Buck circuit question

[Ben]

Hello,

Been working with a buck circuit, today and think I've got in a bit of
mix.

Sorry for the bad ascii art!



cct:

Switch Inductor
---------- \------|----LLLLLLL-----|---------o +ve output 5V DC
| |
| |
/ \ ------
Diode --- ------ Capacitor
| |
| |
| |
---------------------------------------------o ground 0V DC


Diode is a low power shotky switch is actually transistor.

Ok m so when the switch is closed the current flows through the
inductor, creating a magnetic field. The cap charges and the output
voltage of 5V is reached. So the p.d at the output is 5V say at 500mA
to keep things simple.

When the switch is opened, there is a back e.m.f generated by the
inductutor. Which basically means that the p.d at the didoe and
inductior is -5V. The diode is now reverse biased. So current flows
through the inductutor to the cap keeping the output voltage a 5V.

What about the current does it still stay at 500mA? I thought that caps
could not store energy?

The process is then repeated to said f when the switch is again opened
and closed.

Many thanks.

Neb

 

[DJ Delorie]

Ben <nebATwalrus81DOTplusDOTcom> writes:

Switch Inductor
---------- \------|----LLLLLLL-----|---------o +ve output 5V DC
| |
| |
/ \ ------
Diode --- ------ Capacitor
| |
| |
| |
---------------------------------------------o ground 0V DC


What about the current does it still stay at 500mA? I thought that
caps could not store energy?

The inductor acts like a "current smoother", much like a capacitor
"stores" voltage, an inductor "stores" current.

So think of it in terms of current, not voltage:

When the switch is closed, the inductor's current gradually increases.
When it exceeds the current draw on the output, the excess is used to
charge the capacitor, and the voltage increases. When the voltage on
the cap exceeds +5v, the switch is opened.

When the switch is open, the inductor's current gradually decreases
(using the diode). When it no longer exceeds the current draw on the
output, the capacitor provides the additional current, which
discharges the capacitor. When its voltage drops below +5v, the
switch is closed.

The switching frequency is just the fastest that the switch changes
state. Since the switch is either "on" or "off" (and never halfway,
like a resistor), you minimize heat losses there. The inductor and
capacitor are "ideal" in that the only heat loss is from their
inefficiencies and leakages. Most of the power loss is at the diode,
since it has a fixed Vf. Some switchers add a mosfet in parallel with
the diode to reduce those losses also.

 

[Ben]

DJ Delorie wrote:

Ben <nebATwalrus81DOTplusDOTcom> writes:
Switch Inductor
---------- \------|----LLLLLLL-----|---------o +ve output 5V DC
| |
| |
/ \ ------
Diode --- ------ Capacitor
| |
| |
| |
---------------------------------------------o ground 0V DC


What about the current does it still stay at 500mA? I thought that
caps could not store energy?

The inductor acts like a "current smoother", much like a capacitor
"stores" voltage, an inductor "stores" current.

So think of it in terms of current, not voltage:

When the switch is closed, the inductor's current gradually increases.
When it exceeds the current draw on the output, the excess is used to
charge the capacitor, and the voltage increases. When the voltage on
the cap exceeds +5v, the switch is opened.

When the switch is open, the inductor's current gradually decreases
(using the diode). When it no longer exceeds the current draw on the
output, the capacitor provides the additional current, which
discharges the capacitor. When its voltage drops below +5v, the
switch is closed.

The switching frequency is just the fastest that the switch changes
state. Since the switch is either "on" or "off" (and never halfway,
like a resistor), you minimize heat losses there. The inductor and
capacitor are "ideal" in that the only heat loss is from their
inefficiencies and leakages. Most of the power loss is at the diode,
since it has a fixed Vf. Some switchers add a mosfet in parallel with
the diode to reduce those losses also.

Ok,

I think that I've got it sorted now.

So when the switch is closed the inductor provides the current for the
load, which provides the votlage for the load (via the cap) and the
current also.

When the switch is closed the inductor discharges slowly via diode and
once the CURRENT drops the cap takes over with current, once the cap has
been exhausted the voltage drops and an error amp/fb loop turns the
swithc on again.

Or have I got in a slight mess?

 

[Ben]

Forgot to say thanks!

