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Multiplexing LEDs - calculating resistor value

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roxlu
electronics forum beginner

Joined: 01 Jun 2006
Posts: 37

Posted: Mon Jun 26, 2006 8:40 pm Post subject: Note

Multiplexing LEDs - calculating resistor value

Hi all,

My previous post about speeding up multiplexing resulted in a display
with
8x48 LEDs. As I suggested and read, the LEDs are becomming less bright
when more than 2 LEDs are on in a matrx of 8x48 >=.

Can anyone help me to calculate the resistor value I need to get bright
LEDs.
Or, how can I caclulate this? Now I'm using a 100R per column which is
probably
to much.

Here is a small part of the circuit to show the current limiting
resistor of 100R.
http://imagebin.org/5461

Thanks
Greetings

roxlu
electronics forum beginner

Joined: 01 Jun 2006
Posts: 37

Posted: Mon Jun 26, 2006 9:20 pm Post subject: Note

Re: Multiplexing LEDs - calculating resistor value

Spehro Pefhany wrote:

Quote:

On 26 Jun 2006 13:40:26 -0700, the renowned "roxlu"
diederickh@gmail.com> wrote:

Hi all,

My previous post about speeding up multiplexing resulted in a display
with
8x48 LEDs. As I suggested and read, the LEDs are becomming less bright
when more than 2 LEDs are on in a matrx of 8x48 >=.

Can anyone help me to calculate the resistor value I need to get bright
LEDs.
Or, how can I caclulate this? Now I'm using a 100R per column which is
probably
to much.

Here is a small part of the circuit to show the current limiting
resistor of 100R.
http://imagebin.org/5461

Thanks
Greetings

Arrange it so that 48 LEDs are on at once (48 columns and 8 rows). You
want the resistor to pass about 8 times the average current you would
like for the LEDs. eg. if they are reasonably bright at 3mA, you want
about 24mA through the resistor. Just use G.S. Ohm's law, from the 5V
supply minus the LED forward drop (at 24mA, not 3mA) minus two
Vce(sat) drops.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

Hi Spehro,

Thanks for your reply, but thats what I did, and its not giving me good
results. I calculated the value like this: 5v - 1.7v (led) - 1.2 (2x0.6
or the transistors) = 2.1 v
Now I'm using a 100R, so thats around 2.1 / 100 = 21mA. Though they are
still not bright, which is probably due to the 'refresh rate'.

Greetings

Jonathan Kirwan
electronics forum Guru

Joined: 08 May 2005
Posts: 531

Posted: Mon Jun 26, 2006 9:22 pm Post subject: Note

Re: Multiplexing LEDs - calculating resistor value

On 26 Jun 2006 13:40:26 -0700, "roxlu" <diederickh@gmail.com> wrote:

Quote:

One thing that comes to mind is that you should be able to drive any
number of 'columns' at once, but only one 'row.' If you are enabling
two or more rows at once for a given column, then there is a problem.
There are other things I might arrange differently, but it sounds as
though this may be an issue for now.

Jon

Spehro Pefhany
electronics forum Guru

Joined: 01 May 2005
Posts: 2326

Posted: Mon Jun 26, 2006 9:24 pm Post subject: Note

Re: Multiplexing LEDs - calculating resistor value

On 26 Jun 2006 13:40:26 -0700, the renowned "roxlu"
<diederickh@gmail.com> wrote:

Quote:

Arrange it so that 48 LEDs are on at once (48 columns and 8 rows). You
want the resistor to pass about 8 times the average current you would
like for the LEDs. eg. if they are reasonably bright at 3mA, you want
about 24mA through the resistor. Just use G.S. Ohm's law, from the 5V
supply minus the LED forward drop (at 24mA, not 3mA) minus two
Vce(sat) drops.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

Spehro Pefhany
electronics forum Guru

Joined: 01 May 2005
Posts: 2326

Posted: Mon Jun 26, 2006 9:25 pm Post subject: Note

Re: Multiplexing LEDs - calculating resistor value

On Mon, 26 Jun 2006 17:24:57 -0400, Spehro Pefhany
<speffSNIP@interlogDOTyou.knowwhat> wrote:

Quote:

