Connecting LEDs in series.

[C M Baker]

As a novice with electronics, I am constructing an emergency lighting
system using LEDs instead of filament lamps, to minimise battery
requirements and extend the power available. I am using a 12v lead
acid battery as a power source and Hibrite LEDs for illumination. I
can use the formula for calculating the current limiting resistor for
1 LED, but it seems pointless wiring the required number in parallel
and disipating a lot of wasted power as heat from the resistors. Is
there a way of calculating the size of the current limiting resistor
for LEDs in series, I know I can set it up on a breadboard and start
with a high resistance and reduce it until the current in the circuit
matches the design current of the LED, but there must be a formula for
calculating it directly.

I have experimented with 1, 2 and 3 LEDs in series and plotted the
limiting resistors against the supply voltage and the sum voltage for
the number of resistors. I get a straight line graph with equal
spacing between the plots for 1, 2 and 3 LEDs in series, which tells
me (I think?) that there must be a simple link for the formula, but at
57 years, my maths is rather rusty and I can't correlate the observed
data.

Any one out there with the answer, I shall be most pleased to hear
from you.

Many Thanks, Colin Baker.

 

[Costas Vlachos]

"C M Baker" <cmearsbaker@wight365.net> wrote in message
news:551757ce.0403250843.24d92509@posting.google.com...

As a novice with electronics, I am constructing an emergency lighting
system using LEDs instead of filament lamps, to minimise battery
requirements and extend the power available. I am using a 12v lead
acid battery as a power source and Hibrite LEDs for illumination. I
can use the formula for calculating the current limiting resistor for
1 LED, but it seems pointless wiring the required number in parallel
and disipating a lot of wasted power as heat from the resistors. Is
there a way of calculating the size of the current limiting resistor
for LEDs in series, I know I can set it up on a breadboard and start
with a high resistance and reduce it until the current in the circuit
matches the design current of the LED, but there must be a formula for
calculating it directly.

I have experimented with 1, 2 and 3 LEDs in series and plotted the
limiting resistors against the supply voltage and the sum voltage for
the number of resistors. I get a straight line graph with equal
spacing between the plots for 1, 2 and 3 LEDs in series, which tells
me (I think?) that there must be a simple link for the formula, but at
57 years, my maths is rather rusty and I can't correlate the observed
data.

Any one out there with the answer, I shall be most pleased to hear
from you.

Many Thanks, Colin Baker.

Assuming the LEDs need the same forward current, you just sum the forward
voltages and use that in the equation. For example, suppose you have 3
standard LEDs in series, with Vf = 2V and If = 20mA each, powered by a 12V
lead-acid battery. These batteries can go up to 14V with such small loads,
so I'll use 14V as the supply voltage. Then

R = (14V-(2V+2V+2V))/0.02A = (14V-6V)/0.02A = 8V/0.02A = 400 Ohm

So you need only one 400 Ohm resistor and the 3 LEDs, all in series. Notice
that the closer the total Vf gets to the supply (14V in your case), the
smaller R becomes, until R goes to zero at 14V (7 LEDs in series). Don't let
R become too small or zero, because then there is no mechanism for limiting
the current and any fluctuation in battery voltage will kill the LEDs. I
wouldn't put more than 3 LEDs in series in this setup. Then parallel as many
3-LED groups as necessary to achieve the total number of LEDs you want.

cheers,
Costas

 

[Peter Bennett]

On 25 Mar 2004 08:43:21 -0800, cmearsbaker@wight365.net (C M Baker)
wrote:

As a novice with electronics, I am constructing an emergency lighting
system using LEDs instead of filament lamps, to minimise battery
requirements and extend the power available. I am using a 12v lead
acid battery as a power source and Hibrite LEDs for illumination. I
can use the formula for calculating the current limiting resistor for
1 LED, but it seems pointless wiring the required number in parallel
and disipating a lot of wasted power as heat from the resistors. Is
there a way of calculating the size of the current limiting resistor
for LEDs in series, I know I can set it up on a breadboard and start
with a high resistance and reduce it until the current in the circuit
matches the design current of the LED, but there must be a formula for
calculating it directly.

