LED Matrix Questions

[FyberOptic]

Hiya folks. I got into electronics mostly as a result of messing with
digital components, so unfortunately I still find myself confused when
it comes to various analog aspects of circuitry. So I was hoping
someone could help me learn how this works a little better.

Basically, I thought about messing around with an LED matrix.
Preferably small to start out with, like a 4x4. I was going to use a
74LS374 latch to control it, since a.) that's what I already have
laying here, and b.) it's what I was going to use in a different
project soon (one where I'd actually be addressing and writing to the
latch in a microprocessor environment), and thought being able to
operate LEDs with it would be a good test for that when the time comes
(making sure my address decoding worked properly, etc). Anyway, I
know that the LS series can't source much current. About 2.6ma for
this part if I'm reading the datasheet right. But it can sink 24ma
apparently.

The red LEDs I have apparently want 2v at 30ma to power them. A
100ohm resistor would get it down to 30ma if I did my math right (5v
vcc minus 2v forward voltage of LED, divided by 0.030 amps). But now
I'm confused on how many milliamps are going through such a circuit,
even if it's just a simple vcc->resistor->led->gnd setup with nothing
else in the mix. Is it just 30ma total? Or does the LED also have a
resistance that one would combine with the resistor's value to
calculate the amps of the whole circuit?

In the full LED matrix circuit, I'd be using NPN transistors from pins
of the latch to control the voltage going to the rows of LEDs. But
just to clarify, I'd have to be careful of what resistor I used on the
pin coming from the chip to the base of the transistor, right? I'm
assuming that current from the output pin of the IC would mingle with
the current going across the collector and emitter from my power
source, which could possibly result in too much current going through
the LED and/or into the IC pin sinking the other end of the matrix.

I hate that my knowledge of analog electronics leaves me so baffled
sometimes over seemingly simple concepts, but hopefully I'll be able
to cram all of such knowledge into my head eventually!

 

[DJ Delorie]

First off, you probably need only 20mA on the LEDS - 30mA is the *max*
on most LEDS, not the ideal. 20mA is probably plenty bright enough
for you anyway. Read your specs again and be sure.

In a matrix, you normally activate one row/column at a time, and any
LEDs in that row/column. Since the '374 can sink 24mA, use four of
those to connect, say, each column to GND. For the rows, use a PNP or
P-MOSFET transistor for each row, capable of passing 20*4 = 80mA, and
drive the row that way. For the base resistor, you need to do some
math, at least for PNP transistors...

The voltage across the resistor is Vcc - Vol - Vbe. Let's assume Vol
is 0.3v and Vbe is 0.7v, leaving 4v across the resistor.

The Ice current is 80mA.

The base current Ibe is Ice / hFE. For example, if hFE is 100, Ibe =
0.080 / 100 = 800uA

The base resistor is then Vbe / Ibe, or 4 / 0.0008 = 5000 ohms
maximum. I'd go with half that, or 2.2k or so, or even as low as 1k
since the '374 can sink plenty of current.

For the LEDs themselves, you want resistors on the "drive each" edge
(one per column, in this case), not the "drive row/column" edge.
I.e. they go away from the transistors, or the columns in this case.
That way, each resistor is only responsible for one LED at a time.
Size the resistors for Vcc - Vol - Vce - Vled, which is probably about
5 - 0.3 - 0.1 - 2.0 or 2.6v. For 20mA, 2.6/0.02 = 130 ohms.

Note: size resistors based on the *largest* voltage you can calculate
from the specs (i.e. smallest drop across each other device), not the
average.

Also, size the *wattage* of the transistors big enough to handle Ice *
Vce.

If you use a P-MOSFET the math is a little easier. No "base" resistor
is needed, The "Vce" equivalent for math purposes is zero, just choose
one with a big enough Ids for your current needs. The only catch is
that the turn-on voltage is a bit higher than bipolar, so bipolar is
better if Vcc is small enough. Wattage is less of a problem as Vds is
so much smaller, check the specs.

The next thing to consider is that, if you're cycling through the rows
fast enough, each LED is only on for 1/Nth of the time (or not on at
all, of course). You can usually get away with higher currents in the
LEDs and transistors since they're not on all the time.

Size the resistor wattages for the full load, though.

Here's an 8x8 raster module I made with two LEDs in each cell (red and
green), with 16 resistors and the row lines on the back side:

http://www.delorie.com/electronics/usb-gpio/led-raster-1.jpg (or .html)

 

[DJ Delorie]

FyberOptic <fyberoptic@gmail.com> writes:

But if I understand how LEDs work, the voltage isn't what's most
important, it's the amps that affect brightness.

Yup, but if you don't have a current *regulator*, you end up using the
voltage as well as current specs to figure out the load resistor.

So, if you know the LED drops about 2v at 20mA, you figure out what
resistor passes 20mA at whatever voltage is left:

1. Select operating current
2. Look up voltage drop at that current
3. Calculate voltage drop across resistor
4. Choose resistor value according to voltage drop and desired current.