DJ Delorie wrote:

Ben <nebATwalrus81DOTplusDOTcom> writes:
Switch Inductor
---------- \------|----LLLLLLL-----|---------o +ve output 5V DC
| |
| |
/ \ ------
Diode --- ------ Capacitor
| |
| |
| |
---------------------------------------------o ground 0V DC


What about the current does it still stay at 500mA? I thought that
caps could not store energy?

The inductor acts like a "current smoother", much like a capacitor
"stores" voltage, an inductor "stores" current.

So think of it in terms of current, not voltage:

When the switch is closed, the inductor's current gradually increases.
When it exceeds the current draw on the output, the excess is used to
charge the capacitor, and the voltage increases. When the voltage on
the cap exceeds +5v, the switch is opened.

When the switch is open, the inductor's current gradually decreases
(using the diode). When it no longer exceeds the current draw on the
output, the capacitor provides the additional current, which
discharges the capacitor. When its voltage drops below +5v, the
switch is closed.

The switching frequency is just the fastest that the switch changes
state. Since the switch is either "on" or "off" (and never halfway,
like a resistor), you minimize heat losses there. The inductor and
capacitor are "ideal" in that the only heat loss is from their
inefficiencies and leakages. Most of the power loss is at the diode,
since it has a fixed Vf. Some switchers add a mosfet in parallel with
the diode to reduce those losses also.

 

[DJ Delorie]

Ben <nebATwalrus81DOTplusDOTcom> writes:

So when the switch is closed the inductor provides the current for
the load, which provides the voltage for the load (via the cap) and
the current also.

When the switch is closed the inductor discharges slowly via diode
and once the CURRENT drops the cap takes over with current, once the
cap has been exhausted the voltage drops and an error amp/fb loop
turns the swithc on again.

Aside from saying "closed" twice (the second one should be "open"),
yes.

 

[Ben]

Excelent,

thanks, I was focussing on V and then i rather than looking at i. It
makes sense, L charges c abd provides load, then c provides load and v
drops then repeats

thanks very much


DJ Delorie wrote:

Ben <nebATwalrus81DOTplusDOTcom> writes:

So when the switch is closed the inductor provides the current for
the load, which provides the voltage for the load (via the cap) and
the current also.

When the switch is closed the inductor discharges slowly via diode
and once the CURRENT drops the cap takes over with current, once the
cap has been exhausted the voltage drops and an error amp/fb loop
turns the swithc on again.

Aside from saying "closed" twice (the second one should be "open"),
yes.

 

[Jon Slaughter]

"Ben" <nebATwalrus81DOTplusDOTcom> wrote in message
news:13l0lao7rhi06bf@corp.supernews.com...

Hello,

Been working with a buck circuit, today and think I've got in a bit of
mix.

Sorry for the bad ascii art!

Yes, it very bad. You need to use a fixed width font when doing ascii so
each character gets the same space. Else it could make no sense. You could,
for example, use word or notepad and select a fixed width font and then copy
it into the email if you cannot change the font in the email program(you can
do it with outlook if you go into options).

cct:

Switch Inductor
---------- \------|----LLLLLLL-----|---------o +ve output 5V DC
| |
| |
/ \ ------
Diode --- ------ Capacitor
| |
| |
| |
---------------------------------------------o ground 0V DC


Diode is a low power shotky switch is actually transistor.

Ok m so when the switch is closed the current flows through the inductor,
creating a magnetic field. The cap charges and the output voltage of 5V
is reached. So the p.d at the output is 5V say at 500mA
to keep things simple.

When the switch is opened, there is a back e.m.f generated by the
inductutor. Which basically means that the p.d at the didoe and inductior
is -5V. The diode is now reverse biased. So current flows
through the inductutor to the cap keeping the output voltage a 5V.

What about the current does it still stay at 500mA? I thought that caps
could not store energy?

lol, no. Caps do store energy. They are like a tank of water. You can pour
energy in and drain it out later if you want. As long as you pour more than
you take it will always be full(on average). The capacitor there is to act
like a temporary battery while the main battery is disconnected.

Think about it this way, if you charged the capacitor and stuck its leads
across a resistor then current would flow. The big difference between a cap
and a battery is that a battery supplies constant voltage while the voltage
on a cap will decrease over time and it depends on the charge it has stored
and its capacitance. (so it depends on how fast your removing that charge,
i.e., current)

Resistors do no store energy. Inductors and capacitors do. This is reason
you don't see a resistor in there.