As usual, R = E/I, where E is the voltage across the resistor.

For your application:
R = (Esupply - (n * Eled)) / I


--
Peter Bennett VE7CEI
email: peterbb (at) interchange.ubc.ca
GPS and NMEA info and programs: http://vancouver-webpages.com/peter/index.html
Newsgroup new user info: http://vancouver-webpages.com/nnq

 

[C M Baker]

Peter Bennett <peterbb@nowhere.invalid> wrote in message news:<ddh660p2ssedsakpprinuoprir1vq9fgba@4ax.com>...

On 25 Mar 2004 08:43:21 -0800, cmearsbaker@wight365.net (C M Baker)
wrote:

As a novice with electronics, I am constructing an emergency lighting
system using LEDs instead of filament lamps, to minimise battery
requirements and extend the power available. I am using a 12v lead
acid battery as a power source and Hibrite LEDs for illumination. I
can use the formula for calculating the current limiting resistor for
1 LED, but it seems pointless wiring the required number in parallel
and disipating a lot of wasted power as heat from the resistors. Is
there a way of calculating the size of the current limiting resistor
for LEDs in series, I know I can set it up on a breadboard and start
with a high resistance and reduce it until the current in the circuit
matches the design current of the LED, but there must be a formula for
calculating it directly.

As usual, R = E/I, where E is the voltage across the resistor.

For your application:
R = (Esupply - (n * Eled)) / I

Thanks for your reply, but refering to my calcuations and results for
a voltage supply of 12v and 3 LEDs Vf, 2.5v If, 0.025amps, the
calculated resistor was 180ohms but when set up on the breadboard, the
actual resistor value needed was 230ohms. I am informed that LEDs are
non-linear devices and do not obey Ohms Law when in strings, as they
need a trigger voltage of approx 1.5v before they will light. I have
read most of the postings refering to multiple LEDs in series
and note that about half refer to LEDs following Ohms Law and the rest
seem to think that they do not.

Any further information about this subject will be of great interest,
not only to myself, but to many others, judging by the number of
postings on this matter.
Many thanks for taking the time to reply. Colin.

 

[Watson A.Name \"Watt Sun]

Costas Vlachos wrote:

"C M Baker" <cmearsbaker@wight365.net> wrote in message
news:551757ce.0403250843.24d92509@posting.google.com...

As a novice with electronics, I am constructing an emergency lighting
system using LEDs instead of filament lamps, to minimise battery
requirements and extend the power available. I am using a 12v lead
acid battery as a power source and Hibrite LEDs for illumination. I
can use the formula for calculating the current limiting resistor for
1 LED, but it seems pointless wiring the required number in parallel
and disipating a lot of wasted power as heat from the resistors. Is
there a way of calculating the size of the current limiting resistor
for LEDs in series, I know I can set it up on a breadboard and start
with a high resistance and reduce it until the current in the circuit
matches the design current of the LED, but there must be a formula for
calculating it directly.

I have experimented with 1, 2 and 3 LEDs in series and plotted the
limiting resistors against the supply voltage and the sum voltage for
the number of resistors. I get a straight line graph with equal
spacing between the plots for 1, 2 and 3 LEDs in series, which tells
me (I think?) that there must be a simple link for the formula, but at
57 years, my maths is rather rusty and I can't correlate the observed
data.

Any one out there with the answer, I shall be most pleased to hear
from you.

Many Thanks, Colin Baker.

Assuming the LEDs need the same forward current, you just sum the forward
voltages and use that in the equation. For example, suppose you have 3
standard LEDs in series, with Vf = 2V and If = 20mA each, powered by a 12V
lead-acid battery. These batteries can go up to 14V with such small loads,
so I'll use 14V as the supply voltage. Then

R = (14V-(2V+2V+2V))/0.02A = (14V-6V)/0.02A = 8V/0.02A = 400 Ohm

So you need only one 400 Ohm resistor and the 3 LEDs, all in series. Notice
that the closer the total Vf gets to the supply (14V in your case), the
smaller R becomes, until R goes to zero at 14V (7 LEDs in series). Don't let
R become too small or zero, because then there is no mechanism for limiting
the current and any fluctuation in battery voltage will kill the LEDs. I
wouldn't put more than 3 LEDs in series in this setup. Then parallel as many
3-LED groups as necessary to achieve the total number of LEDs you want.

