Introduction: Binary Clock
Here is a simple example how to build cool looking binary 24 hour clock. Red LEDs shows seconds, green LEDs minutes and yellow LEDs hours.
Case contains four buttons to adjust the time. Clock works with 9 volts.
This clock is easy to do and parts cost only few bucks, so it is also cheap to do.
Case contains four buttons to adjust the time. Clock works with 9 volts.
This clock is easy to do and parts cost only few bucks, so it is also cheap to do.
Step 1: Schematic and Parts
I used the blue colored case, because it was cheap and looked good to my eyes.
Parts:
- Clock crystal (Q1) 32.768 kHz.
I think the easiest way to get that crystal is take it from the old wall clock.
- 560pF, 22pF capacitors and one 10M resistor
- 1 x 4060 IC, which is the 14bit ripple counter.
With 32.768 KHz clock crystal this IC gives 2Hz out from the pin number 3
- 3 x 4024 IC
This is 7bit ripple counter
- 2 x 4082 IC
Dual 4-input AND gate
- 1 x 2,1mm plugin
- 17 x led
Red, yellow, green or what ever you like
- 17 x 470 Ohm resistors
I used the 9 Volt supply, so the output from the pins is something around 9V. Typical forward
voltage for these LEDs are about 2 Volts. Let's want that, the current to the LED is something about
0,015 A = 15 mA, then (9-2)V / 0,015A = 466 Ohm -> 470 Ohm is size of resistors.
Now it's time to download 4020 14-stage ripple counter data sheet and we will find that, the
max output current is 4mA =), but it is enough and works anyway.
Parts:
- Clock crystal (Q1) 32.768 kHz.
I think the easiest way to get that crystal is take it from the old wall clock.
- 560pF, 22pF capacitors and one 10M resistor
- 1 x 4060 IC, which is the 14bit ripple counter.
With 32.768 KHz clock crystal this IC gives 2Hz out from the pin number 3
- 3 x 4024 IC
This is 7bit ripple counter
- 2 x 4082 IC
Dual 4-input AND gate
- 1 x 2,1mm plugin
- 17 x led
Red, yellow, green or what ever you like
- 17 x 470 Ohm resistors
I used the 9 Volt supply, so the output from the pins is something around 9V. Typical forward
voltage for these LEDs are about 2 Volts. Let's want that, the current to the LED is something about
0,015 A = 15 mA, then (9-2)V / 0,015A = 466 Ohm -> 470 Ohm is size of resistors.
Now it's time to download 4020 14-stage ripple counter data sheet and we will find that, the
max output current is 4mA =), but it is enough and works anyway.
Step 2: Testing
It is better to test the circuit on bread board before to do the final soldering. When everything works as it should be, it is time to start solder.
HOW IT WORKS:
4060 is 14-bit (/16,384) ripple counter with internal oscillator and it gives with the 32768 Hz crystal 2Hz signal at the last output Q14, which is pin number 3. Then the 2Hz signal goes to 4024, which is also 7-bit (/128) ripple counter.
With the 2Hz clock input, the output Q1(/2) pin number 12 is low one second and high one second. Q2(/4) pin number 11 is low two seconds and then high two seconds. Q3(/8) is low four seconds and then high four seconds. When the last four (most significant digits 111100 = 60) goes to 1, the 4082 dual 4-input AND gate turn its output to 1.
Signal goes to reset pin and the counter starts calculating again from zero to 60 and the same signal also goes to the second 4024 ripple counter clock input. This signal comes to clock input every 60s and it works same way than the first ripple counter, but it calculates minutes.
HOW IT WORKS:
4060 is 14-bit (/16,384) ripple counter with internal oscillator and it gives with the 32768 Hz crystal 2Hz signal at the last output Q14, which is pin number 3. Then the 2Hz signal goes to 4024, which is also 7-bit (/128) ripple counter.
With the 2Hz clock input, the output Q1(/2) pin number 12 is low one second and high one second. Q2(/4) pin number 11 is low two seconds and then high two seconds. Q3(/8) is low four seconds and then high four seconds. When the last four (most significant digits 111100 = 60) goes to 1, the 4082 dual 4-input AND gate turn its output to 1.
Signal goes to reset pin and the counter starts calculating again from zero to 60 and the same signal also goes to the second 4024 ripple counter clock input. This signal comes to clock input every 60s and it works same way than the first ripple counter, but it calculates minutes.
