Maker Pro
Maker Pro

Inductor/diode boost converter question

M

MK

Jan 1, 1970
0
Can someone verify my understanding of a boost converter?

For a basic boost (step-up) converter, once the inductor has been
charged and a magnetic field has formed, the switch to ground is opened
and the charge gets transferred to the capacitor through the inductor.

The inductor voltage spikes because the current cannot change by a
finite amount instantaneously. Once the voltage spike reaches to 0.7 V
(assuming a silicon diode) above the capacitor voltage, the diode
forward biases and transfers the charge to the capacitor.

Thus, the capacitor is only charged each cycle by a voltage slightly
above the diode turn on voltage.

Specifically then besides verifying my above observation, is my idea
about the inductor and diode also valid?

Thanks.
 
J

Jim Thompson

Jan 1, 1970
0
Can someone verify my understanding of a boost converter?

For a basic boost (step-up) converter, once the inductor has been
charged and a magnetic field has formed, the switch to ground is opened
and the charge gets transferred to the capacitor through the inductor.

The inductor voltage spikes because the current cannot change by a
finite amount instantaneously. Once the voltage spike reaches to 0.7 V
(assuming a silicon diode) above the capacitor voltage, the diode
forward biases and transfers the charge to the capacitor.

Thus, the capacitor is only charged each cycle by a voltage slightly
above the diode turn on voltage.

Specifically then besides verifying my above observation, is my idea
about the inductor and diode also valid?

Thanks.

Nope. It's driven by a current... _almost_ current-source-like.

...Jim Thompson
 
M

MK

Jan 1, 1970
0
Sorry, what is driven by a current (what is current source like?) - can
you clarify.
 
M

MK

Jan 1, 1970
0
Sorry, yes, the capacitor is charged by the charge that gets placed on
it from the inductor. But only after the voltage generated by the
inductor forward biases the diode - correct?
 
T

Tim Williams

Jan 1, 1970
0
MK said:
Sorry, yes, the capacitor is charged by the charge that gets placed on
it from the inductor. But only after the voltage generated by the
inductor forward biases the diode - correct?

Well let me put it this way. The complete, utterly horrific detailed story
is:

When the switch opens, there is some current I in the inductor (due to V = L
* dI/dt (inductor equation) integrated over the switch-ON period, minus
series resistance, which causes a slight RL exponential curve). As the
switch turns off (over perhaps 10 to 1000 nanoseconds), voltage rises and
current is transferred to the parasitic capacitance (of the inductor,
switch, diode and anything else, incidental or intentional) according to
dV/dt = I/C (capacitor equation). The capacitance limits rise time to the
shape of the leading edge of a decaying resonant pulse. For slow turn-off
times and fast (i.e., low capacitance) parts, the turn-off curve can control
the rise time instead.

Note that, when using solid state switches (transistor, diode), capacitance
is often a function of voltage; for large capacitances, the rise can start
out slow, speed up in the middle, then slow down again, because capacitance
is highest near one side or the other. I have something of an example here:
http://webpages.charter.net/dawill/Images/Induction721.jpg
This is an inductor switched with IGBTs. Rise/fall time is very quick
(comparable to the listed data of the transistors themselves), but still,
there is a noticable change in slope at the toe and shoulder of the edges.

When voltage reaches the diode turn-on voltage (in perhaps another 10 to
1000 ns), current is transferred from the capacitances to the diode. The
change in current in the diode causes an overshoot in voltage (which can be
seen in the picture as well, as that circuit uses diodes to restrain the
inductive flyback), in proportion to the loop inductance (from capacitor to
diode to inductor; note that two inductances are in series, so in a manner
of speaking, the overshoot is partly error from measuring it from some given
point along that loop!). Voltage remains here until the inductor discharges
(again, dt = L*dI/V). When current reaches zero, the parasitic capacitance
(and also charge stored in the diode, depending on reverse recovery time)
holds voltage for a short time, causing a small negative current in the
inductor. This negative current causes the voltage to alternately swing
down and up, according to the resonant frequency of the inductor vs.
capacitances. If the downward swing is sufficient, voltage may be clipped
by a diode connected in reverse across the switch (usually intrinsic, since
the switch is usually a MOSFET), clipping the peak before it continues to
resonate.