Three red LEDs adds up to about 6V. I wouldn't be concerned about
putting them in series until they add up to 8 or 9V, as long as the max
current, which is usually around 30 or 40 mA, isn't exceeded at max
charginng voltage which is about 15 or 16VDC. Or you can use a constant
current circuit. See the data sheets for the LM317 for an example of a
simple constant current circuit.

cheers,
Costas

 

[Watson A.Name \"Watt Sun]

C M Baker wrote:

Peter Bennett <peterbb@nowhere.invalid> wrote in message news:<ddh660p2ssedsakpprinuoprir1vq9fgba@4ax.com>...

On 25 Mar 2004 08:43:21 -0800, cmearsbaker@wight365.net (C M Baker)
wrote:

As a novice with electronics, I am constructing an emergency lighting
system using LEDs instead of filament lamps, to minimise battery
requirements and extend the power available. I am using a 12v lead
acid battery as a power source and Hibrite LEDs for illumination. I
can use the formula for calculating the current limiting resistor for
1 LED, but it seems pointless wiring the required number in parallel
and disipating a lot of wasted power as heat from the resistors. Is
there a way of calculating the size of the current limiting resistor
for LEDs in series, I know I can set it up on a breadboard and start
with a high resistance and reduce it until the current in the circuit
matches the design current of the LED, but there must be a formula for
calculating it directly.

As usual, R = E/I, where E is the voltage across the resistor.

For your application:
R = (Esupply - (n * Eled)) / I

Thanks for your reply, but refering to my calcuations and results for
a voltage supply of 12v and 3 LEDs Vf, 2.5v If, 0.025amps, the
calculated resistor was 180ohms but when set up on the breadboard, the
actual resistor value needed was 230ohms. I am informed that LEDs are
non-linear devices and do not obey Ohms Law when in strings, as they
need a trigger voltage of approx 1.5v before they will light. I have
read most of the postings refering to multiple LEDs in series
and note that about half refer to LEDs following Ohms Law and the rest
seem to think that they do not.

Red LEDs drop about 2V, green about 2.4V. Did you measue the V drop at
the rated current? If it's 2V, then three LEDs would be 6V, and 7.8V /
..025 A gives 312 ohms. It's a simple matter of addition and division,
no maths that us old farts can't handle. ;-)

If you want to play around, put a 220 ohm resistor and a 500 ohm pot in
series for the resistor, and then adjust until the required current is
flowing. Remember that the DMM will drop about .1 or .2V so when it's
gone, add that v into the mess. Then remove the pot and resistor and
measure it. Then use the next higher standard value resistor.

Any further information about this subject will be of great interest,
not only to myself, but to many others, judging by the number of
postings on this matter.
Many thanks for taking the time to reply. Colin.

 

[C M Baker]

"Robert C Monsen" <rcsurname@comcast.net> wrote in message news:<427ac.129464$1p.1761076@attbi_s54>...

"C M Baker" <cmearsbaker@wight365.net> wrote in message
news:551757ce.0403260349.243c3366@posting.google.com...
Peter Bennett <peterbb@nowhere.invalid> wrote in message
news:<ddh660p2ssedsakpprinuoprir1vq9fgba@4ax.com>...
On 25 Mar 2004 08:43:21 -0800, cmearsbaker@wight365.net (C M Baker)
wrote:

As a novice with electronics, I am constructing an emergency lighting
system using LEDs instead of filament lamps, to minimise battery
requirements and extend the power available. I am using a 12v lead
acid battery as a power source and Hibrite LEDs for illumination. I
can use the formula for calculating the current limiting resistor for
1 LED, but it seems pointless wiring the required number in parallel
and disipating a lot of wasted power as heat from the resistors. Is
there a way of calculating the size of the current limiting resistor
for LEDs in series, I know I can set it up on a breadboard and start
with a high resistance and reduce it until the current in the circuit
matches the design current of the LED, but there must be a formula for
calculating it directly.

