## Re: Newbie LED question

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• Hi Pedro, If you want to run the 16 LED s at 30 mA each, your 12V power supply must have at least 500 mA (16 X 30 mA=480 mA) or better capability. A power
Message 1 of 14 , Jul 31, 2001
Hi Pedro,
If you want to run the 16 LED's at 30 mA each, your 12V power supply
must have at least 500 mA (16 X 30 mA=480 mA) or better capability.
A power supply with 1A capability would run much cooler. Each of the
LED's, if they don't already have a built-in resistor, will need a
series limiting resistor to insure your 30 mA of current is not
exceeded. The forward voltage (VF)drop is not mentioned in your spec,
but the LED's are probably dropping about 2 V, that would mean that
you have about 10 V dropped across your
limiting resistors. Using Ohms Law R = E/I you can calculate the
resistor value to be R = 10V/.03A = 333 ohms. Resistors come in
standard values, the closest value for the above application is a 330
ohm. The power rating of the resistor can be found by using the
formula P = E X I, in this example P = 10V X .03A = 300 milliwatts.
Therefore the 330 ohm resistors should have a power rating of at
least .5 watts (500 milliwatts). If you are desiring to have all 16
LEDs on at the same time, connect each of the LEDs with their
respective resistors in series and then all of them across the 12 V
source. Insure that the polarity is such, that the LED's are forward
biased, negative through the series resistors to the cathodes, and
positive voltage to the anodes. Hope this helps.
John
--- In Electronics_101@y..., "Pedro de-Oliveira" <olive_@h...> wrote:
> Hi all
>
> I am a newbie to the art of electronics so please be gentle with
me :-)
>
> I am trying to run 16 LED's from a single 12v power source. The
LED specs
> are as follows:
>
> Parameter Value Units
> DC Forward Current[1] 30 mA
> Peak Forward Current 100 mA
> Average Forward Current 30 mA
> Power Dissipation 120 mW
> Reverse Voltage (IR = 100 mA) 5 V
> LED Junction Temperature 100 °C
> Operating Temperature Range 40 to +80 °C
> Storage Temperature Range 40 to +100 °C
>
> What I would like to know is what would be the best way to hook
these things
> up and what value resistors I would need to do it.
>
> Cheers
> Pedro
>
>
>
> _________________________________________________________________
http://explorer.msn.com/intl.asp
• It should also be noted that it is preferable to use one (330 ohm) resistor per LED rather than one (e.g. 20 ohm) resistor for the whole lot, because of
Message 2 of 14 , Aug 1, 2001
It should also be noted that it is preferable to use one (330 ohm) resistor
per LED rather than one (e.g. 20 ohm) resistor for the whole lot, because of
internal resistance deviations in each LED, as one device may end up drawing
more current than the rest and cause some damage.

Just my 2 cents ...

Jon

----- Original Message -----
From: <jperrynew@...>
To: <Electronics_101@yahoogroups.com>
Sent: Wednesday, August 01, 2001 4:57 PM
Subject: [Electronics_101] Re: Newbie LED question

Hi Pedro,
If you want to run the 16 LED's at 30 mA each, your 12V power supply
must have at least 500 mA (16 X 30 mA=480 mA) or better capability.
A power supply with 1A capability would run much cooler. Each of the
LED's, if they don't already have a built-in resistor, will need a
series limiting resistor to insure your 30 mA of current is not
exceeded. The forward voltage (VF)drop is not mentioned in your spec,
but the LED's are probably dropping about 2 V, that would mean that
you have about 10 V dropped across your
limiting resistors. Using Ohms Law R = E/I you can calculate the
resistor value to be R = 10V/.03A = 333 ohms. Resistors come in
standard values, the closest value for the above application is a 330
ohm. The power rating of the resistor can be found by using the
formula P = E X I, in this example P = 10V X .03A = 300 milliwatts.
Therefore the 330 ohm resistors should have a power rating of at
least .5 watts (500 milliwatts). If you are desiring to have all 16
LEDs on at the same time, connect each of the LEDs with their
respective resistors in series and then all of them across the 12 V
source. Insure that the polarity is such, that the LED's are forward
biased, negative through the series resistors to the cathodes, and
positive voltage to the anodes. Hope this helps.
John
--- In Electronics_101@y..., "Pedro de-Oliveira" <olive_@h...> wrote:
> Hi all
>
> I am a newbie to the art of electronics so please be gentle with
me :-)
>
> I am trying to run 16 LED's from a single 12v power source. The
LED specs
> are as follows:
>
> Parameter Value Units
> DC Forward Current[1] 30 mA
> Peak Forward Current 100 mA
> Average Forward Current 30 mA
> Power Dissipation 120 mW
> Reverse Voltage (IR = 100 mA) 5 V
> LED Junction Temperature 100 °C
> Operating Temperature Range -40 to +80 °C
> Storage Temperature Range -40 to +100 °C
>
> What I would like to know is what would be the best way to hook
these things
> up and what value resistors I would need to do it.
>
> Cheers
> Pedro
>
>
>
> _________________________________________________________________
http://explorer.msn.com/intl.asp