At this point, inductor current is essentially zero, and the switch can turn
on again, consuming a short peak current (depending on current risetime) due
to capacitance, and a longer current ramp due to the inductor charging.

Tim
 
D

default

Jan 1, 1970
0
Sorry, yes, the capacitor is charged by the charge that gets placed on
it from the inductor. But only after the voltage generated by the
inductor forward biases the diode - correct?

Yes but . . . the collapsing magnetic field is busy producing a high
voltage spike and that spike is allowed to charge the capacitor to a
value higher than the supply (the diode merely subtracts its forward
voltage drop from the total the inductor produces)

The overshoot caused by the collapsing field is high, but only present
for a short period of time - so you run the converter at a high
frequency and use diodes that switch quickly.
 
J

John Larkin

Jan 1, 1970
0
Can someone verify my understanding of a boost converter?

For a basic boost (step-up) converter, once the inductor has been
charged and a magnetic field has formed, the switch to ground is opened
and the charge gets transferred to the capacitor through the inductor.

The inductor voltage spikes because the current cannot change by a
finite amount instantaneously. Once the voltage spike reaches to 0.7 V
(assuming a silicon diode) above the capacitor voltage, the diode
forward biases and transfers the charge to the capacitor.

Thus, the capacitor is only charged each cycle by a voltage slightly
above the diode turn on voltage.

Specifically then besides verifying my above observation, is my idea
about the inductor and diode also valid?

Thanks.

Yup, pretty much true. Note that, since the inductor-diode current is
at a low duty cycle, the peak diode current must be a good deal higher
than the average output current.

There are lots of numerical subtleties, of course: inductor
saturation, copper loss, distributed coil+diode capacitances, stuff
like that. Semiconductor manufacturers' appnotes have lots of examples
that actually work.

John
 
H

Homer J Simpson

Jan 1, 1970
0
Sorry, yes, the capacitor is charged by the charge that gets placed on
it from the inductor. But only after the voltage generated by the
inductor forward biases the diode - correct?

Pretty much. If you want an analogy, think of a pendulum which pushes
against a spring and ratchet on each stroke. It slowly compresses the
spring, but it has to go further each time to add a little more to the
compression.

The pendulum is the inductor, the ratchet is the diode and the spring is the
capacitor.
 
J

Jim Thompson

Jan 1, 1970
0
Sorry, what is driven by a current (what is current source like?) - can
you clarify.


Jim said:
[snip]
The inductor voltage spikes because the current cannot change by a
finite amount instantaneously.
[snip]

Nope. It's driven by a current... _almost_ current-source-like.

...Jim Thompson

You answered your own question. Study it. (And don't top post ;-)

...Jim Thompson
 
J

Jim Thompson

Jan 1, 1970
0
Sorry, what is driven by a current (what is current source like?) - can
you clarify.


Jim said:
[snip]

The inductor voltage spikes because the current cannot change by a
finite amount instantaneously. [snip]

Nope. It's driven by a current... _almost_ current-source-like.

...Jim Thompson

You answered your own question. Study it. (And don't top post ;-)

...Jim Thompson

And take your question to S.E.B

...Jim Thompson
 
M

MK

Jan 1, 1970
0
Thanks for the replies. The concept is quite clear now.


Jim said:
Sorry, what is driven by a current (what is current source like?) - can
you clarify.


Jim Thompson wrote:
[snip]

The inductor voltage spikes because the current cannot change by a
finite amount instantaneously. [snip]

Nope. It's driven by a current... _almost_ current-source-like.

...Jim Thompson

You answered your own question. Study it. (And don't top post ;-)

...Jim Thompson

And take your question to S.E.B

...Jim Thompson
 
Top