Voltage Limiter...

[Robert Baer]

Ricketty C wrote:

On Tuesday, September 1, 2020 at 4:18:09 PM UTC-4, Robert Baer wrote:
Ricketty C wrote:
On Monday, August 31, 2020 at 7:07:22 PM UTC-4, Lasse Langwadt Christensen wrote:
tirsdag den 1. september 2020 kl. 01.02.06 UTC+2 skrev Ricketty C:
On Monday, August 31, 2020 at 5:53:52 PM UTC-4, Robert Baer wrote:
Ricketty C wrote:
I suppose this works like a voltage regulator but it is hard to search those to find which ones will work down to what voltage with minimum drop out. This circuit limits the output voltage to about 12.4 volts and has a minimum drop out of 0.15 volts down to a Vin of around 6 volts. This is powering a motor, so the current probably won\'t change as fast as I\'ve simulated.

This seems to work pretty well without instabilities. The parts cost is minimal even if the parts count is higher than with a regulator.

http://arius.com/temp/MotorOverCurrent_Vlimit_FET.asc

http://arius.com/temp/MotorOverCurrent_Vlimit_FET.plt

I\'m pretty sure there aren\'t any missing models. At least not in LTspice XVII. I wonder why it went from IV to XVII...?

Does anyone know of a regulator that will provide 12.4 volts from 17 volts max with minimum drop out voltage as the input drops to 10 volts? It needs to have a shutdown input as well and draw very low idle current, <100 uA, better to be <50 uA.

That would be VERY interesting..input down to 10V and output staying
near 12.4V...you did imply linear...

I assume there is a joke in there somewhere. You do realize I\'m talking about an output dropping once the input drops below the output plus some drop out. That\'s my point, regulators are spec\'d for a minimum drop out where they meet full specs. I\'m interested in the drop out once they are out of regulation and the pass transistor is turned on fully.

Actually, using a regulator may not be an option anymore. We are combining the input DC @ 15V with the battery voltage through diodes to feed the main power rail. The motor is the only heavy draw. I\'m wondering if it makes sense to use two FETs back to back to prevent reverse current flow between the two circuits. One pair for the main power source and a second pair for the battery resulting in a lower voltage drop so the battery can be used more effectively.

Using a regulator will still require a diode to prevent back current flow to the main power input. So two diodes and a regulator or four FETs?

I actually saw this circuit on a schematic for a Trenz FPGA board, but much lower power levels. Not sure which way I want to go.


you just need a \"diode\" from each, something like this has been used on the raspberry pi to enable using a psu or USB power and not backfeed the USB

https://robotics.ong.id.au/wp-content/uploads/2014/07/rpi_protection.png

I can\'t seem to make that work for me. It is set up to equalize the voltage when passing current in one direction. But it can\'t be used to limit the output voltage so well because the lowest drop you can get is 0.7 volts or so through the parasitic diode. I need to scrub about 2.5 volt from the main power input or a varying amount from the battery, down to as close to zero as possible.

I could just use a FET for the limiter and then use two of these to combine the two supplies. Or maybe only one is needed on the battery. When the power source is unplugged, I\'m not sure it would hurt anything to reverse power the input. I only feeds the battery charger. I guess it would create some additional power drain. There is an OFF state where only the minimal circuitry is powered with the intent to use less power than the battery self drain current of ~100 uA. Heck, the FET diode circuit as drawn would draw <300 uA! I\'d need to knock the resistors up to 1Meg to get the current low enough at 14V. But that sounds like a winner. Thanks.

So the final circuit can be a diode from the DC input (no worry with the drop and it\'s used elsewhere so it\'s on the BoM regardless), this circuit for the battery connection to the common power rail. Then my previous voltage limiter circuit to supply the motor and provide a means of cutting power in case the software runs amuck resulting in an over current.

I\'m not a fan of adding unique parts to the BOM, but I think this one is worth it! Thanks again.

Color me confused; you talk about 12V motor max and yet have a 5V
\"transient protector\" at the output???
And it takes only 250K for those resistors to get a \"leakage\" of 48uA
at 12V in.
Looked at the DMMT5401, a good choice, beta may be a little low at a
measly 100 but seems adequate,giving Vce(sat) <200mV up to 100mA.
Again NO HINT as to motor current.
More info??

Thanks.

Sorry, I\'m confused. Where do you see a 5 volt transient protector??? The diagram you are likely looking at was posted by someone else to illustrate the properties of the circuit, not a suggestion for what I should use. I don\'t think the beta is a critical parameter. The circuit seem to produce a voltage drop of 80 mV used with an AOD4185 in my circuit.