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• Hi and thanks for your help I understood all that your message said and will now get 12x330 Ohm resistors to wire before each of the LED s. There is just a bit
Message 3 of 14 , Aug 1, 2001
Hi and thanks for your help

I understood all that your message said and will now get 12x330 Ohm
resistors to wire before each of the LED's. There is just a bit I
don't understand in your message and this is cut and pasted below:

<SNIP>

>Insure that the polarity is such, that the LED's are forward
> biased, negative through the series resistors to the cathodes, and
> positive voltage to the anodes. Hope this helps.

</SNIP>

What does this mean, do they all have to lie the same way or +ve to
+ve

Thanks again

Pedro
• Hi Pedro, If your LED doesn t produce light, simply reverse the leads and it should be in the forward current direction. Each LED has an anode and cathode end
Message 4 of 14 , Aug 1, 2001
Hi Pedro,
If your LED doesn't produce light, simply reverse the leads and it
should be in the forward current direction. Each LED has an anode and
cathode end and is polarity sensitive.
John

--- In Electronics_101@y..., "Pedro de Oliveira" <olive_@h...> wrote:
> Hi and thanks for your help
>
> I understood all that your message said and will now get 12x330 Ohm
> resistors to wire before each of the LED's. There is just a bit I
> don't understand in your message and this is cut and pasted below:
>
> <SNIP>
>
> >Insure that the polarity is such, that the LED's are forward
> > biased, negative through the series resistors to the cathodes, and
> > positive voltage to the anodes. Hope this helps.
>
> </SNIP>
>
> What does this mean, do they all have to lie the same way or +ve to
> +ve
>
>
> Thanks again
>
> Pedro
• 16 x 30mA = 480mA. Therefore, the 12V source must be able to put out half an amp for the LEDs. half an amp! That s as much as a television draws! Wow...
Message 5 of 14 , Aug 1, 2001
16 x 30mA = 480mA.   Therefore, the 12V source must be able to put out half an amp for the LEDs.

half an amp!  That's as much as a television draws!  Wow...

• I have been experimenting using leds for lighting system on my electric scooter and have built a combined back/brake light with 3 rows of 5 red leds in series
Message 6 of 14 , Aug 1, 2001
I have been experimenting using leds for lighting system on my electric scooter and have built a combined back/brake light with 3 rows of 5 red leds in series operating on 12 volts SLA. The brake light is connected parallel 2 rows of 5 leds in series and the back light is1 row of 5 leds in series. Current draw is approx. 90ma for the brake light and 45ma for the back light (total current draw=135ma). No current limiting resistors were  used and the leds are operating at maximum light intensity. I have used the light for more than 6 months and did not encounter any problem, the circuit diagram is posted below should you be interested.

Bob Wong - Singapore
----- Original Message -----
Sent: Thursday, August 02, 2001 4:08 AM
Subject: Re: [Electronics_101] Newbie LED question

16 x 30mA = 480mA.   Therefore, the 12V source must be able to put out half an amp for the LEDs.

half an amp!  That's as much as a television draws!  Wow...

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• Not Neccesarily .... a Television draws 0.5A @ 110/240V (depending on where you are in the world) so your television is sucking 55/120W (P=VA; 110 x 0.5 = 55)
Message 7 of 14 , Aug 1, 2001
Not Neccesarily .... a Television draws 0.5A @ 110/240V (depending on where you are in the world) so your television is sucking 55/120W (P=VA; 110 x 0.5 = 55)  whereas the LED's running @ 12V are drawing a little under 6W.

