transistor IC

Need to drive three of these with Arduino: Single Hi-Red 4.0 inch CC 7-Seg LED Display Data

I'm thinking use a transistor to convert the Arduino's 5V output to a 10V @ 25mA output going to each display segment. Vs will be 12 volts, so 400 ohm resistor on the Vs line I figure?

I thinking connect digital pin to base, Vs power line to collector, and the LED segment to emitter.

But for 14 segments, it would be better if I could use a couple of IC's versus 14 transistors. Do these exist? The closest I can find is a common-emitter 8x darlington, http://www.mouser.com/ProductDetail/STMicroelectronics/ULN2803A/?qs=sGAEpiMZZMutXGli8Ay4kAHmCxIxFTXSfbpCT5ik4XI%3D

It's a common cathode so that makes it harder to drive.
You need to look for a high sided driver like the Allegro 2981

They make a CA but only in a 5 inch, and I think that would be too big for what I'm doing.

What makes a CC harder to drive than a CA? If it was a CA, could I just write the analog pins to be zero volts, or would the 10V Vs just zap the sockets? (The arduino running on a 5V regulator off same 12V power brick)

So for the driver you specified, pin 9 would be my 10V, the inputs on left woult be the arduino 5V signal, and the outputs would be the 10V going to each segment?

Mouser doesn't seem to have that part number, not sure what to C/R to.

(I prefer ordering from mouser because they're 3 miles away, no shipping, no wait time)

The array I linked to is similar to what you have specified, except the diodes are pointing the opposite direction.

datasheet: http://www.st.com/internet/com/TECHNICAL_RESOURCES/TECHNICAL_LITERATURE/DATASHEET/CD00000179.pdf

How could I modify a circuit about this chip to perform the desired function?

You would need to do it like this:

What makes a CC harder to drive than a CA?

Because with a common anode all you have to do is to pull down to ground with a transistor or transistor array, these are quite common and cheap as you have found.

With common cathode you have to have a source of voltage generated from the arduino. As this is a much rarer situation and needs more electronics to achieve these are more expensive.

Many beginners are in the mindset that common cathode is the sensible straightforward thing to have, in this they are mistaken.

CrossRoads circuit technique is called shunt switching, and consumes more power when the LEDs are off than when they are on.

Grumpy_Mike:

What makes a CC harder to drive than a CA?

Because with a common anode all you have to do is to pull down to ground with a transistor or transistor array, these are quite common and cheap as you have found.

With common cathode you have to have a source of voltage generated from the arduino. As this is a much rarer situation and needs more electronics to achieve these are more expensive.

Many beginners are in the mindset that common cathode is the sensible straightforward thing to have, in this they are mistaken.

CrossRoads circuit technique is called shunt switching, and consumes more power when the LEDs are off than when they are on.

The ULN2003 is 67 cents at mouser for 7 array, the ULN2803 is $1 at mouser for 8 array.

It takes more current because having the arduino output turned on is what turns off the LED segmant?

How many mA are we talking about? This is going on a 12V/1A power brick with voltage regulators so I think I would be fine.

So if I understand that, the darlington array "steals" the ground and re-routes it around the LED segment, so it does not light? Arduino 5V signal goes to base, LED Vs goes to collector, and emitter goes to ground.

It takes more current because having the arduino output turned on is what turns off the LED segmant?

Yes.

the darlington array "steals" the ground and re-routes it around the LED segment, so it does not light?

Yes the darlington shorts out the LEDs.

How many mA are we talking about?

When the LEDs are on there is 20mA going through them. But when the LEDs are off there is:-
12 / 137.5 = 87.2 mA going through the darlington, and it is burning 12 X 0.0872 = 1.05 Watts, so it needs to be a big resistor.

  1. Check newark, digikey, dipmicro.com for prices as well.

  2. Takes more current in the sense that current is always on, vs only being on when the LED is on.

  3. 20-25ma/segment if you are running full brightness. So 175mA per digit to display an 8, 200 with decimal point.

  4. Yes.

The other option is to put a PNP transistor, or P-Channel MOSFET, between +12V and the individual segment Anodes,
i.e. “roll your own” driver.
The Allegro 2981 part does that in an array of 8 drivers; unfortunately it has a 1.8V drop from +12V to the output.

Grumpy_Mike:

It takes more current because having the arduino output turned on is what turns off the LED segmant?

Yes.

the darlington array "steals" the ground and re-routes it around the LED segment, so it does not light?

Yes the darlington shorts out the LEDs.

How many mA are we talking about?