* The posed ciRcuit eXplicitly showed a stupid SMBJ5.0A at the output.
And 250K resistors will give you 48uA \"leakage\"; no need for 1meg;
much faster at power drop.
The higher the beta, the wider the range for Iout/Iin .

To minimize leakage from the battery I used 1 Meg resistors which slow the response to the line power dropping resulting in a half volt drop in voltage for half a millisecond in the simulation. I consider that acceptable. The requirements on this circuit are not severe. It\'s powering a motor, not digital logic. There are also some whopping big caps on this rail to deal with the flyback current from the PWM turning off. Cutting power at this FET is a different matter, so I added a flyback diode and small resistor.

The motor current trips a cut off over 4 amps. Motor rating is 3.5 amps continuous.

* NOW the secret is out!
I suspected that; now you know why i picked out high current transistors.
You may need to use those 4A rated Zetek transistors to carry that
kind of kick-back and maintain <400mV Vo-Vin; prevents the poor
unsuspecting FET from having to handle the motor sass.
High beta here is your friend.

Don\'t know what current it will actually be operated or if any of
these numbers need to be limited instantaneous or average. Not sure how
long it takes to snap the motor shaft when the mechanism binds. They\'ve
broken several Pololu units so far. They are using a better motor now.

The rev 2 board will have the building blocks to manage this no matter what. The current limit will actually be done in the FPGA, so no worry about the software crashing and breaking the hardware.

 

[Ricketty C]

On Wednesday, September 2, 2020 at 2:51:43 AM UTC-4, Robert Baer wrote:

Ricketty C wrote:
On Tuesday, September 1, 2020 at 4:18:09 PM UTC-4, Robert Baer wrote:
Ricketty C wrote:
On Monday, August 31, 2020 at 7:07:22 PM UTC-4, Lasse Langwadt Christensen wrote:
tirsdag den 1. september 2020 kl. 01.02.06 UTC+2 skrev Ricketty C:
On Monday, August 31, 2020 at 5:53:52 PM UTC-4, Robert Baer wrote:
Ricketty C wrote:
I suppose this works like a voltage regulator but it is hard to search those to find which ones will work down to what voltage with minimum drop out. This circuit limits the output voltage to about 12.4 volts and has a minimum drop out of 0.15 volts down to a Vin of around 6 volts. This is powering a motor, so the current probably won\'t change as fast as I\'ve simulated.

This seems to work pretty well without instabilities. The parts cost is minimal even if the parts count is higher than with a regulator.

http://arius.com/temp/MotorOverCurrent_Vlimit_FET.asc

http://arius.com/temp/MotorOverCurrent_Vlimit_FET.plt

I\'m pretty sure there aren\'t any missing models. At least not in LTspice XVII. I wonder why it went from IV to XVII...?

Does anyone know of a regulator that will provide 12.4 volts from 17 volts max with minimum drop out voltage as the input drops to 10 volts? It needs to have a shutdown input as well and draw very low idle current, <100 uA, better to be <50 uA.

That would be VERY interesting..input down to 10V and output staying
near 12.4V...you did imply linear...

I assume there is a joke in there somewhere. You do realize I\'m talking about an output dropping once the input drops below the output plus some drop out. That\'s my point, regulators are spec\'d for a minimum drop out where they meet full specs. I\'m interested in the drop out once they are out of regulation and the pass transistor is turned on fully.

Actually, using a regulator may not be an option anymore. We are combining the input DC @ 15V with the battery voltage through diodes to feed the main power rail. The motor is the only heavy draw. I\'m wondering if it makes sense to use two FETs back to back to prevent reverse current flow between the two circuits. One pair for the main power source and a second pair for the battery resulting in a lower voltage drop so the battery can be used more effectively.

Using a regulator will still require a diode to prevent back current flow to the main power input. So two diodes and a regulator or four FETs?

I actually saw this circuit on a schematic for a Trenz FPGA board, but much lower power levels. Not sure which way I want to go.


you just need a \"diode\" from each, something like this has been used on the raspberry pi to enable using a psu or USB power and not backfeed the USB

https://robotics.ong.id.au/wp-content/uploads/2014/07/rpi_protection..png

I can\'t seem to make that work for me. It is set up to equalize the voltage when passing current in one direction. But it can\'t be used to limit the output voltage so well because the lowest drop you can get is 0.7 volts or so through the parasitic diode. I need to scrub about 2.5 volt from the main power input or a varying amount from the battery, down to as close to zero as possible.

