uuuuu said:

Are you saying that the reverse voltage which is induced on L can be

greater than the forward voltage that was there just before the

switching occured, i.e. greater by 0.6V (diode drop) ?

Sure. For example, if the inductor has 7 volts switched into it, and

5 volts out, there is only 2 volts causing current ot increase when

the switch is on, but when the switch turns off, there is -(5 +.6) or

so across it as the inductor produces enough induced voltage to

forward bias the diode.

Is it possible for the induced voltage to be arbitrarily high ?

Yes. The induced voltage is limited only by the speed that the switch

turns off, and the capacitive load across the inductance and the

parallel resistance across the inductance (that represents the current

induced in the core). If the inductor and switch were perfect and the

stray capacitance approached zero, the induced voltage would approach

infinity (but would last for a time approaching zero) and the inductor

current would quench to zero almost instantaneously.

Also, can you please explain the formula in fig. 28, i.e. how is this

formula arrived at ?

This equation just describes the averaging effect of a low pass filter

with no losses. The average voltage over the time of one pulse cycle

(Tp) of a pulse that spends Ton time at Vpk, and (Tp-Ton) at zero

volts is:

Vout = ((Ton*Vpk + 0*(Tp-Ton))/Tp = (Ton*Vpk)/Tp

Of course, it a real circuit, the off state voltage is one diode drop

negative, and the inductor has some resistive voltage drop, so this

formula overestimates the output voltage.

Also, I believe you did not say why the voltage on L increases and decreases

linearly and produces a saw-tooth wave form, which formula are you using ?

The voltage across L is assumed to have one of two fixed values during

a cycle. when the switch is on, it is assumed to be the unregulated

voltage supply, and when the switch is off, it is assumed to be the

negative of the output voltage plus a diode drop.