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                                                                12v to 5v Buck Converter Circuit


Hey, Guys!  Today I have no place for travelling and sit in front of the computer and thought of a simple question that “what’s 12v to 5v Buck Converter Circuit”.  Then let’s me fix this question below.
 
This is a typical DC-DC buck converter circuit. The core component is LM2576-5, is 5V fixed voltage. 
 
Working principle:
 
12V input voltage through anti-reverse Schottky diode D1, put into the LM2576-5 NO.1 pin (VIN port, also connect with the internal switch). The other route go by R10 and L3 for power indication. LM2576-5 NO.2 pin inside is the emitter switch, the external energy storage capacitor LL1 and Schottky diode D2 connected. LM2576-5 NO.4 pin is the  output voltage samplle side, the internal is a resistor divider, after the sample voltage  being splited then send to the non-inverting input, comparision with the 1.23V that connected by reverse phase, then comparision with the output voltage with the built-in OSC oscillation signal, and again comparision the output signal is again associated with the reset signal and non-retransmission, such as the boost stage, which ultimately controls the operating state of the switching transistor .
 
When 12 voltage in the built-in switch tube conductioned, added to the LL1 and load (R11 and L4), to provide the power to load, meanwhile charging LL1 magnetization, C20, C21 charge (C20 is the filter capacitor, C21 can be considered as the peak absorption capacitor) The LM2576-5  NO.4 pin, take the samples of the load voltage, when the voltage rises to a certain value (threshold), the internal circuit auto dealth with, will turn off the switch, LM2576-5 N0.2 pin without output, due to the inductance LL1,The Schottky diode D2 is turned on and LL1 and C20 continue to supply power to the load. D2 tunred on only in the switch off, for the LL1 circuit back, which is called freewheeling diode. During this time, the load voltage will slowly decrease until below the threshold, the internal circuit again forced the switch to turn on, repeat the aforementioned process.
 
Hence, the load above could get the stable 5V voltage.