For example, if your load is very low and your capacitor is very big then
you don't need an inductor and diode there. The issue here is that you want
to minimize the capacitor size cause you have to quickly fill it up when the
power supply starts. If its to big you could draw excessive current from the
PS and destroy something. But that would mean you could leave the switch off
much longer too. Although it would probably be an extremely noisy method.

The process is then repeated to said f when the switch is again opened and
closed.
Neb

http://www.hills2.u-net.com/electron/smps.htm

 

[John Larkin]

On Fri, 30 Nov 2007 18:24:23 +0000, Ben <nebATwalrus81DOTplusDOTcom>
wrote:

Hello,

Been working with a buck circuit, today and think I've got in a bit of
mix.

Sorry for the bad ascii art!

Looks fine to me.

cct:

Switch Inductor
---------- \------|----LLLLLLL-----|---------o +ve output 5V DC
| |
| |
/ \ ------
Diode --- ------ Capacitor
| |
| |
| |
---------------------------------------------o ground 0V DC


Diode is a low power shotky switch is actually transistor.

Ok m so when the switch is closed the current flows through the
inductor, creating a magnetic field. The cap charges and the output
voltage of 5V is reached. So the p.d at the output is 5V say at 500mA
to keep things simple.

Right. The diode is now reverse biased, off.

When the switch is opened, there is a back e.m.f generated by the
inductutor. Which basically means that the p.d at the didoe and
inductior is -5V. The diode is now reverse biased.

No. When the transistor is off, the diode is forward biased. It keeps
the voltage from swinging very much negative, which it would do if the
diode weren't there. The diode "clamps" the switch node from swinging
negative.


So current flows

through the inductutor to the cap keeping the output voltage a 5V.

Well, it keeps pumping current into the capacitor and the load.

What about the current does it still stay at 500mA? I thought that caps
could not store energy?

Caps do store energy. So do inductors.

The current through the inductor builds up when the transistor is on,
and decays when it's off. Those phases correspond to adding energy to
the inductor, and taking it out. If the L is big enough and the
frequency is high enough, the current through the inductor is *almost*
constant. In real life, there's a pretty-much triangular ripple
current in the inductor, because nobody wants to buy huge inductors.

The ripple current in the inductor in turn causes a bit of ripple
voltage across the capacitor, and the load.

John

 

[John Larkin]

On Fri, 30 Nov 2007 13:49:36 -0600, "Jon Slaughter"
<Jon_Slaughter@Hotmail.com> wrote:

"Ben" <nebATwalrus81DOTplusDOTcom> wrote in message
news:13l0lao7rhi06bf@corp.supernews.com...
Hello,

Been working with a buck circuit, today and think I've got in a bit of
mix.

Sorry for the bad ascii art!

Yes, it very bad. You need to use a fixed width font when doing ascii so
each character gets the same space. Else it could make no sense. You could,
for example, use word or notepad and select a fixed width font and then copy
it into the email if you cannot change the font in the email program(you can
do it with outlook if you go into options).

It looked OK to me, Agent with a fixed-pitch font.

John

 

[Ben]

"The diode "clamps" the switch node from swinging

negative."

?

So I'm a bit confused. So am I correct in thinking that thorught out
the on/off phase, the current always flows from the inductor to the
capacitor. The back e.m.f would go to the switch when turned off if it
were not for the diode?

I thought that the diode would conduct as it were revese biased by the
back e.m.f

I'm all condused again!



John Larkin wrote:

On Fri, 30 Nov 2007 18:24:23 +0000, Ben <nebATwalrus81DOTplusDOTcom
wrote:

Hello,

Been working with a buck circuit, today and think I've got in a bit of
mix.

Sorry for the bad ascii art!




Looks fine to me.


cct:

Switch Inductor
---------- \------|----LLLLLLL-----|---------o +ve output 5V DC
| |
| |
/ \ ------
Diode --- ------ Capacitor
| |
| |
| |
---------------------------------------------o ground 0V DC


Diode is a low power shotky switch is actually transistor.

