at 13:27
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\$\begingroup\$ While convenient from an application perspective to looking at an LED in combination with a resistor as something that you "apply voltage" to, the path to understanding the LED is to consider the current pushed through it. \$\endgroup\$ May 28, 2020 at 14:02
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\$\begingroup\$ The downside is that an LED comes with restrictions of supply voltage and that means being careful when you apply voltage to it. AND then drive it carefully @ChrisStratton - I'm not trying to inform how to drive a LED but I am saying it has to be done carefully. \$\endgroup\$– Andy akaMay 28, 2020 at 14:04
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4\$\begingroup\$ That's really, really looking at it the wrong way, an LED is not a voltage-mode device but a current-mode one, and needs an approximation to a current source, not a voltage source. \$\endgroup\$ May 28, 2020 at 14:04
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1\$\begingroup\$ @ChrisStratton please feel free to make an answer of your own. Like I say you have to be careful when driving an LED and many sincere thanks for the downvote and I look forward to your answer. \$\endgroup\$– Andy akaMay 28, 2020 at 14:10
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\$\begingroup\$ Re "pure resistance": That is a little bit misleading. It does not follow Ohm's law; it is a non-linear resistance (and probably no significant reactive component / reactance of zero at the normal operating frequencies). Perhaps clarify it in your answer? \$\endgroup\$ May 29, 2020 at 18:03
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To add to the existing answers with an explanation from the practical site:
A light bulb is usually manufactured for a specific voltage and then connected to a voltage source.
On the other side, an LED is sensitive to the current. The voltage/current curve varies not only by color, but also by manufacturing tolerances. It is also very steep, so a small overvoltage will cause a huge current and immediately destroy it.
Usually, LEDs are driven by a constant-current source. In simple circuits, this is emulated by a series resistor that limits the current. It needs to be chosen so that the current stays within the specifications even for LED's with particular low drop voltage, or with the resistance decrease that might occur when temperature changes.
So, an LED with a Vdrop varying between 3.2 and 3.4 volt can not reliably driven on a 3.5 V source using a series resistor since the current will vary largely between the allowed Vdrop range. When using a 9 V supply, there is a large enough voltage drop at the resistor to stabilize it even when using UV LEDs with a high Vdrop, like a UV LED reaching 4.4 V Vdrop (the Vdrop increases roughly inversely proportional to the wavelength of the emitted light).
However, as noted, in almost all industry applications using high-power LEDs (so not just as a status LED) they are current driven.
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\$\begingroup\$ A comprehensive answer showing interesting sides of the problem... As I have figuratively said above, we have to absorb the redundant voltage by some "elastic" element (resistor) or by a combination of elements where some of them are "stiff" (diodes)... but at least one of them is "elastic". Now I agree with you that the portion of voltage drop across the latter should be high enough to ensure small current variations when the supplying voltage varies. \$\endgroup\$ May 29, 2020 at 9:29
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\$\begingroup\$ What is the reason for the weird capitalisation of words (not rhetorical question - I am really interested to know why)? E.g., is it instead of bold and italics? Or did an old version of English capitalise common nouns (as in German)? Or something else? \$\endgroup\$ May 29, 2020 at 18:36
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\$\begingroup\$ Peter, Mortensen, some of what you think are words are actually abbreviations, where all the letters are capitalized. For example, LED is an abbreviation for light emitting diode. \$\endgroup\$– user6030May 29, 2020 at 23:18
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2\$\begingroup\$ Part of the capitalisation was indeed weird because english is not my native language, and i was just tired and typing without proof-reading, so thanks for fixing. \$\endgroup\$– BoldarMay 30, 2020 at 2:35
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Both the light bulb and the LED are a non-linear elements and they are differently non-linear. (see here: http://physicsexperiments.eu/2097/light-bulb-current-voltage-characteristic and here https://www.electronics-tutorials.ws/diode/diode_8.html for I/V graphs.)
In particular, the resistance of a diode goes rapidly down when the voltage or current is increased and the resistance of an incandescent filament goes slightly up.
In order to get a stable operation (as in small change of any parameter not to bring large and possibly damaging change of some other parameter) the LED must be fed a more or less constant current and the lightbulb must be fed a constant voltage.
Not that you cannot do the other way 'round.
You can perfectly power a LED with its 3.09 +/- 1% volts (3.09 are typical example for a blue or white LED and the exact number also depends on the temperature of the LED) in order to keep the light output in +/- 50% limits. For the same LED, 3.5 volts are instantly damaging and 2.8 volts are no light at all. Not a deal, is it?
The light bulb when used with a current source instead of a voltage source is more forgiving. It will start rather slowly (say, 1-3 s for high-power lamps) and will get brighter and brighter with aging, leading to much faster aging.
A resistor for the LED is just a simple current source. You can as well use other (more or less constant) current source that doesn't explicitly include a resistor. Power LEDs are used with integrated switching-mode current source. You can also use a single CR2032 battery as a current source with no other elements - it will power a 5mm led for a day or two with 4-10mA current.
