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Bootstrap diode

J

Jon Slaughter

Jan 1, 1970
0
Can someone explain how it works and/or give a schematic/circuit? I just
know they are used in some high-side mosfet drivers for creating the
high-side gate voltage but can't seem to find out any details.
 
J

Jon Slaughter

Jan 1, 1970
0
Jon Slaughter said:
Can someone explain how it works and/or give a schematic/circuit? I just
know they are used in some high-side mosfet drivers for creating the
high-side gate voltage but can't seem to find out any details.

http://www.national.com/an/AN/AN-1317.pdf

Seems to imply that such a configuration cannot work unless the low side is
used often? Probably just as often as the high side?
 
J

Joerg

Jan 1, 1970
0
Jon said:
http://www.national.com/an/AN/AN-1317.pdf

Seems to imply that such a configuration cannot work unless the low side is
used often? Probably just as often as the high side?

Pretty much, no exercise -> no voltage. How often depends on the size of
the cap and the hunger of your high side circuitry. Bootstraps are meant
for complementary circuits where the whole chebang runs at some
reasonable frequency and both high- and low-side are switched.
 
R

Rich Grise

Jan 1, 1970
0
http://www.national.com/an/AN/AN-1317.pdf

Seems to imply that such a configuration cannot work unless the low side is
used often? Probably just as often as the high side?

Yes - ideally, it'd be symmetrical, to avoid current imbalances and stuff.

I kinda redrew the circuit in my head, and the "bootstrap" portion is just
a bent half-wave voltage doubler. :)

Cheers!
Rich
 
J

Jon Slaughter

Jan 1, 1970
0
Joerg said:
Pretty much, no exercise -> no voltage. How often depends on the size of
the cap and the hunger of your high side circuitry. Bootstraps are meant
for complementary circuits where the whole chebang runs at some reasonable
frequency and both high- and low-side are switched.


I guess then I can't use this for a full bridge since the low side drain and
high side source must connected directly. I guess with a motor inbetween the
cap will not be charged properly?
 
L

legg

Jan 1, 1970
0
I guess then I can't use this for a full bridge since the low side drain and
high side source must connected directly. I guess with a motor inbetween the
cap will not be charged properly?
The bootstrap cap is normally referenced to the high-side source. As
long as the source is pulled low, either by the low side fet or the
load, the bootstrap cap will charge.

In ~self-oscillating full-bridge situations, where the drive signals
depend on the output state, care must be taken to ensure that the
bottom fet does in fact turn on at least once before drive to the
upper fet is required.

This may complicate start-up and limiting situations.

RL
 
J

Jon Slaughter

Jan 1, 1970
0
legg said:
The bootstrap cap is normally referenced to the high-side source. As
long as the source is pulled low, either by the low side fet or the
load, the bootstrap cap will charge.

But with a motor the cap will not be pulled to ground or even close.
In ~self-oscillating full-bridge situations, where the drive signals
depend on the output state, care must be taken to ensure that the
bottom fet does in fact turn on at least once before drive to the
upper fet is required.

This may complicate start-up and limiting situations.

My confusion is actually how it accomplishes. In the pdf link it has a
diagram of a basic driver. It shows a level shifter and some logic then an
omp amp.

My guess it that it has some logic to level shift the signal to whatever
floating hi and low it uses. The op amp is for drive capabilities. The op
amp is floating as the cap supplies power to it. Essentially as Rich as
mentioned it is similar to a voltage doubler. The cap is charged on the low
side then "lifted" to the high side giving effectively 2Vcc from ground but
puts Vcc across the op amp to power it and drive the high side gate.

If a load is between the high side source and low side drain then the cap
will not be charged up properly.

I imagine I could create some additional circuitry to charge the cap up on
the low side. Basically disconnect the cap and connect it to ground during
the low side. This is so that the cap can be charged during the load side
but is separate from the load. Of course this creates more problems than
it's worth.
 
L

legg

Jan 1, 1970
0
But with a motor the cap will not be pulled to ground or even close.

If the way you use it prevents it's normal function, you are
misapplying the part. I think this is one of the reasons why app notes
are published.
My confusion is actually how it accomplishes. In the pdf link it has a
diagram of a basic driver. It shows a level shifter and some logic then an
omp amp.

