Series Resister or not

G

GrimReaper

Guest
Would I be right in assuming that if the supply voltage was the same as the
forward voltage of the LED then there is no need to fit a series resister or
do you have to have a higher supply voltage and always fit a series resister
to drop the supply voltage back to the forward voltage of the LED to limit
the current?

GrimReaper


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"Baphomet" <fandaDEATH2SPAMMERS@catskill.net> wrote in message
news:vpau491fi1ee9d@corp.supernews.com...
"GrimReaper" <266@082.com> wrote in message
news:bn3sfr$h5$1@newsg3.svr.pol.co.uk...
Would I be right in assuming that if the supply voltage was the same as
the
forward voltage of the LED then there is no need to fit a series
resister
or
do you have to have a higher supply voltage and always fit a series
resister
to drop the supply voltage back to the forward voltage of the LED to
limit
the current?

GrimReaper

You should always use a higher voltage supply and limit the current to the
led with a series resistor, unless you're using a current limited power
supply.
Click to expand...
Thank you my friend, I now know what to do

Regards
GrimReaper


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"GrimReaper" <266@082.com> wrote in message
news:bn3sfr$h5$1@newsg3.svr.pol.co.uk...
Would I be right in assuming that if the supply voltage was the same as
the
forward voltage of the LED then there is no need to fit a series resister
or
do you have to have a higher supply voltage and always fit a series
resister
to drop the supply voltage back to the forward voltage of the LED to limit
the current?

GrimReaper
Click to expand...
You should always use a higher voltage supply and limit the current to the
led with a series resistor, unless you're using a current limited power
supply.
 
Very simplicticly.
You could imagine a LED as a constant voltage device and given that, the
current will rise to infinity or until the source limits it or the diode
turns to smoking. The resistor on the other hand, drops more volts as the
current rises and soon there is equilibrium.
--
John G

Wot's Your Real Problem?

"GrimReaper" <266@082.com> wrote in message
news:bn3sfr$h5$1@newsg3.svr.pol.co.uk...
Would I be right in assuming that if the supply voltage was the same as
the
forward voltage of the LED then there is no need to fit a series resister
or
do you have to have a higher supply voltage and always fit a series
resister
to drop the supply voltage back to the forward voltage of the LED to limit
the current?

GrimReaper


---
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Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.529 / Virus Database: 324 - Release Date: 16/10/03
Click to expand...
 
"GrimReaper" wrote
Would I be right in assuming that if the supply voltage was the same as the
forward voltage of the LED then there is no need to fit a series resister or
do you have to have a higher supply voltage and always fit a series resister
to drop the supply voltage back to the forward voltage of the LED to limit
the current?
Click to expand...

Yes, use a higher voltage and a series resistor.

This link may help

http://pub40.ezboard.com/fbasicelectronicsfrm5.showMessage?topicID=16.topic
 
"John G" <greentestatoptusnet.com.au> wrote in message
news:3f95c311$0$1089$afc38c87@news.optusnet.com.au...
Very simplicticly.
You could imagine a LED as a constant voltage device and given that, the
current will rise to infinity or until the source limits it or the diode
turns to smoking. The resistor on the other hand, drops more volts as the
current rises and soon there is equilibrium.
--
John G
Click to expand...
While I agree with the sentiment, diodes do actually have a 'resistance' of
sorts; its simply nonlinear, ie, it doesn't obey ohms 'law', so its a bit
more difficult to figure out. The forward voltage is related to the natural
log of the current through it, so the more current, the more voltage across
it, just like a resistor. IIRC, one rule of thumb is that if you increase
the current by 10 times (a decade), the forward voltage will go up something
like 60mV, at room temperature (at least this works for bipolar transistor
diode junctions).

I think this means that the OP can omit the resistor without explosions...

Regards
Bob Monsen
 
"GrimReaper" <266@082.com> wrote in message
news:bn3sfr$h5$1@newsg3.svr.pol.co.uk...
Would I be right in assuming that if the supply voltage was the same as
the
forward voltage of the LED then there is no need to fit a series resister
or
do you have to have a higher supply voltage and always fit a series
resister
to drop the supply voltage back to the forward voltage of the LED to limit
the current?

