# how to connect up bulk leds ?

M

#### mark krawczuk

Jan 1, 1970
0
hi, i want to make my own lead light for workin on cars etc.... i have
ordered 100 , 10 mm white leds.
what is the best way to connect them up ? in series or in paralell ? i will
be using around 70-90 leds.
thanks,

mark k

R

Jan 1, 1970
0
mark said:
hi, i want to make my own lead light for workin on cars etc.... i have
ordered 100 , 10 mm white leds.
what is the best way to connect them up ? in series or in paralell ? i
will be using around 70-90 leds.

As LED's are driven with current rather than voltage, it's generally best to
connect multiple LED's in series, together with a current-limiting
resistor.
The number of LED's you can connect in series depends on the available
supply voltage and the type op LED's used.

A white LED has a forward voltage drop of approximately 3.6 volts, so if you
connect two in series, you need a supply voltage of at least 7.2 volts;
with three in series the required supply voltage is 3 x 3.6 = 10.8 volts
minimum and so on. Also, you have to add a few extra volts for the
current-limiting resistor.

So if you have a 12V supply, and you want to operate each LED at 20mA of
current, these are the calculations:
- You can connect three white LED's in series, resulting in a 10.8 volt
forward voltage drop, with 1.2 volts remaining (12 - 10.8).
- To get 20mA of current with a 1.2 volts voltage requires a resistor of
1.2 / 0.02 = 60 ohms. So you cold use a 68 ohm series resistor.

With 90 LED's in total, this means that you have to make 30 strings, each
composed of three white LED's and a 68 ohm resistor connected in series.
These strings are connected in parallel to the 12V supply, resulting in a
total current of 30 x 0.02 = 0.6A.

For 24 volts, you can double the amount of LED's in each string:
6 x 3.6 = 21.6 volts. To get 20mA you divide once again the remaining
voltage by the desired current: (24 - 21.6) / 0.02 = 120 ohms series
resistor. Now you need only 15 strings of 6 LED's and one resistor each.
The total current drawn is 15 x 0.02 = 0.3A.

I would strongly recommend not raising the voltage any higher. And note that
this is all based on a DC supply. If you have an AC supply, you could
double the amount of LED's per string, with two LED's instead of one per
3.6V voltage step, connected anti-parallel, and approximately half the
resistor value to double the current (as each LED is switched for on only
half the time). Also note that LED's don't take reverse voltages very well.

B

#### [email protected]

Jan 1, 1970
0
As LED's are driven with current rather than voltage, it's generally best to
connect multiple LED's in series, together with a current-limiting
resistor.
The number of LED's you can connect in series depends on the available
supply voltage and the type op LED's used.

A white LED has a forward voltage drop of approximately 3.6 volts, so if you
connect two in series, you need a supply voltage of at least 7.2 volts;
with three in series the required supply voltage is 3 x 3.6 = 10.8 volts
minimum and so on. Also, you have to add a few extra volts for the
current-limiting resistor.

So if you have a 12V supply, and you want to operate each LED at 20mA of
current, these are the calculations:
- You can connect three white LED's in series, resulting in a 10.8 volt
forward voltage drop, with 1.2 volts remaining (12 - 10.8).
- To get 20mA of current with a 1.2 volts voltage requires a resistor of
1.2 / 0.02 = 60 ohms. So you cold use a 68 ohm series resistor.

With 90 LED's in total, this means that you have to make 30 strings, each
composed of three white LED's and a 68 ohm resistor connected in series.
These strings are connected in parallel to the 12V supply, resulting in a
total current of 30 x 0.02 = 0.6A.

For 24 volts, you can double the amount of LED's in each string:
6 x 3.6 = 21.6 volts. To get 20mA you divide once again the remaining
voltage by the desired current: (24 - 21.6) / 0.02 = 120 ohms series
resistor. Now you need only 15 strings of 6 LED's and one resistor each.
The total current drawn is 15 x 0.02 = 0.3A.

I would strongly recommend not raising the voltage any higher. And note that
this is all based on a DC supply. If you have an AC supply, you could
double the amount of LED's per string, with two LED's instead of one per
3.6V voltage step, connected anti-parallel, and approximately half the
resistor value to double the current (as each LED is switched for on only
half the time). Also note that LED's don't take reverse voltages very well..

The catch with this approach is that the "3.6V" drop per white LED is
strictly nominal. The manufacturer's data sheet will also specify
upper and lower limits to the LED voltage drop at 20mA (or whatever is
approprate for the specific LED which may not always be 20mA) at room
temperature - usually (but - not always - 20C).