As usual, R = E/I, where E is the voltage across the resistor.

For your application:
R = (Esupply - (n * Eled)) / I

Thanks for your reply, but refering to my calcuations and results for
a voltage supply of 12v and 3 LEDs Vf, 2.5v If, 0.025amps, the
calculated resistor was 180ohms but when set up on the breadboard, the
actual resistor value needed was 230ohms. I am informed that LEDs are
non-linear devices and do not obey Ohms Law when in strings, as they
need a trigger voltage of approx 1.5v before they will light. I have
read most of the postings refering to multiple LEDs in series
and note that about half refer to LEDs following Ohms Law and the rest
seem to think that they do not.

Any further information about this subject will be of great interest,
not only to myself, but to many others, judging by the number of
postings on this matter.
Many thanks for taking the time to reply. Colin.

You can make a simple 'constant current' source with a single PNP
transistor, a couple of diodes, and a couple of resistors that will keep the
current through your LEDs constant over the voltage range you are interested
in.


12V +------+-------+
| |
| |
V .-.
- | |
| | | 33 ohms (R1)
| '-'
V |
- |
| |
| b |< e
+-----| PNP, like 2N4403
| |\ c
| |
| |
| |
.-.
1k ohm| | 5 LEDs in series
| |
'-' |
| |
| |
=== ===
GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

From 10V to 15V, the current through those 5 LEDs will be quite close to
20mA. Use a larger R1 for less current. Note that the current through the
LEDs will be something like

I = 0.7/R1, so

R = 33 => I = 21mA
R = 47 => I = 15mA
R = 68 => I = 10mA

Diodes are 1N914 or 1N4001, for example. Really, anything, even diode
connected transistors will work in a pinch.

Regards,
Bob Monsen

Thanks Bob, this looks very interesting and I will experiment with the
bread board. I now have 3 emergency light units up and running, with 2
Hibrite white LEDS and 12 Hibrite green LEDs. The combination is for
best light value at lowest cost and illuminates both bedrooms and
office at sufficient levels for safety. Other units now under
construction. Many thanks for your help.

 

[Watson A.Name \"Watt Sun]

Robert C Monsen wrote:

"C M Baker" <cmearsbaker@wight365.net> wrote in message
news:551757ce.0403260349.243c3366@posting.google.com...

Peter Bennett <peterbb@nowhere.invalid> wrote in message

news:<ddh660p2ssedsakpprinuoprir1vq9fgba@4ax.com>...

On 25 Mar 2004 08:43:21 -0800, cmearsbaker@wight365.net (C M Baker)
wrote:

As a novice with electronics, I am constructing an emergency lighting
system using LEDs instead of filament lamps, to minimise battery
requirements and extend the power available. I am using a 12v lead
acid battery as a power source and Hibrite LEDs for illumination. I
can use the formula for calculating the current limiting resistor for
1 LED, but it seems pointless wiring the required number in parallel
and disipating a lot of wasted power as heat from the resistors. Is
there a way of calculating the size of the current limiting resistor
for LEDs in series, I know I can set it up on a breadboard and start
with a high resistance and reduce it until the current in the circuit
matches the design current of the LED, but there must be a formula for
calculating it directly.

As usual, R = E/I, where E is the voltage across the resistor.

For your application:
R = (Esupply - (n * Eled)) / I

Thanks for your reply, but refering to my calcuations and results for
a voltage supply of 12v and 3 LEDs Vf, 2.5v If, 0.025amps, the
calculated resistor was 180ohms but when set up on the breadboard, the
actual resistor value needed was 230ohms. I am informed that LEDs are
non-linear devices and do not obey Ohms Law when in strings, as they
need a trigger voltage of approx 1.5v before they will light. I have
read most of the postings refering to multiple LEDs in series
and note that about half refer to LEDs following Ohms Law and the rest
seem to think that they do not.