My guess it that it has some logic to level shift the signal to whatever
floating hi and low it uses. The op amp is for drive capabilities. The op
amp is floating as the cap supplies power to it. Essentially as Rich as
mentioned it is similar to a voltage doubler. The cap is charged on the low
side then "lifted" to the high side giving effectively 2Vcc from ground but
puts Vcc across the op amp to power it and drive the high side gate.

If a load is between the high side source and low side drain then the cap
will not be charged up properly.

In a full bridge there are four switches. The low side switches in
each case are responsible for refreshing the bootstrap cap specific to
the high-side switch on it's own side of the bridge. The load is
applied between the switch pair junctions on each side of the bridge.

Your reference app note describes a half-bridge driver; one side of a
bridge is used.

An asymmetrical bridge replaces two diagonally situated switches of
the full bridge with rectifiers - it is capable of producing load
current in only one direction.
I imagine I could create some additional circuitry to charge the cap up on
the low side. Basically disconnect the cap and connect it to ground during
the low side. This is so that the cap can be charged during the load side
but is separate from the load. Of course this creates more problems than
it's worth.
A 'leak' resistor between the source and the negative rail could serve
this function, prior to the first high-side gate drive signal. Once
any motor load current is established, its own flyback energy will
ensure that the source is pulled low in the freewheeling interval.

RL
 
J

Joerg

Jan 1, 1970
0
Jon said:
But with a motor the cap will not be pulled to ground or even close.


My confusion is actually how it accomplishes. In the pdf link it has a
diagram of a basic driver. It shows a level shifter and some logic then an
omp amp.

My guess it that it has some logic to level shift the signal to whatever
floating hi and low it uses. The op amp is for drive capabilities. The op
amp is floating as the cap supplies power to it. Essentially as Rich as
mentioned it is similar to a voltage doubler. The cap is charged on the low
side then "lifted" to the high side giving effectively 2Vcc from ground but
puts Vcc across the op amp to power it and drive the high side gate.

If a load is between the high side source and low side drain then the cap
will not be charged up properly.

It can be, depends on the load. If the low side FET conducts then its
drain pulls towards zero. Then the load tugs on the source of the high
side FET and pulls that to ground.

Of course if the load is sluggish this falls apart.

I imagine I could create some additional circuitry to charge the cap up on
the low side. Basically disconnect the cap and connect it to ground during
the low side. This is so that the cap can be charged during the load side
but is separate from the load. Of course this creates more problems than
it's worth.

In that case I'd use a real motor driver or something wit a separate
charge pump.
 
J

Jamie

Jan 1, 1970
0
Jon said:
But with a motor the cap will not be pulled to ground or even close.




My confusion is actually how it accomplishes. In the pdf link it has a
diagram of a basic driver. It shows a level shifter and some logic then an
omp amp.

My guess it that it has some logic to level shift the signal to whatever
floating hi and low it uses. The op amp is for drive capabilities. The op
amp is floating as the cap supplies power to it. Essentially as Rich as
mentioned it is similar to a voltage doubler. The cap is charged on the low
side then "lifted" to the high side giving effectively 2Vcc from ground but
puts Vcc across the op amp to power it and drive the high side gate.

If a load is between the high side source and low side drain then the cap
will not be charged up properly.

I imagine I could create some additional circuitry to charge the cap up on
the low side. Basically disconnect the cap and connect it to ground during
the low side. This is so that the cap can be charged during the load side
but is separate from the load. Of course this creates more problems than
it's worth.
For what its worth..

I see the circuit as a simple method of supplying the HS line, which
is the Vee of the driver of the high side, a "common" point as long as
the charge in the cap sustains..
When the low side output is on, this will then provide the "COMMON"
point for the HS line and also the cap how ever, the high side isn't on
at this time so the only thing that takes place is the recharging of the
cap.

The diode is there only to prevent back drain of this cap and supply
voltage for charging.

For larger caps with slower switching times, an external larger
diode should be used to relieve the strain from the internal one, which
could actually short it. This is something you do not want because it
will then most likely take out the high side output due to insufficient
drive at the gate which will put it into linear state.

