Connecting LED's to direct 120VAC?

[[email protected]]

Is it possible to connect a blue (or white) LED to direct 120VAC using
just a resistor? or world I have to convert the AC to DC? Because I
tore apart a old power strip with a suppose-of surge protector and I
only see a red LED in series with just a resistor, and no rectifier or
diodes. If so, what would the resistor value be? (ohm/watt) Thanks

 

[Pooh Bear]

Is it possible to connect a blue (or white) LED to direct 120VAC using
just a resistor?
No

or world I have to convert the AC to DC?
Yes

Because I
tore apart a old power strip with a suppose-of surge protector and I
only see a red LED in series with just a resistor, and no rectifier or
diodes.

Sounds like a neon lamp.

If so, what would the resistor value be? (ohm/watt) Thanks

For the neon ? About 100kohm in the US and other 120V countries, 220kohm
in Europe and other 230V countries.

Graham

 

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[John Fields]

Is it possible to connect a blue (or white) LED to direct 120VAC using
just a resistor?

---
No. You'll need, as a minimum, another diode.
---

or world I have to convert the AC to DC?
---
No.
---

Because I
tore apart a old power strip with a suppose-of surge protector and I
only see a red LED in series with just a resistor, and no rectifier or
diodes.

---
Then it's either dual LED's in anti-parallel in the same housing or
a neon lamp.
---

If so, what would the resistor value be? (ohm/watt) Thanks

Assuming you want to put a standard 20mA LED across the mains,
you'll need to place either another LED or a diode in anti-parallel
with it and the resistor in series with the parallel pair of diodes,
like his:


120V MAINS>---+
|
[R]
|
+---+---+
|A |K
[LED] [DIODE]
| |
+---+---+
|
120V MAINS>---+


The value of the resistor will be:

Vmains - Vled 120V - 2V
R = --------------- = ----------- = 5900 ohms
Iled 0.02A

The closest standard 5% values available are 5600 ohms and 6200
ohms, so err on the side of caution and choose the 6200 ohms.

The resistor will allow:


Vmains - Vled 120V - 2V
I = --------------- = ----------- = 0.019 amperes
R 6200R

to flow in the circuit, and it will dissipate:


P = (Vmains - Vled) * I = 118V * 0.019A = 2.24 watts.

That's quite a bit of power, but you can make an array of 6200 ohm
resistors which will be able to handle it like this:



MAINS>--+--[6K2]-+-[6K2]-+-[6K2]--+--[1N4001>]--+
| | |
+--[6K2]-+-[6K2]-+-[6K2]--+ |
| | |
+--[6K2]-+-[6K2]-+-[6K2]--+--[<LED]-----+
|
MAINS>------------------------------------------+

If you use 1/2 watt resistors the array will be capable of
dissipating 4.5 watts, so it'll get warm, but it should be OK.

120V mains can kill, so if you're not familiar with working with
120V mains, get someone who is to help you.

 

[John Fields]

well, the LED is a diode in a sence,

---
No, it _is_ a diode.

That's why it's called a "light emitting diode."

forward current
is when the LED is emitting. and backward current 4etf..
beside the fact that it may leak a bit on reverser current.
i don't see a problem there.

---
LOL, with 170V across it when the mains reverses polarity and
back-bias the diode, it'll do a lot more than "leak a bit", it'll
let all the magic smoke out.
---

R := (E-LED_Voltage"1.7")/LED_Current"0.050ma" = 2366 ohms

---
Geez, have you go it messed up or what???

With a diode in anti-parallel with the LED it's just:


E - LED_Voltage
R = -----------------
LED_Current


Where do you get the 1.7 from?
And the 0.050mA?...

for 120V mains and a 20mA LED, which is common, it comes out like
this:


E - LED_Voltage 120V - 2V
R = ----------------- = ----------- = 5900 ohms
LED_Current 0.02A

and the power dissipation would be:


P = LED_Current * E - LED-Voltage = 0.02A * 118V = 2.36 watts

 

[Jamie]

Is it possible to connect a blue (or white) LED to direct 120VAC using
just a resistor? or world I have to convert the AC to DC? Because I
tore apart a old power strip with a suppose-of surge protector and I
only see a red LED in series with just a resistor, and no rectifier or
diodes. If so, what would the resistor value be? (ohm/watt) Thanks

well, the LED is a diode in a sence, forward current
is when the LED is emitting. and backward current 4etf..
beside the fact that it may leak a bit on reverser current.
i don't see a problem there.
R := (E-LED_Voltage"1.7")/LED_Current"0.050ma" = 2366 ohms
W = I*V = ~6 Watts.

not sure if the reverse side of the diooe would cause some
problems. you could always throw a diode across the LED to
force current flow in the diode on the reverser side of the
voltage for the LED>

 

[John Fields]

Is it possible to connect a blue (or white) LED to direct 120VAC using
just a resistor?