When the LEDs are on there is 20mA going through them. But when the LEDs are off there is:-
12 / 137.5 = 87.2 mA going through the darlington, and it is burning 12 X 0.0872 = 1.05 Watts, so it needs to be a big resistor.

The darlington can handle that much current, yes?

150 ohms @ 2 watts: 32 cents

137 Ohms @ 2 watts: 54 cents

The darlington can handle that much current, yes?

Yes they can take 500mA each with a total per package not exceeding about 650mA.

Grumpy_Mike:

The darlington can handle that much current, yes?

Yes they can take 500mA each with a total per package not exceeding about 650mA.

So if I were display a "1", 5 segments would be turned off @ ~95mA (add some overhead) = 475 mA

So between the Arduino, voltage regulators, darlington chips, and LED boards (remember I'm driving 2 boards), if I displayed a 1 on both boards, I'd be pulling about 1 3/8 amps. So I guess a 12VDC @ 2 amp power brick it is.

So I said 2 board thinking I could wire the third differently, but I can't.

I forgot a filtering capacitor and free-wheel diodes on the input lines before the regulators.

This is pretty much what I'm trying to do:

The flip switch makes the first digit either a 5 or an 8

The period is always on (hard wired)

The first 10 position rotary switch makes the second digit be 0 thru 9

The second 10 position switch makes the third digit be 0 thru 9

Okay, that much IO you're gonna need some shift registers.
These are available at mouser

Output shift register with high voltage/high current sink output driver

If you have a part number for the rotary switches, can suggest parts for that too.

Rotary switches like these panel mount parts?

Probably wire the 1-2-4-8 pins in parallel and alternately drive the common pins from unique IO pins to read the state of the switch.

So what happens after you read the switches & display the digits?

CrossRoads:
Okay, that much IO you’re gonna need some shift registers.
These are available at mouser
http://www.ti.com/lit/ds/symlink/tpic6595.pdf
Output shift register with high voltage/high current sink output driver

If you have a part number for the rotary switches, can suggest parts for that too.

Rotary switches like these panel mount parts?
http://www.mouser.com/ProductDetail/NKK-Switches/FR01AR10PB-W-S/?qs=sGAEpiMZZMtHyRFzBQ9JV3U0zgiV7QHI

Probably wire the 1-2-4-8 pins in parallel and alternately drive the common pins from unique IO pins to read the state of the switch.

So what happens after you read the switches & display the digits?

more down the lines of this, but same theory: http://www.mouser.com/ProductDetail/Alpha-Taiwan/SR2512F-0110-19R0B-E9-N-W/?qs=sGAEpiMZZMvNbjZ2WlReYnqYHrQfuERsnKqDF70Kn5Q%3D

So what exactly is the shift register vs the Darlington array?

What do you mean 1-2-4-8 pins?

This is just to print a x.xx format number where the first digit is selectable between 2 known values, and last 2 digits are selectable.

A shift register is a device you can store an 8-bit value in.
The darlington array is just a buffer that can drive more current.
The tcip6595 combines both into 1 part.

If you look at the data sheet for the rotary switch, the pinouts and truth table, you will see the 1-2-4-8 output pins, the switch makes contacts internally to output a binary code of the digit selected: 0000, 0001, 0010, 0011,0100, 0101, 0110, 0111,1000,1001 for 0-9.

What this used for? Showing 5:xx and 8:xx as prices? Just curious.

What do you mean 1-2-4-8 pins?

The pins from your rotary switches are labeled as these if they are binary encoded rotary switches, are they?

That’s over my head. I thought it was like a source switch, where each position is on one at a time, to a common.

CrossRoads:
What this used for? Showing 5:xx and 8:xx as prices? Just curious.

Dial-in's actually.

Just to mess around with something half useful for a change.

http://www.racedigitaldelay.com/dial-megadial-b.html

http://www.biondoracing.com/MegaPanel.shtml

it wont look as pretty, but if I can make it for much cheaper, why not?

Our drag car is 5.55 in the 1/8 mile and 8.67 in the 1/4, so that's how the first digit is chosen. But the 2 decimal places can be anything based on the conditions.

Okay, browsed the links, I don't understand what the tower does with the number.

Your code will read the 4 pins, result is:

digit1 = (8 * 8bit + 4* 4bit + 22bit + 11bit);

so (1001) = 9,
1000 = 8
0111 = 7
0110 = 6
0101 = 5
0100 = 4
0011 = 3
0010 = 2
0001 =1
0000 = 0

Pretty basic stuff.