# LEDs in SERIES

Discussion in 'Electronic Basics' started by chris, Jun 3, 2004.

1. ### chrisGuest

I'm going to be wiring some LEDs in series. they are 1.2v 100ma IR
LEDs.
Let's say I use 4. 4x1.2v = 4.8 volt source. would the total amp draw
still only be 100ma? Is this better than wiring them in parallel as
far as current goes?

2. ### WayneGuest

It depends on your source voltage and current capability.
For instance, for your series circuit, if you have a 4.8V 1A
supply, you will fry your leds unless you limit the current
to a safe level ( 100ma ).

3. ### WongGuest

Think that you will need a current-limit-resistor. For example, IR LED
in series base on your assumption:

4.8 - 1.2 - 1.2 = 2.4 = Voltage drop on resistor.
Since 100 mA is required, therefore 2.4/100m gives you 24, so you
need 24 ohm resistor. In practical, 24 ohm is not common so you may
choose any other values around that.

4. ### AmnonGuest

Hi
Since you want to use four LEDs , each with 1.2V drop, this means a total of
4x1.2=4.8V drop.
Thus your power supply must be a bit larger than 4.8V to allow the addition
of a current limiter serial resistor as well,in serial to these 4 LEDs.
For power supply of Xvolt (where X > 4.8) , the serial resistor value will
have to had a value of R=(X - 4.8)/0.1
to guarantee a limit of 0.1amp (100mA).

Working in parallel means that X can be smaller now (X>1.2v) but it should
supply 4x100mA=400mA, instead of only 100mA, and you will need four
resistors instead of only one. However, if one LED blows, the other will
still lit, where in serial connection all will go off.
Amnon

5. ### dBGuest

No, it isn't a 4.8V source, it's 4.8V across the diodes. The source
voltage will have to be higher and you will have to connect a resistor
in series to linmit the current to your required value.

The same value of current flows out of a series circuit as flows into
it. It will "only" be 100mA if the value of the series resistor has
the appropriate value.

If you run the diodes at 100mA and use a series arrangement than the
total current will be 100mA. If you use the parallel arrangement the
total current will be 400mA. It is up to you to decide which is
"better".

You must get used to thinking of a l.e.d. as a current operated
device, not a voltage operated one.
A l.e.d. requires a current within a certain range to function
correctly. When a current within this range flows through it a
voltage will be developed across it. The value of this voltage will
change very little if the current varies throughout the specified
range.

If you try to operate a l.e.d. from a voltage source you will be
asking for trouble and you shouldn't be surprised if you are rewarded
with it. Your l.e.d. will probably have a very bright life - but a
very short one.

http://p199.ezboard.com/fbasicelectronicsfrm5.showMessage?topicID=16.topic

6. ### chrisGuest

I'm going to be wiring some LEDs in series. they are 1.2v 100ma IR
"Current operated device" - That gets me every time!
I am having trouble getting past the idea the LEDs are not light bulbs
and don't have their own resistance. I'm always thinking voltage and
not current.
I think that since I'm limited by the amount of remaining current I
have available from the transformer, I'll do the series option, using
a resistor to set the current.

7. ### Robert C MonsenGuest

A 1A supply only means the amount of current you can pull from it
without overheating the transformer. He would have to limit the
current using a resistor if he only had a 100mA supply.

The issue is that when the LEDs get hotter, they can pass LOTS more
current. However, resistors don't have this flaw, so using, say, a 6V
supply, with a 12 ohm resistor, would work better.

Another issue is that when manufacturers say 100mA, they probably mean
maximum. The OP might want to limit the current even further to
increase LED life.

Regards

8. ### C M BakerGuest

I asked the same thing a couple of months back and most people
responded with the formula (Supply Voltage minus forward voltage) =
voltage drop.
Voltage drop divided by max forward current = value of limiting
resistor.
This works fine for 1 LED, but as soon as you put two or more in
series, that formula does not work. I tried several ways to work out
what the correct value of the series current limiting resistor should
be and ended up by using a bread board and milliammeter and starting
with a high resistor and reducing the value until the meter showed the
correct forward current for the LEDs used. Even with this method,the
voltage applied to the series is affected by the internal resistance
of the meter, so opt for a series current limiting resistor that is of
the nearest standard value, higher rather than lower, this will extend
the life of your LEDs. I have successfully built several mains failure
emergency backup lights using LEDs in series running off a 12volt
battery. Hope this information helps.

