The output voltage for a buck converter using a resistor divider is given by:

[ \Large V_{\text{out}} = V_{\text{FB}}\left(1 + \frac{R_1}{R_2}\right) ]

For the TPS5430DDA, the typical feedback reference voltage, ( V_{\text{FB}} ), is 0.8 V.

With the resistor values provided:

the calculation is:

[ \Large V_{\text{out}} = 0.8,V \left(1 + \frac{33,kΩ}{10,kΩ}\right) = 0.8,V \times (1 + 3.3) = 0.8,V \times 4.3 = 3.44,V ]

Thus, using the current resistor divider, the output voltage is approximately 3.44 V.

Note:
To obtain a 5 V output, you would need to adjust the resistor values. For example, if ( V_{\text{out}} = 5,V ) is desired with ( V_{\text{FB}} = 0.8,V ), then:

[ \Large 5,V = 0.8,V\left(1+\frac{R_1}{R_2}\right) \quad \Longrightarrow \quad \frac{R_1}{R_2} = \frac{5}{0.8} - 1 = 6.25 - 1 = 5.25 ]

If you set ( R_2 = 10,kΩ ), then ideally:

[ \Large R_1 = 5.25 \times 10,kΩ = 52.5,kΩ ]

You may choose the nearest standard resistor value (e.g., 51 kΩ or 56 kΩ) and recalculate to fine-tune the output voltage.

The TPS5430DDA is typically rated for an output current around 3A. This means that as configured, it is not capable of supporting a continuous 5A load.

For a 5A output, you would need to select a buck converter that is specified to handle currents in excess of 5A or consider paralleling appropriate devices designed for such loads.

Let me know if you need recommendations for alternative components or additional design adjustments.