I'm working on a circuit where I have two 74HC595 in parrallel. (2 serial to parrallel 8bit shift register with latch)
The outputs of the registers are connected to 2 40174 (D type flip flop)
So that allow me to keep the signal up or low until I ask the flip flop to update, even if the register value change and give me an extra control on when I want to apply the changes from the registers.
The flip flops outputs are connected to 4066 switchs. They allow me to "multiplex" 12 data signals.
My question is, can I use a single 40174 pin to control multiple 4066 switchs ?
Do I need to do something special ? (resistor ? capacitor ?) Or can I just wire them as-is ?
40174 Datasheet : http://www.ti.com/lit/ds/symlink/cd40174b.pdf
4066 Datasheet : http://www.ti.com/lit/ds/symlink/cd4066b.pdf
Best Answer
What you're asking about is called fan-out.
You need to compare the output current capability of the 40174 with the input leakage current of the 4066 control inputs.
I looked at datasheets for some 40174 and 4066 parts from NXP and the 4066 had an input leakage spec of 1 uA, while the 40174 had an ouput drive capability (worst case is sinking current for low output) of 0.4 mA. So you can conceivabley drive quite a few 4066 loads from a 40174. But you didn't say exactly which part numbers you're using, so you need to check your own datsheets.
Also be aware that the more you fan out, the slower the switching transient will be, so you also need to consider the capacitive load of your 4066's and your required switching delay. The 4066 load is about 7.5 pF (max). With 0.5 mA drive this gives a 66 V/us slew rate (very roughly), divided by the number of loads. If you want to switch 5 V in less than 1 us, you can't drive more than about 13 loads. If you want to switch very frequently (e.g. 10,000 switches per second or more) you need to start looking at power handling capabilities of the '174's as well.
Edit: looking at the TI datasheets you linked, your parts have similar specs to the NXP parts I looked at.