9. ### C M BakerGuest

See the series of articles 6th April 2004

10. ### Peter BennettGuest

This calculation should work for several LEDs in series, just as well
as it works for single LEDs - but you should use the "typical" forward
voltage drop, not the maximum.

Note that the formula with several LEDs in series is Supply voltage
minus (number of LEDs times typical voltage drop). You should then
use "voltage drop divided by desired current = resistance" (_not_
maximum current!!).

Also, it is rarely necessary to run LEDs at their maximum rated
current - I find most LEDs give quite adequate light with under 10 mA,
although their rated maximum current may be 20 - 30 mA. (and if I plan
on using well under the maximum current, I don't need to bother
checking the specs every time.)

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter

11. ### Dr Engelbert BuxbaumGuest

C M Baker wrote:

It does. You just add the voltage drops of all the LEDs together.

Say, you want to operate a red (1.8 V), green (2.4 V) and blue (3.5 V)
LED in series. The total voltage drop across them would be (1.8 + 2.4 +
3.5) V = 7.7 V (the LED act as resistors in series, the total voltage
dropped is the sum of the individual drop voltages, the current flowing
is identical through all of them).

If you use a 9 V supply, you have to drop 1.3 V across the current
limiting resistor. If you want 20 mA to flow through the diodes, you'd
use 1.3 V / 0.02 A = 65 Ohm, or rather the next higher standard value.
The power dropped over the resistor would be 1.3 V * 0.02 A = 26 mW,
thus a standard 1/8 watt resistor would do just fine.

As always, there is a little trick to save oneself the calculation: If
you use a J-FET instead of a resistor to limit the current, the current
depends only on the amplification of the FET (up to the maximum voltage
of the FET, that is), not on the supply voltage. A BF245A will supply
about 10 mA, a BF245B 15 mA and a BF245C about 18 mA:

o +5..30 V
|
|
| |S
|----
G | BF245
------->|----
| | |D
| |
-------------
|
__|__
<-- \ | /
<-- \|/ LED
-----
|
|
|
--- Gnd

If you have a AC supply, you can use a capacitor as a blind resistor to
limit the current through the LED. Since current and voltage are out of
phase, little power is dropped across the capacitor, limiting heat
developed. This is usefull as "power on" indicator in mains circuits:

o 230 V, 50 Hz
|
|
----- 100 nF, 1000 V =, 630 V ~
-----
|
|
---
| |
| | 10 Ohm, 1/4 W
| |
---
|
|
-----------
__|__ | 1N4148
<-- \ | / -----
<-- \|/ /|\
----- / | \
LED | --|--
-----------
|
|
o neutral

Note that in this circuit, all components have contact to mains, thus
appropriate care is required. The capacitor needs to be a type certified
for use across mains. For countries using other mains voltages or
frequencies the value of the capacitor needs recalculation.

Shouldn't we put this in a FAQ somewhere, I recall answering this
question a couple of times.

12. ### C M BakerGuest

As I sit here at my computer, I have 6 LEDs, specification of which
is: Forward Voltage 1.7v, Max If 100ma, running from a 13.7v supply,
in series. According to calculations, current limiting resistor should
be, (13.7 - (6 x 1.7)) = 3.5v using V=IR, 3.5/.1 = 35 ohms. To limit
the current to approx 100ma, I have had to put a 100 ohm & 120 ohm in
parallel,(to limit heat buildup in the resistors) giving a calculated
resistance of 54.5 ohms. At switch on, this gave a current of 96.5ma
which has gradually risen to 100.5ma during the last hour.
My 'research' and my friendly man at the local MAPLINS store say that
as LEDs are not straight forward resistive components like light
bulbs, they don't follow Ohms Law directly, and all my testing on a
bread board also indicates the same. There appears to be two 'camps'
concerning series LEDs, one saying yes they do follow Ohms Law and one
saying no they don't. Someone out there must surely have the
definitive answer or formula. I am not trying to be rude to those
emminent people who say they do follow Ohms Law, but why do they do it
for you and not for me?

