Triac, Transistor, or relay?

B

Baronvonrex420

Guest
I want to use the outputs from a 7 seg driver (74LS47 OR 7447) to light a LED
display AND to send a larger current (12vdc) to an LED array. Each seg of the
array is comprised of 6 LEDS, 2 strings of 3 in series. I would assume I
couldn't do this with the driver IC alone.

What is the best way to accomplish this? Triac, Transistor? Cost is a major
factor.

Also, please explain how I would route the pins, for instance, base to driver
emitter to ground, collector to power? Any additional components required such
as resistors or caps?

I need to get this project completed, so apologies in advance for the
newbieness!

Thanks,

Greg
http://www.rts-group.com
 
Baronvonrex420 wrote:
I want to use the outputs from a 7 seg driver (74LS47 OR 7447) to light a LED
display AND to send a larger current (12vdc) to an LED array. Each seg of the
array is comprised of 6 LEDS, 2 strings of 3 in series. I would assume I
couldn't do this with the driver IC alone.

What is the best way to accomplish this? Triac, Transistor? Cost is a major
factor.

Also, please explain how I would route the pins, for instance, base to driver
emitter to ground, collector to power? Any additional components required such
as resistors or caps?

I need to get this project completed, so apologies in advance for the
newbieness!

Thanks,

Greg
http://www.rts-group.com
------------------
Click to expand...
Parallel the LED with an optocoupler, or a PNP transistor base.

-Steve
--
-Steve Walz rstevew@armory.com ftp://ftp.armory.com/pub/user/rstevew
Electronics Site!! 1000's of Files and Dirs!! With Schematics Galore!!
http://www.armory.com/~rstevew or http://www.armory.com/~rstevew/Public
 
I want to use the outputs from a 7 seg driver (74LS47 OR 7447) to light a LED
display AND to send a larger current (12vdc) to an LED array. Each seg of
the
array is comprised of 6 LEDS, 2 strings of 3 in series. I would assume I
couldn't do this with the driver IC alone.

What is the best way to accomplish this? Triac, Transistor? Cost is a major
factor.

Also, please explain how I would route the pins, for instance, base to driver
emitter to ground, collector to power? Any additional components required
such
as resistors or caps?

I need to get this project completed, so apologies in advance for the
newbieness!

Thanks,

Greg
Click to expand...
The ULN2003 is made for your application. It's a 16 pin IC with 7 darlington
drivers, transistors and resistors built in. If you put a logic 1 at the
input, the darlington output goes on and is capable of sinking enough current
to drive your LED displays. The IC is relatively cheap, and commonly
available. This will certainly be a better solution than discrete components,
even if you cost the transistors and resistors at zero. Make sure you use
current limiting resistors on your LEDs.

http://www.allegromicro.com/datafile/2001.pdf

Good luck.
Chris
 
You didn't specify what the total current per segment would be. I'm assuming
that its over 24mA (SN74ls47) and 40mA (SN7447). If not, then you can use
the driver directly as the outputs can handle up to 15V.

If you exceed the segment current, then there are a variety of ways to go.
If the total segment current is less than 120mA then you can use 3 SN7447
chips. One for the 7-seg display, and the other two for each of the strings
of 3 leds. Or you can use the ULN2003 driver (as suggested by another).
However, don't forget to switch over to the SN7448 or SN47LS48 as it will
drive the segment inputs of the ULN2003 correctly.

First, answer the questions, and we can then proceed.

Dana Frank Raymond



"Baronvonrex420" <baronvonrex420@aol.com> wrote in message
news:20030928202952.10934.00000255@mb-m12.aol.com...
I want to use the outputs from a 7 seg driver (74LS47 OR 7447) to light a
LED
display AND to send a larger current (12vdc) to an LED array. Each seg of
the
array is comprised of 6 LEDS, 2 strings of 3 in series. I would assume I
couldn't do this with the driver IC alone.

What is the best way to accomplish this? Triac, Transistor? Cost is a
major
factor.

Also, please explain how I would route the pins, for instance, base to
driver
emitter to ground, collector to power? Any additional components required
such
as resistors or caps?

I need to get this project completed, so apologies in advance for the
newbieness!

Thanks,

Greg
http://www.rts-group.com
Click to expand...
 