Ok m so when the switch is closed the current flows through the
inductor, creating a magnetic field. The cap charges and the output
voltage of 5V is reached. So the p.d at the output is 5V say at 500mA
to keep things simple.

Right. The diode is now reverse biased, off.

When the switch is opened, there is a back e.m.f generated by the
inductutor. Which basically means that the p.d at the didoe and
inductior is -5V. The diode is now reverse biased.


No. When the transistor is off, the diode is forward biased. It keeps
the voltage from swinging very much negative, which it would do if the
diode weren't there. The diode "clamps" the switch node from swinging
negative.


So current flows
through the inductutor to the cap keeping the output voltage a 5V.

Well, it keeps pumping current into the capacitor and the load.



What about the current does it still stay at 500mA? I thought that caps
could not store energy?

Caps do store energy. So do inductors.

The current through the inductor builds up when the transistor is on,
and decays when it's off. Those phases correspond to adding energy to
the inductor, and taking it out. If the L is big enough and the
frequency is high enough, the current through the inductor is *almost*
constant. In real life, there's a pretty-much triangular ripple
current in the inductor, because nobody wants to buy huge inductors.

The ripple current in the inductor in turn causes a bit of ripple
voltage across the capacitor, and the load.

John

 

[Jamie]

Ben wrote:

"The diode "clamps" the switch node from swinging
negative."

?

So I'm a bit confused. So am I correct in thinking that thorught out
the on/off phase, the current always flows from the inductor to the
capacitor. The back e.m.f would go to the switch when turned off if it
were not for the diode?

I thought that the diode would conduct as it were revese biased by the
back e.m.f

I'm all condused again!

Yes, the diode will short the back EMF, Also called wheeling voltage
and thus a wheeling diode.
Polarity is reversed on the EMF.

Which leads to some interesting facts here. Since this is true, the
polarity on the other end must be (+) in which case can increase the
voltage on the cap. So the current cycle when the switch is on will
most likely not be on long enough to introduce the full voltage
on the output side of the inductor because, when it shorts via the
diode. Additional voltage will appear at the output. The regulating
circuit must be in tuned to this process of course.

I may have over looked something but that is just about the gist of
it.




--
"I'd rather have a bottle in front of me than a frontal lobotomy"
http://webpages.charter.net/jamie_5

 

[John Larkin]

On Fri, 30 Nov 2007 20:43:08 +0000, Ben <nebATwalrus81DOTplusDOTcom>
wrote:

"The diode "clamps" the switch node from swinging
negative."

?

So I'm a bit confused. So am I correct in thinking that thorught out
the on/off phase, the current always flows from the inductor to the
capacitor. The back e.m.f would go to the switch when turned off if it
were not for the diode?

Current always flows through the inductor [1] in the direction of the
load. If the diode weren't there, terrible things would happen when
the transistor turns off. The voltage at the switch node would try to
swing to negative infinity, the transistor would break down from
overvoltage, and the transistor would probably die. Even if the
transistor didn't die, it would get very hot and the thing wouldn't
act like a switching regulator.

I thought that the diode would conduct as it were revese biased by the
back e.m.f

"Back emf" is a fuzzy and hazardous term. I only use it with reference
to DC motors, where it actually means something.

When the transistor is off, current flows from ground, *forward*
through the diode, through the inductor, into the capacitor. In that
state, the energy stored in the inductor drives the load.


Redraw it as...


SPDT
Switch Inductor
--------o\
\o----------LLLLLLL------+---------o +ve output 5V DC
o |
| ------
| ------ Capacitor
| |
| |
| |
--------+--------------------------+---------o ground 0V DC


and wiggle the switch up and down, so that the inductor is connected
to Vin half the time and ground half the time.

This is a "synchronous switcher."


Oh your ascii art had hard tabs embedded, which is why some people
didn't see it right. Spaces are safer, because some weird people don't
use 8-space tabs.

John


[1] as long as the mess is operating in continuous mode.

 

[John Popelish]

Ben wrote:

Hello,

Been working with a buck circuit, today and think I've got in a bit of
mix.

(snip)

Diode is a low power shotky switch is actually transistor.

Ok m so when the switch is closed the current flows through the
inductor, creating a magnetic field. The cap charges and the output
voltage of 5V is reached. So the p.d at the output is 5V say at 500mA
to keep things simple.