---
No. You'll need, as a minimum, another diode.
---

or world I have to convert the AC to DC?
---
No.
---

Because I
tore apart a old power strip with a suppose-of surge protector and I
only see a red LED in series with just a resistor, and no rectifier or
diodes.

---
Then it's either dual LED's in anti-parallel in the same housing or
a neon lamp.
---

If so, what would the resistor value be? (ohm/watt) Thanks

Assuming you want to put a standard 20mA LED across the mains,
you'll need to place either another LED or a diode in anti-parallel
with it and the resistor in series with the parallel pair of diodes,
like his:


120V MAINS>---+
|
[R]
|
+---+---+
|A |K
[LED] [DIODE]
| |
+---+---+
|
120V MAINS>---+


The value of the resistor will be:

Vmains - Vled 120V - 2V
R = --------------- = ----------- = 5900 ohms
Iled 0.02A

The closest standard 5% values available are 5600 ohms and 6200
ohms, so err on the side of caution and choose the 6200 ohms.

The resistor will allow:


Vmains - Vled 120V - 2V
I = --------------- = ----------- = 0.019 amperes
R 6200R

to flow in the circuit, and it will dissipate:


P = (Vmains - Vled) * I = 118V * 0.019A = 2.24 watts.

That's quite a bit of power, but you can make an array of 6200 ohm
resistors which will be able to handle it like this:



MAINS>--+--[6K2]-+-[6K2]-+-[6K2]--+--[1N4001>]--+
| | |
+--[6K2]-+-[6K2]-+-[6K2]--+ |
| | |
+--[6K2]-+-[6K2]-+-[6K2]--+--[<LED]-----+
|
MAINS>------------------------------------------+

If you use 1/2 watt resistors the array will be capable of
dissipating 4.5 watts, so it'll get warm, but it should be OK.

120V mains can kill, so if you're not familiar with working with
120V mains, get someone who is to help you.

---
One thing I forgot to mention was that if you use a high-efficiency
LED like an HLMP4700, which only needs 2mA to work, then the
resistor can go up to 62K and the dissipation will drop to 0.224
watts. That means you can use a single 62k 5% 1/2 watt resistor for
the job. Much better, huh?

 

[Chris]

Is it possible to connect a blue (or white) LED to direct 120VAC using
just a resistor? or world I have to convert the AC to DC? Because I
tore apart a old power strip with a suppose-of surge protector and I
only see a red LED in series with just a resistor, and no rectifier or
diodes. If so, what would the resistor value be? (ohm/watt) Thanks

Hi, Phillip. Graham and Mr. Fields have given good advice. Whether
you're using an LED or neon bulb, though, remember neither is rated to
isolate line voltage. Make sure to use a cover of some kind (like the
diffusing cap you have over the neon bulb in the power strip) to ensure
that no one can touch the surface of the element when it's at line
voltage potential.

Good luck, and play safe
Chris

 

[Pooh Bear]

Now wait a minute, since you have picked out a mistake i made i will also
pick out one that i think your making.!. first of all you did say 170 V
which is peek, now your using 120 volts here and the above math looks
like it
assumes DC voltage ? you are using 20 ma Leds for your case.
last time i remember P = I*V*cos(angle). in cases where we are dealing
with Sine wave energy.

Good Lord !

You don't understand how to use peak and RMS either ?

John is correct.

Graham

 

[Jamie]

I got to say, its a good thing the readers here have
you around to set them straight.!

---
LOL, with 170V across it when the mains reverses polarity and
back-bias the diode, it'll do a lot more than "leak a bit", it'll
let all the magic smoke out.
---

Yeah ok, you got me there, i forgot that,. i was thinking RMS etc..

---
Geez, have you go it messed up or what???