GrimReaper
Click to expand...
As a general rule, you should use a series resistor and a slightly higher
(+0.5-1V) voltage. However a LED is not a constant voltage device and most
LEDs have noticeable resistances that allows them to operate quite well
ballanced at a constant voltage. Note nevertheless that the difference in
the forward voltage rises very slowly on substantial current increase due to
a LED's nonlinearity. This means that your power supply voltage should be
_really_ constant, small changes may become lethal for the LED. Never rely
on a 'trafo, diode and cap' style of power supply for a voltage stability.
Note also that an LED is a temperature-sensitive device. When operating at a
constant voltage it should be kept away from high temperatures and mounted
in a case that can provide slight heatsink capabilities. If you plan to
operate the LED with batteries, their internal resistance can help making a
series resistor unnecessary, but batteries change condition easier than one
can sometimes think, so I would classify a device of this sort as 'not
reliable'.
 
Thanks for all your info. I want to connect 21 Blue LED's as part of a Coral
Tank lighting system. The LED's have a forward voltage of 3.4v. A current of
20ma. Based on your comments I was going to buy a 12 volt stabilised power
supply and put 3 LED's in series with a 91 ohm resistor and repeat this
series configuration 7 times. Does that make sense?
No good at drawing that here. I am a marine biologist and what I know about
electronics I have gained from here and Google searches - that's it.


Regards
GrimReaper


---
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Checked by AVG anti-virus system (http://www.grisoft.com).
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"GrimReaper" <266@082.com> wrote in message
news:bn9b10$dcb$1@news6.svr.pol.co.uk...
Thanks for all your info. I want to connect 21 Blue LED's as part of a
Coral
Tank lighting system. The LED's have a forward voltage of 3.4v. A current
of
20ma. Based on your comments I was going to buy a 12 volt stabilised power
supply and put 3 LED's in series with a 91 ohm resistor and repeat this
series configuration 7 times. Does that make sense?
No good at drawing that here. I am a marine biologist and what I know
about
electronics I have gained from here and Google searches - that's it.


Regards
GrimReaper
Click to expand...
That'll do nicely GR... now using P=I˛R, calulate the power rating of the
resistor you
need....Oh, use the nearest standard value resistor, it isn't at all
critical. Don't forget
to add a little extra power rating for slop margin.

You have 11.2 volts across the LEDs, leaving .8 volts across the resistor,
when supplied
with 12 volts.
 
On Thu, 23 Oct 2003 20:43:31 +0100, "GrimReaper" <266@082.com> wrote:

Thanks for all your info. I want to connect 21 Blue LED's as part of a Coral
Tank lighting system. The LED's have a forward voltage of 3.4v. A current of
20ma. Based on your comments I was going to buy a 12 volt stabilised power
supply and put 3 LED's in series with a 91 ohm resistor and repeat this
series configuration 7 times. Does that make sense?
Click to expand...
---
Yes.

The difference between the supply voltage and the sum of the LED forward
voltages divided by the LED current is 90 ohms and you have chosen the
nearest standard 5% value, 91 ohms. Well done.

All that's left to do is determine the wattage required for the resistor
to dissipate, and that'll be the difference between the supply voltage
and the sum of the LED forward voltages _multiplied_ by the LED
current,

(12V-10.2V)*0.02A = 0.036 watts

so a 1/4 watt resistor will be fine.
---

No good at drawing that here. I am a marine biologist and what I know about
electronics I have gained from here and Google searches - that's it.
Click to expand...
---
So far so good!

--
John Fields
 
"John Fields" <jfields@austininstruments.com> wrote in message
news:29dgpvka10cf2sk8oke8ktcu6ljfh2a74o@4ax.com...
On Thu, 23 Oct 2003 20:43:31 +0100, "GrimReaper" <266@082.com> wrote:

Thanks for all your info. I want to connect 21 Blue LED's as part of a
Coral
Tank lighting system. The LED's have a forward voltage of 3.4v. A current
of
20ma. Based on your comments I was going to buy a 12 volt stabilised
power
supply and put 3 LED's in series with a 91 ohm resistor and repeat this
series configuration 7 times. Does that make sense?

---
Yes.

The difference between the supply voltage and the sum of the LED forward
voltages divided by the LED current is 90 ohms and you have chosen the
nearest standard 5% value, 91 ohms. Well done.

All that's left to do is determine the wattage required for the resistor
to dissipate, and that'll be the difference between the supply voltage
and the sum of the LED forward voltages _multiplied_ by the LED
current,

(12V-10.2V)*0.02A = 0.036 watts

so a 1/4 watt resistor will be fine.
---

No good at drawing that here. I am a marine biologist and what I know
about
electronics I have gained from here and Google searches - that's it.

