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Boost energy & power flow (i/p to o/p)

R

reggie

Jan 1, 1970
0
Hi all,

I am trying to understand the flow of energy an d thus power through a
boost converter. I understand how the circuit works, but am a little
confused on some concepts proposed by A.Presmans book switching power
supply design second edition.

(For those of you with the book it’s the equation 1.13 on page 26 that
I am not sure about)

The books is describing the power throughput of a discontinuous boost
converter with an input full wave rectified line voltage of Vdc,

The equation that I am struggling with its meaning is:

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

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. This equation seems to supply
energy from the input during the fet off time because of the Vdc terms
inclusion in this equation.

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?

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.

Can someone please explain the energy and power transfer for a
discontinuous boost converter from input to output in words.

Thanks in advance,

Reggie.
 
R

reggie

Jan 1, 1970
0
Hi Reggie,


"The equation that I am struggling with its meaning is:

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 theload
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 theload
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

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Thanks Joel,

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..

Thanks Joel for this. What you have said above makes perfect sense to
me.
The way I'd think of it is...

-- The input voltage is assumed to be constant, Vdc
Agree

-- The average current pulled from the power supply is the same as the average
current in the inductor, Ip/2
Agree

-- Hence -- while the FET is off -- the power supply delivers Vdc*Ip/2 watts.

Not so sure, Average current Iout = area under curve/length of base.
Area of decaying current triangle with base of Tr and height of Ip;

area = 1/2Tr*Ip and length of base is Tr so

Average current Iout (over Tr not T) = (1/2Tr*Ip)/ Tr = Ip/2

This has a time element included, which makes me think that the Vdc
needs to be multiplied by a time element (the Tr/T).
-- Over a complete switching cycle, the FET is off for Tr/T ( * 100%) of the
time.

Not so sure as described above
-- Hence, on average the power supply delivers Vdc * Ip/2 * Tr/T watts to the
load (without taking into account what the inductor does).

Agree, I know I am nit picking, and the result is the same but its an
understanding I am after, how would I construct those formulas without
the book is what I am trying to achieve…

So is this statement inaccurate

Pdc = [(average input voltage during tr) X (Average current during
tr)]
+ (what the inductor does)

Pdc = (average power during tr only as when fet is on Vdc is not
supplying load)
+(1/2.L.Ip.Ip/T)

I am now not sure why (1/2 .L.Ip.Ip/T) there is a T and not a Ton term
as the inductor is filled up when during ton not the whole cycle T.

Does what I am saying make sense to you?

Am I way off in my thoughts and grasp of what these equations mean?

The problem is if I cannot remember equations only derivations

Thanks again,

Reggie..
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