Controlling 10V 7 Segment LEDs

R

Robert Monsen

Guest
Ok, I've got a simple scheme for controlling these using an 'unnamed 5V
device' :)

The problem is that I need to be able to switch the 10V common anode, and
ground the a-g inputs, all using 5V outputs from a shift register. An open
collector buffer like a 7407 works for the latter.

http://home.comcast.net/~rcmonsen/steve/steve.html

Page down to the second section. It shows a schematic for powering the
common anode using two transistors and three resistors.

However, this is a pretty common thing to want to do. I'm guessing there is
an IC that can do it for me without the added transistors/resistors. I spent
a few moments in the philips device tree looking for something, but didn't
find anything that would work.

I looked at analog switches, but their 'on' voltage is about 7V for a Vcc of
10V.

I looked at CMOS buffers, but they have the same problem.

Is there a part that is cheaper than the 6 times (two transistors/3
resistors/extra solder and wire runs)?

(Of course, the other way is simply to drive each 7 segment from its own
shift register. However, that means 6 shift registers, and 8 open collector
buffers, and their wire runs)

Thanks,
Bob Monsen
 
"Robert Monsen" <postmaster@BulkingPro.com> wrote in message
news:F9mlb.837748$YN5.940299@sccrnsc01...
Ok, I've got a simple scheme for controlling these using an 'unnamed 5V
device' :)

The problem is that I need to be able to switch the 10V common anode, and
ground the a-g inputs, all using 5V outputs from a shift register. An open
collector buffer like a 7407 works for the latter.

http://home.comcast.net/~rcmonsen/steve/steve.html

Page down to the second section. It shows a schematic for powering the
common anode using two transistors and three resistors.

However, this is a pretty common thing to want to do. I'm guessing there
is
an IC that can do it for me without the added transistors/resistors. I
spent
a few moments in the philips device tree looking for something, but didn't
find anything that would work.

I looked at analog switches, but their 'on' voltage is about 7V for a Vcc
of
10V.

I looked at CMOS buffers, but they have the same problem.

Is there a part that is cheaper than the 6 times (two transistors/3
resistors/extra solder and wire runs)?

(Of course, the other way is simply to drive each 7 segment from its own
shift register. However, that means 6 shift registers, and 8 open
collector
buffers, and their wire runs)

Thanks,
Bob Monsen
Click to expand...

You might use a 7406 inverting OC buffer on the digit lines. When the
shift register output goes low, the buffer output will be open collector.
From there you can pull it up to the digit supply voltage with a single
resistor per digit.
 
"Lord Garth" <LGarth@Tantalus.net> wrote in message
news:SAmlb.1125$G16.155@newssvr24.news.prodigy.com...
You might use a 7406 inverting OC buffer on the digit lines. When the
shift register output goes low, the buffer output will be open collector.
From there you can pull it up to the digit supply voltage with a single
resistor per digit.
Click to expand...
Yes, that would work. Lets see:

Assume a 12V supply, I need something like 50mA through the LEDs, so I drop
the voltage to ~10V using a 47 ohm resistor. Then, that 40 ohms goes across
the entire 12V, so it dissipates 300mW, so a 1/2W resistor will work. Also,
I can use the same part for both the anode and cathode. I think this will
work.

Its even nicer that I can use the same part for both the anode and cathode.
I'll remember this trick.

Thanks for the help...

Regards,
Bob Monsen
 
"Robert Monsen" <postmaster@BulkingPro.com> wrote in message
news:51plb.850379$uu5.152700@sccrnsc04...
"Lord Garth" <LGarth@Tantalus.net> wrote in message
news:SAmlb.1125$G16.155@newssvr24.news.prodigy.com...

You might use a 7406 inverting OC buffer on the digit lines. When the
shift register output goes low, the buffer output will be open
collector.
From there you can pull it up to the digit supply voltage with a single
resistor per digit.


Yes, that would work. Lets see:

Assume a 12V supply, I need something like 50mA through the LEDs, so I
drop
the voltage to ~10V using a 47 ohm resistor. Then, that 40 ohms goes
across
the entire 12V, so it dissipates 300mW, so a 1/2W resistor will work.
Also,
I can use the same part for both the anode and cathode. I think this will
work.

Its even nicer that I can use the same part for both the anode and
cathode.
I'll remember this trick.

Thanks for the help...

Regards,
Bob Monsen
Click to expand...
Not. Posted too quickly.

Its actually 300mA through the resistor, which means the dissipation is .3 *
12= 3.6W. Ouch

P = 50mA * V^2/(V - 10)... so there is a minima for P at 20V, and thus the
absolute minimum for this technique, given the preconditions, is 2W.

It still might be ok, but there also might be another way. Also, the 7406
would have to sink 100mA. Only one pin is on at a time, though, so this
should be OK.

Regards,
Bob Monsen
 
On Wed, 22 Oct 2003 07:06:03 GMT, "Robert Monsen"
<postmaster@BulkingPro.com> wrote:


Not. Posted too quickly.

Its actually 300mA through the resistor, which means the dissipation is .3 *
12= 3.6W. Ouch

P = 50mA * V^2/(V - 10)... so there is a minima for P at 20V, and thus the
absolute minimum for this technique, given the preconditions, is 2W.

It still might be ok, but there also might be another way. Also, the 7406
would have to sink 100mA. Only one pin is on at a time, though, so this
should be OK.
Click to expand...
Use the 7406 outputs to drive PNP transistors. You will need a
resistor from base to +10 to turn the transistor off when the 7406 is
OC, and a resistor from base to 7406 output to limit base current when
the 7406 is on.



--
Peter Bennett VE7CEI
GPS and NMEA info and programs: http://vancouver-webpages.com/peter/index.html
Newsgroup new user info: http://vancouver-webpages.com/nnq
 
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