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I want to design a Switch Mode Li-ion charger which uses 5V as input and charges a single cell Li-ion cell.
I am referring this AppNote

On page 4 they have given the equation to find inductor value i.e. $$ L = (V_i – V_{sat} – V_o) \times \frac{T_{on}}{2 \times I_{max}} $$ If I use 5V Vin and want to charge at 1A current, the inductor value comes out to be very low. I assume Vo=4.2V. Ton=1.32\$\mu\$s (50% duty cycle) for 376KHz PWM.

So I am confused as to how they use switch mode chargers in mobile phones.

This is the schematic of typical charger:

schematic

simulate this circuit – Schematic created using CircuitLab

so I was wondering with 5V input and voltage drops at MOSFET and schottky diode would I have enough voltage to charge the battery?

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    \$\begingroup\$ What is the problem? Why would that not work for phones? One of the main points of modern switch mode converter design is to minimise inductor value, as otherwise they are bulky and heavy among other things. Don't forget that a 1H inductor is enormous and weighs a ton. \$\endgroup\$ – CL22 Apr 7 '15 at 11:27
  • \$\begingroup\$ @Jodes I have added my doubt in the question. \$\endgroup\$ – Sajid Apr 8 '15 at 7:12
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The inductor in the circuit can be used to create a higher voltage than you had to start with. It stores energy when you put a current through it. When you remove that current, a voltage will appear across its terminals. This can be very much higher than the original voltage.

Circuits which provide a higher DC voltage from the original are known as step-up DC to DC converters. There are also step-down converters.

A feature of switch-mode supplies (and chargers) is they are quite efficient when converting the voltage (up or down), often 90% efficiency or more.

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