Simple FET Driver...

[Rick C]

I have a circuit with a pair of pFETs to switch power and it gets some shoot through during the switching. Mostly it\'s a concern because of the slow turn off using a resistive pullup. Then there\'s the issue of the two power supplies being somewhat different voltages. The high drive has to turn off the FETs completely. The modes are for one or the other to be on or for both to be off.

I\'m wondering if a digital logic device can be used to drive this circuit. 4000 series CMOS can handle the 15V max levels but I don\'t know if it is capable of adequate drive to switch a 10 amp FET. I guess it can\'t be much worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS device. I\'d still be concerned about the inputs though. A low input voltage could provide too little drive for the CMOS input. So everything would have to be referenced to the diode routed power which might draw too much current from the battery when it\'s all off.

Any suggestions on a driver that isn\'t high quiescent drain? Most of what I find are bipolar. It should be very cheap too.

--

Rick C.

- Get 1,000 miles of free Supercharging
- Tesla referral code - https://ts.la/richard11209

 

[Edward Rawde]

\"Rick C\" <gnuarm.deletethisbit@gmail.com> wrote in message
news:e2a6129f-a999-4b9d-b079-35fb2a4cc290n@googlegroups.com...
I have a circuit with a pair of pFETs to switch power and it gets some
shoot through during the switching. Mostly it\'s a concern because of the
slow turn off using a resistive pullup. Then there\'s the issue of the two
power supplies being somewhat different voltages. The high drive has to
turn off the FETs completely. The modes are for one or the other to be on
or for both to be off.

I\'m wondering if a digital logic device can be used to drive this circuit.
4000 series CMOS can handle the 15V max levels but I don\'t know if it is
capable of adequate drive to switch a 10 amp FET. I guess it can\'t be much
worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS
device. I\'d still be concerned about the inputs though. A low input
voltage could provide too little drive for the CMOS input. So everything
would have to be referenced to the diode routed power which might draw too
much current from the battery when it\'s all off.

After reading that I am unable to get a clear picture of what you\'re trying
to do. I would have posted a reference to a scan of the intended circuit,
complete with labels showing what power is coming from where and going where
and when and why.

Any suggestions on a driver that isn\'t high quiescent drain? Most of what
I find are bipolar. It should be very cheap too.

I don\'t see anything wrong with using slow switching FETs to switch power
rails, provided it\'s not so slow that the FET heats up too much.
Maybe you need a transistor or two to get the FET gate drive right but hard
to tell without a schematic.

--

Rick C.

 

[bitrex]

On 12/1/2020 6:31 PM, Rick C wrote:

I have a circuit with a pair of pFETs to switch power and it gets some shoot through during the switching. Mostly it\'s a concern because of the slow turn off using a resistive pullup. Then there\'s the issue of the two power supplies being somewhat different voltages. The high drive has to turn off the FETs completely. The modes are for one or the other to be on or for both to be off.

I\'m wondering if a digital logic device can be used to drive this circuit. 4000 series CMOS can handle the 15V max levels but I don\'t know if it is capable of adequate drive to switch a 10 amp FET. I guess it can\'t be much worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS device. I\'d still be concerned about the inputs though. A low input voltage could provide too little drive for the CMOS input. So everything would have to be referenced to the diode routed power which might draw too much current from the battery when it\'s all off.

Any suggestions on a driver that isn\'t high quiescent drain? Most of what I find are bipolar. It should be very cheap too.

Isolated gate drivers make high-side interfacing so much easier, and
they\'re so relatively inexpensive now that it\'s hard to justify not
using them most times.

Please see figure 3.1:

<https://www.mouser.com/datasheet/2/368/Si8751_2-1397983.pdf>

these are about a buck in small quantity.

 

[bitrex]

On 12/1/2020 11:10 PM, bitrex wrote:

On 12/1/2020 6:31 PM, Rick C wrote:
I have a circuit with a pair of pFETs to switch power and it gets some
shoot through during the switching.  Mostly it\'s a concern because of
the slow turn off using a resistive pullup.  Then there\'s the issue of
the two power supplies being somewhat different voltages.  The high
drive has to turn off the FETs completely.  The modes are for one or
the other to be on or for both to be off.

I\'m wondering if a digital logic device can be used to drive this
circuit.  4000 series CMOS can handle the 15V max levels but I don\'t
know if it is capable of adequate drive to switch a 10 amp FET.  I
guess it can\'t be much worse than the 10k ohm pull up I\'m using now.

Not sure how to power this.  Could use diodes to route power to the
CMOS device.  I\'d still be concerned about the inputs though.  A low
input voltage could provide too little drive for the CMOS input.  So
everything would have to be referenced to the diode routed power which
might draw too much current from the battery when it\'s all off.

Any suggestions on a driver that isn\'t high quiescent drain?  Most of
what I find are bipolar.  It should be very cheap too.


Isolated gate drivers make high-side interfacing so much easier, and
they\'re so relatively inexpensive now that it\'s hard to justify not
using them most times.

Please see figure 3.1:

https://www.mouser.com/datasheet/2/368/Si8751_2-1397983.pdf

these are about a buck in small quantity.

Only downside is they\'re very moisture-sensitive so if you want to
experiment either buy a development board (they\'re cheap) or buy in
small quantity and mount them to a SMT adapter straight out of the bag
when they come in to you from the warehouse unless you have some kind of
humidity-controlled storage for the parts.

 

[Rick C]

On Tuesday, December 1, 2020 at 9:56:01 PM UTC-5, Edward Rawde wrote:

\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:e2a6129f-a999-4b9d...@googlegroups.com...
I have a circuit with a pair of pFETs to switch power and it gets some
shoot through during the switching. Mostly it\'s a concern because of the
slow turn off using a resistive pullup. Then there\'s the issue of the two
power supplies being somewhat different voltages. The high drive has to
turn off the FETs completely. The modes are for one or the other to be on
or for both to be off.

