LED voltage drop

Hi,

If you connect a 9v battery across a red LED it blows.

The voltage drop across an LED is fixed at about 1.4V so what happens
to the other 7.6v??

Gareth

 

A battery as well as a LED have an internal resistance. As the internal
resistance of the LED is very small, most of the voltage drop comes across
the internal resistance of the battery. So you will see the battery voltage
fall way below 9V. The voltage across the LED will rise but not too much
until the LED blows. Be aware that a lot of LEDs have a voltage drop of
around 2V these days. To be sure, you'll need to look at the datasheet.

petrus bitbyter

 

Well, the led burns up and then you see the full 9 volts across its
terminals.
So the 7.6 volts didn't disappear.
The question is, what happens in that split second between 1.4 and 9?
Is there really a discontinuity, i.e. an instantaneous change from 1.4
to 9;
or would you be able to see intermediate voltage values if you had
fast enough test gear?
I suspect you would see intermediate values.

 

Well, the led burns up and then you see the full 9 volts across its
terminals.
So the 7.6 volts didn't disappear.
The question is, what happens in that split second between 1.4 and 9?
Is there really a discontinuity, i.e. an instantaneous change from 1.4
to 9;
or would you be able to see intermediate voltage values if you had
fast enough test gear?
I suspect you would see intermediate values.

 

Funny you mention that - IR leds? I just got some tricolor leds that
are specified at 2V red and 3.4 V green and blue - and they drop 2V
red, ~1.7 green, and 1.1 V blue (!)

I've checked my meter several times and even went so far as to put a
"normal" white LED in series and got 3.5 across the white and 1.1
across the tricolor blue. Different meter same thing.

It isn't "possible," but I'm inclined to believe what the meter is
telling me. (particularly since I can light the silly thing on an old
alkaline cell with a 1.1V drop 3 ma and with a 6V supply and the drop
is still 1.1 and at 20 ma)

--

 

Others have answered correctly that the "other 7.6V" is dropped
across the internal resistance of the battery.

Let's look at a proper circuit. The right way to reduce
the voltage from the battery's 9 volts to the ~1.4 volts
that the LED needs is by dropping that ~ 7.6 volts in
a controlled manner selected by the circuit designer.
Usually, it is done by placing a resistor in series with
the LED. The value of the resistor is determined
using ohm's law where voltage (E) equals current (I) times
resistance (R) or E=IR. Assuming a typical LED, the
current (I) should be limited to about .02 amps. So,
since you want to drop ~ 7.6 volts, plug the numbers into
the equation and solve: 7.6 = .02 R. Therfore R = 7.6/.02 or
about 380 ohms.

The internal resistance of a battery is extremely low compared
to 380 ohms, so it can be ignored. And a standard value 390
ohm resistor is close enough to 380 ohms that it can be used.
So you end up with a circuit that looks like this:

--------
+-----|390 ohms|-----+
| -------- |
| 0 LED
| |
| +----------------+
| |
-----
| 9V |
| |
| |
-----

Incidently, the 1.4 volt figure is usually closer to
1.8 volts for a typical red LED. The actual voltage
depends on the individual LED of course, and the
amount of current through it.

Ed

 

Single alkaline cell lights a blue LED?

Your RGB LED is probably a poorly documented one made from mix-and-match
parts, including whatever goes into those blue ones with built-in boost
converters. Some Lumex blue and white ones that work from single alkaline
cells were available from Digi-Key for a few years.

- Don Klipstein ()

 

The LED has resistance, usaully a few to around 10 ohms. The 9V battery
has a couple to a few ohms of resistance (more when in weak condition).

Connect a good alkaline 9V battery across an LED, and for the sake of
argument (numbers picked to make calculations easier):

The LED has an inherent voltage drop of 2 volts (usually changing by .1
volt per factor of 10 change in current, now saying 2.15 volts for
~30x rated current),

The fresh 9V alkaline battery has an open circuit voltage of 9.65V,

This fresh battery has a resistance of 3 ohms (fresh 9V batteries usually
deliver about 8 volts to #93 automotive lamps and close to 6 volts for
#1156 ones),

and the LED has a resistance of 9.5 ohms, for a total of 12.5 ohms,

then:

9.65 volts minus 2.15 volts means 7.5 volts across the total resistance
of 12.5 ohms, for a current of .6 amp.

LED inherent voltage drop at .6 amp is 2.15 volts
LED internal resistance of 9.5 ohms * .6 amp = 5.7 volts
LED total voltage is 7.85 volts
LED power = 7.85 volts * .6 amp = 4.71 watts

With LED current about 20 times maximum rating of most 3 and 5 mm LEDs
and LED power 30-50 times the maximum of most such LEDs, the LED will
probably completely fail within a second, with a fair chance of cracking
and production of foul-smelling smoke.

Battery internal resistance of 3 ohms * .6 amp = 1.8 volt drop,
from 9.65 volts = 7.85 volts

If the battery has more internal resistance than the 3 ohms that I
picked, then current will be less, and the LED voltage (which equals the
battery terminal voltage here) will be less.

================================================

For another set of numbers, the same hypothetical LED has inherent
voltage drop at ~10x rated current being 2.1 volts, same internal
resistance of 9.5 ohms. The battery is a weaker one with open circuit
voltage of 8.75 volts and internal resistance of 14.25 ohms.
Although these numbers were chosen for easier calculations, they are
close to actual of a somewhat-well-used 9V battery that I actually have at
this moment. That battery delivers approx. 3 volts to a #93 automotive
lamp and approx. 1.3 volts to an 1156 one.

Total resistance is 23.75 ohms. 8.75 volts minus 2.1 volts is 6.65
volts across this resistance.

Current is 6.65 / 23.75, which is .28 amp.

Multiply .28 amp by 9.5 ohm LED resistance, and taht is 2.66 volts in
addition to the 2.1 volts unrelated to resistance, for a total of 4.76
volts.

4.76 volts times .28 amp is about 1.33 watts.

.28 amp and 1.33 watts exceed the current and power ratings of most 3 mm
and 5 mm (T1 and T1-3/4) LEDs by about an order of magnitude. The LED is
likely to suffer significant permanent degradation in anywhere from a
fraction of a second to a few seconds, and is likely to completely fail
anywhere from within a second to having endured several seconds of such
abuse.

Multiply .28 amp by the 14.25 ohm battery resistance, and that is 3.99
volts to subtract from the 8.75 battery open circuit voltage, resulting in
4.76 volt battery terminal voltage.

=======================================

If you have any old 9V batteries that you are tempted to connect LEDs
directly across, beware - the internal resistance of batteries is very
non-constant. Though tending to vary inversely with temperature, it tends
to vary directly with usage in the battery's recent history - the
resistance increases with use and decreases with rest, even when usage
heats the battery.
Old 9V batteries do occaisionally damage LEDs connected directly across
them. If you have a temptation to connect LEDs directly across old 9V
batteries, that's one purpose of 150 ohm resistors. 150 ohms won't mean
much when the battery resistance gets to several hundred ohms.

- Don Klipstein ()

 

Hi all,

Thanks for all the responses!! Makes perfect sense now - as the
internal resistance of things is usually ignored I completely forgot
about it.

Gareth