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I am using this DC-DC buck converter.

Schematics :

enter image description here

Buck Converter Specifications :

  • Input Voltage - 18V to 32V
  • Switching Frequency - 300kHz
  • Output Voltage 9V
  • Load Current - 0mA to 200mA Maximum.

I was just curious to understand what will happen to the buck converter when I provide an input voltage of less than 9V.

So, I started to increase the input voltage from 0V. I noticed that I didn't get any output voltage below 5V.

Only when the input was 5V, I started getting some output voltage. I provided an input voltage of 5V, 9V, 10V and increased.

Below are the waveforms of the input voltage, output voltage and switching frequency of the buck converter and I provided a load of 0.1mA at the output.

Input Voltage of 5V :

enter image description here

Input Voltage of 9V :

enter image description here

Input Voltage of 10V :

enter image description here

Input Voltage of 14V :

enter image description here

My questions :

  1. Can someone tell me why I was not getting any output voltage when I provided an input voltage less than 5V?

  2. When I provided an input voltage less than 9V, the duty cycle of the switching frequency is 97% and the switching frequency is around 37kHz. I understand that DC-DC switching converter architecture is totally different from linear regulator, but why doesn't the input travel directly to the output when the input voltage is less than the 9V? From the datasheet, the oscillating frequency (switching frequency) is only between 270kHz to 330kHz. But how am I getting a value of 37kHz when I provide input of less than 9V? Can someone explain what is happening internally during this low input voltage state?

  3. When I provide an input voltage of exactly 10V, we can see then the switching frequency is halved from what we desired. We get a switching frequency of around 150kHz and there is some sort of glitch or inverted triangle between the switching pulses. Can someone tell me why is that?

After I increase above 10V, the waveforms of input, output and switching frequency are as expected, except the ringing.

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  • \$\begingroup\$ If your actual minimum input voltage is 18V, why do you care about any of this out of spec behaviour? \$\endgroup\$
    – Reinderien
    Nov 5, 2020 at 4:48
  • \$\begingroup\$ As I mentioned in the question, just curious to understand the behaviour during low voltage input conditions to understand what will happen and how is it happening. \$\endgroup\$
    – Newbie
    Nov 5, 2020 at 6:16

2 Answers 2

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  1. It does not work with such low voltages. It even has an under-voltage lockout to specifically prevent that, so it requires more than 4.5V to start.

  2. While some regulators have that feature, this regulator is not built to pass input to output. There is a NMOS doing the high side switching and it cannot turn on unless the gate voltage is made by switching and the boost cap, so this sets the max duty cycle. What happens is the chip knows it should be in MaxDuty mode where frequency is divided by 8, so 37 kHz is spot-on correct.

  3. It might alternate between MaxDuty and Normal modes or do other kind of pulse skipping if there is light load. Components behave weirdly when put into weird circumstances. Especially if the external components are not specifically selected to provide good operation over the intended range.

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  • \$\begingroup\$ Thank you for the answer. \$\endgroup\$
    – Newbie
    Nov 5, 2020 at 9:21
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The inductance is way too high compared to suggested values. Refer to Fig.32 at p.17. A 22uH inductor is fine, but you placed 470uH. I know, the values suggested are given as "minimum", but remember that increasing the inductance is not always the preferred way. A very large inductor may result in unexpected behavior.

I'll answer only the 1st question as the other two may be solved by decreasing the inductor to 22uH:

  1. Note-6 at datasheet p.5: Voltage more than 4.65V is necessary for IC start. The IC can operate to 4.5V after IC start.
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