I\'m wondering if a digital logic device can be used to drive this circuit.
4000 series CMOS can handle the 15V max levels but I don\'t know if it is
capable of adequate drive to switch a 10 amp FET. I guess it can\'t be much
worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS
device. I\'d still be concerned about the inputs though. A low input
voltage could provide too little drive for the CMOS input. So everything
would have to be referenced to the diode routed power which might draw too
much current from the battery when it\'s all off.
After reading that I am unable to get a clear picture of what you\'re trying
to do. I would have posted a reference to a scan of the intended circuit,
complete with labels showing what power is coming from where and going where
and when and why.
Any suggestions on a driver that isn\'t high quiescent drain? Most of what
I find are bipolar. It should be very cheap too.
I don\'t see anything wrong with using slow switching FETs to switch power
rails, provided it\'s not so slow that the FET heats up too much.
Maybe you need a transistor or two to get the FET gate drive right but hard
to tell without a schematic.

It\'s not switching in the sense of motor control or a DC converter. It is a switch over from line to battery when input power fails. Each switch is two pFETs to prevent back feed of current. The battery is 12V nominal SLA and the input is a 12V DC for running a motor with 10 amp peaks and 4 amp nominal. I use a comparator on the input DC to detect when it is absent which switches between the two power sources. Using transistors with pull ups means the pull up is slow and there are some 50 us where current passes through both switches between the two power sources. Maybe this isn\'t much of a real problem, but 15 or more amps seems like something to minimize the time duration of if the current itself can\'t be minimized.

If logic gates are used instead of the transistors with pullups the transition time will be much shorter and the issue of shoot through current will be less significant. But I\'m probably showing my ignorance of what is needed for this circuit to work well. One guy on the team has picked a LT part at $4 to control the pFETs. It\'s a nice part, but when was an LT part ever cheap? It just doesn\'t seem like a hard circuit for an experienced power designer to pull off.

--

Rick C.

+ Get 1,000 miles of free Supercharging
+ Tesla referral code - https://ts.la/richard11209

 

[Rick C]

On Tuesday, December 1, 2020 at 11:10:23 PM UTC-5, bitrex wrote:

On 12/1/2020 6:31 PM, Rick C wrote:
I have a circuit with a pair of pFETs to switch power and it gets some shoot through during the switching. Mostly it\'s a concern because of the slow turn off using a resistive pullup. Then there\'s the issue of the two power supplies being somewhat different voltages. The high drive has to turn off the FETs completely. The modes are for one or the other to be on or for both to be off.

I\'m wondering if a digital logic device can be used to drive this circuit. 4000 series CMOS can handle the 15V max levels but I don\'t know if it is capable of adequate drive to switch a 10 amp FET. I guess it can\'t be much worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS device. I\'d still be concerned about the inputs though. A low input voltage could provide too little drive for the CMOS input. So everything would have to be referenced to the diode routed power which might draw too much current from the battery when it\'s all off.

Any suggestions on a driver that isn\'t high quiescent drain? Most of what I find are bipolar. It should be very cheap too.

Isolated gate drivers make high-side interfacing so much easier, and
they\'re so relatively inexpensive now that it\'s hard to justify not
using them most times.

Please see figure 3.1:

https://www.mouser.com/datasheet/2/368/Si8751_2-1397983.pdf

these are about a buck in small quantity.

The price is still higher than I\'d like since two of them are now $2.50 which isn\'t much cheaper than the $4 part. I also am not following how these work. Is power being driven across the barrier? A barrier is absolutely not needed in this case.

--

Rick C.

-- Get 1,000 miles of free Supercharging
-- Tesla referral code - https://ts.la/richard11209

 

[server]

On Tue, 1 Dec 2020 21:55:48 -0500, \"Edward Rawde\"
<invalid@invalid.invalid> wrote:

\"Rick C\" <gnuarm.deletethisbit@gmail.com> wrote in message
news:e2a6129f-a999-4b9d-b079-35fb2a4cc290n@googlegroups.com...
I have a circuit with a pair of pFETs to switch power and it gets some
shoot through during the switching. Mostly it\'s a concern because of the
slow turn off using a resistive pullup. Then there\'s the issue of the two
power supplies being somewhat different voltages. The high drive has to
turn off the FETs completely. The modes are for one or the other to be on
or for both to be off.

I\'m wondering if a digital logic device can be used to drive this circuit.
4000 series CMOS can handle the 15V max levels but I don\'t know if it is
capable of adequate drive to switch a 10 amp FET. I guess it can\'t be much
worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS
device. I\'d still be concerned about the inputs though. A low input
voltage could provide too little drive for the CMOS input. So everything
would have to be referenced to the diode routed power which might draw too
much current from the battery when it\'s all off.

After reading that I am unable to get a clear picture of what you\'re trying
to do. I would have posted a reference to a scan of the intended circuit,
complete with labels showing what power is coming from where and going where
and when and why.

It\'s amazing how one sketch can clarify a huge messy muddle of words.



--

John Larkin Highland Technology, Inc

The best designs are necessarily accidental.

 

[bitrex]

On 12/2/2020 12:01 AM, Rick C wrote:

On Tuesday, December 1, 2020 at 11:10:23 PM UTC-5, bitrex wrote:
On 12/1/2020 6:31 PM, Rick C wrote:
I have a circuit with a pair of pFETs to switch power and it gets some shoot through during the switching. Mostly it\'s a concern because of the slow turn off using a resistive pullup. Then there\'s the issue of the two power supplies being somewhat different voltages. The high drive has to turn off the FETs completely. The modes are for one or the other to be on or for both to be off.

I\'m wondering if a digital logic device can be used to drive this circuit. 4000 series CMOS can handle the 15V max levels but I don\'t know if it is capable of adequate drive to switch a 10 amp FET. I guess it can\'t be much worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS device. I\'d still be concerned about the inputs though. A low input voltage could provide too little drive for the CMOS input. So everything would have to be referenced to the diode routed power which might draw too much current from the battery when it\'s all off.

Any suggestions on a driver that isn\'t high quiescent drain? Most of what I find are bipolar. It should be very cheap too.

Isolated gate drivers make high-side interfacing so much easier, and
they\'re so relatively inexpensive now that it\'s hard to justify not
using them most times.

Please see figure 3.1:

https://www.mouser.com/datasheet/2/368/Si8751_2-1397983.pdf

these are about a buck in small quantity.

The price is still higher than I\'d like since two of them are now $2.50 which isn\'t much cheaper than the $4 part. I also am not following how these work. Is power being driven across the barrier? A barrier is absolutely not needed in this case.

