# LED Multiplexing Power Consumption

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#### Suraj143

##### Active Member
I have a LED matrix 8 rows & 24 Columns.Scanning rate is 1/8th rate.& designed it for 4mA average current.

My question is
When the whole circuit power from a 5V regulator what will be current taken?
Is it 4mA X 24 = 96mA ?
or
is it 4mA X 24 X 8 = 768mA?

Can I measure the current by using a digital multi meter? I heard telling that you cannot measure current in multiplexing circuits using a digital multimeter...!!!

#### ronsimpson

##### Well-Known Member
4mA X 24 X 8 = 768mA
Yes.
It is hard to know what is really happening.
If the LED is on 1/8 the time and the average current is 4mA then the LED must be on at 32mA for 1/8 of the time.
When all the LEDs are on; at any point in time there will be 24 LEDs on each at 32mA which equals 768mA.

What I did was use the "4mA average" X 24 X8 = 768
AND
"32mA peak" X 24 = 768. There are only 24 LEDs on at a time.
Both ways doing the math gives us the same "768mA".

#### JimB

##### Super Moderator
I have a LED matrix 8 rows & 24 Columns.Scanning rate is 1/8th rate.& designed it for 4mA average current.
Just to clarify, do you mean that:
Each LED has 4mA when it is conducting?
You illuminate one row of 24 LEDs at a time?
At any one time, up to 24 LEDs will be illuminated?

If that is the case, the maximum current will be 4mA x 24 = 96mA

Can I measure the current by using a digital multi meter? I heard telling that you cannot measure current in multiplexing circuits using a digital multimeter...!!!
The current drawn by the LEDs will vary depending how many LEDs in each row are illuminated.

A DMM will have a difficult time to display a rapidly varying value.
If the DMM has a peak-hold facility, it may be able to catch the maximum current.

You could use a good old analogue meter, but the mechanical inertial of the meter movement will make it display the average current over the mux cycle.

For your testing, you could arrange that all the LEDs are illuminated in all rows, then the current should be more or less constant, give or take any glitches when switching between rows.

JimB

#### gophert

##### Well-Known Member
Peak for the panel will be 96 mA. You should have no trouble finding an AC adapter that can accommodate - most cell phone chargers - even old ones - can handle 5v up to 500 mA (if it has a USB connector. Most newer ones have 2.1 A or more.

If you are doing batteries, and you want to guess/estimate load and battery lifetime., then an analog ammeter as suggested above or add a big capacitor across the supply rails and use s digital meter. Be sure to disconnect one battery lead and put your AmMeter in Series with your supply (all power runs through the meter for ammeter)., if you connect it in parallel with your device power supply, you will creat a dead short and pop your meter (or your fuse).

Cheers.

#### Suraj143

##### Active Member
Hi,
I use one row lit at a time.1/8th scan rate..

I think Ron is correct.The peak power to the LEDs will be 32mA. So the average will be 4mA.
Current drawn from the regulator 32mA X 24 = 768mA.

I cannot touch my 7805.It heats too much.(12V -5V) X 768mA = 5.376Watts

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#### Pommie

##### Well-Known Member
I'd get a heatsink on that 7805 or switch to a switching regulator. An alternative maybe a 6Ω 5W resistor before the regulator.

Mike.

#### gophert

##### Well-Known Member
Hi,
I use one row lit at a time.1/8th scan rate..

I think Ron is correct.The peak power to the LEDs will be 32mA. So the average will be 4mA.
Current drawn from the regulator 32mA X 24 = 768mA.

I cannot touch my 7805.It heats too much.(12V -5V) X 768mA = 5.376Watts

I'm sorry but the math on this post doesn't make sense.

What else is powered on this circuit?

#### Pommie

##### Well-Known Member
The maths makes perfect sense to me. 24 LEDs on at any one time with a current of 32mA means if all LEDs are lit (192) then the current draw will be 768mA. Or did I miss something?

Mike.