I could just use a FET for the limiter and then use two of these to combine the two supplies. Or maybe only one is needed on the battery. When the power source is unplugged, I\'m not sure it would hurt anything to reverse power the input. I only feeds the battery charger. I guess it would create some additional power drain. There is an OFF state where only the minimal circuitry is powered with the intent to use less power than the battery self drain current of ~100 uA. Heck, the FET diode circuit as drawn would draw <300 uA! I\'d need to knock the resistors up to 1Meg to get the current low enough at 14V. But that sounds like a winner. Thanks.

So the final circuit can be a diode from the DC input (no worry with the drop and it\'s used elsewhere so it\'s on the BoM regardless), this circuit for the battery connection to the common power rail. Then my previous voltage limiter circuit to supply the motor and provide a means of cutting power in case the software runs amuck resulting in an over current.

I\'m not a fan of adding unique parts to the BOM, but I think this one is worth it! Thanks again.

Color me confused; you talk about 12V motor max and yet have a 5V
\"transient protector\" at the output???
And it takes only 250K for those resistors to get a \"leakage\" of 48uA
at 12V in.
Looked at the DMMT5401, a good choice, beta may be a little low at a
measly 100 but seems adequate,giving Vce(sat) <200mV up to 100mA.
Again NO HINT as to motor current.
More info??

Thanks.

Sorry, I\'m confused. Where do you see a 5 volt transient protector??? The diagram you are likely looking at was posted by someone else to illustrate the properties of the circuit, not a suggestion for what I should use. I don\'t think the beta is a critical parameter. The circuit seem to produce a voltage drop of 80 mV used with an AOD4185 in my circuit.
* The posed ciRcuit eXplicitly showed a stupid SMBJ5.0A at the output.
And 250K resistors will give you 48uA \"leakage\"; no need for 1meg;
much faster at power drop.
The higher the beta, the wider the range for Iout/Iin .

You must have gotten out of bed on the wrong side these days. What is going on with you??? No one \"posed\" a circuit, \"eXplicitly\" or otherwise. It was an example of usage showing the concept... as I just explained. Do you not understand the difference?

I don\'t want 48 uA of leakage. Better to have lower leakage and the speed is not important here. Optimization is about improving the circuit for the particular application. The need here is to keep the vampire drain as low as practical so the self drain rate is not significantly increased. There are other parts of the circuit that are drawing current when off, so no reason to increase this drain without purpose.

To minimize leakage from the battery I used 1 Meg resistors which slow the response to the line power dropping resulting in a half volt drop in voltage for half a millisecond in the simulation. I consider that acceptable.. The requirements on this circuit are not severe. It\'s powering a motor, not digital logic. There are also some whopping big caps on this rail to deal with the flyback current from the PWM turning off. Cutting power at this FET is a different matter, so I added a flyback diode and small resistor.

The motor current trips a cut off over 4 amps. Motor rating is 3.5 amps continuous.
* NOW the secret is out!
I suspected that; now you know why i picked out high current transistors.
You may need to use those 4A rated Zetek transistors to carry that
kind of kick-back and maintain <400mV Vo-Vin; prevents the poor
unsuspecting FET from having to handle the motor sass.
High beta here is your friend.

Sorry, I don\'t know what Zetek transistors you are talking about. The simulation of this circuit is showing less than 100 mV drop out from the AOD4185. Why do I care about beta in a MOSFET?

This circuit is not directly driving the motor. There is an H-bridge in the path and 2,000 uF of capacitor on the power to it. I don\'t think there is going to be too much sass from the motor, but if there is a diode takes care of that. No?

The bipolar transistors don\'t carry the motor current at all. Why do they need to be high current? Unless you are saying something very different. A picture is worth a thousand words.

--

Rick C.

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

 

[Robert Baer]

Ricketty C wrote:

On Wednesday, September 2, 2020 at 2:51:43 AM UTC-4, Robert Baer wrote:
Ricketty C wrote:
On Tuesday, September 1, 2020 at 4:18:09 PM UTC-4, Robert Baer wrote:
Ricketty C wrote:
On Monday, August 31, 2020 at 7:07:22 PM UTC-4, Lasse Langwadt Christensen wrote:
tirsdag den 1. september 2020 kl. 01.02.06 UTC+2 skrev Ricketty C:
On Monday, August 31, 2020 at 5:53:52 PM UTC-4, Robert Baer wrote:
Ricketty C wrote:
I suppose this works like a voltage regulator but it is hard to search those to find which ones will work down to what voltage with minimum drop out. This circuit limits the output voltage to about 12.4 volts and has a minimum drop out of 0.15 volts down to a Vin of around 6 volts. This is powering a motor, so the current probably won\'t change as fast as I\'ve simulated.