When the switch is opened, there is a back e.m.f generated by the
inductutor. Which basically means that the p.d at the didoe and
inductior is -5V. The diode is now reverse biased. So current flows
through the inductutor to the cap keeping the output voltage a 5V.

What about the current does it still stay at 500mA? I thought that caps
could not store energy?

The process is then repeated to said f when the switch is again opened
and closed.

I'm jumping in late, but see you are still having a little
difficulty with the basic concepts. This circuit puts the
inductor in one of two situations. When the switch is on,
there is a voltage drop across the inductor that is the
difference between the input voltage connected through the
switch and the output voltage (across the capacitor and
load). This fixed voltage causes the inductor current to
ramp steadily up.

This comes out of the basic formula that relates voltage and
current for an inductor. V=L*(di/dt) or voltage across the
inductor is equal to the inductance times the rate of change
of current. V is in volts, L in henries and di/dt is in
amperes per second.

In the switch on condition, V is (Vin - Vout). So the
current ramps up at that voltage divided by the inductance.

Lets assume that this ramp averages the load current, so
that during the early part of the ramp, the capacitor
voltage is sagging a little as it provides the current
deficit, and near the end of the ramp, when the inductor
current is more than the load current, the capacitor absorbs
the excess, as its voltage rises.

Then the switch opens and blocks inductor current. So the
inductor produces whatever negative voltage it takes to suck
the current (the inductor was passing, right before the
switch opened) up from ground, through the diode. This
means that the input end of the inductor drives a diode drop
more negative than ground. This requires that the inductor
voltage drop be the output voltage plus the forward bias
diode drop. So that new V (negative, this time, since the
inductor is the energy source, not an energy absorber)
produces a negative rate of change of the current through
the inductor. That means the current starts where it was
when the switch opened and ramps steadily down toward zero.
And again, the ramp is assumed to average about equal to
the load current, so at the beginning of this ramp down, the
capacitor is still absorbing excess current ands its voltage
is still creeping up. But the moment the inductor current
ramp falls below the load current, the capacitor starts
supplying the current deficit to the load, and its voltage
starts to decay a little.

Before the current reaches zero, I am assuming the switch
closes again, and reverses the voltage across the inductor,
starting another increasing current ramp, and the cycle repeats.

So the inductor current is a triangle wave between some
minimum and some maximum current whose average is equal to
the load current. The output voltage has a ripple as the
capacitor sucks up current at the high parts of the inductor
current triangle, and supplies current at the low parts of
the inductor current triangle.

The larger the inductor, the flatter the current triangle,
so the less the capacitor must absorb and contribute each
cycle.

The higher the switch frequency, the closer the peak current
of the triangle is to the minimum current of the triangle.

The larger the capacitor the less its voltage swings for a
given current ripple it must absorb.

--
Regards,

John Popelish

 

[Ben]

OK,

Thanks for all the replies,

I think I've got it clear now. Switch closed creates a positive current
ramp, that intially charges cap and load, with the current increasing to
an excessive ammount to bring V out up to max. Diode is "off"

When switch is closed, diode is on, current still flows in the same
direction, ie from ground to inductor to charge c, only becuase the
diode is "on" due to being revese biased, by "back e.m.f". However the
current starts at an excess, (the max point reached during on time) so
Vout is up to max. As load drains the current the excess is drawn from
inductor then cap and once excess is depleted the V out drops. At this
point the switch turns on again.

Thanks.

John Popelish wrote:

Ben wrote:
Hello,

Been working with a buck circuit, today and think I've got in a bit
of mix.

(snip)
Diode is a low power shotky switch is actually transistor.

Ok m so when the switch is closed the current flows through the
inductor, creating a magnetic field. The cap charges and the output
voltage of 5V is reached. So the p.d at the output is 5V say at 500mA
to keep things simple.

When the switch is opened, there is a back e.m.f generated by the
inductutor. Which basically means that the p.d at the didoe and
inductior is -5V. The diode is now reverse biased. So current flows
through the inductutor to the cap keeping the output voltage a 5V.

What about the current does it still stay at 500mA? I thought that
caps could not store energy?

The process is then repeated to said f when the switch is again opened
and closed.