---
So far so good!

--
John Fields
Click to expand...
Oops John caught me !

Seems I flubbed that 3.4 times three ....back to second grade I go!!!
 
"Lord Garth" <LGarth@Tantalus.net> wrote in message
news:XUWlb.1010$sc3.860@newssvr23.news.prodigy.com...
"John Fields" <jfields@austininstruments.com> wrote in message
news:29dgpvka10cf2sk8oke8ktcu6ljfh2a74o@4ax.com...
On Thu, 23 Oct 2003 20:43:31 +0100, "GrimReaper" <266@082.com> wrote:

Thanks for all your info. I want to connect 21 Blue LED's as part of a
Coral
Tank lighting system. The LED's have a forward voltage of 3.4v. A
current
of
20ma. Based on your comments I was going to buy a 12 volt stabilised
power
supply and put 3 LED's in series with a 91 ohm resistor and repeat this
series configuration 7 times. Does that make sense?

---
Yes.

The difference between the supply voltage and the sum of the LED forward
voltages divided by the LED current is 90 ohms and you have chosen the
nearest standard 5% value, 91 ohms. Well done.

All that's left to do is determine the wattage required for the resistor
to dissipate, and that'll be the difference between the supply voltage
and the sum of the LED forward voltages _multiplied_ by the LED
current,

(12V-10.2V)*0.02A = 0.036 watts

so a 1/4 watt resistor will be fine.
---

No good at drawing that here. I am a marine biologist and what I know
about
electronics I have gained from here and Google searches - that's it.

---
So far so good!

--
John Fields

Oops John caught me !

Seems I flubbed that 3.4 times three ....back to second grade I go!!!



John / Lord Garth
Click to expand...
Thank you for that. I had not thought about the power dissipated in the
resistor


Kind Regards
GrimReaper


---
All Outgoing mail is checked by AVG Anti Virus
Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.530 / Virus Database: 325 - Release Date: 22/10/03
 
"GrimReaper" <266@082.com> wrote in message
news:bn9h3r$cg8$1@news8.svr.pol.co.uk...
"Lord Garth" <LGarth@Tantalus.net> wrote in message
news:XUWlb.1010$sc3.860@newssvr23.news.prodigy.com...

"John Fields" <jfields@austininstruments.com> wrote in message
news:29dgpvka10cf2sk8oke8ktcu6ljfh2a74o@4ax.com...
On Thu, 23 Oct 2003 20:43:31 +0100, "GrimReaper" <266@082.com> wrote:

Thanks for all your info. I want to connect 21 Blue LED's as part of
a
Coral
Tank lighting system. The LED's have a forward voltage of 3.4v. A
current
of
20ma. Based on your comments I was going to buy a 12 volt stabilised
power
supply and put 3 LED's in series with a 91 ohm resistor and repeat
this
series configuration 7 times. Does that make sense?

---
Yes.

The difference between the supply voltage and the sum of the LED
forward
voltages divided by the LED current is 90 ohms and you have chosen the
nearest standard 5% value, 91 ohms. Well done.

All that's left to do is determine the wattage required for the
resistor
to dissipate, and that'll be the difference between the supply
voltage
and the sum of the LED forward voltages _multiplied_ by the LED
current,

(12V-10.2V)*0.02A = 0.036 watts

so a 1/4 watt resistor will be fine.
---

No good at drawing that here. I am a marine biologist and what I know
about
electronics I have gained from here and Google searches - that's it.

---
So far so good!

--
John Fields

Oops John caught me !

Seems I flubbed that 3.4 times three ....back to second grade I go!!!



John / Lord Garth
Thank you for that. I had not thought about the power dissipated in the
resistor


Kind Regards
GrimReaper
Click to expand...
That's certainly one of the things that makes this such a powerful medium. A
final solution becomes a collaborative effort, each one seeing a possible
solution from a slightly different perspective.
 