This seems to work pretty well without instabilities. The parts cost is minimal even if the parts count is higher than with a regulator.

http://arius.com/temp/MotorOverCurrent_Vlimit_FET.asc

http://arius.com/temp/MotorOverCurrent_Vlimit_FET.plt

I\'m pretty sure there aren\'t any missing models. At least not in LTspice XVII. I wonder why it went from IV to XVII...?

Does anyone know of a regulator that will provide 12.4 volts from 17 volts max with minimum drop out voltage as the input drops to 10 volts? It needs to have a shutdown input as well and draw very low idle current, <100 uA, better to be <50 uA.

That would be VERY interesting..input down to 10V and output staying
near 12.4V...you did imply linear...

I assume there is a joke in there somewhere. You do realize I\'m talking about an output dropping once the input drops below the output plus some drop out. That\'s my point, regulators are spec\'d for a minimum drop out where they meet full specs. I\'m interested in the drop out once they are out of regulation and the pass transistor is turned on fully.

Actually, using a regulator may not be an option anymore. We are combining the input DC @ 15V with the battery voltage through diodes to feed the main power rail. The motor is the only heavy draw. I\'m wondering if it makes sense to use two FETs back to back to prevent reverse current flow between the two circuits. One pair for the main power source and a second pair for the battery resulting in a lower voltage drop so the battery can be used more effectively.

Using a regulator will still require a diode to prevent back current flow to the main power input. So two diodes and a regulator or four FETs?

I actually saw this circuit on a schematic for a Trenz FPGA board, but much lower power levels. Not sure which way I want to go.


you just need a \"diode\" from each, something like this has been used on the raspberry pi to enable using a psu or USB power and not backfeed the USB

https://robotics.ong.id.au/wp-content/uploads/2014/07/rpi_protection.png

I can\'t seem to make that work for me. It is set up to equalize the voltage when passing current in one direction. But it can\'t be used to limit the output voltage so well because the lowest drop you can get is 0.7 volts or so through the parasitic diode. I need to scrub about 2.5 volt from the main power input or a varying amount from the battery, down to as close to zero as possible.

I could just use a FET for the limiter and then use two of these to combine the two supplies. Or maybe only one is needed on the battery. When the power source is unplugged, I\'m not sure it would hurt anything to reverse power the input. I only feeds the battery charger. I guess it would create some additional power drain. There is an OFF state where only the minimal circuitry is powered with the intent to use less power than the battery self drain current of ~100 uA. Heck, the FET diode circuit as drawn would draw <300 uA! I\'d need to knock the resistors up to 1Meg to get the current low enough at 14V. But that sounds like a winner. Thanks.

So the final circuit can be a diode from the DC input (no worry with the drop and it\'s used elsewhere so it\'s on the BoM regardless), this circuit for the battery connection to the common power rail. Then my previous voltage limiter circuit to supply the motor and provide a means of cutting power in case the software runs amuck resulting in an over current.

I\'m not a fan of adding unique parts to the BOM, but I think this one is worth it! Thanks again.

Color me confused; you talk about 12V motor max and yet have a 5V
\"transient protector\" at the output???
And it takes only 250K for those resistors to get a \"leakage\" of 48uA
at 12V in.
Looked at the DMMT5401, a good choice, beta may be a little low at a
measly 100 but seems adequate,giving Vce(sat) <200mV up to 100mA.
Again NO HINT as to motor current.
More info??

Thanks.

Sorry, I\'m confused. Where do you see a 5 volt transient protector??? The diagram you are likely looking at was posted by someone else to illustrate the properties of the circuit, not a suggestion for what I should use. I don\'t think the beta is a critical parameter. The circuit seem to produce a voltage drop of 80 mV used with an AOD4185 in my circuit.

* Have yet to see the rest of it; the AOD4185 is way more beefy than
needed for a measly 4A motor.
In fact, that DMG2305UX will do the job nicely.

* The posed ciRcuit eXplicitly showed a stupid SMBJ5.0A at the output.
And 250K resistors will give you 48uA \"leakage\"; no need for 1meg;
much faster at power drop.
The higher the beta, the wider the range for Iout/Iin .

