hello,
https://pdf1.alldatasheet.com/datasheet-pdf/view/26845/TI/CD4013B.html
I want to know few points in this datasheet.
♣ In 1st page> features > 6th point > Max. input current of 1uA at 18V
what is this current stands for ?
Thats the maximum current drawn from the driving circuit.
♣ why they use drain to source voltage terminology , when it is clearly supply voltage.
Because CD4000 is CMOS technology
Vdd [ drain] Vss [source]
♣ Max. current that this IC can give to output is 15mA max (both Iol and Ioh), which can not light up an LED, then i suppose it will need a driver circuit to do this.
Relook at the tables for Source and Sinking current, its much lower than 15mA
♣ Vol is 0.05V Max., it means this voltage will always be deducted from calculated output voltage. and how will one calculate output voltage. without using test circuits.
This is the LOW state voltage of the output
♣ Iin , if i use this IC how can one measure current which is to be given at Input.
without using test circuits.
Its not required to measure the input current, its stated as above 1uA at 18V
♣ what is load capacitance ? according to fig5 if i use 5V supply , where should i look in graph for what will be my propogation delay, because i dont know whats my load capacitance.
Load capacitance is the 'capacitance' that a following circuit places on the output
At 5V if you assume the max load capacitance of 100pF, you will get ~170nSec propagation delay