In article <woqlb.604238$Oz4.625427@rwcrnsc54>,
postmaster@BulkingPro.com mentioned...
"John G" <greentestatoptusnet.com.au> wrote in message
news:3f95c311$0$1089$afc38c87@news.optusnet.com.au...
Very simplicticly.
You could imagine a LED as a constant voltage device and given that, the
current will rise to infinity or until the source limits it or the diode
turns to smoking. The resistor on the other hand, drops more volts as the
current rises and soon there is equilibrium.
--
John G


While I agree with the sentiment, diodes do actually have a 'resistance' of
sorts; its simply nonlinear, ie, it doesn't obey ohms 'law', so its a bit
more difficult to figure out. The forward voltage is related to the natural
log of the current through it, so the more current, the more voltage across
it, just like a resistor. IIRC, one rule of thumb is that if you increase
the current by 10 times (a decade), the forward voltage will go up something
like 60mV, at room temperature (at least this works for bipolar transistor
diode junctions).

I think this means that the OP can omit the resistor without explosions...
Click to expand...
"Just like a resistor"? You're saying that the resistor, which obeys
ohm's law, obeys the natural log relatonship? I don't think so.

And there is a resistive component to the current/voltage
relationship, just that it's not a predominant component. There is
the bond wire resistance, the internal resistance of the chip itself,
etc. But as the LED gets hotter, the V drop gets lower, so putting
the LED on a constant voltage supply means that it will draw excessive
current as it warms up. This could let the smoke out.

Regards
Bob Monsen
Click to expand...

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
 
"Watson A.Name - Watt Sun" <alondra101@hotmail.com> wrote in message
news:MPG.1a04bea074a3ad84989828@news.dslextreme.com...
In article <woqlb.604238$Oz4.625427@rwcrnsc54>,
postmaster@BulkingPro.com mentioned...
"John G" <greentestatoptusnet.com.au> wrote in message
news:3f95c311$0$1089$afc38c87@news.optusnet.com.au...
Very simplicticly.
You could imagine a LED as a constant voltage device and given that,
the
current will rise to infinity or until the source limits it or the
diode
turns to smoking. The resistor on the other hand, drops more volts as
the
current rises and soon there is equilibrium.
--
John G


While I agree with the sentiment, diodes do actually have a 'resistance'
of
sorts; its simply nonlinear, ie, it doesn't obey ohms 'law', so its a
bit
more difficult to figure out. The forward voltage is related to the
natural
log of the current through it, so the more current, the more voltage
across
it, just like a resistor. IIRC, one rule of thumb is that if you
increase
the current by 10 times (a decade), the forward voltage will go up
something
like 60mV, at room temperature (at least this works for bipolar
transistor
diode junctions).

I think this means that the OP can omit the resistor without
explosions...

"Just like a resistor"? You're saying that the resistor, which obeys
ohm's law, obeys the natural log relatonship? I don't think so.
Click to expand...
No, I'm not saying that. You apparently misread my post...

Regards,
Bob Monsen
 
In article <3f980a8f_1@news.arcor-ip.de>, dklingb@cs.tu-berlin.de
mentioned...
"GrimReaper" <266@082.com> wrote in message
news:bn3sfr$h5$1@newsg3.svr.pol.co.uk...
Would I be right in assuming that if the supply voltage was the same as
the
forward voltage of the LED then there is no need to fit a series resister
or
do you have to have a higher supply voltage and always fit a series
resister
to drop the supply voltage back to the forward voltage of the LED to limit
the current?

GrimReaper

As a general rule, you should use a series resistor and a slightly higher
(+0.5-1V) voltage. However a LED is not a constant voltage device and most
LEDs have noticeable resistances that allows them to operate quite well
ballanced at a constant voltage. Note nevertheless that the difference in
the forward voltage rises very slowly on substantial current increase due to
a LED's nonlinearity. This means that your power supply voltage should be
_really_ constant, small changes may become lethal for the LED.
Click to expand...
It looks like you have never measured the voltage across a LED while
it's operating. If you had, you would know that it has a negative
tempco. As the LED gets warmer, it drops less voltage. This is
really pronounced on big LEDs such as the Luxeon Star. If you connect
a LED to a constant voltage, it draws more current as it gets hotter,
which causes it to draw more current and get even hotter, in a vicious
circle which will destroy the LED if it doesn't shorten its life
considerably. So I'd say your advice is bogus.

