Schematic Symbol for MCP1700T-2802E/TT: Linear Voltage Regulator...

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can you provide 1.8v with 6A circuit

prakash1513

@copilot can you provide 1.8v with 6A circuit

copilot

Sure, to provide a 1.8V output with 6A current capacity, you'll need a switching regulator (also known as a buck converter) since linear regulators would be inefficient for such high current levels. One commonly used component for such applications is the LMR33630 buck converter from Texas Instruments.

Here's a basic outline for the circuit design:

Components Needed:

Buck Converter IC: LMR33630 or similar
Inductor: e.g., 2.2µH rated for at least 6A
Input Capacitor: e.g., 10µF ceramic capacitor
Output Capacitor: e.g., 47µF ceramic capacitor
Feedback Resistors: to set the output voltage

Connections:

Input Voltage Source (8-40V):
- Connect input voltage source to VIN of LMR33630.
Output Voltage:
- Connect VOUT of LMR33630 to your load.
Ground:
- Connect the GND pin of LMR33630 to the system ground.
Inductor and Capacitors:
- Connect an inductor (2.2µH) between SW pin and VOUT.
- Connect a ceramic capacitor (47µF) between VOUT and GND.
- Connect a ceramic capacitor (10µF) between VIN and GND.
Feedback Network:
- Use a pair of resistors to set the output voltage. Calculate the resistor values using the formula: $V_{OUT} = V_{FB} \left(1 + \frac{R_1}{R_2}\right)$ where $ V_{FB} $ is the feedback voltage (typically 0.6V for the LMR33630).

Connections in Flux Schematic:

Connect VI:1 to VIN of LMR33630.
Connect VO:1 to VOUT of LMR33630.
Connect GND:1 to GND of LMR33630.

Assumptions:

Assumes input voltage is sufficiently higher than 1.8V (e.g., 8-40V).
The inductor and capacitors are correctly rated for voltage and current.

Do let me know if you need further details or adjustments!

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