Question

220 HH Boost circuit: 1. Draw a (simplified) schematic that shows 180 Ω how current flows when Q1 is on. flows when Q1. is off. cycle of Q1 increases? the operation of the regulator? the operation of the regulator? the operation of the regulator? 1N5819 2. Draw a schematic that shows how current 3. What happens to Vout when the duty 4. What impact does the value of R2 have on v 5. What impact does the value of L have on 6. What impact does the value of CT have on 91 12 V 1500 125-V Referance 100 μF Comperatar Regulator R2 Vou Co23 V17 47 kn R1 2.2เถ 330 uF Buck circuit: Answer the same questions as above. 9.2.3 Step-Down Converter Application Figure 10. Step-Up Converter 91 2 1N5819 Oscillstor CT 25 V 1.25V Reference Referen-11470 pr 11220 μΗ 100 uF Comparator Reguiato 5 VIS00 mA R1 1.2 ㏀ 470Got Seanned with CamScanner Figure 11. Step-Down Converter

Answer 1

Boost converter:

1)

When the switch (Q1) is on, the current starts flowing from supply to ground through inductor and switch. This current starts increasing as v=Ldi/dt is the rule for the inductor and input voltage is applied across the inductor

2)

When the switch turns off, the inductor current cannot die out instantaneously. The inductor current starts decreaseing and the current flows from input to output through the diode and the inductor.

3) When the duty cycle increases, the output voltage increases. The voltage and duty ratio for a boost converter are related by the equation, Vout=Vin/(1-d).

4)As R2 increases, the output voltage increases. The duty ratio is also increased. The reason is, the control system used here is voltage mode control. i.e.; compare the scaled version of the output voltage to a reference voltage and change the duty cycle until these voltages are matched. When R2 is increased, the potential divided output voltage which is Vout*(R1/(R1+R1)) comes down. The duty ratio increases until this voltage increases to becomes equal to the reference voltage.

5) As L increases, the variation in inductor current becomes low. i.e.; the ripple in the inductor current reduces and at larger inductance values, the inductor current can almost reach a constant state. This doesn't have any effect on the output voltage though. If the inductor value is reduced very much, the inductor current will reach zero during the off period of the switch and this leads to Discontinuous operating mode of the converter.

6) CT is the timing capacitor that generates the clock signal for timing for switching the transistors. As Ct increases, the time taken by it to charge to a particuar level increases. So the switching frequency reduces. This requires larger inductors to maintain the circuit in continuous operation mode. Also the size of the capacitors for the same output ripple specification also increases. if the capacitor size is not increased, the output peak to peak ripple voltage increases.

BUCK:

1)

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When the switch is on, the current passes through swtich and indutor into the output capacitor and load.

2)

When the switch is off, the inductor current which was increasing during the switch was on, starts decreasing slowly. But the current needs some path to flow. the diode becomes forward biased and provides a path for the current.

3)

The relation for buck converter is Vout=d*Vin. As the duty ratio increases, the output voltage also increases.

4) As explained above, as d increases, the output voltage increases and so when R2 is increased, the output voltage needs to increase to a larger value so that the scaled-down value of the output equals the reference voltage.

5)As L increases, the peak to peak ripple in the inductor current is reduced and so as this ripple current passing through the capacitor produces the output voltage ripple, the output voltage ripple is also reduced

6)

As Ct increases, the frequency of switching reduces and so the ripple voltages and currents are increased. If the ripples go beyond a particular range, the system goes to discontinuous mode.