If you get a batch of LEDs with a lower forward drop, 68R may give you
rather more current than you expect.

There is also the problem that the forward voltage across an LED is
temperature dependent and will fall by 2mV per degree Celcius increase
in junction temperature. This means that you can get more current
flowing through the LED as it warms up, and - potentially, higher
power dissipation in the LED (if the voltage drop across the resistor
is less than the voltage drop across the LEDs).

For sufficiently long strings of LED's and sufficiently small
resistors, the process can run away, melting your LEDs. A properly
designed constant current driver can be immune to this problem.

R

Jan 1, 1970
0
As LED's are driven with current rather than voltage, it's generally best
to connect multiple LED's in series, together with a current-limiting
resistor.
[snip]

The catch with this approach is that the "3.6V" drop per white LED is
strictly nominal. The manufacturer's data sheet will also specify
upper and lower limits to the LED voltage drop at 20mA (or whatever is
approprate for the specific LED which may not always be 20mA) at room
temperature - usually (but - not always - 20C).

If you get a batch of LEDs with a lower forward drop, 68R may give you
rather more current than you expect.

There is also the problem that the forward voltage across an LED is
temperature dependent and will fall by 2mV per degree Celcius increase
in junction temperature. This means that you can get more current
flowing through the LED as it warms up, and - potentially, higher
power dissipation in the LED (if the voltage drop across the resistor
is less than the voltage drop across the LEDs).

For sufficiently long strings of LED's and sufficiently small
resistors, the process can run away, melting your LEDs. A properly
designed constant current driver can be immune to this problem.

Indeed, with a very small voltage overhead (say less than 5%) and the
resulting small resistors, these things must be taken into account.
But as soon as you have some 10% or more of the total supply voltage across
the current-limiting resistor, things will be fine in the vast majority of
cases. But if you get a batch of unknown LED's, it can't hurt all the same
to take a handful of LED's and measure the actual forward voltage drop at
the desired current, and recalculate the series resistor accordingly.

About temperature: with 25mA through a white LED, the internal power
dissipation (~0.1 watts) is hardly capable of raising the temperature more
than a few degrees. The increase in forward voltage drop versus increase in
current (iow: the dynamic resistance) is usually much larger than the
influence of temperature, and prevents runaway effects -- e.g. for a
typical white LED, I found a forward voltage drop of 3.6 volts at 13mA; at
20mA, the voltage drop had increased to 4.2 volts. Only when heating up the
LED to approximately 70 degrees Celsius was I able to nibble off 0.1 volt.

In over 25 years of using LED's, I have never blown even one using series
resistors (I *did* destroy some by accidentally hooking 'em up in reverse,
though). And several people I know even say they hooked up white and blue
LED's directly to 5V power supplies without failure -- although I
definitely wouldn't recommend this.

D

#### D from BC

Jan 1, 1970
0
hi, i want to make my own lead light for workin on cars etc.... i have
ordered 100 , 10 mm white leds.
what is the best way to connect them up ? in series or in paralell ? i will
be using around 70-90 leds.
thanks,

mark k

Perhaps like this:

--line AC---Rectifier---cap---Pot---LED1...LED?---Com

An Idea for an Experimental Method
0) Set Pot to 0ohm
1) Make an LED string that adds up to the peak voltage for a safe peak
current.
2) Power it up and adjust the pot to compensate for self heating.
Quickly...
2) Raise temperature of all the LEDs in an oven for the greatest
ambient temp. 40C ??
3) Adjust the pot resistance to maintain a safe peak current.

Even trying to dodge a crapload of math it's still tricky

D from BC
British Columbia

J

#### Jon Slaughter

Jan 1, 1970
0
As LED's are driven with current rather than voltage, it's generally best
to
connect multiple LED's in series, together with a current-limiting
resistor.
The number of LED's you can connect in series depends on the available
supply voltage and the type op LED's used.

A white LED has a forward voltage drop of approximately 3.6 volts, so if
you
connect two in series, you need a supply voltage of at least 7.2 volts;
with three in series the required supply voltage is 3 x 3.6 = 10.8 volts
minimum and so on. Also, you have to add a few extra volts for the
current-limiting resistor.

So if you have a 12V supply, and you want to operate each LED at 20mA of
current, these are the calculations:
- You can connect three white LED's in series, resulting in a 10.8 volt
forward voltage drop, with 1.2 volts remaining (12 - 10.8).
- To get 20mA of current with a 1.2 volts voltage requires a resistor of
1.2 / 0.02 = 60 ohms. So you cold use a 68 ohm series resistor.