P.S. Each of you row drivers is going to have to handle 48 * 24mA =
1.2A in the example I gave above (at 1/8 duty cycle). That will also
be the total maximum current draw of the array. Your PNP BJTs will
probably not be saturated at that current without a lot more base
current than ~4mA. Or just use 8 P-channel MOSFETs for the source
drivers.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

Jonathan Kirwan
electronics forum Guru

Joined: 08 May 2005
Posts: 531

Posted: Mon Jun 26, 2006 9:37 pm Post subject: Note

Re: Multiplexing LEDs - calculating resistor value

On Mon, 26 Jun 2006 17:24:57 -0400, Spehro Pefhany
<speffSNIP@interlogDOTyou.knowwhat> wrote:

Quote:

Yes, but didn't he also say "becoming less bright?" Sizing that 100
ohm resistor in the collector leg assuming 8 LED rows being on will
brighten the case where only one LED row is used, yes?

Jon

Spehro Pefhany
electronics forum Guru

Joined: 01 May 2005
Posts: 2326

Posted: Mon Jun 26, 2006 9:37 pm Post subject: Note

Re: Multiplexing LEDs - calculating resistor value

On 26 Jun 2006 14:20:28 -0700, the renowned "roxlu"
<diederickh@gmail.com> wrote:

Quote:

Spehro Pefhany wrote:
On 26 Jun 2006 13:40:26 -0700, the renowned "roxlu"
diederickh@gmail.com> wrote:

Hi all,

My previous post about speeding up multiplexing resulted in a display
with
8x48 LEDs. As I suggested and read, the LEDs are becomming less bright
when more than 2 LEDs are on in a matrx of 8x48 >=.

Can anyone help me to calculate the resistor value I need to get bright
LEDs.
Or, how can I caclulate this? Now I'm using a 100R per column which is
probably
to much.

Here is a small part of the circuit to show the current limiting
resistor of 100R.
http://imagebin.org/5461

Thanks
Greetings

Arrange it so that 48 LEDs are on at once (48 columns and 8 rows). You
want the resistor to pass about 8 times the average current you would
like for the LEDs. eg. if they are reasonably bright at 3mA, you want
about 24mA through the resistor. Just use G.S. Ohm's law, from the 5V
supply minus the LED forward drop (at 24mA, not 3mA) minus two
Vce(sat) drops.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

Hi Spehro,

Thanks for your reply, but thats what I did, and its not giving me good
results. I calculated the value like this: 5v - 1.7v (led) - 1.2 (2x0.6
or the transistors) = 2.1 v
Now I'm using a 100R, so thats around 2.1 / 100 = 21mA. Though they are
still not bright, which is probably due to the 'refresh rate'.

Greetings

Above the (visual) fusion frequency, brightness is mostly a function
of average current. See my PS. BTW, 1.7V sounds a bit low for the LED,
even for red, but 0.6V is high for Vce(sat) if the transistor is
actually saturated.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

roxlu
electronics forum beginner

Joined: 01 Jun 2006
Posts: 37

Posted: Mon Jun 26, 2006 9:38 pm Post subject: Note

Re: Multiplexing LEDs - calculating resistor value

Spehro Pefhany wrote:

Quote:

On 26 Jun 2006 14:20:28 -0700, the renowned "roxlu"
diederickh@gmail.com> wrote:

Spehro Pefhany wrote:
On 26 Jun 2006 13:40:26 -0700, the renowned "roxlu"
diederickh@gmail.com> wrote:

Hi all,

My previous post about speeding up multiplexing resulted in a display
with
8x48 LEDs. As I suggested and read, the LEDs are becomming less bright
when more than 2 LEDs are on in a matrx of 8x48 >=.

Can anyone help me to calculate the resistor value I need to get bright
LEDs.
Or, how can I caclulate this? Now I'm using a 100R per column which is
probably
to much.

Here is a small part of the circuit to show the current limiting
resistor of 100R.
http://imagebin.org/5461

Thanks
Greetings

Arrange it so that 48 LEDs are on at once (48 columns and 8 rows). You
want the resistor to pass about 8 times the average current you would
like for the LEDs. eg. if they are reasonably bright at 3mA, you want
about 24mA through the resistor. Just use G.S. Ohm's law, from the 5V
supply minus the LED forward drop (at 24mA, not 3mA) minus two
Vce(sat) drops.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

Hi Spehro,

Thanks for your reply, but thats what I did, and its not giving me good
results. I calculated the value like this: 5v - 1.7v (led) - 1.2 (2x0.6
or the transistors) = 2.1 v
Now I'm using a 100R, so thats around 2.1 / 100 = 21mA. Though they are
still not bright, which is probably due to the 'refresh rate'.

