0
\$\begingroup\$

I am hobbyist and am looking to reduce signal frequency using D type flip flops. I have a circuit diagram for a 4013 IC but I would like to use a 74HC74 instead and I do not understand if it is possible.

From wikipedia:

  • 7474 are dual D positive edge triggered flip-flop, asynchronous preset and clear.
  • 7479 is a dual D flip-flop

Likewise the 4000 series:

  • 4013 is a Dual D Type Flip Flop

I wasn't able to find a 74HC79 components however I was able to find 74HC74 and a 4013 IC.

From looking at the functional diagram in the specifications, they are essentially the same other then that one has CD [asynchronous clear-direct input (active HIGH)] and the other has RD [asynchronous reset-direct input (active LOW)] on the diagram.

I want to construct the following diagram using 74HC74 instead of 4013: enter image description here

Could this be done using 74HC74? How would such a circuit be different? or can RD be connected to the negative wire just like CD?

\$\endgroup\$

2 Answers 2

Reset to default
3
\$\begingroup\$

Yes, you can use a 7474 to divide by two.

BUT the set and reset inputs are active low in the 7474, whereas they are active high in the 4013.

This means that to divide by 2, the set and clear must be pulled up or connected to the rail, not to ground.

One further point, with CMOS, inputs should not be left floating, BAD things can happen. Take all inputs on unused sections of the device to rail or ground, it doesn't matter which, obviously take all used inputs to the correct logic level. Outputs should be left unconnected if not used.

Download a data sheet for both parts and study them carefully for pinout and pin function, IIRC the packages are not pin-compatible, so you will have to change the pin numbers as well.

\$\endgroup\$
4
  • \$\begingroup\$ Thank you! To confirm does "Take all inputs on unused sections of the device to rail or ground" mean that all pins which are not part of the main circuit should be connected to either ground or rail? That is, all I/O of the second flip flop should be connected to ground or rail? \$\endgroup\$
    – Greg
    Commented Feb 8, 2017 at 8:24
  • 1
    \$\begingroup\$ No, only all the inputs of the circuitry should be connected to either rail or ground. So Q and Q_bar should be left floating for the unused flipflop. \$\endgroup\$
    – Douwe66
    Commented Feb 8, 2017 at 8:42
  • 1
    \$\begingroup\$ Confirming Douwe, only the inputs, don't connect any outputs to ground or rail. All the device inputs should be connected to a valid logic levels, even on the unused elements. If they float to a middle voltage, the chip can draw excessive supply current and overheat. \$\endgroup\$
    – Neil_UK
    Commented Feb 8, 2017 at 9:16
  • \$\begingroup\$ Given that the asynchronous clear-direct input (CD) and the asynchronous reset-direct input (RD) are active low and require rails to function, are there any power saving benefits of connecting the unused inputs to ground vs rails? \$\endgroup\$
    – Greg
    Commented Feb 8, 2017 at 9:22
2
\$\begingroup\$

The difference between the 4000 series and 7400 series is amongst other things the voltage rating:

CD4013: 3.0V - 15V power supply
74HC74: 2.0V - 6.0V power supply

Furthermore the pinout differs, so take a look at the datasheet for the right location of the pins!

The other difference you already mentioned are the set and reset pin. For the 74HC74 it is active low, this means that you need to connect them to VDD if you want the flip flop to operate. When reset is active the output pins (Q and Qbar) have a fixed state.

Noting the above mentioned points, you could easily interchange the two flipflops. The functionality is the same after all.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.