With 90 LED's in total, this means that you have to make 30 strings, each
composed of three white LED's and a 68 ohm resistor connected in series.
These strings are connected in parallel to the 12V supply, resulting in a
total current of 30 x 0.02 = 0.6A.

For 24 volts, you can double the amount of LED's in each string:
6 x 3.6 = 21.6 volts. To get 20mA you divide once again the remaining
voltage by the desired current: (24 - 21.6) / 0.02 = 120 ohms series
resistor. Now you need only 15 strings of 6 LED's and one resistor each.
The total current drawn is 15 x 0.02 = 0.3A.

I would strongly recommend not raising the voltage any higher. And note
that
this is all based on a DC supply. If you have an AC supply, you could
double the amount of LED's per string, with two LED's instead of one per
3.6V voltage step, connected anti-parallel, and approximately half the
resistor value to double the current (as each LED is switched for on only
half the time). Also note that LED's don't take reverse voltages very
well.

He might use a smaller series resistance and then add a variable pot and
"extra bright" switch inline with all the other members(or even a 3-way
switch). This will give him the option to increase the brightness if he
needs(But he would need to know it won't last as long). The pot will allow
for brightness controll. he could then, if he wanted, add a fan and/or temp
sensor with slow shut off or reduced brightness.

It would really suck to hook all those LED's up and turn on the power and 2

B

#### [email protected]

Jan 1, 1970
0
mark krawczuk wrote:
hi, i want to make my own lead light  for workin on cars etc.... i have
ordered  100 ,  10 mm white leds.
what is the best way to connect them up ?  in series or in paralell? i
will be using around 70-90 leds.
As LED's are driven with current rather than voltage, it's generally best
to connect multiple LED's in series, together with a current-limiting
resistor.
[snip]

The catch with this approach is that the "3.6V" drop per white LED is
strictly nominal. The manufacturer's data sheet will also specify
upper and lower limits to the LED voltage drop at 20mA (or whatever is
approprate for the specific LED which may not always be 20mA) at room
temperature - usually (but - not always - 20C).
If you get a batch of LEDs with a lower forward drop, 68R may give you
rather more current than you expect.
There is also the problem that the forward voltage across an LED is
temperature dependent and will fall by 2mV per degree Celcius increase
in junction temperature. This means that you can get more current
flowing through the LED as it warms up, and - potentially, higher
power dissipation in the LED (if the voltage drop across the resistor
is less than the voltage drop across the LEDs).
For sufficiently long strings of LED's and sufficiently small
resistors, the process can run away, melting your LEDs. A properly
designed constant current driver can be immune to this problem.

Indeed, with a very small voltage overhead (say less than 5%) and the
resulting small resistors, these things must be taken into account.
But as soon as you have some 10% or more of the total supply voltage across
the current-limiting resistor, things will be fine in the vast majority of
cases. But if you get a batch of unknown LED's, it can't hurt all the same
to take a handful of LED's and measure the actual forward voltage drop at
the desired current, and recalculate the series resistor accordingly.

About temperature: with 25mA through a white LED, the internal power
dissipation (~0.1 watts) is hardly capable of raising the temperature more
than a few degrees.

The Agilent HLMP-CW18 5mm white LED has a thermal resistance from
junction to ambient of 240C/watt, which means that the junction is
running 24C above ambient when dissipating 100mW.

The numbers are usually available if you can be bothered to look. If
they aren't, you'd better look for a more reputable manufacturer.
The increase in forward voltage drop versus increase in
current (iow: the dynamic resistance) is usually much larger than the
influence of temperature, and prevents runaway effects -- e.g. for a
typical white LED, I found a forward voltage drop of 3.6 volts at 13mA; at
20mA, the voltage drop had increased to 4.2 volts. Only when heating up the
LED to approximately 70 degrees Celsius was I able to nibble off 0.1 volt.

For that particular LED. The HLMP-CW18 has a typical forward voltage
of 3.4V at 20mA and a maximum of 4.0V, so your parts would seem to
have been some kind of cheap rubbish.
In over 25 years of using LED's, I have never blown even one using series
resistors (I *did* destroy some by accidentally hooking 'em up in reverse,
though). And several people I know even say they hooked up white and blue
LED's directly to 5V power supplies without failure -- although I
definitely wouldn't recommend this.

Very wise of you.

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