Circuit Connection Guidance

connect these components correctly to the chip based on the application circuit

Thought for 50 sec

Below is a typical connection diagram for U1 with the surrounding components in your design:

Diagram

Connections Summary

VIN pin to [VBAT], with [C1] (10µF) and [C3] (0.1µF) both placed from VIN to ground.
LX pin to [L1], then from the other side of [L1] to the output node where [C2] (33µF) also connects to ground.
FB pin to the junction of [R1] and [R2]; [R1] goes from the output node to FB, [R2] goes from FB to ground.
EN pin can be tied to VIN (or controlled externally) for enable.
GND pin to ground.

You can now wire these nets in the schematic to match the above connections. If you need help with the wiring interface, use @help.

Add decoupling capacitors

Verify inductor specifications

List all components used

Sign up to send

Files

Pins

Symbol

Layout

Preview

are there any issues in this project?

hasofukirsch

@copilot are there any issues in this project?

1y

copilot

Based on the textual description of your project, here are a few observations and suggestions:

Feedback Network: The feedback network formed by R1 and R2 looks standard. However, no information on the desired output voltage is provided, which is needed to ensure the correct resistor values. The output voltage $V_{OUT}$ for a typical buck converter can be calculated using the formula:

$V_{OUT} = V_{REF} \left(1 + \frac{R1}{R2}\right)$

where $V_{REF}$ is the reference voltage of the RT8059GJ5, usually around 0.6V. With R1 = 3.4kΩ and R2 = 1kΩ, this gives approximately 2.64V as $V_{OUT}$.
Placement Suggestion: The note suggests placing C2 close to L1 to minimize loop area. This is crucial for stability and noise reduction in switching regulators.
Enable Pin: U1's EN (Enable) pin is tied to Net1. Make sure it is connected to the correct logical level to enable the converter. If connected directly to VIN, it enables automatically, otherwise consider adding a resistor divider if you want to control it.
Power Input: Ensure the power input VBAT is within the acceptable range specified by the RT8059GJ5 datasheet to avoid damage.
Bypass/Decoupling: C3 is used for bypassing/decoupling. Ensure it's placed physically close to U1's VIN pin for optimal performance.

If you have specific parameters or operation conditions in mind, consider comparing them against the provided values I mentioned and adjust accordingly. Would you like any further insight on a specific part or connection?

1y

hasofukirsch

check the voltage divider again

1y

copilot

Sure! Let's revisit the voltage divider calculation for the desired output voltage.

For the feedback network of a buck converter:

R1 = 3.4 kΩ (top resistor)
R2 = 1 kΩ (bottom resistor)
Typical reference voltage, $V_{REF}$, for the RT8059GJ5 is approximately 0.6V (please verify with the datasheet).

The formula for the output voltage, $V_{OUT}$, is:

V_{OUT} = V_{REF} \left(1 + \frac{R1}{R2}\right)

Plugging in the values:

V_{OUT} = 0.6 \, \text{V} \times \left(1 + \frac{3.4 \, \text{k}\Omega}{1 \, \text{k}\Omega}\right)

V_{OUT} = 0.6 \, \text{V} \times \left(1 + 3.4\right)

V_{OUT} = 0.6 \, \text{V} \times 4.4

V_{OUT} = 2.64 \, \text{V}

Thus, the calculated output voltage is approximately 2.64V, assuming the reference voltage is 0.6V. Please verify the reference voltage in your RT8059GJ5 datasheet to ensure accuracy. Let me know if this aligns with your design requirements!

1y

hasofukirsch

is the reference voltage the same as the output voltage? i set the output voltage to 3.3V

1y

copilot

The reference voltage $V_{REF}$ is typically not the same as the output voltage $V_{OUT}$ in a buck converter. Instead, $V_{REF}$ is used in conjunction with the feedback network (voltage divider) to set the desired $V_{OUT}$.

If you want the output voltage to be 3.3V, you can use the formula to determine the necessary resistor values or verify if the provided resistor values work with your setup.

The formula for $V_{OUT}$ is:

V_{OUT} = V_{REF} \left(1 + \frac{R1}{R2}\right)

Given:

Desired $V_{OUT}$ = 3.3V
$V_{REF}$ for the RT8059GJ5 = 0.6V
R2 = 1 kΩ

You can solve for R1:

3.3 = 0.6 \left(1 + \frac{R1}{1 \, \text{k}\Omega}\right)

Let's rearrange and solve for R1:

\frac{3.3}{0.6} = 1 + \frac{R1}{1 \, \text{k}\Omega}

\frac{3.3}{0.6} - 1 = \frac{R1}{1 \, \text{k}\Omega}

\frac{R1}{1 \, \text{k}\Omega} = 4.5

R1 = 4.5 \, \text{k}\Omega

The calculated value for R1 should be approximately 4.5 kΩ. Since R1 is currently set to 3.4 kΩ in your design, you'll need to adjust it to achieve 3.3V at the output.

Let me know if you need further assistance!

1y

hasofukirsch

do a final check on the voltage divider circuit. is it suitable for 3.3V output?

1y

copilot

Let's run through the calculations one more time to verify if the current resistor values in your voltage divider circuit are appropriate for achieving a 3.3V output.

Given:

$V_{REF}$ = 0.6V for the RT8059GJ5.
Desired $V_{OUT}$ = 3.3V.
R1 = 4.5 kΩ.
R2 = 1 kΩ.

The formula for the output voltage $V_{OUT}$ is:

V_{OUT} = V_{REF} \left(1 + \frac{R1}{R2}\right)

Substitute the given values:

V_{OUT} = 0.6 \, \text{V} \left(1 + \frac{4.5 \, \text{k}\Omega}{1 \, \text{k}\Omega}\right)

V_{OUT} = 0.6 \, \text{V} \left(1 + 4.5\right)

V_{OUT} = 0.6 \, \text{V} \times 5.5

V_{OUT} = 3.3 \, \text{V}

Therefore, with R1 set to 4.5 kΩ and R2 set to 1 kΩ, the voltage divider is indeed suitable to achieve the desired 3.3V output. Your current setting for R1 is correct at 4.5 kΩ to get 3.3V output, considering the reference voltage is 0.6V. Let me know if there's anything else you'd like to clarify!

1y

1

hasofukirsch1_04 Connecting Components Done

No files yet