Greetings

Above the (visual) fusion frequency, brightness is mostly a function
of average current. See my PS. BTW, 1.7V sounds a bit low for the LED,
even for red, but 0.6V is high for Vce(sat) if the transistor is
actually saturated.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

Oke thanx, I measured the 1.7, and the 0.6 is told to me. Someone told
me that when a column is on 1/48 of the time, the current which is
pushed through the LED must be 48 times the normal value.. well.. this
can't be used in this case.. But what if I push 50mA trhough the LEDs
will that be oke?

Greetings

Spehro Pefhany
electronics forum Guru

Joined: 01 May 2005
Posts: 2326

Posted: Mon Jun 26, 2006 10:04 pm Post subject: Note

Re: Multiplexing LEDs - calculating resistor value

On 26 Jun 2006 14:38:43 -0700, the renowned "roxlu"
<diederickh@gmail.com> wrote:

Quote:

Spehro Pefhany wrote:
On 26 Jun 2006 14:20:28 -0700, the renowned "roxlu"
diederickh@gmail.com> wrote:

Spehro Pefhany wrote:
On 26 Jun 2006 13:40:26 -0700, the renowned "roxlu"
diederickh@gmail.com> wrote:

Hi all,

My previous post about speeding up multiplexing resulted in a display
with
8x48 LEDs. As I suggested and read, the LEDs are becomming less bright
when more than 2 LEDs are on in a matrx of 8x48 >=.

Can anyone help me to calculate the resistor value I need to get bright
LEDs.
Or, how can I caclulate this? Now I'm using a 100R per column which is
probably
to much.

Here is a small part of the circuit to show the current limiting
resistor of 100R.
http://imagebin.org/5461

Thanks
Greetings

Arrange it so that 48 LEDs are on at once (48 columns and 8 rows). You
want the resistor to pass about 8 times the average current you would
like for the LEDs. eg. if they are reasonably bright at 3mA, you want
about 24mA through the resistor. Just use G.S. Ohm's law, from the 5V
supply minus the LED forward drop (at 24mA, not 3mA) minus two
Vce(sat) drops.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

Hi Spehro,

Thanks for your reply, but thats what I did, and its not giving me good
results. I calculated the value like this: 5v - 1.7v (led) - 1.2 (2x0.6
or the transistors) = 2.1 v
Now I'm using a 100R, so thats around 2.1 / 100 = 21mA. Though they are
still not bright, which is probably due to the 'refresh rate'.

Greetings

Above the (visual) fusion frequency, brightness is mostly a function
of average current. See my PS. BTW, 1.7V sounds a bit low for the LED,
even for red, but 0.6V is high for Vce(sat) if the transistor is
actually saturated.

Oke thanx, I measured the 1.7, and the 0.6 is told to me. Someone told
me that when a column is on 1/48 of the time, the current which is
pushed through the LED must be 48 times the normal value..

Greetings

I suggest 1/8 duty cycle not 1/48.

Quote:

well.. this
can't be used in this case.. But what if I push 50mA trhough the LEDs
will that be oke?

Maybe. Check the specs. Provided the mux doesn't stop for too long.
Also, each LED will only have an average of < 1mA current, so it may
not be bright enough.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

John Fields
electronics forum Guru

Joined: 24 Mar 2005
Posts: 3260

Posted: Mon Jun 26, 2006 10:47 pm Post subject: Note

Re: Multiplexing LEDs - calculating resistor value

On 26 Jun 2006 14:38:43 -0700, "roxlu" <diederickh@gmail.com> wrote:

Quote:

---
It depends on the LEDs.

There should be some sort of pulse rating given on the data sheet,
and you'll need a separate resistor for each LED if you intend for
them not to interact with each other.

Also, you'd be able to get full apparent brightness out of the LEDs
if you interchanged rows and columns. That is, if you strobed the
rows instead of the columns. That way you'd only have to increase
the current to the LEDs 8 times since you'd be multiplexing them 8:1

The down side is that if you're using 20mA LEDs which can take 160mA
for 1/8 of the time, your row drivers would have to be able to
switch 160mA * 48 lamps ~ 8 amperes. That's not bad at all and if
you use N channel MOSFETs as the row witches and get some devices
with lowish Rds(on) 100mV, say, then they'll only be dissipating

P = EIT = 0.1V * 7.68A * 0.125 = 0.096 watts,

so they won't even get warm.

