rockycheng
Member level 5
It is said that the error amplifier ouput Veao is a constant value, but I have some doubt. Let's consider the load regulation case as an example.
Assuming the transient load resistance R decreases, then the average inductance current Iav shall increase to make Vo(=R*Iav) remain the same. On the other hand, the duty cycle D(=Vo/Vin) is not changed, since both Vo and Vin are contants. And both the upwards and downwards current rates (m1 and m2, respectively) are not changed, for the same reason. So, in order to make all the requirements above satisfied simultaneously, the only way seems to move the whole current curve up, while keeping D, m1, m2 the same. This way, Veao (no matter with or without the ramp compensation) has to be increased to an appropriate value, and keep there, in the following steady states. Since the gain of the error amplifier is quite large, the voltage of the inverting input node could still be considered equal to Vref.
Assuming the transient load resistance R decreases, then the average inductance current Iav shall increase to make Vo(=R*Iav) remain the same. On the other hand, the duty cycle D(=Vo/Vin) is not changed, since both Vo and Vin are contants. And both the upwards and downwards current rates (m1 and m2, respectively) are not changed, for the same reason. So, in order to make all the requirements above satisfied simultaneously, the only way seems to move the whole current curve up, while keeping D, m1, m2 the same. This way, Veao (no matter with or without the ramp compensation) has to be increased to an appropriate value, and keep there, in the following steady states. Since the gain of the error amplifier is quite large, the voltage of the inverting input node could still be considered equal to Vref.