No, the point is the secondary side generates its own gate-drive boost
voltage internally so it can drive the high-side PFET hard on and off
using a complementary-type driver with no pullups or such.

\"So everything would have to be referenced to the diode routed power
which might draw too much current from the battery when it\'s all off.\"

And the isolation means you just drive the primary like any
ground-referenced CMOS or optocoupler input you don\'t have to worry
about referencing shit.

 

[boB]

On Tue, 1 Dec 2020 21:01:04 -0800 (PST), Rick C
<gnuarm.deletethisbit@gmail.com> wrote:

On Tuesday, December 1, 2020 at 11:10:23 PM UTC-5, bitrex wrote:
On 12/1/2020 6:31 PM, Rick C wrote:
I have a circuit with a pair of pFETs to switch power and it gets some shoot through during the switching. Mostly it\'s a concern because of the slow turn off using a resistive pullup. Then there\'s the issue of the two power supplies being somewhat different voltages. The high drive has to turn off the FETs completely. The modes are for one or the other to be on or for both to be off.

I\'m wondering if a digital logic device can be used to drive this circuit. 4000 series CMOS can handle the 15V max levels but I don\'t know if it is capable of adequate drive to switch a 10 amp FET. I guess it can\'t be much worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS device. I\'d still be concerned about the inputs though. A low input voltage could provide too little drive for the CMOS input. So everything would have to be referenced to the diode routed power which might draw too much current from the battery when it\'s all off.

Any suggestions on a driver that isn\'t high quiescent drain? Most of what I find are bipolar. It should be very cheap too.

Isolated gate drivers make high-side interfacing so much easier, and
they\'re so relatively inexpensive now that it\'s hard to justify not
using them most times.

Please see figure 3.1:

https://www.mouser.com/datasheet/2/368/Si8751_2-1397983.pdf

these are about a buck in small quantity.

The price is still higher than I\'d like since two of them are now $2.50 which isn\'t much cheaper than the $4 part. I also am not following how these work. Is power being driven across the barrier? A barrier is absolutely not needed in this case.

You could look at the TI series UCC27533DBV. Low power too.

For high side you will need to keep it powered if you want it to stay
pulled up for any length of time over a few milliseconds.

The complexity of that may depend on how high of voltage you want to
switch.

 

[bitrex]

On 12/2/2020 1:04 AM, boB wrote:

On Tue, 1 Dec 2020 21:01:04 -0800 (PST), Rick C
gnuarm.deletethisbit@gmail.com> wrote:

On Tuesday, December 1, 2020 at 11:10:23 PM UTC-5, bitrex wrote:
On 12/1/2020 6:31 PM, Rick C wrote:
I have a circuit with a pair of pFETs to switch power and it gets some shoot through during the switching. Mostly it\'s a concern because of the slow turn off using a resistive pullup. Then there\'s the issue of the two power supplies being somewhat different voltages. The high drive has to turn off the FETs completely. The modes are for one or the other to be on or for both to be off.

I\'m wondering if a digital logic device can be used to drive this circuit. 4000 series CMOS can handle the 15V max levels but I don\'t know if it is capable of adequate drive to switch a 10 amp FET. I guess it can\'t be much worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS device. I\'d still be concerned about the inputs though. A low input voltage could provide too little drive for the CMOS input. So everything would have to be referenced to the diode routed power which might draw too much current from the battery when it\'s all off.

Any suggestions on a driver that isn\'t high quiescent drain? Most of what I find are bipolar. It should be very cheap too.

Isolated gate drivers make high-side interfacing so much easier, and
they\'re so relatively inexpensive now that it\'s hard to justify not
using them most times.

Please see figure 3.1:

https://www.mouser.com/datasheet/2/368/Si8751_2-1397983.pdf

these are about a buck in small quantity.

The price is still higher than I\'d like since two of them are now $2.50 which isn\'t much cheaper than the $4 part. I also am not following how these work. Is power being driven across the barrier? A barrier is absolutely not needed in this case.


You could look at the TI series UCC27533DBV. Low power too.

For high side you will need to keep it powered if you want it to stay
pulled up for any length of time over a few milliseconds.

The complexity of that may depend on how high of voltage you want to
switch.

There\'s a variant of the chip I posted that has a pin for programming
the high-side current consumption when in the on state, obviously with
the trade off being switching speed vs. lower power

 

[Edward Rawde]

\"Rick C\" <gnuarm.deletethisbit@gmail.com> wrote in message
news:02352733-1a66-48f7-a9b5-29b8c46cab13n@googlegroups.com...
On Tuesday, December 1, 2020 at 9:56:01 PM UTC-5, Edward Rawde wrote:

\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:e2a6129f-a999-4b9d...@googlegroups.com...
I have a circuit with a pair of pFETs to switch power and it gets some
shoot through during the switching. Mostly it\'s a concern because of the
slow turn off using a resistive pullup. Then there\'s the issue of the two
power supplies being somewhat different voltages. The high drive has to
turn off the FETs completely. The modes are for one or the other to be on
or for both to be off.

I\'m wondering if a digital logic device can be used to drive this
circuit.
4000 series CMOS can handle the 15V max levels but I don\'t know if it is
capable of adequate drive to switch a 10 amp FET. I guess it can\'t be
much
worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS
device. I\'d still be concerned about the inputs though. A low input
voltage could provide too little drive for the CMOS input. So everything
would have to be referenced to the diode routed power which might draw
too
much current from the battery when it\'s all off.
After reading that I am unable to get a clear picture of what you\'re
trying
to do. I would have posted a reference to a scan of the intended circuit,
complete with labels showing what power is coming from where and going
where
and when and why.
Any suggestions on a driver that isn\'t high quiescent drain? Most of what
I find are bipolar. It should be very cheap too.
I don\'t see anything wrong with using slow switching FETs to switch power
rails, provided it\'s not so slow that the FET heats up too much.
Maybe you need a transistor or two to get the FET gate drive right but
hard
to tell without a schematic.

It\'s not switching in the sense of motor control or a DC converter. It is
a switch over from line to battery when input power fails.

That is what I assumed. Most circuits I\'ve ever designed which did that
didn\'t need a switch at all because the battery was already there on the
rail, ready to take over the load current in the absence of input power.