You must have gotten out of bed on the wrong side these days. What is going on with you??? No one \"posed\" a circuit, \"eXplicitly\" or otherwise. It was an example of usage showing the concept... as I just explained. Do you not understand the difference?

I don\'t want 48 uA of leakage.

* First you said 100uA would be nice, and indicated that 50uA would be
better.
Now you are bitching that 48uA is too much.
With higher beta, 400 or more, you could have a 10uA drive/\"leakage\"..

Better to have lower leakage and the speed is not important here. Optimization is about improving the circuit for the particular application. The need here is to keep the vampire drain as low as practical so the self drain rate is not significantly increased. There are other parts of the circuit that are drawing current when off, so no reason to increase this drain without purpose.


To minimize leakage from the battery I used 1 Meg resistors which slow the response to the line power dropping resulting in a half volt drop in voltage for half a millisecond in the simulation. I consider that acceptable. The requirements on this circuit are not severe. It\'s powering a motor, not digital logic. There are also some whopping big caps on this rail to deal with the flyback current from the PWM turning off. Cutting power at this FET is a different matter, so I added a flyback diode and small resistor.

* Then use 2 FETs back-to-back instead, making for a more symmetrical
fancy voltage follower.

The motor current trips a cut off over 4 amps. Motor rating is 3.5 amps continuous.
* NOW the secret is out!
I suspected that; now you know why i picked out high current transistors.
You may need to use those 4A rated Zetek transistors to carry that
kind of kick-back and maintain <400mV Vo-Vin; prevents the poor
unsuspecting FET from having to handle the motor sass.
High beta here is your friend.

Sorry, I don\'t know what Zetek transistors you are talking about. The simulation of this circuit is showing less than 100 mV drop out from the AOD4185. Why do I care about beta in a MOSFET?

* Idiot! I was obviously talking abut the transistor pair, and being
used as a \"fancy voltage follower\".

This circuit is not directly driving the motor. There is an H-bridge in the path and 2,000 uF of capacitor on the power to it. I don\'t think there is going to be too much sass from the motor, but if there is a diode takes care of that. No?

* The circuit, as shown without the FET and a motor (load acting as a
generator,can feed current into this \"fancy voltage follower\" will dump
that current into U14b.
That current flows thru the resistors and out U14a; granted not 3A,
but near the 48uA input current.
Using 2 FETs back-to-back allows that \"back current\" to flow to the
input.

The bipolar transistors don\'t carry the motor current at all. Why do they need to be high current? Unless you are saying something very different. A picture is worth a thousand words.

And you are tossing ultra-beefy FETs and H-bridges and other unknown
parts in a scramble that nobody knows about.
Perhaps a circuit different than the \"fancy voltage follower\" will do
what is needed, whatever that is.

 

[Ricketty C]

On Monday, August 31, 2020 at 7:07:22 PM UTC-4, Lasse Langwadt Christensen wrote:

tirsdag den 1. september 2020 kl. 01.02.06 UTC+2 skrev Ricketty C:
On Monday, August 31, 2020 at 5:53:52 PM UTC-4, Robert Baer wrote:
Ricketty C wrote:
I suppose this works like a voltage regulator but it is hard to search those to find which ones will work down to what voltage with minimum drop out. This circuit limits the output voltage to about 12.4 volts and has a minimum drop out of 0.15 volts down to a Vin of around 6 volts. This is powering a motor, so the current probably won\'t change as fast as I\'ve simulated.

This seems to work pretty well without instabilities. The parts cost is minimal even if the parts count is higher than with a regulator.

http://arius.com/temp/MotorOverCurrent_Vlimit_FET.asc

http://arius.com/temp/MotorOverCurrent_Vlimit_FET.plt

I\'m pretty sure there aren\'t any missing models. At least not in LTspice XVII. I wonder why it went from IV to XVII...?

Does anyone know of a regulator that will provide 12.4 volts from 17 volts max with minimum drop out voltage as the input drops to 10 volts? It needs to have a shutdown input as well and draw very low idle current, <100 uA, better to be <50 uA.

That would be VERY interesting..input down to 10V and output staying
near 12.4V...you did imply linear...

I assume there is a joke in there somewhere. You do realize I\'m talking about an output dropping once the input drops below the output plus some drop out. That\'s my point, regulators are spec\'d for a minimum drop out where they meet full specs. I\'m interested in the drop out once they are out of regulation and the pass transistor is turned on fully.