16 x 30mA = 480mA.   Therefore, the 12V source must be able to put out half an amp for the LEDs.

half an amp!  That's as much as a television draws!  Wow...
• Robert, Depending on the type of LED you are using, let s assume that Vf is 2V - so when you put 5 of them in series, you have an actual voltage drop of 10V,
Message 8 of 14 , Aug 1, 2001
Robert,

Depending on the type of LED you are using, let's assume that Vf is 2V - so when you put 5 of them in series, you have an actual voltage drop of 10V, leaving you 2V in excess ... so in actual fact you are over-driving your LED's by 0.4V (say 20% per each) for short periods of time - no big deal, it will reduce the life of your LED's somewhat, but probably not enough to notice. I will agree that it is a slightly more efficient usage of current, in that you are not dissapating 16 seperate lots of heat - certainly a potential issue when you are running on limited supply source.

If you are using it on a scooter another thing to point out would be that if you are running it on the main drive batteries, when in use the voltage at the terminals will actually be a lot less that 12V, if you are running it on a seperate battery, as the battery discharges you will be running closer to the 10V mark too, and so getting closer to the LED's capabilities.

l8r

Jon
----- Original Message -----
Sent: Thursday, August 02, 2001 8:17 AM
Subject: Re: [Electronics_101] Newbie LED question

I have been experimenting using leds for lighting system on my electric scooter and have built a combined back/brake light with 3 rows of 5 red leds in series operating on 12 volts SLA. The brake light is connected parallel 2 rows of 5 leds in series and the back light is1 row of 5 leds in series. Current draw is approx. 90ma for the brake light and 45ma for the back light (total current draw=135ma). No current limiting resistors were  used and the leds are operating at maximum light intensity. I have used the light for more than 6 months and did not encounter any problem, the circuit diagram is posted below should you be interested.

Bob Wong - Singapore
----- Original Message -----
Sent: Thursday, August 02, 2001 4:08 AM
Subject: Re: [Electronics_101] Newbie LED question

16 x 30mA = 480mA.   Therefore, the 12V source must be able to put out half an amp for the LEDs.

half an amp!  That's as much as a television draws!  Wow...

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• Jon..... Honestly, I don t even know what type of LED I have and bought these because of their ultra-bright characteristics from a surplus electronics
Message 9 of 14 , Aug 2, 2001
Jon.....

Honestly, I don't even know what type of LED I have and bought these because of their ultra-bright characteristics from a surplus electronics component shop at US\$ 1.20 for 30 pieces.  Overdriving the  LED is done on purpose to give the high visibility and intensity that I am looking for.  How long these LED will last is another story. I would expect the LED on the brake light will last longer as these are only switch on intermitently during braking while the back light is switch on all the time. As the system is of an experimental project, I might have to make changes to ensure more reliability over time.

My scooter is running with 2x12volt/12ah SLA battery connected in series for 24volt operation. Perhaps, an additional 12 volt voltage regulator circuit stepping down from the 24 voltage system would help to stabilise voltage.

Bob Wong - Singapore
----- Original Message -----
Sent: Thursday, August 02, 2001 8:56 AM
Subject: Re: [Electronics_101] Newbie LED question

Robert,

Depending on the type of LED you are using, let's assume that Vf is 2V - so when you put 5 of them in series, you have an actual voltage drop of 10V, leaving you 2V in excess ... so in actual fact you are over-driving your LED's by 0.4V (say 20% per each) for short periods of time - no big deal, it will reduce the life of your LED's somewhat, but probably not enough to notice. I will agree that it is a slightly more efficient usage of current, in that you are not dissapating 16 seperate lots of heat - certainly a potential issue when you are running on limited supply source.

If you are using it on a scooter another thing to point out would be that if you are running it on the main drive batteries, when in use the voltage at the terminals will actually be a lot less that 12V, if you are running it on a seperate battery, as the battery discharges you will be running closer to the 10V mark too, and so getting closer to the LED's capabilities.

l8r

Jon
----- Original Message -----
Sent: Thursday, August 02, 2001 8:17 AM
Subject: Re: [Electronics_101] Newbie LED question

I have been experimenting using leds for lighting system on my electric scooter and have built a combined back/brake light with 3 rows of 5 red leds in series operating on 12 volts SLA. The brake light is connected parallel 2 rows of 5 leds in series and the back light is1 row of 5 leds in series. Current draw is approx. 90ma for the brake light and 45ma for the back light (total current draw=135ma). No current limiting resistors were  used and the leds are operating at maximum light intensity. I have used the light for more than 6 months and did not encounter any problem, the circuit diagram is posted below should you be interested.