I'm jumping in late, but see you are still having a little difficulty
with the basic concepts. This circuit puts the inductor in one of two
situations. When the switch is on, there is a voltage drop across the
inductor that is the difference between the input voltage connected
through the switch and the output voltage (across the capacitor and
load). This fixed voltage causes the inductor current to ramp steadily up.

This comes out of the basic formula that relates voltage and current for
an inductor. V=L*(di/dt) or voltage across the inductor is equal to the
inductance times the rate of change of current. V is in volts, L in
henries and di/dt is in amperes per second.

In the switch on condition, V is (Vin - Vout). So the current ramps up
at that voltage divided by the inductance.

Lets assume that this ramp averages the load current, so that during the
early part of the ramp, the capacitor voltage is sagging a little as it
provides the current deficit, and near the end of the ramp, when the
inductor current is more than the load current, the capacitor absorbs
the excess, as its voltage rises.

Then the switch opens and blocks inductor current. So the inductor
produces whatever negative voltage it takes to suck the current (the
inductor was passing, right before the switch opened) up from ground,
through the diode. This means that the input end of the inductor drives
a diode drop more negative than ground. This requires that the inductor
voltage drop be the output voltage plus the forward bias diode drop. So
that new V (negative, this time, since the inductor is the energy
source, not an energy absorber) produces a negative rate of change of
the current through the inductor. That means the current starts where
it was when the switch opened and ramps steadily down toward zero. And
again, the ramp is assumed to average about equal to the load current,
so at the beginning of this ramp down, the capacitor is still absorbing
excess current ands its voltage is still creeping up. But the moment
the inductor current ramp falls below the load current, the capacitor
starts supplying the current deficit to the load, and its voltage starts
to decay a little.

Before the current reaches zero, I am assuming the switch closes again,
and reverses the voltage across the inductor, starting another
increasing current ramp, and the cycle repeats.

So the inductor current is a triangle wave between some minimum and some
maximum current whose average is equal to the load current. The output
voltage has a ripple as the capacitor sucks up current at the high parts
of the inductor current triangle, and supplies current at the low parts
of the inductor current triangle.

The larger the inductor, the flatter the current triangle, so the less
the capacitor must absorb and contribute each cycle.

The higher the switch frequency, the closer the peak current of the
triangle is to the minimum current of the triangle.

The larger the capacitor the less its voltage swings for a given current
ripple it must absorb.

 

[Dan Coby]

"Ben" <nebATwalrus81DOTplusDOTcom> wrote in message news:13l20c3nh79236e@corp.supernews.com...

I think I've got it clear now. Switch closed creates a positive current ramp, that intially
charges cap and load, with the current increasing to an excessive ammount to bring V out up to
max. Diode is "off"

Yes.

When switch is closed,

I assume that you mean 'when the switch is open'.

.... diode is on, current still flows in the same direction, ie from
ground to inductor to charge c, only becuase the diode is "on" due to being revese biased, by
"back e.m.f".

No. The diode is forward biased. Diodes will only conduct current when
they are forward biased. As another poster has said, the term 'back emf'
is confusing in this context.

.... However the current starts at an excess, (the max point reached during on time) so
Vout is up to max. As load drains the current the excess is drawn from inductor then cap and once
excess is depleted the V out drops.

The current through the inductor will start to drop once the switch is opened.
The left side of the inductor will be at 0.7 volts (the forward drop across
the diode), the right side of the inductor will be at Vout. Thus the voltage
across the induction will be Vout - 0..7 volts. This voltage will be trying to
reduce the inductor current.

Another way to look at what is happening with the inductor current is to look
at that is happening with the inductor's energy. An inductor stores energy in
its magnetic field. The energy in an inductor is given by

energy = (1/2) * L * i ** 2
or
energy = (1/2) * L * i squared

When the switch is closed, the battery is supplying energy to both the output
and to the inductor. As the inductor's current is increased, the energy stored
in the inductor is increasing.

When the switch is opened, the battery is no longer supplying energy. Instead
the inductor starts to provide energy into the capacitor and the load. This
results in the inductor current dropping. When the inductor current becomes
less than the load current, the output voltage will start to drop. (Note: The
output voltage was still increasing for a short time after when the switch opened
while the inductor current was greater than the load current. During this time,
the excess current from the inductor was continuing to charge the capacitor.)