Why PNP transistors instead of NPN?

Because the '374 can sink 24mA but only source 2mA.

So, you can use the '374 to drive the GND side of the LEDs directly,
but you need a transistor to drive the Vcc side.

Couldn't that lead to some potential problems at the start of the
system?

Normally in an MCU-driven system, one of the first things you do is
initialize all your I/O ports. Also, normally you'd tie nOE high to
disable the outputs until the MCU overrides it. Besides, you
shouldn't rely on ANY specific "safe" power-on state. Use the nOE
pins to keep the outputs off until you latch in safe values.

Now that I think of it, though, the LS version has a Voh low enough
that it might (will?) trigger the PNP transistors even when high. I'm
used to using CMOS parts, where the output is symmetrical so it
doesn't matter which way the transistors go. Try it and see.

Of course, you could use an NPN transistor to drive the PNP
transistor. Or an open collector inverter/buffer.

With today's MCUs, the outputs tend to be CMOS so Voh is as close to
VDD as Vol is to GND, and they default to High-Z on reset, so they're
safe and more flexible.

The voltage across the resistor is Vcc - Vol - Vbe.  Let's assume Vol
is 0.3v and Vbe is 0.7v, leaving 4v across the resistor.

What was the Vol, exactly?

V = voltage for logic LOW output
OL

V = voltage for logic HIGH output
OH

 

[David L. Jones]

On Sep 16, 11:17 am, FyberOptic <fyberop...@gmail.com> wrote:

Hiya folks. I got into electronics mostly as a result of messing with
digital components, so unfortunately I still find myself confused when
it comes to various analog aspects of circuitry. So I was hoping
someone could help me learn how this works a little better.

Basically, I thought about messing around with an LED matrix.
Preferably small to start out with, like a 4x4. I was going to use a
74LS374 latch to control it, since a.) that's what I already have
laying here, and b.) it's what I was going to use in a different
project soon (one where I'd actually be addressing and writing to the
latch in a microprocessor environment), and thought being able to
operate LEDs with it would be a good test for that when the time comes
(making sure my address decoding worked properly, etc). Anyway, I
know that the LS series can't source much current. About 2.6ma for
this part if I'm reading the datasheet right. But it can sink 24ma
apparently.

The red LEDs I have apparently want 2v at 30ma to power them.

LED's are current driven devices, and the voltage drop changes with
the current through it.
See here:
http://cq.cx/pics/int-led-vi.png

You don't "power a LED from 2V", you power a LED with X amount of
current.

Depending on the LED, 30mA is most likely the absolute maximum current
rating. Normally you would not go that high.
Most ordinary LED's will work on anywhere from a few mA to say 20mA,
with brightness that is fairly linear with current. With your
particular LED you may find that it is plenty bright enough at say
5mA.

A
100ohm resistor would get it down to 30ma if I did my math right (5v
vcc minus 2v forward voltage of LED, divided by 0.030 amps).

That is correct for a 5V supply.

But now
I'm confused on how many milliamps are going through such a circuit,
even if it's just a simple vcc->resistor->led->gnd setup with nothing
else in the mix. Is it just 30ma total? Or does the LED also have a
resistance that one would combine with the resistor's value to
calculate the amps of the whole circuit?

The LED does not have any extra resistance, it's just the voltage drop
for a given current.
In the case you have given, assuming a 2V drop with a 5V supply, and
100ohms, the current will be 30mA.

Dave.

 

[FyberOptic]

On Sep 15, 11:23 pm, DJ Delorie <d...@delorie.com> wrote:

First off, you probably need only 20mA on the LEDS - 30mA is the *max*
on most LEDS, not the ideal.  20mA is probably plenty bright enough
for you anyway.  Read your specs again and be sure.

Yeah the "DC forward current" rating was in a box with max specs, it
seems. So "forward voltage" is 2v but at 20mA, not 30, according to
this. 3v max. But if I understand how LEDs work, the voltage isn't
what's most important, it's the amps that affect brightness.

In a matrix, you normally activate one row/column at a time, and any
LEDs in that row/column.  Since the '374 can sink 24mA, use four of
those to connect, say, each column to GND.  For the rows, use a PNP or
P-MOSFET transistor for each row, capable of passing 20*4 = 80mA, and
drive the row that way.  For the base resistor, you need to do some
math, at least for PNP transistors...

Why PNP transistors instead of NPN? Couldn't that lead to some
potential problems at the start of the system? I mean, if the latch
pins are all '0' by default, then wouldn't the PNP transistor would be
active at startup and allow the current to pass into the LEDs by
default?. As in, every row and column would be activated at once
until the 4 output latch pins were set to '1'? If all four LEDs were
active in a column at once, wouldn't the IC pin then be sinking too
much current (the whole 80mA), possibly killing that pin or the whole
IC?

The voltage across the resistor is Vcc - Vol - Vbe.  Let's assume Vol
is 0.3v and Vbe is 0.7v, leaving 4v across the resistor.

What was the Vol, exactly?

Thanks for the response!