[snip]

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
 
In article <2GWlb.1007$B%2.35@newssvr23.news.prodigy.com>,
LGarth@Tantalus.net mentioned...
"GrimReaper" <266@082.com> wrote in message
news:bn9b10$dcb$1@news6.svr.pol.co.uk...
Thanks for all your info. I want to connect 21 Blue LED's as part of a
Coral
Tank lighting system. The LED's have a forward voltage of 3.4v. A current
of
20ma. Based on your comments I was going to buy a 12 volt stabilised power
supply and put 3 LED's in series with a 91 ohm resistor and repeat this
series configuration 7 times. Does that make sense?
No good at drawing that here. I am a marine biologist and what I know
about
electronics I have gained from here and Google searches - that's it.


Regards
GrimReaper

That'll do nicely GR... now using P=I˛R, calulate the power rating of the
resistor you
need....Oh, use the nearest standard value resistor, it isn't at all
critical. Don't forget
to add a little extra power rating for slop margin.

You have 11.2 volts across the LEDs, leaving .8 volts across the resistor,
when supplied
with 12 volts.
Click to expand...
Eh? 3.4V times 3 is 10.2V. That leaves 1.8V across the res. There
will be some variation in the voltage drop, but it shouldn't be too
much. That can be measured with a DMM. If it's too much, change the
LED.

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
 
In article <29dgpvka10cf2sk8oke8ktcu6ljfh2a74o@4ax.com>,
jfields@austininstruments.com mentioned...
On Thu, 23 Oct 2003 20:43:31 +0100, "GrimReaper" <266@082.com> wrote:

Thanks for all your info. I want to connect 21 Blue LED's as part of a Coral
Tank lighting system. The LED's have a forward voltage of 3.4v. A current of
20ma. Based on your comments I was going to buy a 12 volt stabilised power
supply and put 3 LED's in series with a 91 ohm resistor and repeat this
series configuration 7 times. Does that make sense?

---
Yes.

The difference between the supply voltage and the sum of the LED forward
voltages divided by the LED current is 90 ohms and you have chosen the
nearest standard 5% value, 91 ohms. Well done.
Click to expand...
If you can't get that relatively hard-to-find value, use two 47 ohms
(much more common) in series, or 100 ohms will work okay with no
perceptible difference in brightness.

All that's left to do is determine the wattage required for the resistor
to dissipate, and that'll be the difference between the supply voltage
and the sum of the LED forward voltages _multiplied_ by the LED
current,

(12V-10.2V)*0.02A = 0.036 watts

so a 1/4 watt resistor will be fine.
Click to expand...
I bought a couple dozen blue LEDs and I've found that every one of
them that I've taken out of the package is intermittent. And it's
really strange, too. I crank the PS up to 1.5V and the LED starts to
draw current, maybe a few mA, but the current is really erratic, I
can't get a stable reading. If I go up to 3.5V at 25 mA, sometimes
the LED will light steady, sometimes it will blink, maybe a few times
a minute, maye less, once every few minutes I've checked at least 6,
probably more, and they're all like that. Bad news. Sucks.


---

No good at drawing that here. I am a marine biologist and what I know about
electronics I have gained from here and Google searches - that's it.

---
So far so good!
Click to expand...

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
 
In article <MPG.1a0561a02a6c440c989831@news.dslextreme.com>, Watson
A.Name - "Watt Sun" wrote:
I bought a couple dozen blue LEDs and I've found that every one of
them that I've taken out of the package is intermittent. And it's
really strange, too. I crank the PS up to 1.5V and the LED starts to
draw current, maybe a few mA, but the current is really erratic, I
can't get a stable reading. If I go up to 3.5V at 25 mA, sometimes
the LED will light steady, sometimes it will blink, maybe a few times
a minute, maye less, once every few minutes I've checked at least 6,
probably more, and they're all like that. Bad news. Sucks.
Click to expand...
Maybe it's static damage. Most blue LEDs are static-sensitive.

- Don Klipstein (don@misty.com)
 
In an LED VI Curve, the independant variable is current, and the
dependant variable is voltage.
Use the resistor. You can series a number of leds with a higher
voltage supply and one resistor, or use a current scource, if you want
to get sophisticated.
Dan

On Tue, 21 Oct 2003 19:04:46 +0100, "GrimReaper" <266@082.com> wrote:

Would I be right in assuming that if the supply voltage was the same as the
forward voltage of the LED then there is no need to fit a series resister or
do you have to have a higher supply voltage and always fit a series resister
to drop the supply voltage back to the forward voltage of the LED to limit
the current?

GrimReaper


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Colorado Springs, CO
My advice may be worth what you paid for it.
 
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