Step 3: Finalizing
Next we drill holes for the LEDs. My LEDs were 5mm so I used the 5mm drill. LED stays tight in that hole and no glue is needed. I cutted the board, so it fits perfectly to the bottom of the box.
I left the LED wires on purpose that long, so the LEDs are easier fit to their correct places.
I left the LED wires on purpose that long, so the LEDs are easier fit to their correct places.
Step 4: Setting the Time
I drilled three holes to left side of box for time setting buttons. Hours, minutes and seconds. There is also one button on other side, which is set-button.
When I put the power plug in the LEDs start blinking. Then I press the set-button to down and keep it down. Same time I adjust the right time to the clock with the other side buttons. When the time is correct, it is time to release the set-button.
When I put the power plug in the LEDs start blinking. Then I press the set-button to down and keep it down. Same time I adjust the right time to the clock with the other side buttons. When the time is correct, it is time to release the set-button.
Step 5: How to Read It?
Binary clock is easy to read. It only need little bit simple math.
Okay, If we want to set 11:45:23 to our clock
It is easier convert binary to decimal than decimal to binary. I try to explain both ways.
Base number is 2
Here is the key numbers: 1 2 4 8 16 32 64 128,...
Our decimal number is 11 and that we are converting to binary. Let's find out the smallest number, which is smaller than our number from the key number list. It is 8, Let's reduce that number from our number 11-8=3. It goes to our number one time so let's put the number 1 up.
Now our number is 3 (11-8=3). Now we have to take number which is next to that number what we just used. It was 8, so the next is 4. Let's do the same thing, how many times 4 goes to 3 ? zero! Let's put the 0 number up.
Next on list is after 4 is 2. How many times 2 goes to 3 ? one time! Ok, number 1 to up.
There is one number left and our number is 3-2=1 and the last number on that list is 1 and it goes to 1 one time and that's it no numbers left. Because it goes the one time our last marked number is 1.
What we have: 1011
So the number 11 with four bits is 1011, with five bits 01011, six bits 001011, seven 0001011 etc.
Okay, let's convert it back to decimal. It is easier anyway.
Our binary number is 1011.
And our magiz numbers =) is 1 2 4 8 16, ...
Let's put our binary numbers under the magiz numbers. We have to start read from least significant digit, so that's why the counting is from right to left
8 4 2 1
1 0 1 1
Now we have to do summation with the numbers which are over the every 1 number. There are 1, 2 and 8, right?
1+2+8 = 11
Rest numbers are 45 and 23.
45 is 101101
23 is 10111 with six bits it's 010111
11:45:23 is 01011:101101:010111
Easy? =)
Okay, If we want to set 11:45:23 to our clock
It is easier convert binary to decimal than decimal to binary. I try to explain both ways.
Base number is 2
Here is the key numbers: 1 2 4 8 16 32 64 128,...
Our decimal number is 11 and that we are converting to binary. Let's find out the smallest number, which is smaller than our number from the key number list. It is 8, Let's reduce that number from our number 11-8=3. It goes to our number one time so let's put the number 1 up.
Now our number is 3 (11-8=3). Now we have to take number which is next to that number what we just used. It was 8, so the next is 4. Let's do the same thing, how many times 4 goes to 3 ? zero! Let's put the 0 number up.
Next on list is after 4 is 2. How many times 2 goes to 3 ? one time! Ok, number 1 to up.
There is one number left and our number is 3-2=1 and the last number on that list is 1 and it goes to 1 one time and that's it no numbers left. Because it goes the one time our last marked number is 1.
What we have: 1011
So the number 11 with four bits is 1011, with five bits 01011, six bits 001011, seven 0001011 etc.
Okay, let's convert it back to decimal. It is easier anyway.
Our binary number is 1011.
And our magiz numbers =) is 1 2 4 8 16, ...
Let's put our binary numbers under the magiz numbers. We have to start read from least significant digit, so that's why the counting is from right to left
8 4 2 1
1 0 1 1
Now we have to do summation with the numbers which are over the every 1 number. There are 1, 2 and 8, right?
1+2+8 = 11
Rest numbers are 45 and 23.
45 is 101101
23 is 10111 with six bits it's 010111
11:45:23 is 01011:101101:010111
Easy? =)