With a diode in anti-parallel with the LED it's just:


E - LED_Voltage
R = -----------------
LED_Current


Where do you get the 1.7 from?
And the 0.050mA?...

And you haven't seen LED's at these ratings?

for 120V mains and a 20mA LED, which is common, it comes out like
this:


E - LED_Voltage 120V - 2V
R = ----------------- = ----------- = 5900 ohms
LED_Current 0.02A

and the power dissipation would be:

Now wait a minute, since you have picked out a mistake i made i will also
pick out one that i think your making.!. first of all you did say 170 V
which is peek, now your using 120 volts here and the above math looks
like it
assumes DC voltage ? you are using 20 ma Leds for your case.
last time i remember P = I*V*cos(angle). in cases where we are dealing
with Sine wave energy.

P = LED_Current * E - LED-Voltage = 0.02A * 118V = 2.36 watts

gee. thanks, coming from you i take that as a complement! :=)

btw, i have seen many simple LED light systems from the
120 service using nothing more than a single resistor which
worked fine..
we have some LED upgrades for incandescent lamp holders for
control panels which you can clearly see that they are a
LED and Resistor combo for a 120 vac fixture.(no back diodes).

oh well, happier days are coming.

 

[Michael A. Terrell]

you make me laugh.
you leave out the most relevant part of the quote just to
re'engineer another one and there by, miss lead others into
your little poke.
John is a popular guy but, he isn't perfect nor is he always
correct! but i must say that at least he tries to keep it
professional.

btw, because i decided to use a 50 ma Led and 1.7 variety does
that mean something ?, it was only an example! and yes i did over
look over the Peak voltage which i did admit doing but i guess you just
didn't want to include that because that would have spoiled your fun?

and please do not reply to this, it most likely will be crap i have
heard in the profession since i started 30+ years ago.

have a good time..

If you haven't learned how to power an LED in thirty years its time
to quit.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

 

[Jamie]

Jamie wrote:




Good Lord !

You don't understand how to use peak and RMS either ?

John is correct.

Graham

you make me laugh.
you leave out the most relevant part of the quote just to
re'engineer another one and there by, miss lead others into
your little poke.
John is a popular guy but, he isn't perfect nor is he always
correct! but i must say that at least he tries to keep it
professional.

btw, because i decided to use a 50 ma Led and 1.7 variety does
that mean something ?, it was only an example! and yes i did over
look over the Peak voltage which i did admit doing but i guess you just
didn't want to include that because that would have spoiled your fun?

and please do not reply to this, it most likely will be crap i have
heard in the profession since i started 30+ years ago.

have a good time..

 

[John Fields]

I got to say, its a good thing the readers here have
you around to set them straight.!

---
Not the readers, the writers, sometimes.
---

Yeah ok, you got me there, i forgot that,. i was thinking RMS etc..

---
RMS, peak, it doesn't matter.

With 120V across it in reverse it would still pop.
---

And you haven't seen LED's at these ratings?

---
OK, I get it; you're saying the drop across the LED is 1.7V with 50
microamperes through it?

While the LED may have 1.7V across it with 50 microamps through it,
most LEDs are rated for 20 milliamperes of current.

In any case, you got the calculation wrong because what you want to
do is drop the _difference_ between the mains voltage and the LED
voltage across the resistor, as I've shown below.
---

Now wait a minute, since you have picked out a mistake i made i will also
pick out one that i think your making.!. first of all you did say 170 V
which is peek, now your using 120 volts here and the above math looks
like it assumes DC voltage ?

---
A 120VRMS sine wave will have voltage peaks at 170V and -170V and is
what you have to consider when you're dealing with the reverse
voltage across the diode, which is given in volts, DC. The RMS
voltage of the sine wave will cause the equivalent heating that
120VDC will, so since the resistor (and the LED) will be dissipating
power and getting hot, we use the RMS value of the sine wave to
figure out what value the reistor needs to be/
---

you are using 20 ma Leds for your case.
last time i remember P = I*V*cos(angle). in cases where we are dealing
with Sine wave energy.

---
Then why didn't you take that into consideration in your
calculation?
---

gee. thanks, coming from you i take that as a complement! :=)

---
Compliment, not complement.
---

btw, i have seen many simple LED light systems from the
120 service using nothing more than a single resistor which
worked fine..

