I am using this DC-DC buck converter.
Schematics :
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 :
Input Voltage of 9V :
Input Voltage of 10V :
Input Voltage of 14V :
My questions :
Can someone tell me why I was not getting any output voltage when I provided an input voltage less than 5V?
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?
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.