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I want to make a voltage down converter to power my controller (Atmega328P).

Input = 12V / Output = 5V (3A). As a converter, I use ST1S40.

My connection diagram

enter image description here

I took the circuit from datasheet, changed the values ​​of resistors R1, R2. The rest is unchanged. I did not find the connection scheme of this converter with an output voltage of + 5V.

Why did I choose him? To connect it does not require a lot of components. Unfortunately, I did not find other options.

Kevin Reid
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Delta
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  • Your EN(able) pin isn't wired? – Tyler Dec 14 '18 at 14:24
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    A lot of regulators have the EN pin pulled up internally. This one has it pulled down internally, so if you let it float your supply is disabled. – Cristobol Polychronopolis Dec 14 '18 at 14:29
  • For clarity, it is good practice to draw power flowing in from the left and exiting from the right (as opposed to a symbol with pin-pin correspondence with footprint) – BB ON Dec 14 '18 at 14:41
  • In addition to what @BBON said, here are [guidelines for drawing readable schematics](https://electronics.stackexchange.com/q/28251/7036). – Nick Alexeev Dec 15 '18 at 05:57

1 Answers1

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If you want a 5 volt output then you need to recalculate R1 and R2. At the moment it will regulate to 6.8 volts and not 5 volts. This is because the FB pin needs to "see" 0.8 volts and with 5 volts at the output, it will see 0.588 volts and, as a result, the duty cycle will increase until it sees 0.8 volts and that occurs when Vout = 6.8 volts.

As others have mentioned in comments you need to have the enable input pulled high.

You also need to ensure that the thermal pad beneath your chip is connected to ground and that you have applied enough copper on the PCB to ensure the junction temperature does not exceed 150 degC. Read section 6.4 of the data sheet.

Section 6.5 also draws your attention to correct PCB layout and, at the bottom also reminds you about the thermal pad: -

Thanks to the exposed pad of the device, the ground plane helps to reduce the thermal resistance junction to ambient; so a large ground plane, soldered to the exposed pad, enhances the thermal performance of the converter allowing high power conversion.

Andy aka
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  • 20 / (105 + 20) * 5 = 0.8 I'm doing wrong? – Delta Dec 14 '18 at 15:04
  • Yes, that would work OK. – Andy aka Dec 14 '18 at 15:05
  • Then R1 (105k) and R2 (20k) have correct values? – Delta Dec 14 '18 at 15:12
  • You have calculated the values correctly. – Andy aka Dec 14 '18 at 15:16
  • Ok, is there any dependency on other elements (inductors and capacitors)? I collect it for the first time and do not really understand yet. :( – Delta Dec 14 '18 at 15:43
  • It turns out I placed the wrong resistor on the circuit, you were right. 150k i need to replace with 105k. – Delta Dec 14 '18 at 15:50
  • This is a simple question and answer site. You asked and I provided feedback. This site does not seek to provide an expose on the ins and outs of every detail of the chip. Read the data sheet is my advice and if you have a new question then please raise it. All the components chosen should be as those receommended in the data sheet. Any discepency or inconsistency would make a good question but you have to find it first. – Andy aka Dec 14 '18 at 15:51
  • @Алекс Гарисон no problem – Andy aka Dec 14 '18 at 15:55