Circuit Diagram: https://components101.com/articles/boost-converter-basics-working-design
Good day Fellow Electrical/Electronic Engineers I am having a conceptual challenge here when deriving the DC Current transfer functions (i.e Relationship between Input Current (Ii) from Unregulated DC input power supply and Load Current (Io) drawn by the load from the regulated Boost Converter operating in Continuous Conduction mode). So the challenge is this, when i use the Volt-Second balance equations and derive the ON-time and OFF-time ratios which are equal to the Voff and Von voltage ratio....i can actually replace the ON-time with Voff, and OFF-time with Von in the Duty cycle equation to get a relationship between Vin (unregulated DC voltage input) and Vo (Regulated DC voltage output) connected by the Duty cycle for switching ON and OFF the MOSFET or BJT switches. Using the Power Balance equation, i.e Pin=Pout, which implies 100% efficiency, assuming no power losses in the rectifier diode when forward biased and the BJT in saturation mode, i then found that Ii=Io/(1-D). Now, if i try to derive this expression, i.e Ii=Io/(1-D), using basic concepts i get stuck. WHY? In continuous conduction mode, the Inductor current of the Boost regulator is not allowed to discharge till the Inductor current reaches 0A or NO conduction in the OFF-state. This means the inductor does not completely discharge before the next ON-state cycle begins. Now, during the OFF-state of the BJT or MOSFET switch, the inductor current discharges and the current waveform is a negative gradient moving from the peak inductor current to some arbitrary current below the average inductor current=input current (Ii). Literature seems to show the average load current as Io=Ii*Toff/T=Ii*(1-D)...which implies Ii=Io/(1-D). However, when i analyze this i actually get confused because then it implies that the total charge discharged during the OFF-state is Ii*Toff and that charge which is discharged during the OFF-state actually charges the output Capacitor of the Boost Regulator with Io*Ton charge and also supplies the load with Io*Toff charge which implies that during the OFF-state the inductor charges the Capacitor with enough charge that it, the Capacitor, can supply the load Io current for Ton duration during the ON-state when the inductor is cut-off from the output circuit by the rectifier diode. It also means that the inductor supplies Io current for Toff duration during the OFF-state.. However, where does this charge come from....If Ii*Toff=Io*Ton+Io*Toff=Io*(Ton+Toff)=Io*T...this implies that the charge to charge the capacitor and supply the load during the OFF-state is actually the total charge, or Area, under the current discharge waveform of the Inductor in the OFF-state which contradicts the Continuous Conduction Mode philosophy. Who can assist?
It introduces right half plane zero in control to input transfer function.Which means change in duty ratio of switch will first dip the output voltage and than rise to steady dtate.
Consider driving a car without hill start support on a sloped road. Even though you want to go forward, there is a situation where the car slides backwards to a certain extent. For the boost converter, this creates an effect that delays the system reaching stability.
In control systems, this situation is known as undershoot and is caused by a positive zero in the denominator of the system's transfer function (zero in the right half plane).