If you kept your PNP's for column drivers they'd only have to pass
160mA, so with a saturation voltage somewhere around 0.3V they'd
dissipate 48 milliwatts while they were on, but they'd only be on
1/48th of the time, so they'd really only be dissipating a
milliwatt!

If you're using a 5V supply, calculate the LED resistors like this:

Vsupply - Vce(sat) - (Rds(on)* Iled) - Vled
R = ---------------------------------------------
Iled

5V - 0.3V - 0.016V - 1.2V
= ------------------------------ = 21.7 ohms
0.16A

The closest standard 5% value is 22 ohms, so that's what I'd use.

The resistor would have to dissipate:

P = I²RT = 0.16A² * 22R * 0.125 = 0.07 watts,

so a 1/2 watt resistor would be fine.

--
John Fields
Professional Circuit Designer

pbdelete@spamnuke.ludd.lu
electronics forum Guru Wannabe

Joined: 15 Sep 2005
Posts: 273

Posted: Mon Jun 26, 2006 11:03 pm Post subject: Note

Re: Multiplexing LEDs - calculating resistor value

roxlu <diederickh@gmail.com> wrote:

Quote:

My previous post about speeding up multiplexing resulted in a display with
8x48 LEDs. As I suggested and read, the LEDs are becomming less bright
when more than 2 LEDs are on in a matrx of 8x48 >=.

Can anyone help me to calculate the resistor value I need to get bright LEDs.
Or, how can I caclulate this? Now I'm using a 100R per column which is
probably to much.

I have found many multiplexed led displays in public have a way too low
update frequency. Which cause headache all sorts of annoyances..

John Fields
electronics forum Guru

Joined: 24 Mar 2005
Posts: 3260

Posted: Mon Jun 26, 2006 11:08 pm Post subject: Note

Re: Multiplexing LEDs - calculating resistor value

On Mon, 26 Jun 2006 17:47:40 -0500, John Fields
<jfields@austininstruments.com> wrote:

Quote:

---
Oops...

If you do 8:1 multiplexing (see below) you'll only need one resistor
per column since there'll never be more than one LED on, per column,
at any given time, and its power dissipation (since it could be on
100% of the time) will be:

P = I²R = 0.16A² * 22R = 0.56 watts,

so a 22 ohm 1 watt resistor would be a good choice
---

Quote:

Also, you'd be able to get full apparent brightness out of the LEDs
if you interchanged rows and columns. That is, if you strobed the
rows instead of the columns. That way you'd only have to increase
the current to the LEDs 8 times since you'd be multiplexing them 8:1

The down side is that if you're using 20mA LEDs which can take 160mA
for 1/8 of the time, your row drivers would have to be able to
switch 160mA * 48 lamps ~ 8 amperes. That's not bad at all and if
you use N channel MOSFETs as the row witches and get some devices
with lowish Rds(on) 100mV, say, then they'll only be dissipating

P = EIT = 0.1V * 7.68A * 0.125 = 0.096 watts,

so they won't even get warm.

If you kept your PNP's for column drivers they'd only have to pass
160mA, so with a saturation voltage somewhere around 0.3V they'd
dissipate 48 milliwatts while they were on, but they'd only be on
1/48th of the time, so they'd really only be dissipating a
milliwatt!

If you're using a 5V supply, calculate the LED resistors like this:

Vsupply - Vce(sat) - (Rds(on)* Iled) - Vled
R = ---------------------------------------------
Iled

5V - 0.3V - 0.016V - 1.2V
= ------------------------------ = 21.7 ohms
0.16A

The closest standard 5% value is 22 ohms, so that's what I'd use.

The resistor would have to dissipate:

P = I²RT = 0.16A² * 22R * 0.125 = 0.07 watts,

so a 1/2 watt resistor would be fine.

