1.)MATLAB PID auto tuner doesn’t work for a power converter,because PID Tuner needs a linear model to design a PID controller

Continuous conduction mode

case-1: When switch S is ON

When switch in ON for a time t_on, the diode will be open circuited since it does not allow currents in reverse direction from input to output. Hence the Buck Boost converter can be redrawn as follows

During this state the inductor charges and the inductor current increases. The current through the inductor is given as

Assume that prior to the opening of switch the inductor current is I’_{L, off}. Since the input voltage is constant

Assume the switch is open for ton seconds which is given by D*T_s where D is duty cycle and T_s is switching time period. The current through the inductor at the end of switch on state is given as

                     I_{L, on} = (1/L) *V_in*D*T_s + I’_{L, on} (equation 1)

Hence ΔI_L,on = (1/L)*V_in*D*T_s.

case 2: When switch is off

When switch in OFF the diode will be forward biased as it allows current from output to input (p to n terminal) and the Buck Boost converter circuit can be redrawn as follows

The inductor now discharges through the diode and RC combination. Assume that prior to the closing of switch the inductor current is I’’_{L, off}. The current through the inductor is given as     

Note the negative sign at the front end of equation signifies that the inductor is discharging. Assume the switch is open for t_off seconds which is given by (1-D)*T_s where D is duty cycle and T_s is switching time period. The current through the inductor at the end of switch off state is given as

                                I’’’_{L, off} = -(1/L) *V_out*(1-D)*T_s + I’’_{L, off} (equation 2)

In steady state condition as the current through the inductor does not change abruptly, the current at the end of switch on state and the current at the end of switch off state should be equal. Also the currents at the start of switch off state should be equal to current at the end of switch on state. Hence

                                             I’’’_{L, off} =I_{L, on} also I’_{L, off} =I’’_{L, off}

Using the equations 1 and 2 we get

(1/L) *V_in*D*T_s =  (1/L) *V_out*(1-D)*T_s

V_in*D =V_out*(1-D) 

V_out/V_in = D/(1-D)

Since D < 1>out can be greater than or less than V_in.For D>0.5 the Buck boost converter acts as boost converter with V_out >V_in.For D<0>out >V_in.

Assuming no losses in the circuit and applying the law of conservation of energy

                                                            V_out*I_out = V_in*I_in

This implies I_out/I_in = (1-D)/D, Thus I_out > I_in for D<0>out < I>in for D<0>

2.)importance of ADC and DAC blocks in closed loop control of power converter. 

similarly ADC And  DAC helps to convert the signal from continuous to discrete and vice versa.

 3.)