On the first page of the datasheet we are having this statement.
In the equations table with designe procedure we will have this formula.
And all the online calculation tools tells me that I cannot get more than 250 mA of output current without 1.5 A Ipk limit, for Vin = 4.8 and Vout = 12 V. Trying to recalculate it by datasheet's records seems to not going well.
So what the difference between Ipk and Iout? What the actuall transformation ratio? What the maximum output power of this converter? The last question, maybe better describes it's abilities, rather that current.
Best Answer
In a boost converter the switch has to handle the average INPUT current plus half the inductor current ripple.
Note that when the switch turns on, the inductor is carrying the average input current minus half the peak to peak inductor current:
Once the switch is closed the inductor current ramps up to the average input current PLUS one half the inductor p-p ripple current. That value must be below the peak switch current limit.
Boost switch, diode and inductor currents:
Output power will be equal to input power minus losses, so Iout = efficiency * Iin*Vin/Vout.
So for example if you have 5V in and 10V out with 80% efficiency (neglecting inductor ripple) and a switch peak current limit of 1A:
Max input power possible: 5V*1A = 5W. Max output current possible 5W/10V = 0.5A, 0.5A*efficiency = 450mA.
For more accurate calculations you have to take the inductor ripple current into account as explained in the data sheet.
So it all depends on your ratio of Vin to Vout as to how much current/power you can get out of the device.