#### gophert

##### Well-Known Member
The maths makes perfect sense to me. 24 LEDs on at any one time with a current of 32mA means if all LEDs are lit (192) then the current draw will be 768mA. Or did I miss something?

Mike.
Where do you (and the OP) suddenly pull 32mA from? The current per LED is listed as 4mA. Each row is 24 LEDs. Since only one of those rows is on at a time, all others must be off so, at any given time, the max per row will be 96mA.

Said another way, no need to multiply by 8 rows to get 768 mA) since only one row is on at a time.

#### ronsimpson

##### Well-Known Member
1/8th rate.& designed it for 4mA average current.
Where do you (and the OP) suddenly pull 32mA from? The current per LED is listed as 4mA.
4mA average = 32mA peak

#### Pommie

##### Well-Known Member
Which part of the maths don't you understand? The 32mA comes from the fact that the average current per LED is 4mA and it's only on 1/8th of the time therefore the actual current must be 32mA (4=32/8). Simples.

Mike.

#### gophert

##### Well-Known Member
I think the information given is vague and subjust to mix-interpretation.

There are two ways of looking at "average" on a multiplexed display.

Case One, as Ron and Pommie, assuming all LEDs are on all the time and then divide by 8 (because there are 8 matrix3d rows). Or conversely, multiply the average current (4mA) by 8 to get peak power.

Case Two, a matrix display is rarely illuminating all 24 LEDs in a given row so the max current of each LED may be 10ma but only 40% of the 24 LEDs each row may be on at any one time so the AvERAGE for the row is 4mA per LED.

ronsimpson and Pommie are assuming Case 1, I am assuming Case 2. I still don't have the clear information to decide which case is correct.

#### Pommie

##### Well-Known Member
We're assuming the worst case which is sensible. Do you assume that only 50% of your circuit will be powered so you only need a small power supply?

Mike.

#### gophert

##### Well-Known Member
We're assuming the worst case which is sensible. Do you assume that only 50% of your circuit will be powered so you only need a small power supply?

Mike.

I would say Case 2 is closer to being correct based on this statement from the OP.
.
I cannot touch my 7805.It heats too much.(12V -5V) X 768mA = 5.376Watts
There is no way an LM7805 is surviving 5.3 watts.

It will however get quite hot at 0.7 watts = 7v x 0.096 amps.

#### JimB

##### Super Moderator
I think the information given is vague and subjust to mix-interpretation.
Which is why I asked earlier:
Just to clarify, do you mean that:
Each LED has 4mA when it is conducting?
You illuminate one row of 24 LEDs at a time?
At any one time, up to 24 LEDs will be illuminated?
So far Suraj has not replied.*

But as this ie ETO we can have endless speculation and conjecture based on complete information.

JimB

*On edit: I see that he has, but insists on talking about 1/8 timing and "average" current which I believe to be confusing the issue.

#### Suraj143

##### Active Member
Hi sorry for the lack of details.

I illuminate one row of LEDs at a time (24 Leds).
At any time 24 LEDs will be illuminated.
So when I do this process one at a time speedily for all 8 rows it will display a picture.I call this multiplexing.

My LEDs lights up brightly when I apply 4mA continuos current.So when I multiplexind @ 1/8th rate the brightness decreased by 8 times.So what I do is I increase the current by 8 times.So now my LED gets 32mA peak current and the average will be 4mA.

#### JimB

##### Super Moderator
Thank you for the clarification.

In that case I agree with ronsimpson, the current will be 32 x 24 = 768mA.

JimB

#### Suraj143

##### Active Member
The only problem is 7805 is getting too hot. (12-5) * 768mA = 5.376Watts

Here unable to find modern switchmode converters.

#### JimB

##### Super Moderator
The only problem is 7805 is getting too hot. (12-5) * 768mA = 5.376Watts
Yes, that can be a bit of a problem!

Is there a lower volts supply available in the equipment?
If not, and you are stuck with using 12v, then you could try using a bigger regulator like an LM317 on a big heat sink.
You could also use an external resistor to drop some of the 7v before the 7805. But beware of going below the minimum input voltage of the 7805.

JimB