Answer :
Final answer:
To design a Buck Boost Converter from 6V to 48V, one must consider components like inductors, diodes, transistors, and capacitors, and aim for an efficiency of at least 85%. A current mode controller is essential for constant current operation, and design details require EDA software for schematics and PCB layout.
Explanation:
Designing a Buck Boost Converter
To design a Buck Boost Converter that steps up the voltage from 6V to 48V using the constant current (CC) method, you need to consider the power requirement and the conversion efficiency. The power output is given as 18W for a 3V output, which infers a current of 6A. To step up to 48V, we need to ensure that the converter can handle the same power level or higher. Given losses in the converter, let's aim for an efficiency of at least 85%.
The key components of a buck-boost converter include an inductor, a diode, a switch (usually a transistor), and a capacitor. For a constant current design, a current mode controller should be used, which uses an external signal to control the duty cycle of the switch and maintain a constant output current. The duty cycle 'D' can be roughly estimated by the formula, D=1-(Vin/Vout), which for our case of 6V to 48V would be 1-(6/48), giving 0.875 or 87.5%. However, the actual circuit design will need to accommodate the dynamic response and load variations to maintain stable output.
