Note Vout75V State the type of switching regulator shown in

Note: Vout=7.5V

State the type of switching regulator shown in FIGURE 2, and briefly explain its operation. In FIGURE 2, R_1 = 10 k omega, V_ref = 1.25 V and the duty cycle D is 30%. Determine: the required value of R_2 [select the nearest preferred value from the E24 series]. The value of V_IN.

Solution

Ans)

a)

The switching regulator shown in figure2 is Buck Regulator/Converter(Step down Converter)

The basic operation is to steps down voltage from its input (supply) to its output (load)

The TR1 --Transistor shown in the figure acts as switch i.e it will on and off with certain frequency and duty cylce.Whenever switch is on and off you will get discontinuos voltage of Vin and 0 depending upon the duration of on period you will get different output voltages ,the inductor and capcitor acts as a filter to smoothen the discontinuos voltage to constant voltage

b)

The resistors R1 and R2 combination sets the required duty cycle which in turn sets the output voltage as

Output voltage Vo=D*Vin where D= duty cyle and Vin = input voltage

D=Ton/T where swith on time Ton and T is the total time period of oscillator

Vref of 1.25 corresponds to Full output voltage=input voltage so in order to set 30% duty cyle R1 and R2 combination should give 30% of Vref

so 30% of 1.25=1.25*30/100=0.375 V =Voltage across R2(Vr2)

output voltage Vo=7.5 V given

Voltage across R1= Vr1=Vo-Vr2=7.5-0.375=7.125 V

I=Vr1/R1=7.125/10K=0.7125 mA

R2=Vr2/I=0.375/0.7125m=526.32 ohms

So required R2=526.32 ohms

Either 510 or 560 ohms from E24 series by rounding to higher value you can go for 560 ohms ,but you will

get slightly higher output voltage

ii) D=30%=0.3

Vin=Vo/D=7.5/0.3=25 V

So Vin=25 V