On 29 Sep 2003 00:29:52 GMT, baronvonrex420@aol.com (Baronvonrex420)
wrote:

I want to use the outputs from a 7 seg driver (74LS47 OR 7447) to light a LED
display AND to send a larger current (12vdc) to an LED array. Each seg of the
array is comprised of 6 LEDS, 2 strings of 3 in series. I would assume I
couldn't do this with the driver IC alone.

What is the best way to accomplish this? Triac, Transistor? Cost is a major
factor.

Also, please explain how I would route the pins, for instance, base to driver
emitter to ground, collector to power? Any additional components required such
as resistors or caps?

I need to get this project completed, so apologies in advance for the
newbieness!
Click to expand...
---
Assuming 20mA red LEDs:




+12
|
|
E
LS47 SEG OUT->---[1K0]---B 2N4403
C
|
+----->>------+------+
| | |
[270] [510R] [510R]
| | |
[LED] [LED] [LED]
|K |K |K
| [LED] [LED]
| |K |K
GND GND GND

The 270 ohm resistor will dissipate about 10V*20mA = 200mW, so you'd be
OK with a 1/4 watt resistor, but you might want to use a 1/2 watter just
to keep its temperature down. The 510 ohm resistors will dissipate
about 8V*20mA = 160mW, so 1/4 watters will be fine.


If you wanted to use high-efficiency 2mA LED's like HLMP-4700's, you
could get rid of the transistor and the base resistor and just do this:

+12
|
+----->>------+------+
| | |
[5k1] [3k1] [3k1]
| | |
[LED] [LED] [LED]
|K |K |K
| [LED] [LED]
| |K |K
| [LED] [LED]
| |K |K
LS47 SEG OUT->---+----->>------+------+

All the resistors can be 1/4 watters, no problem.

--
John Fields
 
On Mon, 29 Sep 2003 12:07:53 -0500, John Fields
<jfields@austininstruments.com> wrote:


---
Assuming 20mA red LEDs:




+12
|
|
E
LS47 SEG OUT->---[1K0]---B 2N4403
C
|
+----->>------+------+
| | |
[270] [510R] [510R]
| | |
[LED] [LED] [LED]
|K |K |K
| [LED] [LED]
| |K |K
GND GND GND

---
Click to expand...
Oops, kinda...

It'll work, but I forgot to keep the base from floating.

This is better:


+12 +12
| |
[10k] |
| E
LS47 SEG OUT->-+-[1K0]---B 2N4403
C
|
+----->>------+------+
| | |
[270] [510R] [510R]
| | |
[LED] [LED] [LED]
|K |K |K
| [LED] [LED]
| |K |K
GND GND GND


--
John Fields
 
On Mon, 29 Sep 2003 13:24:50 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Mon, 29 Sep 2003 12:07:53 -0500, John Fields
jfields@austininstruments.com> wrote:


---
Assuming 20mA red LEDs:




+12
|
|
E
LS47 SEG OUT->---[1K0]---B 2N4403
C
|
+----->>------+------+
| | |
[270] [510R] [510R]
| | |
[LED] [LED] [LED]
|K |K |K
| [LED] [LED]
| |K |K
GND GND GND

---
Oops, kinda...

It'll work, but I forgot to keep the base from floating.

This is better:


+12 +12
| |
[10k] |
| E
LS47 SEG OUT->-+-[1K0]---B 2N4403
C
|
+----->>------+------+
| | |
[270] [510R] [510R]
| | |
[LED] [LED] [LED]
|K |K |K
| [LED] [LED]
| |K |K
GND GND GND
Click to expand...
---
Hmm...
Maybe the third time will be the charm.

+12 +12
| |
[10K] |
| E
LS47 SEG OUT->-+-[2K]----B 2N4403
C
|
+----->>------+------+
| | |
[510] [300] [300]
| | |
[LED] [LED] [LED]
|K |K |K
GND [LED] [LED]
|K |K
[LED] [LED]
|K |K
GND GND


--
John Fields
 
Thanks all! I have a couple of great plans and a couple of good backups to
work with, I always appreciate a "Plan B". I can get this project off the
ground now!

Greg

baronvonrex420@aol.com
 
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