I was looking at the charge pump regulator CAT3200TDI-GT3 datasheet that used a voltage doubler inside. The efficiency is 80 %.

If I have input 4 V and I need output 5 V with load current 80 mA for example. Does it mean I need more then 80 mA on the input side ?

We can not have higher output voltage and higher output current with low input voltage and low input current, right ?

 

Calculate the output power.
Divide by 0.8 to give the input power (because the efficiency is 0.8)
Input current = power divided by input voltage.

 

By my calculation 125mA input current. I would allow for at least 150mA since the 80% is probably best case.

 

That is correct. The simple rule is that power out will always be less than power in. Working backwards:

\( 5\text{ V}\times80\text{ mA}\;=\;400\text{ mW, Output power} \)

\( \cfrac{400\text{ mW}}{80\%}\;=\;500\text{ mW, Input power} \)

\( \cfrac{500\text{ mW}}{4\text{ V}}\;=\;125\text{ mA, Minimum input current} \)