These types of drivers are only good for PWM system or some kind of
pulsed system.. Most of the time, at idle, you keep the low side on to
keep the cap charged. I guess the ideal thing would be to run it into an
inductive load or have a inductor in series and have the circuit
generate a very narrow PWM to maintain the cap. With the inductor in
line, you shouldn't see much DC at the other end.

If you were to combine two of these for a full bridge, you'd keep both
low side outputs on at idle. Using the load isn't going to help much
here since it may be reactive or a some what higher level of R>

That's my take of it for what its worth!..
 
J

Jon Slaughter

Jan 1, 1970
0
Jamie said:
For what its worth..

I see the circuit as a simple method of supplying the HS line, which is
the Vee of the driver of the high side, a "common" point as long as the
charge in the cap sustains..
When the low side output is on, this will then provide the "COMMON"
point for the HS line and also the cap how ever, the high side isn't on
at this time so the only thing that takes place is the recharging of the
cap.

The diode is there only to prevent back drain of this cap and supply
voltage for charging.

For larger caps with slower switching times, an external larger
diode should be used to relieve the strain from the internal one, which
could actually short it. This is something you do not want because it will
then most likely take out the high side output due to insufficient
drive at the gate which will put it into linear state.

These types of drivers are only good for PWM system or some kind of
pulsed system.. Most of the time, at idle, you keep the low side on to
keep the cap charged. I guess the ideal thing would be to run it into an
inductive load or have a inductor in series and have the circuit generate
a very narrow PWM to maintain the cap. With the inductor in line, you
shouldn't see much DC at the other end.

If you were to combine two of these for a full bridge, you'd keep both
low side outputs on at idle. Using the load isn't going to help much here
since it may be reactive or a some what higher level of R>

That's my take of it for what its worth!..

I just see it as using the cap for temporary power to the op amp. It charges
the cap on the low side(obvious when the low side mosfet is shorted) and
supplies power to the op amp on when the low side is off.


I have a similar one, the MAX5062(I think). It says it can be used for motor
control but all the circuits show it as a half bridge. I've tried using it
once and could run a small motor but trying to run a large one caused the
IC's to burn up. Not sure why the motor size had anything to do with the
IC's burning up... maybe inductive kickback somehow screwed it up?
 
J

Jon Slaughter

Jan 1, 1970
0
Joerg said:
It can be, depends on the load. If the low side FET conducts then its
drain pulls towards zero. Then the load tugs on the source of the high
side FET and pulls that to ground.

Of course if the load is sluggish this falls apart.

if you use a resistive load I don't see how this can happen except at the
moment of switching. If the low side switch is on and the high side is off
then the cap will charge up through the resistor which means it is not
grounded... but if the cap is small enough it might approach ground. The
higher the load the better as it will approach ground more. But then it also
depends on the switching times and requires a dead time.

I could use a small low side mosfet or maybe even a bjt that alternates with
the high side. This allows the cap to charge every time the high side is
used. Then just use the drivers for high side and use something else for a
low side. This might cause some problems with inductive loads. Also if the
low-side small mosfet is ever turned on without the motor's low side then it
will take all the current and burn up rather quickly. (this shouldn't happen
if everything is consistent though)


In that case I'd use a real motor driver or something wit a separate
charge pump.

They seem to be hard to find compared to these "half-bridges".
 
J

Joerg

Jan 1, 1970
0
Jon said:
if you use a resistive load I don't see how this can happen except at the
moment of switching. If the low side switch is on and the high side is off
then the cap will charge up through the resistor which means it is not
grounded... but if the cap is small enough it might approach ground. The
higher the load the better as it will approach ground more. But then it also
depends on the switching times and requires a dead time.

I could use a small low side mosfet or maybe even a bjt that alternates with
the high side. This allows the cap to charge every time the high side is
used. Then just use the drivers for high side and use something else for a
low side. This might cause some problems with inductive loads. Also if the
low-side small mosfet is ever turned on without the motor's low side then it
will take all the current and burn up rather quickly. (this shouldn't happen
if everything is consistent though)

Well, of course you can't leave both switches on for too long. Then you
are better off with a real motor driver.
They seem to be hard to find compared to these "half-bridges".

Not really:

http://www.freescale.com/files/analog/doc/data_sheet/MC33486.pdf

But I don't know what it exactly is that you want to do with it. This is
an example for high power, there's cheaper ones for less amps.
 