Each switch is two pFETs to prevent back feed of current. The battery is
12V nominal SLA and the input is a 12V DC for running a motor with 10 amp
peaks and 4 amp nominal. I use a comparator on the input DC to detect when
it is absent which switches between the two power sources. Using
transistors with pull ups means the pull up is slow and there are some 50
us where current passes through both switches between the two power
sources.

Active pull ups come to mind but without seeing a schematic I don\'t think it
will help if I say any more.

Maybe this isn\'t much of a real problem, but 15 or more amps seems like
something to minimize the time duration of if the current itself can\'t be
minimized.

If logic gates are used instead of the transistors with pullups the
transition time will be much shorter and the issue of shoot through current
will be less significant. But I\'m probably showing my ignorance of what
is needed for this circuit to work well. One guy on the team has picked a
LT part at $4 to control the pFETs. It\'s a nice part, but when was an LT
part ever cheap? It just doesn\'t seem like a hard circuit for an
experienced power designer to pull off.

--

Rick C.

 

[Rick C]

On Wednesday, December 2, 2020 at 12:59:24 AM UTC-5, bitrex wrote:

On 12/2/2020 12:01 AM, Rick C wrote:
On Tuesday, December 1, 2020 at 11:10:23 PM UTC-5, bitrex wrote:
On 12/1/2020 6:31 PM, Rick C wrote:
I have a circuit with a pair of pFETs to switch power and it gets some shoot through during the switching. Mostly it\'s a concern because of the slow turn off using a resistive pullup. Then there\'s the issue of the two power supplies being somewhat different voltages. The high drive has to turn off the FETs completely. The modes are for one or the other to be on or for both to be off.

I\'m wondering if a digital logic device can be used to drive this circuit. 4000 series CMOS can handle the 15V max levels but I don\'t know if it is capable of adequate drive to switch a 10 amp FET. I guess it can\'t be much worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS device. I\'d still be concerned about the inputs though. A low input voltage could provide too little drive for the CMOS input. So everything would have to be referenced to the diode routed power which might draw too much current from the battery when it\'s all off.

Any suggestions on a driver that isn\'t high quiescent drain? Most of what I find are bipolar. It should be very cheap too.

Isolated gate drivers make high-side interfacing so much easier, and
they\'re so relatively inexpensive now that it\'s hard to justify not
using them most times.

Please see figure 3.1:

https://www.mouser.com/datasheet/2/368/Si8751_2-1397983.pdf

these are about a buck in small quantity.

The price is still higher than I\'d like since two of them are now $2.50 which isn\'t much cheaper than the $4 part. I also am not following how these work. Is power being driven across the barrier? A barrier is absolutely not needed in this case.

No, the point is the secondary side generates its own gate-drive boost
voltage internally so it can drive the high-side PFET hard on and off
using a complementary-type driver with no pullups or such.
\"So everything would have to be referenced to the diode routed power
which might draw too much current from the battery when it\'s all off.\"
And the isolation means you just drive the primary like any
ground-referenced CMOS or optocoupler input you don\'t have to worry
about referencing shit.

So how does the device know *how much* boost to use to drive the secondary side? Doesn\'t really matter. It\'s too expensive really.

--

Rick C.

-+ Get 1,000 miles of free Supercharging
-+ Tesla referral code - https://ts.la/richard11209

 

[Rick C]

On Wednesday, December 2, 2020 at 1:10:18 AM UTC-5, Edward Rawde wrote:

\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:02352733-1a66-48f7...@googlegroups.com...
On Tuesday, December 1, 2020 at 9:56:01 PM UTC-5, Edward Rawde wrote:
\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:e2a6129f-a999-4b9d...@googlegroups.com...
I have a circuit with a pair of pFETs to switch power and it gets some
shoot through during the switching. Mostly it\'s a concern because of the
slow turn off using a resistive pullup. Then there\'s the issue of the two
power supplies being somewhat different voltages. The high drive has to
turn off the FETs completely. The modes are for one or the other to be on
or for both to be off.

I\'m wondering if a digital logic device can be used to drive this
circuit.
4000 series CMOS can handle the 15V max levels but I don\'t know if it is
capable of adequate drive to switch a 10 amp FET. I guess it can\'t be
much
worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS
device. I\'d still be concerned about the inputs though. A low input
voltage could provide too little drive for the CMOS input. So everything
would have to be referenced to the diode routed power which might draw
too
much current from the battery when it\'s all off.
After reading that I am unable to get a clear picture of what you\'re
trying
to do. I would have posted a reference to a scan of the intended circuit,
complete with labels showing what power is coming from where and going
where
and when and why.
Any suggestions on a driver that isn\'t high quiescent drain? Most of what
I find are bipolar. It should be very cheap too.
I don\'t see anything wrong with using slow switching FETs to switch power
rails, provided it\'s not so slow that the FET heats up too much.
Maybe you need a transistor or two to get the FET gate drive right but
hard
to tell without a schematic.

It\'s not switching in the sense of motor control or a DC converter. It is
a switch over from line to battery when input power fails.
That is what I assumed. Most circuits I\'ve ever designed which did that
didn\'t need a switch at all because the battery was already there on the
rail, ready to take over the load current in the absence of input power.

We have a 12 volt supply which is boosted to charge the battery. When needed the battery drives the 12 volt rail from it\'s initial level of ~13V to the cut out voltage of 10 volts. If the battery was \"on the rail\" the input power source would need to be run through a buck boost to provide the full 10 amps needed. Not very practical in this case.

Each switch is two pFETs to prevent back feed of current. The battery is
12V nominal SLA and the input is a 12V DC for running a motor with 10 amp
peaks and 4 amp nominal. I use a comparator on the input DC to detect when
it is absent which switches between the two power sources. Using
transistors with pull ups means the pull up is slow and there are some 50
us where current passes through both switches between the two power
sources.
Active pull ups come to mind but without seeing a schematic I don\'t think it
will help if I say any more.

Ok, here\'s the LT schematic... or one of them. I keep going back and forth between using transistors and logic. The logic models don\'t really provide analog like simulation though. I\'m pretty sure everything in the design is from the LTspice library. If you don\'t have it you might be working with the old version of the program, I\'m using the new.

