Understanding the LM2576 Step-Down Regulator

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The LM2576 step-down regulator is designed to convert 15V DC to 5V DC efficiently, featuring a built-in current limit. Its operation involves feedback from the output voltage to a voltage divider, which reduces the voltage to a reference level of 1.23V. A fixed gain error amplifier amplifies the difference between this reference and the attenuated load voltage, while a comparator compares this error voltage with a 52kHz triangle wave oscillator. The comparator's output controls the duty cycle of the switching transistor, adjusting the inductor current to maintain the desired output voltage. Understanding these interactions is crucial for practical applications in electronics.
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Hi,

I am working on a little electronics project where I need to convert 15V DC to 5V DC. I was hinted at the LM2576 (datasheet) because it is easy to use and has a built in current limit which is just what I need.

The functional block diagram (p12) has a nice display of the internals of the regulator:

201291213024302.jpg


I am aware of the basic principle underlying the switch regulator, however I do not understand the way the parts do interact (amp, comparator and reset). Could you briefly explain to me the principle of operation of this particular device with respect to the instances shown in the block diagram? I am trying to get some working knowledge in practical electronics and every insight by an electrical engineer would be a big help to me.
Thank you.
 
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Welcome to PF. Sorry about the delayed answer.
The output voltage to the “Load” is fed back to pin 4 where it is reduced by the R1&R2 voltage divider to about 1.23V. The “Fixed Gain Error Amplifier” amplifies the difference voltage between attenuated load voltage and 1.23V reference. The “comparator” then compares the error voltage with the voltage of the “52kHz (triangle wave) Oscillator”. The comparator digital output switches at 52kHz but the duty cycle changes according to the error voltage. From there on the signal is digital with fast switching edges through to the output switch transistor.
If the error voltage is low then the driver turns on for a greater proportion of the cycle, which increases the L1 inductor current and so maintains output load voltage.

Is that the sort of answer you wanted ?

Edit: take a look at the waveforms in this application note.
http://cds.linear.com/docs/en/application-note/an35f.pdf
 
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