Bob Wong - Singapore
----- Original Message -----
Sent: Thursday, August 02, 2001 4:08 AM
Subject: Re: [Electronics_101] Newbie LED question

16 x 30mA = 480mA.   Therefore, the 12V source must be able to put out half an amp for the LEDs.

half an amp!  That's as much as a television draws!  Wow...

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• Thanks to you all for your excellent suggestions. The reverse voltage on these LED s is 5V. I take this to mean that they want 5V supply voltage to them. I
Message 10 of 14 , Aug 2, 2001
Thanks to you all for your excellent suggestions.

The reverse voltage on these LED's is 5V. I take this to mean that
they "want" 5V supply voltage to them.

I have therefore decided (due to your help) to run eight parallel
arrangements of two LED's in series from the 12V supply. My LED's are
30mA and so I would need a 0.5A power supply. I am using a 30A supply
so this should not be a problem (even after distributing power to the
other components).

Since these LED's are 5V, I assume I need a resistor to drop the rest
of the 12V supply (i.e. 2V) so about 66 Ohm (or closest available). I
will connect these to each series array so I will need eight. I am
thinking of connecting the resistors to the negative side of LED.

Does this sound convincing or have I got it all in a mess????

Pedro de Oliveira
• Pedro, I think you will find that this is the Maximum Reverse Voltage on the leds (also known as the reverse breakdown voltage), which means that if there is
Message 11 of 14 , Aug 2, 2001
Pedro,
I think you will find that this is the Maximum Reverse Voltage on the
leds (also known as the reverse breakdown voltage), which means that if
there is any more than 5 volts applied across the leads in reverse, the
diode will breakdown and probably start smoking and sputtering and do all
sorts of nasty stuff ...

What you need to look at is the Forward Voltage (denoted as Vf), typical
standard LED's have a Vf of about 2V (between 1.9 & 2.1V). The actual
formula for calculating the resistance based on forward voltage drop, supply
voltage, and maximum continuous forward current is as follows:

R=(E-Vf) x 1000 / I

Where:
R = Resistance needed (Ohms)
E = DC Supply Current (Volts)
I = Total LED current draw (mA)

I've attached a circuit of what I'm pretty sure you are trying to do;
calculated on the values I could work out ... based on a Vf of 2V,
Continuous Forward Current of 30mA, with a supply voltage of 12VDC, which
should be pretty safe ... you can recalculate if need be to accomodate a
higher forward voltage.

Regards,

Jonathan

----- Original Message -----
From: "Pedro de Oliveira" <olive_@...>
To: <Electronics_101@yahoogroups.com>
Sent: Friday, August 03, 2001 12:43 AM
Subject: [Electronics_101] Re: Newbie LED question

> Thanks to you all for your excellent suggestions.
>
> The reverse voltage on these LED's is 5V. I take this to mean that
> they "want" 5V supply voltage to them.
>
> I have therefore decided (due to your help) to run eight parallel
> arrangements of two LED's in series from the 12V supply. My LED's are
> 30mA and so I would need a 0.5A power supply. I am using a 30A supply
> so this should not be a problem (even after distributing power to the
> other components).
>
> Since these LED's are 5V, I assume I need a resistor to drop the rest
> of the 12V supply (i.e. 2V) so about 66 Ohm (or closest available). I
> will connect these to each series array so I will need eight. I am
> thinking of connecting the resistors to the negative side of LED.
>
> Does this sound convincing or have I got it all in a mess????
>
> Cheers for your help again.
>
> Pedro de Oliveira
• Hello All. I m trying to make a custom wireless interface for a video game I m making. I m familiar with using a keyboard encoder, but I wouldn t know how to
Message 12 of 14 , Aug 17, 2001
Hello All.

I'm trying to make a custom wireless interface for a video game I'm
making. I'm familiar with using a keyboard encoder, but I wouldn't
know how to make one wireless. Does anyone know of a company that
sells wireless keyboard encoders, or can point me to somewhere for
modifying a non wireless one?

Thanks

Mark
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