Actually, using a regulator may not be an option anymore. We are combining the input DC @ 15V with the battery voltage through diodes to feed the main power rail. The motor is the only heavy draw. I\'m wondering if it makes sense to use two FETs back to back to prevent reverse current flow between the two circuits. One pair for the main power source and a second pair for the battery resulting in a lower voltage drop so the battery can be used more effectively.

Using a regulator will still require a diode to prevent back current flow to the main power input. So two diodes and a regulator or four FETs?

I actually saw this circuit on a schematic for a Trenz FPGA board, but much lower power levels. Not sure which way I want to go.


you just need a \"diode\" from each, something like this has been used on the raspberry pi to enable using a psu or USB power and not backfeed the USB

https://robotics.ong.id.au/wp-content/uploads/2014/07/rpi_protection.png

In a simulation I see one potential problem. I have a battery where the USB jack is and the +5 is our +15 supply after passing through a real diode. When the voltage from supply is coming up as it crosses the 14 volts (or whatever) from the battery there is a range of about 0 to 100 mV across the FET with the gate still driven on. I\'m seeing a peak of about 2.5 amps into the battery lasting for about a quarter ms. I suppose this is not going to hurt anything unless the PSU has something that is tripped by this current added to the normal use current.

My concern is that with the battery potentially being charged up to near the supply voltage minus the diode, we may see this state persist for some long time. 15V - 5% tolerance = 14.25. Battery at 14.15 and we will see 2 amps flowing into the battery. Or is this current surge just a transient effect from the large resistors and the gate capacitance? Yep, slowing the transition makes it go away. Since the brief transient will have no effect on anything this should work fine.

I\'ve balanced up the differential amp and I think the circuit is ready to go.

--

Rick C.

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

 

[whit3rd]

On Monday, August 31, 2020 at 4:02:06 PM UTC-7, Ricketty C wrote:

>... I\'m talking about an output dropping once the input drops below the output plus some drop out. That\'s my point, regulators are spec\'d for a minimum drop out where they meet full specs. I\'m interested in the drop out once they are out of regulation and the pass transistor is turned on fully.

If one has a curve tracer, it\'s easy to resistor-load a three terminal regulator and
test it just as any two-terminal device. Rising ramp and flat (fixed current after
regulation is achieved) out to the full V rating are what you\'d
be looking for.

Lots of old 7805s work just like that (I don\'t recall any oscillation problems, but
that\'s something to consider).

 

[Ricketty C]

On Friday, September 4, 2020 at 2:16:17 PM UTC-4, whit3rd wrote:

On Monday, August 31, 2020 at 4:02:06 PM UTC-7, Ricketty C wrote:

... I\'m talking about an output dropping once the input drops below the output plus some drop out. That\'s my point, regulators are spec\'d for a minimum drop out where they meet full specs. I\'m interested in the drop out once they are out of regulation and the pass transistor is turned on fully.

If one has a curve tracer, it\'s easy to resistor-load a three terminal regulator and
test it just as any two-terminal device. Rising ramp and flat (fixed current after
regulation is achieved) out to the full V rating are what you\'d
be looking for.

Lots of old 7805s work just like that (I don\'t recall any oscillation problems, but
that\'s something to consider).

Yeah, I understand the concept. I was thinking someone might know of some parts that do this rather than my having to test them myself. Doesn\'t matter at this point. The circuit is designed, simulated and seems to work very well. The P-channel FET, two 2N7002\'s and a third to provide shutdown. Drop out is no more than 150 mV at 4 amps. Works for me. The same pFET is being used in the pair of low voltage drop diode circuits that will be used. One on the battery voltage and another on a 5V power source used for debugging/programming without the main power supply. Seems the MCU can be powered and programmed over a USB port with no other power on the board.

I\'m really getting tired of the project though. I can see so many issues that are being half thought out with virtually no planning. This made me think of another one. The power will be the 12V rail which feeds a 5V linear and a pair of 3.3V linears. The 5V and one of the 3.3V linears can be disabled by the CPU. There are some heavier loads on the 5V rail, 100+ mA. The MCU will be powered on the always on 3.3V rail, but I\'ll bet no one has looked at the details to make sure the USB port will be functional with the other supplies turned off. If they need to be on for any reason, the supply in the USB programmer probably won\'t be sufficient. So the always on supply has to power enough of the circuitry so the MCU can be programmed.

Oh well...

--

Rick C.

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