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can you answer me if my adjustable is gonna work with this feedback, including u...
rafaellbx

rafaellbx

@copilot can you answer me if my adjustable is gonna work with this feedback, including use this values for resistor
2y

copilot

Yes, let's analyze your adjustable setup based on the provided circuit information and resistance values you mentioned for R1 in your project, which is an adjustable resistor (potentiometer) specified as 3296Y-1-103LF with an initial resistance of 10kΩ. You have included specific resistance values in your excerpts, but let's focus on the feedback network which typically involves R1 and R2 in configurations involving adjustable voltage regulators such as the LM2596DSADJR4G in your project.
For an adjustable voltage regulator like the LM2596, the output voltage is usually set using a resistor divider network connected to the feedback pin (FB). The general formula for calculating the output voltage ((V_{out})) for such regulators is given by:
[V_{out} = V_{ref}(1 + \frac{R1}{R2}) + I_{adj}R1]
where:
  • (V_{ref}) is the reference voltage of the regulator.
  • (R1) is the resistance connected between the output and the feedback pin.
  • (R2) is the resistance connected between the feedback pin and ground.
  • (I_{adj}) is the adjustment current, which is usually very small and often negligible for the purpose of initial calculations.
Given that the LM2596 has a typical (V_{ref}) of 1.23V (you should confirm the exact value in your specific regulator's datasheet), and you're using R1 (3296Y-1-103LF) with a resistance value that can be adjusted up to 10kΩ. Let's consider R2 in your project has a fixed resistance of 330Ω according to your project description.
Assuming you do not mention specific desired output voltage and considering (I_{adj}) is negligible, let's calculate a scenario using the standard values you've provided. Without exact values for how you've set R1, I'll give you an example calculation if R1 were at its maximum value of 10kΩ:
[V_{out} = 1.23V \times (1 + \frac{10kΩ}{330Ω})]
This simplified calculation doesn't account for (I_{adj}) due to lack of specific current values, but from this, you can see that adjusting R1 will have a significant effect on (V_{out}), given the large ratio margin with R2's 330Ω. The exact (V_{out}) will depend on R1's setting.
The feedback network involving R1 and R2 is crucial for setting the output voltage, and yes, using an adjustable resistor like R1 allows you to vary (V_{out}). You'll want to adjust R1 within safe operational boundaries suggested by the LM2596's datasheet, ensuring that the maximum output voltage does not exceed what your circuit components can handle. Keep in mind, too high a voltage can damage sensitive components downstream of the regulator.
Please confirm the desired output voltage and we can adjust the calculation to fit your specific needs more closely, or if you have set specific values for R1 that differ from the maximum, let me know those values for a more accurate calculation.
2y
can you answer me if my feedback to control d eoutput is correct?
rafaellbx

rafaellbx

@copilot can you answer me if my feedback to control d eoutput is correct?
2y

copilot

To determine if the feedback network to control the output voltage of your project is correctly configured, let's analyze the components and connections related to the feedback mechanism in your project, particularly around the LM2596 voltage regulator (U1).
The LM2596 (U1) is a step-down (buck) regulator, and its feedback (FB) pin is critical for setting and regulating the output voltage. The regulation is achieved by comparing the output voltage (after it's been divided down) to a reference voltage inside the regulator. The feedback components typically include resistors (to set the output voltage) and could include capacitors for stability or noise reduction.
In your project, the feedback network involves these components and connections:
  • Net c793f39162e3e762e8a6f1cb4875dc8f connects U1:(FB), R2:(P2), and R1:(3).
Considering the typical application circuit for the LM2596 requires a voltage divider from the output to the FB pin to set the output voltage, we will analyze your design based on this principle. The reference voltage for the LM2596's FB pin is around 1.23V. The output voltage is set by the resistors in a voltage divider.
Assumptions Made:
  • R1 is likely part of the voltage divider meant to scale down the output voltage to an appropriate level for feedback. With R1 specified as 10kΩ (from the part number 3296Y-1-103LF), it's a variable resistor/potentiometer, which implies you can adjust the voltage division and thus the output voltage.
  • R2 is also part of the feedback network and is listed with a resistance of 330Ω.
Given these components, the output voltage ((V_{out})) formula with feedback involved is:
Vout=Vref(1+R1R2)V_{out} = V_{ref} \left(1 + \frac{R1}{R2}\right)
where (V_{ref} = 1.23V) (typical for the LM2596), (R1) is the resistance from the wiper of the potentiometer to the ground (variable), and (R2 = 330Ω).
The configuration involving R1, R2, and the connection to U1:FB is conceptually correct for adjusting the output voltage. However, the precise voltage you will get depends on the position of the wiper on R1 (since it's a potentiometer). The variable nature of R1 allows you to fine-tune the output voltage as needed.
Important Note: Ensure that the range of adjustment for R1 keeps the voltage at U1:FB around 1.23V for stable operation. Also, consider the stability and transient response, as adding a capacitor at the FB pin (not present in your current configuration) could improve performance based on the LM2596 datasheet recommendations.
Please adjust the resistance values as needed based on your specific output voltage requirement, considering the formula shared.
2y
1
R2
Resistance
330 Ω
L1
Inductance
33u H
R1
Resistance
10k Ω
PS1
U1
D1
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