J

Jon Slaughter

Jan 1, 1970
0
legg said:
If the way you use it prevents it's normal function, you are
misapplying the part. I think this is one of the reasons why app notes
are published.


In a full bridge there are four switches. The low side switches in
each case are responsible for refreshing the bootstrap cap specific to
the high-side switch on it's own side of the bridge. The load is
applied between the switch pair junctions on each side of the bridge.

Your reference app note describes a half-bridge driver; one side of a
bridge is used.

An asymmetrical bridge replaces two diagonally situated switches of
the full bridge with rectifiers - it is capable of producing load
current in only one direction.

Ok? For bi-directional application you use two asymmetrical half-bridges...
I have done that with the MAX5062 but with a much larger motor it causes
them to burn up. The MAX5062 seems to be designed similar to the app note.
In the datasheet for the MAX5062 it says it can be used for Motor control. I
can't think of any other way than to use it asymetrically. It uses a boot
cap and high side sensing and shows circuits similar to other half bridge
drivers. it does not show one for a motor though.

It works fine on a small 12V DC motor(<1A) but the chips burn up very
quickly on a 15A 12V DC motor. I have no clue why this happens. I get bi-dir
and speed control(I use a pot and a pic to determine the speed/dir then
setup the pwm to feed the max's) just fine with small motor. I get dead IC's
when connectin the large one. (even with added diode supression across the
mosfets)


This was about a year ago and I got tired of going wasting so many chips I
just gave up.

A 'leak' resistor between the source and the negative rail could serve
this function, prior to the first high-side gate drive signal. Once
any motor load current is established, its own flyback energy will
ensure that the source is pulled low in the freewheeling interval.

This might cause some problems with the gate drive since the HS source is no
longer floating. One would need to match the resistance to the capacitance
so that it can be charged fast enough yet not pull the source down to ground
to much reducing the HS gate drive.
 
J

Jon Slaughter

Jan 1, 1970
0
Jon Slaughter said:
Ok? For bi-directional application you use two asymmetrical
half-bridges... I have done that with the MAX5062 but with a much larger
motor it causes them to burn up. The MAX5062 seems to be designed similar
to the app note. In the datasheet for the MAX5062 it says it can be used
for Motor control. I can't think of any other way than to use it
asymetrically. It uses a boot cap and high side sensing and shows circuits
similar to other half bridge drivers. it does not show one for a motor
though.

It works fine on a small 12V DC motor(<1A) but the chips burn up very
quickly on a 15A 12V DC motor. I have no clue why this happens. I get
bi-dir and speed control(I use a pot and a pic to determine the speed/dir
then setup the pwm to feed the max's) just fine with small motor. I get
dead IC's when connectin the large one. (even with added diode supression
across the mosfets)


This was about a year ago and I got tired of going wasting so many chips I
just gave up.



This might cause some problems with the gate drive since the HS source is
no longer floating. One would need to match the resistance to the
capacitance so that it can be charged fast enough yet not pull the source
down to ground to much reducing the HS gate drive.

was just looking at

http://www.irf.com/product-info/datasheets/data/ir2011.pdf

which shows what I did but my load was a motor. I used two of them
configured "asymmetrically" as to get bi-dir. I think I even tried
disconnecting one side with similar results.
 
J

Jon Slaughter

Jan 1, 1970
0
Joerg said:
Well, of course you can't leave both switches on for too long. Then you
are better off with a real motor driver.

Not really:

http://www.freescale.com/files/analog/doc/data_sheet/MC33486.pdf

But I don't know what it exactly is that you want to do with it. This is
an example for high power, there's cheaper ones for less amps.

But these have built in mosfets ;/ I've seen a few of these. I need more
than 10A ;/ Why can't they just have pins for external mosfets? This one has
the 2 low side gate drivers but has built in high side. I've thought about
trying to parallel such devices because they have everything else I could
want. But I think paralleling is probably a bad idea. If only I could use my
own high side drivers then I'd have all my problems solved ;/

Maybe if I can find those in 20A+ from 15-30V then I could use them too. Of
course I would still like to use my own outboard mosfets or at least have
the option.

