Hello Everyone !!!
I am trying to make the parallel in parallel out shift register MC14035BC working as a parallel in serial out register. Does anyone has experience on that circuit ?? I have searched all the internet and I haven't find sufficient information neither the datasheet says something similar . Does anyone has application schematics on that circuit ??

Thank you very much.
George

 

When searching for MC14xxx chips you can also search for CD4xxx chips. Same thing.
MCxxxx is the part number of Motorola which is now ON Semiconductor. CD4xxx IC designation from Harris, RCA, and National Semiconductor is now acquired by Texas Instruments.

Here is the datasheet. What else would you like to know about this IC?

http://www.ti.com/lit/ds/symlink/cd4035b.pdf

http://www.electroniccircuits.gr/files/CD4035.pdf

 

Which direction do you want to shift? Do you only want to shift 4 times?

 

Which direction do you want to shift? Do you only want to shift 4 times?

I want to make it work as an parallel in serial out register , I do not mind about the direction but I would like to know how and if you can switch it's direction........

 

I do not mind about the direction but I would like to know how and if you can switch it's direction

Shift right is easy.

After parallel loading, a LOW on P/S# will shift right; output is from Q3. You need to take the output from Q3 before you begin clocking for shifting. After 3 clocks, serial data will be taken from the J/K# inputs.

EDIT: Data from the serial input starts entering the shift register on the first clock.

Shift left is more work.

You need to feed the outputs Q3-Q1 to the corresponding parallel input for the previous flip flop DP2-DP0. Output is taken from Q0 and serial input is from DP3. Shifting left requires setting P/S# to HIGH; so some multiplexing with the existing parallel input data circuitry is needed.

EDIT: changed DP3-DP0 above to be DP2-DP0.

For parallel load with bi-directional shift, MC14194 would be more convenient.

 

Shift right is easy.

After parallel loading, a LOW on P/S# will shift right; output is from Q3. You need to take the output from Q3 before you begin clocking for shifting. After 3 clocks, serial data will be taken from the J/K# inputs.

EDIT: Data from the serial input starts entering the shift register on the first clock.

Shift left is more work.

You need to feed the outputs Q3-Q1 to the corresponding parallel input for the previous flip flop DP2-DP0. Output is taken from Q0 and serial input is from DP3. Shifting left requires setting P/S# to HIGH; so some multiplexing with the existing parallel input data circuitry is needed.

EDIT: changed DP3-DP0 above to be DP2-DP0.

For parallel load with bi-directional shift, MC14194 would be more convenient.

Thank you but it my circuit doesn't seems to work....... I send you a picture of what I have done so that you can tell me what I have done wrong. The upper button is the clock and the the other the reset . The circuit is in complementary and parallel mode. And I have put a "1" to the PI1

Thank you very much
George

 

Thank you but it my circuit doesn't seems to work....... I send you a picture of what I have done so that you can tell me what I have done wrong.

Could you post a schematic? I don't have the patience to trace breadboards if they're not mine.

What do you expect the circuit to do and what is it doing?

 

Could you post a schematic? I don't have the patience to trace breadboards if they're not mine.

What do you expect the circuit to do and what is it doing?

I expect to take that input which is 0010 and shift the compliment in parallel and then by switching to serial , to shift those values serially........ In few words I struggle to make a parallel to serial shift register........
I post you a schematic which I made in paper brevity.

Regards

George

 

You have a lot of floating inputs; they need to be tied HIGH or LOW.

Reset is HIGH, so the shift register is always reset.

T/C# is HIGH, so you won't get complemented output.

 

This is what you should have:

The simulator I used doesn't have resistors, so I used switches. The LEDs are strange in that they don't need a connection to ground. The switches are strange in that when opened, they provide a LOW.

The dark nets are logic HIGH, gray are logic LOW.

It also didn't have a CD4035/MC14035, so I had to create it.

 

View attachment 124899

You have a lot of floating inputs; they need to be tied HIGH or LOW.

Reset is HIGH, so the shift register is always reset.

T/C# is HIGH, so you won't get complemented output.

Thank you very much ,Low means putting the input to the ground ???

 

Low means putting the input to the ground ???

Yes; using a switch, resistor, another gate, transistor, etc.

I just noticed that the dots in your schematic are the switches. So the shift register can function when its switch is opened.