Very simple question about LED

Can I just connect for example 9V 50mA LED to 9V battery and expect it to work flawlessly?

No, you cannot. An LED is a diode that generates light under a specified current while the LED voltage under that current varies by color and type.
Check LED specifications and use a series resistor to set the nominal LED current.
For instance, LED indicators typically need 5...10 mA, and a red LED has typically 1.5... 2 VDC across its terminals under such current. For 9V source (a battery) you will need e.g. 500-600 Ohms series resistor. Blue and white LEDs have >2 V voltage under ~10 mA, so you can set the current with ~300 Ohm resistor.
To increase LED brightness, the source is often pulsed and in short pulses LED allows up to 0.1...1 A pulse current. But LED life can be limited by such overload.

Read about LED lighting design on Wikipedia or recent magazine papers. Many LED manufacturers give recommendations in their specifications.

And the same goes to incandescent light bulb, right?

Hi

Can I just connect for example 9V 50mA LED to 9V battery and expect it to work flawlessly?

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if it is specified to work with 9V it should be possible. Then i´d expect that there is a built in resistor.

Best way is to look into the datasheet.

Klaus

The SIMPLE answer is , unless specifed, NO.

But if you asked about 12V LEDs on Reels, I would say YES. (R limit. by design)


all else ...

RTFS read the fine spec.

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LEDs like all diodes are somewhat like batteries. They come in a range of different voltages depending on chemistry which controls wavelength and voltage, longer wavelength = lower energy=lower voltage. The major factor is number of devices in series, which is common practice now with power LEDs in the range 3~300W.

As in batteries, the bigger power they are, the lower ESR and more current when the voltage varies from nominal (within each chemistry )

For white LEDs typically ESR is approx. 1 Ohm per Watt. e.g. 100mW part is around 10Ω

This means the current will be limited by Ohm's Law between the applied voltage and the threshold voltage of the LED which is not typically given, but you can extract from a straight line approximation on VI curves.

This means if you put a 2V RED 20mA LED on 12V you must add a series voltage drop of 10V and/or add more LEDs in series. (10V/20mA=~520 Ω)

An incandescent light bulb IS ALREADY a resistor so if the voltage is correct for it then another resistor is not needed. The light bulb works on AC or DC.
A resistor is needed in series with an LED to limit the current. An LED is polarized so it works only on DC.

An incandescent Light is almost opposite to LED. It has a surge current vs LED starts slow then surges with overvoltage.

Tungsten filaments have an ESR 10x lower when cold or 10% of hot power rating thus draws up to 10x peak current as a resistive load. We call this a Pos. Temp coeff. or PTC. As temperature rises on the filament, resistance rises and current drops until steady state in fraction of a second. Self heating of the ambient has a longer time constant that slowly rises with the fixture time constant. this affects reliability of the tungsten greatly.

Every notice why some bulbs don't seem to last long??

A common fixture , a small globe will reduce a 1500 hr 60W light bulb to a few hundred hours as Arrhenius Theorem applies with 1/2 the Life time expected for every 10'C rise in rated ambient... The ambient around a light bulb can go from 20 deg C to 50'C in a small glass fixture. (bad, if the fixture is too hot, it will reduce life time!)

The LED on the other hand, has near infinite ESR below threshold except for leakage and then conducts current like a brick wall at the "Zener voltage" and is limited by the ESR of each part in series with the voltage source.

Although incandescent radiates about >80% of its power as invisible heat or IR, the LED emitter doesn't emit InfraRed (IR) from the light but emits Infrared from the conductor losses like crazy. ( So don't let anyone fool you that LED's don't emit IR. They have IR heat from Conduction losses, not from electro-optic emissions.)

The Amp_squared-Ohm power loss, P=I²R of the junction ( metallic semiconductor ESR property) This gets hot when the power dissipated [W] * THermal Resistance [°C/W]= Temperature Rise [°C] and that part radiates InfraRed like crazy if you apply to much voltage which results in excess current. If it's too hot to touch, it will reduce in life expectancy.

Also remember Epoxy is BOTH a Thermal and Electrical insulator. It the lens is hot. THe junction is much much hotter. Cooling is often conducted thru the negative side ( Cathode) which has a lower Thermal resistance than the Anode ( which is connected by a micron size gold wire bond ) The wirebond also can fuse open with excess current (I²t fuse rating)

This is why the best Power LED regulators use heat for feedback , not current, assuming know how to measure temperature

Anyone can measure case temperature with a thermistor, but you can measure true diode junction temperature using the threshold voltage and very low current measured quickly with ADC.)

Most people rely on Constant current (CC) sources or Resistive Ohm's Law drop current limiters from Constant Voltage (CV) sources.

You cannot neglect the Zener curve unless there is a resistor added to the LED to be used as fixed voltage indicator. If you expect AC, use two in parallel with reverse polarity.

I love it when people ask simple questions.