---
Nope. you need at least an antiparallel diode across the LED or two
LEDs in antiparallel in order to keep the reverse voltage from
killing the single LED.
---

we have some LED upgrades for incandescent lamp holders for
control panels which you can clearly see that they are a
LED and Resistor combo for a 120 vac fixture.(no back diodes).

---
Nope. See above.
---

oh well, happier days are coming.

Maybe.

BTW, I saved the best part for last...

The reason you don't need to use P = EI cos(phi) in this case is
because there are essentially no reactive elements in the circuit.
However, since voltage and current are in phase, if you want to, you
can write:

P = EI cos(phi)

since with the phase angle (phi) being equal to zero, what will its
cosine be? (Please leave your answer on the desk on your way out.)

 

[John Fields]

i don't know what to tell you, we must be using magic LED's
then.

---
Nope, there are no magic LEDs. Might seem like magic to you,
though, if you don't know how they work.
---

last night i set up a test on the bench using a zip cord
and a network of resistors operating a Jumbo, a small RED led
and a green one that i had in my junk box. , no back diodes
on any of them. as i sit and type this i can look over to the bench
and they are operating just fine.

---
If that's the case, AND there's no diode external to the LED, AND
you've got LEDs designed to work with an AC supply, then what you
have is probably two anti-parallel LEDs on the same lead frame.
---

we have been slowly updating our panel lights that are in a range
of 6,12, 24 and 120 volt ac types with LED's with nothing but a
Resistor in series. as far as i know they are all still working accept
for maybe one of them that needed replacing due to a electrician putting
a 6 volt set up in a 24 volt fixture.
i'll leave it as this point, maybe we are using special LED's that i
am not aware of. who knows. all i can say is they work..

---
If you don't know how they work, and you don't know the difference
between RMS and peak, and you don't know the difference between
Volt-Amperes and watts, and you don't know how to figure out the
value of the current limiting resistor, then you shouldn't be giving
advice, you should be _seeking_ it.

Also, it would be nice if you'd be considerate enough to at least
punctuate properly.

 

[Jamie]

--- {lots of crap removed}
Nope. you need at least an antiparallel diode across the LED or two
LEDs in antiparallel in order to keep the reverse voltage from
killing the single LED.

i don't know what to tell you, we must be using magic LED's
then. last night i set up a test on the bench using a zip cord
and a network of resistors operating a Jumbo, a small RED led
and a green one that i had in my junk box. , no back diodes
on any of them. as i sit and type this i can look over to the bench
and they are operating just fine.
we have been slowly updating our panel lights that are in a range
of 6,12, 24 and 120 volt ac types with LED's with nothing but a
Resistor in series. as far as i know they are all still working accept
for maybe one of them that needed replacing due to a electrician putting
a 6 volt set up in a 24 volt fixture.
i'll leave it as this point, maybe we are using special LED's that i
am not aware of. who knows. all i can say is they work..

 

[Pooh Bear]

Also, it would be nice if you'd be considerate enough to at least
punctuate properly.

You just can't get the staff these days, John !

Graham

 

[Jasen Betts]

i don't know what to tell you, we must be using magic LED's
then. last night i set up a test on the bench using a zip cord
and a network of resistors operating a Jumbo, a small RED led
and a green one that i had in my junk box. , no back diodes
on any of them. as i sit and type this i can look over to the bench
and they are operating just fine.
we have been slowly updating our panel lights that are in a range
of 6,12, 24 and 120 volt ac types with LED's with nothing but a
Resistor in series. as far as i know they are all still working accept
for maybe one of them that needed replacing due to a electrician putting
a 6 volt set up in a 24 volt fixture.
i'll leave it as this point, maybe we are using special LED's that i
am not aware of. who knows. all i can say is they work..

I saw it analysed once, the led conducts some current in the reverse
direction (but always less than in the forwards direction) and as long as
the current isn't enough to overheat the LED little damage is
done. given that the reverse breakdown voltage may be much higher than the
forwards voltage there can be much more heating during reverse conduction.

In a thread in another newsgroup I ran a LED directly off a 5V supply for
an hour (but it didn't last much longer than that). Going outside
manufacturers specifications can give unpredictable results.

I'd still reccomend putting a diode in there to stop the led conducting
backwards. If nothing else it'll make the led last longer or allow you
to run it brighter.

Bye.
Jasen