13. ### Rich GriseGuest

It depends what you mean by "follow ohm's law." For any given voltage,
at some current, there's an equation R = E * I, so, _at that particular
voltage and current_, the component is acting like it has a resistance
of R ohms. But that value changes for different voltages and currents
in an LED. It's said to be non-linear. So, the camp that says, "It
doesn't follow Ohm's law" just mean that R in that equation is not
constant, i.e. it doesn't behave like a plain "ohmic" resistor.

The reason your current increased is that Vf decreases as they heat
up. This also changes the effective resistance, even at a constant
current.

Hope This Helps!
Rich

14. ### John FieldsGuest

---
As Rich pointed out, your LEDs have a non-linear voltage VS current
characteristic, as well as a negative temperature coefficient of
resistance.

Unfortunately, the friendly man at your local MAPLINS store misled you
with the part about the light bulb being a straightforward resistive
component. It isn't, and because of the positive tempco of tungsten,
a plain old 120V 100W lamp will look like about 14 ohms when it's cold
and rise to about 144 ohms when it's hot.

15. ### Peter BennettGuest

LEDs (and other diodes) do not follow Ohm's Law - that is, the voltage
across the LED does not vary directly with current like a resistor.
The voltage across a forward-biased LED is fairly constant, and is
determined by the chemical composition of the LED (which also
determines its colour). The voltage will increase slightly with
increased current.

value, and may vary slightly between production lots.

16. ### John PopelishGuest

Ohmic resistance has a voltage drop proportional to the current
through the resistance, and that factor of proportionality (volts per
ampere) is called ohms. Nothing is precisely ohmic (perfectly fixed
proportion between voltage and current) regardless of the current.
Some materials are pretty good over a temperature range that is
useful, and they make resistors out of these materials. Any PN
junction (silicon, germanium or exotic materials like used in LEDs)
have a completely different relation between the voltage drop and the
current. Current through these devices is more accurately described
as an exponential function of voltage drop (or conversely, the voltage
drop is related to the logarithm of current) but this effect is also
dependent on temperature (as you found as your LEDs warmed up from the
I*V power they consumed minus the energy they emitted).

Here is a data sheet for an LED that is perhaps similar to the ones
you are using:
http://www.vishay.com/docs/81009/81009.pdf

Figure 4 describes the typical voltage current relationship at room
temperature. Note that one of the scales is logarithmic and one is
linear. If a simple logarithmic relation between current and voltage
drop existed, this graph would be a straight line on this kind of
graph, and it is pretty close for very small currents up to about
rated current of 10^2 milliamps. For higher currents, the series
resistive drops of the wire bonds and the material on each side of the
junction starts to show itself, and the graph heads off toward a more
resistive line.

Figure 5 shows how the voltage drop associated with one point on the
line from figure 4 (the 10 milliamp point) varies with temperature.
You can assume that this variation is reflected all along the current
line (with small error).

17. ### Robert C MonsenGuest

Everybody has the answer (all correct), but not the formula. Here
it is:

If
--- = exp(Vf/(m*Vt)) - 1
Is

If is the forward current, and Vf is the forward voltage.

'Is' is the reverse leakage current, and depends on the device, and
also the temperature. Vt is a combination of terms (it equals kT/q,
where k is Boltzmann's constant, q is the fundamental electric charge,
and T is the temperature in degrees K) which turns out to equal 25.3mV
at 20C. m is a 'fudge factor' which is between 1 and 2.

'Is' is usually quite small, and also depends on temperature. Its not
easy to figure it out for a discrete device, which makes this
relationship somewhat unusable except for doing things like predicting
that increasing the voltage by 60mV will increase the current 10x,
since in the ratio, 'Is' is divided out.

This relation is called the Ebers-Moll equation.

Regards,
Bob Monsen

18. ### C M BakerGuest

Many thanks to all those who have replied, especially Bob Monsen,