Adding the 0.1uF caps across each IC has caused it not to work.
There's the occasional flicker of a digit when you either connect
or disconnect the power supply, but the rest of the time, at least the display is off.

In the diagram on the website, the 4050 ICs are powered by 12 Volts.
Is this correct? I don't see any other way to power the displays.
The displays are common cathode, not common anode.

 

Ok, I added 470uF caps at the input of each reg, and 47uF caps at the outputs.
I connected the 4050 ICs to the 7812 regulator again.
The result is the same as before.

The micro is definitely working but the display is only lit for a moment when
the power is disconnected.

If I used a pot at the power supply again it would probably work.
If not, at least I can undo this stuff to get it back to a working state.
I would have thought this would be much more simple.

Not that it reveals much detail, but...

 

Are you using the 7809 or 7812?

What is the input voltage?

What is the output voltage on both the 7805 and LED regulator when no display is showing, but it is powered up?

 

I'm currently using a 7812.
The input voltage for testing is around 14 Volts from my vehicle battery where the ignition is turned off.

Unfortunately hobby time is up, and it's bed time (work tomorrow),
but I will try to answer that tomorrow. It would be great to get to the bottom of it.

 

A 7812 would need a 14.5V input before it would output any voltage.

You would either need to look into a low dropout 12V regulator, or a 7809 to work with standard automotive voltages.

 

Ok, thanks, I checked, and my vehicle battery is only 13 Volts when not running.
I will bypass the 7812 for now. I hope you'll check back...
I just don't get a lot of time for this stuff during the week

 

You said that you were using 120 ohm resistors for the LED's. Consider that 9V-2V = 7V, and 7V/120 = 0.058A (58mA). If all 3 displays have all segments on, that's 21 x 58mA = 1.2A (can your 9V regulator supply this amount of current).

If only one segment is on, then the PS input resistor would drop 58ma * 10 ohms = 0.58V. Adding the input diode drop and the resisitor drop, you cannot use a 12V regulator. Consider that if half the total segments are on (~10), then you will be consuming about 10 x 58mA = 580mA. So to drop only 1V at this current, the PS resistor value would be 1V/580mA = 1.7 ohms. The voltage drop across the input resistor is killing the headroom.

However, then you multiply this current times the voltage drop across the regulator to find its dissipation, the regulator is going to dissipate ~12V-9V = 3V, and 3V x 580mA = ~1.75 Watts with only half the segments on, twice that if all segments are on.

 

Currently, the regulator for the LEDs is not even in the equation (I've bypassed it).
Since the circuit as a whole is still not working,
I don't know why I will need to put the pot back on the power supply to get it to work.
The 5 Volt regulator I assume has always done it's job in keeping the microcontroller running.

I can't say I've given this any time since last posting though,
but I will def look into it after the weekend and test the voltages as suggested.

Merry Christmas everyone
Art.

 

GOOD AFTERNOON ENGLISH COMPANION ALSO GOOD BUT NOT WRITE THIS IS NOT THE CASE, LOOK I AM MAKING A SPEEDOMETER AND YOU TE MIRO WENT TO EXCELLENT AND WANTED TO KNOW IF YOU CAN SHARE YOUR WORK WITH ME WHAT YOU APPRECIATE ...
SPEEDOMETER this is my email estevans787@gmail.com
GRASS-COLOMBIA GREETINGS

 

Old thread, but it occurs to me now that I never followed up!

http://www.youtube.com/watch?v=LLZGanh06Ss

It works with the same code, but different driver circuit for the seven segs.