Version 4
SHEET 1 1752 916
WIRE 352 -688 320 -688
WIRE -480 -608 -544 -608
WIRE -288 -608 -480 -608
WIRE -16 -608 -288 -608
WIRE 128 -608 80 -608
WIRE 176 -608 128 -608
WIRE 320 -608 320 -688
WIRE 320 -608 272 -608
WIRE 448 -608 320 -608
WIRE 592 -608 544 -608
WIRE 640 -608 592 -608
WIRE 944 -608 736 -608
WIRE 1232 -608 1200 -608
WIRE 1232 -560 1232 -608
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WINDOW 0 0 56 VBottom 2
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SYMATTR Value 10K
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SYMATTR InstName M1
SYMATTR Value 2N7002
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SYMATTR InstName D1
SYMATTR Value MBRS140
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL schottky 656 -160 R0
SYMATTR InstName D2
SYMATTR Value MBRS140
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL schottky -176 16 R90
WINDOW 0 0 32 VBottom 2
WINDOW 3 32 32 VTop 2
SYMATTR InstName D3
SYMATTR Value MBRS140
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL res 1008 -96 R270
WINDOW 0 32 56 VTop 2
WINDOW 3 0 56 VBottom 2
SYMATTR InstName R100
SYMATTR Value 10
SYMBOL res -352 128 R270
WINDOW 0 32 56 VTop 2
WINDOW 3 0 56 VBottom 2
SYMATTR InstName R1
SYMATTR Value 470K
TEXT 816 -496 Left 3 !.tran 1sec
TEXT -472 -664 Left 2 ;Nominal range\\n11.4 < Vin < 12.6

 

[bitrex]

On 12/2/2020 3:49 AM, Rick C wrote:

On Wednesday, December 2, 2020 at 12:59:24 AM UTC-5, bitrex wrote:
On 12/2/2020 12:01 AM, Rick C wrote:
On Tuesday, December 1, 2020 at 11:10:23 PM UTC-5, bitrex wrote:
On 12/1/2020 6:31 PM, Rick C wrote:
I have a circuit with a pair of pFETs to switch power and it gets some shoot through during the switching. Mostly it\'s a concern because of the slow turn off using a resistive pullup. Then there\'s the issue of the two power supplies being somewhat different voltages. The high drive has to turn off the FETs completely. The modes are for one or the other to be on or for both to be off.

I\'m wondering if a digital logic device can be used to drive this circuit. 4000 series CMOS can handle the 15V max levels but I don\'t know if it is capable of adequate drive to switch a 10 amp FET. I guess it can\'t be much worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS device. I\'d still be concerned about the inputs though. A low input voltage could provide too little drive for the CMOS input. So everything would have to be referenced to the diode routed power which might draw too much current from the battery when it\'s all off.

Any suggestions on a driver that isn\'t high quiescent drain? Most of what I find are bipolar. It should be very cheap too.

Isolated gate drivers make high-side interfacing so much easier, and
they\'re so relatively inexpensive now that it\'s hard to justify not
using them most times.

Please see figure 3.1:

https://www.mouser.com/datasheet/2/368/Si8751_2-1397983.pdf

these are about a buck in small quantity.

The price is still higher than I\'d like since two of them are now $2.50 which isn\'t much cheaper than the $4 part. I also am not following how these work. Is power being driven across the barrier? A barrier is absolutely not needed in this case.

No, the point is the secondary side generates its own gate-drive boost
voltage internally so it can drive the high-side PFET hard on and off
using a complementary-type driver with no pullups or such.
\"So everything would have to be referenced to the diode routed power
which might draw too much current from the battery when it\'s all off.\"
And the isolation means you just drive the primary like any
ground-referenced CMOS or optocoupler input you don\'t have to worry
about referencing shit.

So how does the device know *how much* boost to use to drive the secondary side? Doesn\'t really matter. It\'s too expensive really.

Actually I\'m mistaken, my bad. This type of isodriver does transfer
power across the isolation barrier. There\'s no supply input on the
secondary.

 

[bitrex]

On 12/2/2020 4:12 AM, bitrex wrote:

On 12/2/2020 3:49 AM, Rick C wrote:
On Wednesday, December 2, 2020 at 12:59:24 AM UTC-5, bitrex wrote:
On 12/2/2020 12:01 AM, Rick C wrote:
On Tuesday, December 1, 2020 at 11:10:23 PM UTC-5, bitrex wrote:
On 12/1/2020 6:31 PM, Rick C wrote:
I have a circuit with a pair of pFETs to switch power and it gets
some shoot through during the switching. Mostly it\'s a concern
because of the slow turn off using a resistive pullup. Then
there\'s the issue of the two power supplies being somewhat
different voltages. The high drive has to turn off the FETs
completely. The modes are for one or the other to be on or for
both to be off.

I\'m wondering if a digital logic device can be used to drive this
circuit. 4000 series CMOS can handle the 15V max levels but I
don\'t know if it is capable of adequate drive to switch a 10 amp
FET. I guess it can\'t be much worse than the 10k ohm pull up I\'m
using now.

Not sure how to power this. Could use diodes to route power to the
CMOS device. I\'d still be concerned about the inputs though. A low
input voltage could provide too little drive for the CMOS input.
So everything would have to be referenced to the diode routed
power which might draw too much current from the battery when it\'s
all off.

Any suggestions on a driver that isn\'t high quiescent drain? Most
of what I find are bipolar. It should be very cheap too.

Isolated gate drivers make high-side interfacing so much easier, and
they\'re so relatively inexpensive now that it\'s hard to justify not
using them most times.

Please see figure 3.1:

https://www.mouser.com/datasheet/2/368/Si8751_2-1397983.pdf

these are about a buck in small quantity.

The price is still higher than I\'d like since two of them are now
$2.50 which isn\'t much cheaper than the $4 part. I also am not
following how these work. Is power being driven across the barrier?
A barrier is absolutely not needed in this case.

No, the point is the secondary side generates its own gate-drive boost
voltage internally so it can drive the high-side PFET hard on and off
using a complementary-type driver with no pullups or such.
\"So everything would have to be referenced to the diode routed power
which might draw too much current from the battery when it\'s all off.\"
And the isolation means you just drive the primary like any
ground-referenced CMOS or optocoupler input you don\'t have to worry
about referencing shit.