They do have

http://www.freescale.com/files/analog/doc/data_sheet/MC33883.pdf

Which I used two MAX5062's to get the same effect but the drivers burned up
on a large motor(everything worked fine for a small <1A one). But if a
charge pump is different than the boot strap then maybe thats the reason.
Even though the MAX5062's datasheet say they can be used for motor control.
 
J

Jon Slaughter

Jan 1, 1970
0
Joerg said:
Well, of course you can't leave both switches on for too long. Then you
are better off with a real motor driver.


Not really:

http://www.freescale.com/files/analog/doc/data_sheet/MC33486.pdf

But I don't know what it exactly is that you want to do with it. This is
an example for high power, there's cheaper ones for less amps.

Thanks though, I've seen some of hte "pre-drivers" which is what I'm
basically looking for. Again I've done this with the MAX5062's and can't see
why such a device would burn up just because I'm using a larger motor. The
external mosfets are the only ones that should care about the side of the
motor and they were just fine(even after the drivers burned up).

I'm going to try some of those full-bridge pre-drivers though and see what I
can get.
 
J

Joerg

Jan 1, 1970
0
Jon said:
Thanks though, I've seen some of hte "pre-drivers" which is what I'm
basically looking for. Again I've done this with the MAX5062's and can't see
why such a device would burn up just because I'm using a larger motor. The
external mosfets are the only ones that should care about the side of the
motor and they were just fine(even after the drivers burned up).

I'm going to try some of those full-bridge pre-drivers though and see what I
can get.

I never use Maxim so I can't say. One scenario for a burn-up is stray
spikes finding their way into the chip. Traditionally such spikes will
have 10 times the gusto when you go from a 1A motor to a 10A motor :)
 
B

boB

Jan 1, 1970
0
sure you can, in a full bridge the motor doesnt go between the H-
source and L-drain on the same side.
it goes between the connected H-source and L-drain on one side of the
bridge and
the connected H-source and L-drain on the other side of the bridge


+--------------+
| |
1 |- -| 3
------|<+ +>|------
|-| |-|
| |
+-----( M )----+
| |
2 |- -| 4
------|<+ +>|------
|-| |-|
| |
+--------------+


when 2 is on boost cap for 1 is charging, when 4 is on boostcap for 3
is charging.

right side is driven inverted from left side. Maximum dutycycle,
minimum switching frequency,
and the current consumption of the high side driver must all fit
together with boost caps size and
charging current to avoid the high side drive voltage drooping too
much during high side on time and/or not being fully charged for the
minium lowside on time

-Lasse


In addition to this, if there was current in the motor inductance when
the top FET turned OFF, the the corresponding bottom FET doesn't even
have to come on in order for the boost cap to charge..... There will be
a path to ground through the bottom FET's intrinsic, or additional,
Drain-Source diode.

boB
 
J

Jamie

Jan 1, 1970
0
Jon said:
I just see it as using the cap for temporary power to the op amp. It charges
the cap on the low side(obvious when the low side mosfet is shorted) and
supplies power to the op amp on when the low side is off.


I have a similar one, the MAX5062(I think). It says it can be used for motor
control but all the circuits show it as a half bridge. I've tried using it
once and could run a small motor but trying to run a large one caused the
IC's to burn up. Not sure why the motor size had anything to do with the
IC's burning up... maybe inductive kickback somehow screwed it up?
Since the HS line is most likely connected directly to the load, having
inductive collapse would simply put a (-), not common, potential on
the HS line which will increase the overall voltage supplied to the
driver inside the chip. (Vee and HS). to solve that, put a heavy clamp
like a
zener/transorb across the Vcc and HS line. This should absorb the over
voltage being generated.

The other problem could be that you are not properly latching the
low side on in time to clamp this energy. If you are attempting to
do so, maybe you need a small cap to handle the clamp while the low
side fet is clamping.
If you are free running the circuit at times, then you are in trouble.
Induction will kill your driver chip on the high side if you don't have
some kind of clamp there.

In most digital drives, especially AC inverters, you have the option
of using a DB resistor which ties to the DC bus and goes through a
switch that clamps the bus while the motor is not under load, other wise,
you'll get over voltage on the DC bus due to the motor regenerating
energy that adds to the bus through the protective circuits in the bridge..


Something to ponder on.
 
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