LMR14050-Q1: How to caluculate Minimum Vin at 2MHz operation

Hi,
5V/5A @ 2MHzapplication is too aggresive, it will produce lots of heat. Suggest the frequency no higher than 500kHz for so large current output. For the dropout condition, it varies with different circuit configurations, the best way is to evaluate that on bench practically according your design circuit.

I explain that Loads are LED and other circuit.
(1)
LED On = 0.3A(for other circuit) + pk4.7A (for LED) 200us
LED Off = 0.3A(for other circuit) 70us
This ON/Off repeat 12times (270us x 12 times = 3.24ms)

(2)
Break time = 0.3A(for other circuit) 30ms

They repeat (1) and (2).

So Time of 5A is very short.They consider heat is small and no problem.

They would like to know how to caluculate Minimum Vin at following condition.
Fsw:2MHz
Vout:5V
Iout:5A
We would like to know VIN condition (device does continue to SW at 2MHz,and doesn't enter dropout condition(decrease SW frequency).Please give us calculating formula.

Regards,

Kura,
Below is a formula to rough estimation.
Dropout=Vin-Vo=Io*(Rds_on+RL+Rpcb)+(1-Dmax)*(Vd+Io*RL+Io*Rpcb+Vo)/Dmax
Dmax=1-Toff_min/Ts, Toff_min is about 90ns for 2.2Mhz frequency.
Rds_on: High side FET on resistance
RL: indictor resistance
Vd: diode forward voltage
Rpcb: PCB trace resistance
All the parameter should take temperature coefficient into consideration if junction temp varies.

Luke-san

Thank you for reply.

We got additional question from customer.

Please let us know  worst value of Toff_min @ 2.0MHz.

Regards,