I am going to incorporate this 4017 in a circuit and I have downloaded its datasheet,but I can't understand some parts of it.
What is the max Vdd current? And the max current & voltage of the CLOCK pin?
Here is the data sheet.
The max Clock current is ±10mA, as that is the maximum input current on any input pin. The max Clock Voltage is VDD + 0.5V.
The max VDD current is not given in current, but power. It is 500 mW for the package. Individual output transistor power is limited to 100 mW or ±6.8 mA Typical at 25˚C. So max you should source for the whole IC is less than 68 mA.
The clock is an input. The input current of a CMOS part is composed of a leakage current and a dynamic current due to the input capacitance. The typical leakage current at room temperature is very low- the data sheet says 10^-5 uA or 10pA. Maximum is +/- 1uA at 125 degrees C. The input capacitance is typically 5pF.
The supply current is composed of a quiescent leakage current (typically 40nA at 10V), plus a dynamic current proportional to frequency and supply voltage (often specified as 'power dissipation capacitance - the NXP 74HC4017 specifies 35pF). The current into the Vdd pin will also include the total current sourced by all the outputs. It should be mentioned that if the inputs are not at the supply voltages the supply current can be much higher than the stated quiescent current- mA even if you hold an input at just the right level. That's because both the n and p channel transistors turn on somewhat at the same time.
In general, at low frequencies, inputs at CMOS levels, reasonable temperatures, and if you are running from a mains supply, the supply current is negligible except for whatever you are driving, and the input current is also minuscule. If you are trying to run for years from a small battery then uA count (but the difference between typical and maximum specs is enormous).