Re: how to control LED array? (follow-up)

On Fri, 27 May 2005 18:27:05 -0500, Michael Noone
<mnoone.uiuc.edu@xxxxxxxxx> wrote:

>Hi - this is a follow-up to this thread:
>
>http://groups-
>beta.google.com/group/sci.electronics.basics/browse_frm/thread/679679383
>cd7eab0/cb7e229d9bca99d5
>
>I've been working on finalizing the design. I've drawn up a schematic in
>Eagle of what I think should work:
>
>https://netfiles.uiuc.edu/mnoone/www/Electronics/LED-Array.jpg
>
>This is essentially identical to what John Fields was suggesting, except
>I've modified some resistor values as I have now chosen LEDs:
>
>http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=7513499883 (3.3v
>forward voltage, 20-30ma typical current)
>
>The design of this array calls for only one column to be on at any given
>time. The columns are active high, and the rows are active low. The goal
>is to saturate the transistors. The value for R1-R16 was chosen by
>dividing the typical base-emitter saturation voltage by the base
>current, thus 0.85/.015 = 56.67, or about 56 ohms. These values were
>found on page two of both the 2N4401 and 2N4403 datasheets:
>
>http://www.fairchildsemi.com/ds/2N/2N4401.pdf
>http://www.fairchildsemi.com/ds/2N/2N4403.pdf
>
>The value for R16-R24 was chosen by taking (V+ - 2*VCesat - Vled)/ILED =
>(5 - 2*0.4 - 3.3)/.03 = .9/.03 = 30 ohms. A current of 30ma was chosen
>because only one led on each row will be on at any given time, and as
>each column will only be on for 1/8 of the time - I feel it is best to
>use the maximum current allowed.

---
If you want to drive the LEDs at 30mA, then the column driver needs to
pass 240mA with all the LEDs in that column ON.

For a forced beta of 10, Fairchild spec's the 2N4401 Vbe(sat) at 0.9V
with about a 300mA load, so the base resistor would be:

Vcc - Vbe 4.1V
R = ----------- = -------- ~ 171 ohms
Ib 0.024A

Since there's nothing critical about that forced beta and you've got
an awful lot of gain available, I'd round up to the next standard 5%
value, which is 180 ohms.

With about 24mA being pushed through the resistor it'll be
dissipating:

E² 16.8
P = --- = ------ ~ 0.09W
R 180R

but since you'll be doing 8:1 multiplexing it'll only be dissipating
1/8 of that, or about 12 milliwatts, so 1/4 watt thru-hole resistors
will be fine if you're going that route.

With a collector current of about 300mA, Fairchild spec's Vce(sat) at
about 0.2V for a 2N4401, so the multiplexed transistors will be
dissipating:

(Vcc - Vce) Ic 4.8V * 0.24A
P = ---------------- = -------------- = 0.144 watts
8 8

Which the 2N4401s can easily handle.


Looking at the row drivers; even if you're multiplexing, all the LEDs
in a row could be on one right after the other, so each of the 2N4403s
will have to handle a maximum of 30mA of collector current. With 30
mA of collector current and a forced beta of 10, Fairchild spec's the
2N4403's Vbe(sat) at about 0.8V, so the resistance of the base
resistor would be:

Vcc - Vbe 4.2V
R = ----------- = -------- = 1400 ohms
Ib 0.003A

Then, since we've got a drop across the LED of about 3.3V at 30mA, a
drop of about 0.05V across the row driver at 30mA, and a drop of about
0.2V across the column driver at 240mA, the LED's current limiting
resistor needs to be:


Vcc - (Vled + Vce1 + Vce2) 1.45V
R = ---------------------------- = ------- ~ 48.3 ohms
Iled 0.03A

With standard 5% values of 47 ohms and 51 ohms available, it might be
tempting to go for 47 ohms, but in many cases that 30mA current
specified for white LEDs is the _absolute maximum_, so it would be
prudent to round up to 51 ohms.

Also, not knowing what your I/Os look like, I've made the assumption
that the drive from your µC is rail-to-rail and haven't considered the
drop which will be encountered when the ports actually have to supply
current into the row and column driver bases, particularly the column
drivers, so you may want to rework the numbers with that in mind.
>
>Oh - lastly, if anyone is curious as to the odd placement of gnd on the
>row connector and +5v on the column connector (as the layout would
>probabaly be nicer the other way around) - I did this as a reminder that
>the rows are active low and the columns are active high.
>
>So how does everything look? Some of my calculations and final numbers
>differ from the original thread - thus I thought I should check back to
>see if I'm doing something wrong. Thanks so much,

---
See above, and you're welcome. :-)

--
John Fields
Professional Circuit Designer
.