Any further information about this subject will be of great interest,
not only to myself, but to many others, judging by the number of
postings on this matter.
Many thanks for taking the time to reply. Colin.

You can make a simple 'constant current' source with a single PNP
transistor, a couple of diodes, and a couple of resistors that will keep the
current through your LEDs constant over the voltage range you are interested
in.

12V +------+-------+
| |
| |
V .-.
- | |
| | | 33 ohms (R1)
| '-'
V |
- |
| |
| b |< e
+-----| PNP, like 2N4403
| |\ c
| |
| |
| |
.-.
1k ohm| | 5 LEDs in series
| |
'-' |
| |
| |
=== ===
GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

The one problem I see is that almost as much current is wasted thru the
1k as thru the LEDs themselves. Something like 10 or more mA. I'd try
to increase the 1k to something that's not quite so wasteful, especially
if you're putting the resistor inside of something that also has the
LEDs in it. Why? Because of the additional heat.

I'll have to post my version later.

From 10V to 15V, the current through those 5 LEDs will be quite close to
20mA. Use a larger R1 for less current. Note that the current through the
LEDs will be something like

I = 0.7/R1, so

R = 33 => I = 21mA
R = 47 => I = 15mA
R = 68 => I = 10mA

Diodes are 1N914 or 1N4001, for example. Really, anything, even diode
connected transistors will work in a pinch.

Regards,
Bob Monsen

 

[Watson A.Name - \"Watt Su]

"Watson A.Name "Watt Sun - the Dark Remover"" <NOSPAM@dslextreme.com>
wrote in message news:c4cosk$6h565$1@hades.csu.net...
[snip]

You can make a simple 'constant current' source with a single PNP
transistor, a couple of diodes, and a couple of resistors that will
keep the
current through your LEDs constant over the voltage range you are
interested
in.

12V +------+-------+
| |
| |
V .-.
- | |
| | | 33 ohms (R1)
| '-'
V |
- |
| |
| b |< e
+-----| PNP, like 2N4403
| |\ c
| |
| |
| |
.-.
1k ohm| | 5 LEDs in series
| |
'-' |
| |
| |
=== ===
GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

My version is similar but NPN intead of PNP. It has a better current
limiting characteristic. See graph of current for the LEDs in the above
and below schems in alt.binaries.schematics.electronic. BTW, connecting
5 LEDs in series will add up to about 10V. With the 3 to 4V drop needed
for the circuit itself, that leaves no room for a discharged battery.
Soon after the battery discharges below 12V, the current will drop
quickly to where the LEDs don't light. It would be best to use 4 or
less LEDs.



+--------------------+------- Positive
| | Supply V.
| |
| --- LED
10k \ \ / =====>
to 47k / ===
ohms \ |
/ |
| / Q1
| | / Gen'l
| | Purp NPN
+----------------| 2N3904 or
| | 2N2222A
| | \ E
Q2 \ \
Gen'l \ | |
Purp NPN | 470 ohms |
2N3904 or |------/\/\/\----+
2N2222A | |
E / | |
/ \
| 33 ohms /
| for \
| 20 mA /
| |
| |
| |
+---------------------+-------
Negative Supply V.

The one problem I see is that almost as much current is wasted thru
the
1k as thru the LEDs themselves. Something like 10 or more mA. I'd
try
to increase the 1k to something that's not quite so wasteful,
especially
if you're putting the resistor inside of something that also has the
LEDs in it. Why? Because of the additional heat.

I'll have to post my version later.

From 10V to 15V, the current through those 5 LEDs will be quite
close to
20mA. Use a larger R1 for less current. Note that the current
through the
LEDs will be something like

I = 0.7/R1, so

R = 33 => I = 21mA
R = 47 => I = 15mA
R = 68 => I = 10mA

Diodes are 1N914 or 1N4001, for example. Really, anything, even
diode
connected transistors will work in a pinch.

Regards,
Bob Monsen