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PIC18F + 9V 20 mA 4 in. 7-segment display problem

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Hello,

I'm facing a problem here. I have four 7-segment displays, 4 inches each. There are common cathode.
I'm trying to connect them in a PIC18F4550, but I'm failing in build a interface between the microcontroller and the displays (because the voltage level).

Can someone help me?
 

Hi,

A bit more information could help.

Where do you see the problem?
* Finding a 9V high side driver
* multiplexing
* using a current limit circuit...

Best if you can show us your ideas..schematic, description picture...


Klaus
 
Hello Klaus.
I'm building a digital signboard. So I have four 7-segment displays. I'll do the multiplexing using a ULN2003. My problem is to the interface between the PIC and the display. The PIC's output is 5 V, but I need a 9 V output to turn on the segment. I need to convert the PIC's output (5 V) into a 9 V output in order to activate the segments of the display.

I need a IC or circuit that makes this:
circuit.jpg
 

Your block diagram is insufficient.

Each digit needs to drive 4x160mA with constant voltage ( either common Anode or COmmon Cathode) which is multiplexed to give 1/4 of the average max current of 160mA per digit. Then each segment must be current limited by a resistor drop including ESR of driver, Vf of LED @20mA and Vdrop of digit driver ( pref fixed low RdsOn FET.~< 100 mOhm.
Got it? then use Ohm's Law to calculate Rs for each shared segment line. It can be done with 3.3V or 5V.

9V is very inefficient and will not last long with 640mA on 8888. 3.7V LiPo is best bet with 3.7V float charger..

Try to define power again and choice of display CC or CA.
 

Hi SunnySkyguy, thank you for your reply.

I didn't put the resistors in order to simplify the circuit (but I know that I should -- and I'll -- use them). I have no choice in relation of CC or CA, I have four common cathode displays and I should use them. Each segment of display has five internal LEDs, so I need 9 V due the voltage drop along the segment.
My doubt is how can I "transform" 5 V (PIC's output) in 9 V to feed each segment. In addition, I need a safe way to provide the 160 mA when I have the worst case (all segments turned on).

I think that I need a level shifter + current driver, but I don't know if there is a IC that could do this. Or a circuit using transistors :(
 
Last edited:

Hi Klaus. Do you know a IC part number?
 

Hi,

Finding a 9V high side driver
Click to expand...
Look for "high side driver" or "high side switch"

Klaus

- - - Updated - - -

Hi,

Farnell has hundreds..

Choose voltage, current, price, on resistance, speed, features, package.... it's your turn

Klaus
 
YOu cant use ULN200x since they sink 350mA and you need 640mA max.
So choose Nch FETs with logic level input and RdsON for 0.1V drop or 0.1V/640mA =150 mΩ, pref < 100mΩ

For Anode segments you can use a Lithium 9V cell with Pch MOSFET segment drivers or "HIGH SIDE: automotive switch." CUrrent limiting will be not included and determined by internal ESR of Each LED which is approx 16 Ohms per Led or 64 Ohms for the string.( with an unknown variance) above the Red or Yellow THreshold voltage (not Forward voltage)
If you use 12V supply then add Rs for each segment to drop 3V@20mA*4 or 3V/80mA= 38 Ohms. (remember each segment requires 80mA to give 20mA at 25% duty cycled due to MUX. (check spec that this does not exceed ABS MAX.)
 

I would like to thank you for all the help received here :D

Is the UDN2982A sufficient to solve my problem? It's a expensive part :???:
 

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