So how does the device know *how much* boost to use to drive the
secondary side?   Doesn\'t  really matter.  It\'s too expensive really.


Actually I\'m mistaken, my bad. This type of isodriver does transfer
power across the isolation barrier. There\'s no supply input on the
secondary.

So basically on the voltage-controlled devices there\'s a pin to set the
drive power via a set resistor, and the current-controlled devices it\'s
set by how hard you drive the \"optocopuler emulator.\"

 

[Rick C]

On Wednesday, December 2, 2020 at 1:10:18 AM UTC-5, Edward Rawde wrote:

\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:02352733-1a66-48f7...@googlegroups.com...
On Tuesday, December 1, 2020 at 9:56:01 PM UTC-5, Edward Rawde wrote:
\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:e2a6129f-a999-4b9d...@googlegroups.com...
I have a circuit with a pair of pFETs to switch power and it gets some
shoot through during the switching. Mostly it\'s a concern because of the
slow turn off using a resistive pullup. Then there\'s the issue of the two
power supplies being somewhat different voltages. The high drive has to
turn off the FETs completely. The modes are for one or the other to be on
or for both to be off.

I\'m wondering if a digital logic device can be used to drive this
circuit.
4000 series CMOS can handle the 15V max levels but I don\'t know if it is
capable of adequate drive to switch a 10 amp FET. I guess it can\'t be
much
worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS
device. I\'d still be concerned about the inputs though. A low input
voltage could provide too little drive for the CMOS input. So everything
would have to be referenced to the diode routed power which might draw
too
much current from the battery when it\'s all off.
After reading that I am unable to get a clear picture of what you\'re
trying
to do. I would have posted a reference to a scan of the intended circuit,
complete with labels showing what power is coming from where and going
where
and when and why.
Any suggestions on a driver that isn\'t high quiescent drain? Most of what
I find are bipolar. It should be very cheap too.
I don\'t see anything wrong with using slow switching FETs to switch power
rails, provided it\'s not so slow that the FET heats up too much.
Maybe you need a transistor or two to get the FET gate drive right but
hard
to tell without a schematic.

It\'s not switching in the sense of motor control or a DC converter. It is
a switch over from line to battery when input power fails.
That is what I assumed. Most circuits I\'ve ever designed which did that
didn\'t need a switch at all because the battery was already there on the
rail, ready to take over the load current in the absence of input power.
Each switch is two pFETs to prevent back feed of current. The battery is
12V nominal SLA and the input is a 12V DC for running a motor with 10 amp
peaks and 4 amp nominal. I use a comparator on the input DC to detect when
it is absent which switches between the two power sources. Using
transistors with pull ups means the pull up is slow and there are some 50
us where current passes through both switches between the two power
sources.
Active pull ups come to mind but without seeing a schematic I don\'t think it
will help if I say any more.

Just to show what I\'m thinking by using the logic devices here is a schematic of that. It uses the CD4000 library. I don\'t know where you might track that down. I get sick and tired of trying to remember the 5 billion things you have to remember to run LTspice. I have the symbols, but they don\'t seem to simulate. I think I need to tell LTspice where to find the library. It\'s too stupid to just look in the directory where everything else is.

Version 4
SHEET 1 1752 916
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FLAG -288 -112 0
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FLAG 352 -688 V_main
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WINDOW 0 0 56 VBottom 2
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SYMBOL pmos 736 -560 M270
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WINDOW 0 2 93 VBottom 2
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WINDOW 39 30 96 Left 2
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SYMATTR SpiceLine Rser=0.1
SYMATTR Value PULSE(0 12.5 0.2s 0.1s 0.1s 0.3s)
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SYMBOL TLV333 -320 16 M180
WINDOW 0 24 41 Left 2
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SYMBOL res 1216 -576 R0
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SYMBOL res 800 -432 R90
WINDOW 0 0 56 VBottom 2
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SYMBOL schottky 704 -272 R0
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SYMATTR Type diode
SYMBOL schottky 560 -272 R0
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SYMATTR Description Diode
SYMATTR Type diode
SYMBOL cap 944 224 R90
WINDOW 0 0 32 VBottom 2
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SYMBOL voltage 144 256 R0
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SYMBOL schottky -176 0 R90
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WINDOW 0 0 32 VBottom 2
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SYMBOL res -352 112 R270
WINDOW 0 32 56 VTop 2
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SYMATTR InstName R1
SYMATTR Value 470K
SYMBOL CD4000\\\\CD4093B -32 -480 R0
WINDOW 0 -16 16 Left 2
WINDOW 3 -32 112 Left 1
SYMATTR InstName U2A
SYMATTR SpiceLine VDD=12 SPEED=1.0 TRIPDT=5e-9
SYMATTR SpiceModel V12_always 0
SYMBOL res -128 -576 R0
SYMATTR InstName R2
SYMATTR Value 10K
SYMBOL CD4000\\\\CD4093B 128 -320 R0
WINDOW 0 -16 16 Left 2
WINDOW 3 -32 112 Left 1
SYMATTR InstName U2B
SYMATTR SpiceLine VDD=12 SPEED=1.0 TRIPDT=5e-9
SYMATTR SpiceModel V12_always 0
SYMBOL CD4000\\\\CD4093B 448 -304 R0
WINDOW 0 -16 16 Left 2
WINDOW 3 -32 112 Left 1
SYMATTR InstName U2C
SYMATTR SpiceLine VDD=12 SPEED=1.0 TRIPDT=5e-9
SYMATTR SpiceModel V12_always 0
SYMBOL CD4000\\\\CD4093B 672 160 R0
WINDOW 0 -16 16 Left 2
WINDOW 3 -32 112 Left 1
SYMATTR InstName U3A
SYMATTR SpiceLine VDD=3.3 SPEED=1.0 TRIPDT=5e-9
SYMATTR SpiceModel V3.3_always 0
SYMBOL CD4000\\\\CD4093B 768 96 R180
WINDOW 0 -16 16 Left 2
WINDOW 3 -32 112 Left 1
SYMATTR InstName U3B
SYMATTR SpiceLine VDD=3.3 SPEED=1.0 TRIPDT=5e-9
SYMATTR SpiceModel V3.3_always 0
SYMBOL res 560 96 M0
SYMATTR InstName R3
SYMATTR Value 1Meg
SYMBOL res 992 0 R90
WINDOW 0 2 93 VBottom 2
WINDOW 3 1 29 VBottom 2
SYMATTR InstName R4
SYMATTR Value 1Meg
TEXT 816 -496 Left 3 !.tran 1sec
TEXT -472 -664 Left 2 ;Nominal range\\n11.4 < Vin < 12.6
TEXT 96 136 Left 2 ;Low True TriState Control\\nFPGA - Normally hi-Z, activate \\nby pulling high 20 ms, then low
TEXT 976 136 Left 2 ;Low True TriState Control\\nMCU - Normally hi-Z, activate \\nby pulling high 5 ms, then low

