The 4511 has a blanking input. How about decoding 0000 from your counter, and using that to blank the display?

At half past five in the morning I did not have too much time to think about it.. Thank you for remember me about BI..

 

Hi, I am here again.
The last solution with the diode was not correct.

I have updated my project and I need a last solution or suggestion for the clock, and I got 2 possible solutions in my hands. I have attacched a zip file with the images.

In the first one I putted 2 transistor wich the base of one is connected to an RC and the other base to the switch and R, with the output to the clock.

In the second, I have connected the switch to an RC of 10k & 10uF to the clock.
In the simulator seems it works good, but I would like to have a suggestion for it.

Honestly, I am not sure about how much these solutions are good, or if yes, which is the best one, but I am more convinced about the second..

I still have not putted resistors to the display.. 500 Ohm should be fine for 10mA..

Thank you for your suggestions.

simo_x

 

Did you try my suggestion?

Originally Posted by Ron H
The 4511 has a blanking input. How about decoding 0000 from your counter, and using that to blank the display?

 

Yes,as you can see from the both images.

 

If I'm looking at your schematic correctly, when C3 is finally charged, it will keep Q1 in an ON state. When you close your "ROLL" switch, the output of the 555 will send pulses to Q2, turning it on and off. It seems to me that the result will be shorting the power rails through the two transistors. What am I missing?

 

For calculating your LED current limiting resistors, subtract the typical segment Vf from the supply voltage, and divide the result by the desired current, then use the closest value resistor.

If your typical segment Vf = 2v, your supply is 9v, and you want 10mA current per segment, then:
Rlimit >= (9v-2v) / 10mA = 7/0.01 = 700 Ohms; 750 Ohms is the closest standard value >= 700 Ohms. 680 Ohms would also work.
7v/680 Ohms = 10.3mA

 

Referring to project1.bmp:
R2 at 250 Ohms is too small. 1k Ohms is about as small as you should go with a 9v supply. [eta] This is because pin 7 of the 555 gets pulled to ground to discharge the timing cap; if that resistor is too small, you will have excessive power dissipation as heat in the 555 timer.
So, R1 needs to be increased to 2k Ohms, and C1 reduced to 22uF to keep your timing about the same.

Q2 has no base current limiting resistor. This will be very hard on the 555 timer and Q2. I suggest 470 to 1k Ohms as a base resistor.

BillB3857,
When Q2 turns on, C3 will rapidly drain via Q1's BE junction. However, it would be a good idea to have a resistor (1k-10k) between Q1's collector and +9v.

 

If I'm looking at your schematic correctly, when C3 is finally charged, it will keep Q1 in an ON state. When you close your "ROLL" switch, the output of the 555 will send pulses to Q2, turning it on and off. It seems to me that the result will be shorting the power rails through the two transistors. What am I missing?

Nothing. It's a little detail I have noticed later..

 

Hi SgtWookie, thank you for suggestion. Hower, probably the solution with the two transistor maybe is not the best.

Thank you for all. Now, I am waiting what my professor will tell me about my dices. I hope the best, but if I will need I will be back
You are very gentle.

Again, thank you.

 

Referring to the circuit "project2" from the last zip file I attached here, the professor told me that it will not work..
If it's all bad, how can I give one shot of CLK when power is supplied, taking it connected to the button switch and an RC too?

Or really the circuit with the two transistors should be right?
If it is, I thought to add a R of 10 K to Q1 base and between Q1 and Q2 to prevent the returning 0 V to R3 and C3.
Well, the resistor between Q1 and Q2 should be enough.
So the circuit is as I attacched now.