--
John Fields
Professional Circuit Designer

nospam
electronics forum addict

Joined: 14 Feb 2005
Posts: 86

Posted: Mon Jun 26, 2006 11:15 pm Post subject: Note

Re: Multiplexing LEDs - calculating resistor value

"roxlu" <diederickh@gmail.com> wrote:

Quote:

My previous post about speeding up multiplexing resulted in a display
with
8x48 LEDs. As I suggested and read, the LEDs are becomming less bright
when more than 2 LEDs are on in a matrx of 8x48 >=.

Here is a small part of the circuit to show the current limiting
resistor of 100R.
http://imagebin.org/5461

It sounds like you are still driving the matrix wrong. With the circuit you
linked only one bit in the row shift register should be on so only one LED
connected to that 100R resistor will be on at a time and that LED gets the
whole 21 mA.

When you start driving the matrix the right way remember all 48 LEDs could
be on and the row drivers will have to source just over 1 Amp which the
transistors you chose may not be capable of.

--

John Fields
electronics forum Guru

Joined: 24 Mar 2005
Posts: 3260

Posted: Mon Jun 26, 2006 11:17 pm Post subject: Note

Re: Multiplexing LEDs - calculating resistor value

On Mon, 26 Jun 2006 17:47:40 -0500, John Fields
<jfields@austininstruments.com> wrote:

Quote:

The down side is that if you're using 20mA LEDs which can take 160mA
for 1/8 of the time, your row drivers would have to be able to
switch 160mA * 48 lamps ~ 8 amperes. That's not bad at all and if
you use N channel MOSFETs as the row witches and get some devices
with lowish Rds(on) 100mV, say, then they'll only be dissipating

P = EIT = 0.1V * 7.68A * 0.125 = 0.096 watts,

so they won't even get warm.

---
Aarrgghhhhh!!!

Lowish Rds(on) 100 milli _ohms_,

P = I²RT = 7.68A² * 0.1R * 0.125 = 0.737 watts,

so it'll get warm enough to need a heat sink.

--
John Fields
Professional Circuit Designer

roxlu
electronics forum beginner

Joined: 01 Jun 2006
Posts: 37

Posted: Mon Jun 26, 2006 11:28 pm Post subject: Note

Re: Multiplexing LEDs - calculating resistor value

John Fields wrote:

Quote:

On 26 Jun 2006 14:38:43 -0700, "roxlu" <diederickh@gmail.com> wrote:

Oke thanx, I measured the 1.7, and the 0.6 is told to me. Someone told
me that when a column is on 1/48 of the time, the current which is
pushed through the LED must be 48 times the normal value.. well.. this
can't be used in this case.. But what if I push 50mA trhough the LEDs
will that be oke?

---
It depends on the LEDs.

There should be some sort of pulse rating given on the data sheet,
and you'll need a separate resistor for each LED if you intend for
them not to interact with each other.

Also, you'd be able to get full apparent brightness out of the LEDs
if you interchanged rows and columns. That is, if you strobed the
rows instead of the columns. That way you'd only have to increase
the current to the LEDs 8 times since you'd be multiplexing them 8:1

The down side is that if you're using 20mA LEDs which can take 160mA
for 1/8 of the time, your row drivers would have to be able to
switch 160mA * 48 lamps ~ 8 amperes. That's not bad at all and if
you use N channel MOSFETs as the row witches and get some devices
with lowish Rds(on) 100mV, say, then they'll only be dissipating

P = EIT = 0.1V * 7.68A * 0.125 = 0.096 watts,

so they won't even get warm.

If you kept your PNP's for column drivers they'd only have to pass
160mA, so with a saturation voltage somewhere around 0.3V they'd
dissipate 48 milliwatts while they were on, but they'd only be on
1/48th of the time, so they'd really only be dissipating a
milliwatt!

If you're using a 5V supply, calculate the LED resistors like this:

Vsupply - Vce(sat) - (Rds(on)* Iled) - Vled
R = ---------------------------------------------
Iled

5V - 0.3V - 0.016V - 1.2V
= ------------------------------ = 21.7 ohms
0.16A

The closest standard 5% value is 22 ohms, so that's what I'd use.

The resistor would have to dissipate:

P = I²RT = 0.16A² * 22R * 0.125 = 0.07 watts,

so a 1/2 watt resistor would be fine.

--
John Fields
Professional Circuit Designer

So, I should go fo rthe 22ohms when using 8x48? And if so, can I use
simple
NPN transistors to sink to GND, like this: http://imagebin.org/5461

Greetings.

Google

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