 

[Edward Rawde]

\"Rick C\" <gnuarm.deletethisbit@gmail.com> wrote in message
news:6fce9360-b9e9-450f-b078-02ff237ed2d9n@googlegroups.com...
On Wednesday, December 2, 2020 at 1:10:18 AM UTC-5, Edward Rawde wrote:

\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:02352733-1a66-48f7...@googlegroups.com...
On Tuesday, December 1, 2020 at 9:56:01 PM UTC-5, Edward Rawde wrote:
\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:e2a6129f-a999-4b9d...@googlegroups.com...
I have a circuit with a pair of pFETs to switch power and it gets some
shoot through during the switching. Mostly it\'s a concern because of
the
slow turn off using a resistive pullup. Then there\'s the issue of the
two
power supplies being somewhat different voltages. The high drive has to
turn off the FETs completely. The modes are for one or the other to be
on
or for both to be off.

I\'m wondering if a digital logic device can be used to drive this
circuit.
4000 series CMOS can handle the 15V max levels but I don\'t know if it
is
capable of adequate drive to switch a 10 amp FET. I guess it can\'t be
much
worse than the 10k ohm pull up I\'m using now.

Not sure how to power this. Could use diodes to route power to the CMOS
device. I\'d still be concerned about the inputs though. A low input
voltage could provide too little drive for the CMOS input. So
everything
would have to be referenced to the diode routed power which might draw
too
much current from the battery when it\'s all off.
After reading that I am unable to get a clear picture of what you\'re
trying
to do. I would have posted a reference to a scan of the intended
circuit,
complete with labels showing what power is coming from where and going
where
and when and why.
Any suggestions on a driver that isn\'t high quiescent drain? Most of
what
I find are bipolar. It should be very cheap too.
I don\'t see anything wrong with using slow switching FETs to switch
power
rails, provided it\'s not so slow that the FET heats up too much.
Maybe you need a transistor or two to get the FET gate drive right but
hard
to tell without a schematic.

It\'s not switching in the sense of motor control or a DC converter. It is
a switch over from line to battery when input power fails.
That is what I assumed. Most circuits I\'ve ever designed which did that
didn\'t need a switch at all because the battery was already there on the
rail, ready to take over the load current in the absence of input power.
Each switch is two pFETs to prevent back feed of current. The battery is
12V nominal SLA and the input is a 12V DC for running a motor with 10 amp
peaks and 4 amp nominal. I use a comparator on the input DC to detect
when
it is absent which switches between the two power sources. Using
transistors with pull ups means the pull up is slow and there are some 50
us where current passes through both switches between the two power
sources.
Active pull ups come to mind but without seeing a schematic I don\'t think
it
will help if I say any more.

Just to show what I\'m thinking by using the logic devices here is a
schematic of that. It uses the CD4000 library. I don\'t know where you
might track that down. I get sick and tired of trying to remember the 5
billion things you have to remember to run LTspice. I have the symbols,
but they don\'t seem to simulate. I think I need to tell LTspice where to
find the library. It\'s too stupid to just look in the directory where
everything else is.

Ok well leaving aside the million questions I\'d have brought to the design
review meeting (3 months ago) and assuming you really do need two power
switches with bidirectional isolation then I\'d use the circuit shown on the
following page with suitable FETs and component values and watch out for
your FET\'s maximum gate-source voltage rating.
https://electronic-products-design.com/geek-area/electronics/mosfets/using-mosfets-as-general-switches
Which was quickly found like this:
https://www.google.com/search?q=mosfet+general+switch&tbm=isch
I might add some low value gate series resistors mounted close to the FETs.

The transistor, with suitable base resistors, can be driven from 3.3V logic
and if the logic power is always present then do all the logic in logic
unless it\'s cheaper to add some more transistors and /or diodes to do it.
Another transistor with the collector connected to the existing transistor\'s
collector will make an OR gate.
An open collector comparator can likely be connected directly to the FET
gates or to the transistor base drive circuit as necessary.
I can see no need to want to do logic functions at gate drive level.
What\'s D3 for? Some op amps don\'t like to work as comparators but I\'ve no
idea whether TLV333 cares.

 

[Rick C]

On Thursday, December 3, 2020 at 11:57:05 AM UTC-5, Edward Rawde wrote:

\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:5f1f4d13-a97e-45c6...@googlegroups.com...
On Wednesday, December 2, 2020 at 10:48:03 PM UTC-5, Edward Rawde wrote:
jla...@highlandsniptechnology.com> wrote in message
news:u3ngsfhkn7u43qs1v...@4ax.com...
On Wed, 2 Dec 2020 17:12:04 -0500, \"Edward Rawde\"
inv...@invalid.invalid> wrote:

\"Edward Rawde\" <inv...@invalid.invalid> wrote in message
news:rq8q5r$l3r$1...@gioia.aioe.org...
\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:6fce9360-b9e9-450f...@googlegroups.com...
On Wednesday, December 2, 2020 at 1:10:18 AM UTC-5, Edward Rawde
wrote:
\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:02352733-1a66-48f7...@googlegroups.com...
On Tuesday, December 1, 2020 at 9:56:01 PM UTC-5, Edward Rawde wrote:
\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:e2a6129f-a999-4b9d...@googlegroups.com...
I have a circuit with a pair of pFETs to switch power and it gets
some
shoot through during the switching.

[...]

Also I don\'t see how any harm could come to the op amp working as a
comparator (some of them don\'t like that) if D3 is replaced by a short
circuit.
If D3 is really needed then MBRS140 seems overkill.

