While going through internal structure of a buck regulator, there was a feedback loop connected from the voltage divider circuit therein connected to a comparator and the input fed to the SR flipflop R end. How does this feedback control the duty cycle ?
The buck converter does not absolutely need a pulse generator to maintain its operation. The comparator (a) turns on when it senses a drop in output voltage, then (b) turns off when sufficient juice has been admitted to raise the voltage.
The result is a pulse train which operates the buck converter. Voltage regulation is pretty much automatic with this method.
This method requires a certain amount of hysteresis, achieved by a resistive or capacitive network. The aim is to provide snap action. High gain is able to achieve this. A plain logic gate can do the job as well as the comparator and flip-flop.
I made a simulation where two transistors (as a sziklai pair) drive a buck converter.
A single transistor might be persuaded to oscillate if its gain is sufficiently high. However a condition can happen where it stays on constantly at some middle extent, operating as resistive drop instead of On-Off fashion.
Switch regulator feedback circuits may differ greatly in details, but they all function similarly:
If the output voltage is to high, they adjust the PWM duty-cycle to reduce the voltage.
If the output voltage is to low, they adjust the PWM duty-cycle to increase the voltage.
If the output voltage is just right, the the PWM duty-cycle is not changed.
You have to examine the circuit design to determine how that is done for a particular circuit.
Typically it's done with a comparator the has the error voltage as one input and a sawtooth fed to the other input.
Hi all Thanks for the replies;
please find the attachment,
the regulator is operated in current mode, it consists of an internal clock generator, error amplifier and current sense.
I have 2 questions
The error signal is compared with the current slope ramp signal and that is given to the reset input of the sr filpflop. why is this done?
Also why not connect the error signal directly to the reset input? (can this method also control the duty cycle?)
Don't know what you mean with "current slope ramp". The error signal is primarily compared with current measurement, the current measurement is optionally modified by slope compensation signal. To understand the working of this topology, start with the simple case without slope compensation. Then learn when slope compensation is necessary.
Also why not connect the error signal directly to the reset input? (can this method also control the duty cycle?)
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the regulator is operated in current mode, it consists of an internal clock generator, error amplifier and current sense.
I have 2 questions
The error signal is compared with the current slope ramp signal and that is given to the reset input of the sr filpflop. why is this done?
Also why not connect the error signal directly to the reset input? (can this method also control the duty cycle?)
If the error signal is fed directly to reset input of the SR flip flop, then peak current limiting would not be achieved. Comparing the error signal with the measured current inferred voltage, allows for current limiting to be achieved. The error signal input of the comparator always have a maximum voltage of some sort that switches the pwm when the current sense voltage gets to that value.
The slope compensation signal is used in situations where duty cycle is greater than 50% to modify the slope of the current sense signal basically for effective control.
The pwm can still be controlled without current sense but you'll have to use the voltage mode control scheme.