Hi Reggie,
Pdc=Vdc.(Ip/2)(Tr/T)
Pdc= power delivered to load during off time Vdc= input voltage Ip= average inductor current over Tr which equals Ip/2 during Tr Tr= time for current in inductor to decay from peak to zero (reset time) Tr/T = the duty factor of reset time"
OK.
"What I am struggling with is what it means. I thought a boost converter filled op the inductor during the fet on time and emptied the inductor during the fet off time."
Correct.
"This equation seems to supply energy from the input during the fet off time because of the Vdc terms inclusion in this equation."
Yes, this is correct also. Perhaps the way to think of this is... the DC supply provides some energy to the load, and it really is the inductor that provides the "boost!" Consider: If the inductor were very small, when the FET is off you'd still have pretty much a direct connection between the power supply and the load. That "direct connection" results in a power to the load given by the equation above. But since the inductor provides a "boost" as well, you have to account for power, which Pressman does in equation 1.12, and adds it to your equation above to get the total power delivered to the load.
"Does this equation state Pdc = average input voltage during tr. Average current during tr ? As
1) Input voltage during tr = Vdc(Tr/T) ?
2) Ip/2 = average inductor current during tr?"
The way I'd think of it is...
-- The input voltage is assumed to be constant, Vdc
-- The average current pulled from the power supply is the same as the average current in the inductor, Ip/2
-- Hence -- while the FET is off -- the power supply delivers Vdc*Ip/2 watts.
-- Over a complete switching cycle, the FET is off for Tr/T ( * 100%) of the time.
-- Hence, on average the power supply delivers Vdc * Ip/2 * Tr/T watts to the load (without taking into account what the inductor does).
"I know the current through the inductor during the fet off time is the input current and the load and capacitor current combined but the Vdc term is confusing me."
All the power supply "knows" is that there's something out there (the inductor in this case) pulling current from it, so since it's trying to present a fixed DC output voltage, it must be providing power to whatever that "thing" is. Since an ideal inductor doesn't dissipate any power, the power must end up being dissipated in your load.
"Can someone please explain the energy and power transfer for a discontinuous boost converter from input to output in words."
Briefly: During the FET on time, the power supply stores energy in the inductor. During the FET off time, the power supply is connected to the load in series with the inductor. Hence, you end up with both the power supply and the inductor providing energy to the load.
I'll mention that when I was first designing switching power supply, one of the concepts that took awhile to sink in is that you have to be watchful for what's really "controlling" various currents and voltages: If you have an inductor with a certain current in it and then suddenly change the circuit topology, that inductor is going to still force the same current at "t-zero-plus" to be the same as it was at "t-zero-minus." I know this is gone over in EE courses in school, but it takes awhile to get an intuitive "feel" for what's going on. Just as capacitors soak up charge and present a voltage that takes effort to change (i.e., it acts like a little battery), inductors soak up current (well, flux, really) and then present some loop with a current that takes effort to change (i.e., it acts like a little constant current source).
---Joel