Your TLV333 will probably be fine as a comparator but in cases where it
isn\'t, the advice is clearly to use a comparator if you need a comparator:
https://e2e.ti.com/support/amplifiers/f/14/t/684067

Another LTspice cock up. I didn\'t intend to use an opamp. It was supposed to be an open collector comparator. I copied the part from another design and didn\'t notice it was not a comparator. Thanks for that one. That explains a lot with the slow edge rate I\'m seeing.

The harm is if the power input is reversed. The other components in the circuit will be fine, but the op amp^H^H^H^H^H^H comparator would be damaged.

The design review is coming up. We\'ll see what others think of the LT oriented design.

--

Rick C.

--+ Get 1,000 miles of free Supercharging
--+ Tesla referral code - https://ts.la/richard11209

 

[Rick C]

On Thursday, December 3, 2020 at 11:39:15 AM UTC-5, Edward Rawde wrote:

\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:5f1f4d13-a97e-45c6...@googlegroups.com...
On Wednesday, December 2, 2020 at 10:48:03 PM UTC-5, Edward Rawde wrote:
jla...@highlandsniptechnology.com> wrote in message
news:u3ngsfhkn7u43qs1v...@4ax.com...
On Wed, 2 Dec 2020 17:12:04 -0500, \"Edward Rawde\"
inv...@invalid.invalid> wrote:

\"Edward Rawde\" <inv...@invalid.invalid> wrote in message
news:rq8q5r$l3r$1...@gioia.aioe.org...
\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:6fce9360-b9e9-450f...@googlegroups.com...
On Wednesday, December 2, 2020 at 1:10:18 AM UTC-5, Edward Rawde
wrote:
\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:02352733-1a66-48f7...@googlegroups.com...
On Tuesday, December 1, 2020 at 9:56:01 PM UTC-5, Edward Rawde wrote:
\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:e2a6129f-a999-4b9d...@googlegroups.com...
I have a circuit with a pair of pFETs to switch power and it gets
some
shoot through during the switching.

One further thought.
If the problem is what happens when both switches are on then control
them
independently and arrange for that not to be the case.
In other words you need a break before make switchover.
Hang sufficient capacitance on the rail to the motor control circuit so
it
doesn\'t see the join.


How about using two diodes? Too easy I guess.
Yes a couple of diodes producing Vmain would stop Vin and Vbatt ever
meeting
each other.
Yes, they will do that. But they won\'t allow the rest of the circuit to
work and the reason why has already been covered. It\'s also very obvious
if you look at what the circuit is doing.

In fact, the diodes are already there for the low current always on path..
They work like a champ!

Actually, the predecessor circuit to this one had a FET based diode pair to
combine the input and battery circuits. But that was with a higher voltage
input and now we are using a 12 volt input. Use diodes and the battery
will always be powering the circuit until it runs down.
I think some of us are having trouble fully understanding your system
topology. A block diagram may or may not help. In any case, why does it
matter if both switches are on for say 50uS, what is that going to hurt?
If it could possibly hurt anything at all then add a diode so it can\'t.

Why would you use diodes and switches??? I don\'t know if 40 us of shoot through is an issue. I thought I made that clear in the original descriptions.

The input powers a buck/boost charger for the battery and powers the switch selecting the battery or the input power. Is that what you were missing? You talked about running the entire system off the charging circuit so I thought you must have understood that, no?

--

Rick C.

-+- Get 1,000 miles of free Supercharging
-+- Tesla referral code - https://ts.la/richard11209

 

[Edward Rawde]

\"Rick C\" <gnuarm.deletethisbit@gmail.com> wrote in message
news:7fc9a590-39c8-401a-a877-85f9c5598bfbn@googlegroups.com...
On Thursday, December 3, 2020 at 11:57:05 AM UTC-5, Edward Rawde wrote:

\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:5f1f4d13-a97e-45c6...@googlegroups.com...
On Wednesday, December 2, 2020 at 10:48:03 PM UTC-5, Edward Rawde wrote:
jla...@highlandsniptechnology.com> wrote in message
news:u3ngsfhkn7u43qs1v...@4ax.com...
On Wed, 2 Dec 2020 17:12:04 -0500, \"Edward Rawde\"
inv...@invalid.invalid> wrote:

\"Edward Rawde\" <inv...@invalid.invalid> wrote in message
news:rq8q5r$l3r$1...@gioia.aioe.org...
\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:6fce9360-b9e9-450f...@googlegroups.com...
On Wednesday, December 2, 2020 at 1:10:18 AM UTC-5, Edward Rawde
wrote:
\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:02352733-1a66-48f7...@googlegroups.com...
On Tuesday, December 1, 2020 at 9:56:01 PM UTC-5, Edward Rawde
wrote:
\"Rick C\" <gnuarm.del...@gmail.com> wrote in message
news:e2a6129f-a999-4b9d...@googlegroups.com...
I have a circuit with a pair of pFETs to switch power and it
gets
some
shoot through during the switching.

[...]

Also I don\'t see how any harm could come to the op amp working as a
comparator (some of them don\'t like that) if D3 is replaced by a short
circuit.
If D3 is really needed then MBRS140 seems overkill.

Your TLV333 will probably be fine as a comparator but in cases where it
isn\'t, the advice is clearly to use a comparator if you need a comparator:
https://e2e.ti.com/support/amplifiers/f/14/t/684067

Another LTspice cock up. I didn\'t intend to use an opamp. It was supposed
to be an open collector comparator. I copied the part from another design
and didn\'t notice it was not a comparator. Thanks for that one. That
explains a lot with the slow edge rate I\'m seeing.

The harm is if the power input is reversed. The other components in the
circuit will be fine, but the op amp^H^H^H^H^H^H comparator would be
damaged.

The design review is coming up. We\'ll see what others think of the LT
oriented design.

I likely wouldn\'t think LTspice for this kind of circuit. If there was any
doubt about correct operation then by now I\'d have soldered the parts
together and tested it in a lab, or equivalent parts with suitable legs.

If there\'s a risk of the power input being reversed I\'d probably put a
series shottky diode in the power line. Reverse power might damage more than
your comparator.

--

Rick C.