X capacitor, power off and brown-out condition

Thread Starter

Goxeman

Joined Feb 28, 2017
77
Hello,

I present the next circuit:
- 3.3VDC supplied by a DC-DC converter, main power is from a battery of a car (12V or 24V)
- In the output of DC-DC, within other smaller capacitors, I have a 1000uF or 2200uF (like an X capacitor), which I want to use as EMI suppression capacitor and reservoir, basically to give protection and stability to the circuit
- In this PCB there are different IC and one MCU which power is taken from the 1000uF or 2200uF capacitor line

The problem is that after powering off the PCB, the 1000uF or 2200uF capacitor is fully loaded and that load will be going away veeeeeery slowly so the circuit would enter in a brown-out condicion. Basically I want to discharge the capacitors after power down, I am thinking about using a bleeder resistor to discharge this large capacitor in like a hundred miliseconds

What do you think of this? What would you do to discharge the circuit after powering down safeliy the circuit?
 

AlbertHall

Joined Jun 4, 2014
9,843
If you are drawing a low current such that it takes a long time for those capacitors to discharge then you may be able to use smaller capacitors.
A bleeder resistor will work but it will be drawing current all the time and needlessly drawing power from the battery.
 

DickCappels

Joined Aug 21, 2008
6,384
What is the problemwith thie ---what kind of situation are you trying to avoid? Is there anything on the PCB that does not tolerate this situation well?
 

ebeowulf17

Joined Aug 12, 2014
3,274
As the others have said, you can probably ignore the slow shutdown and/or reduce the cap size quite a bit.

Having said that, if you really want a quick power down, you could do something draconian like put a MOSFET in as a switch between the filter caps and the load.

I've uploaded a very rough sketch. Forgive the crude quality - I draw circuits almost exclusively on computers, but I'm not near one now.

The idea is that when 24V supply is present, current fled through R1 and NPN base, activating NPN and pulling MOSFET gate low, turning it on... so when the power is on, the MOSFET is almost transparent, assuming it has a low enough Rds-on spec compared to the current draw.

Then, when 24V supply voltage is cut the NPN stops pulling the MOSFET gate low, so R2 pulls the gate up, turning off the MOSFET almost instantly. Charge stored in the filter caps will eventually disappear through various leakage paths, but won't power the load for indeterminate durations like it currently does.
B666A3C1-F9CA-4750-858F-397CD0D9F45E.jpeg
I haven't simulated or otherwise tested this circuit, and I'm still in many ways a newbie, so don't just take my word for it, but l *think* this would give you an instant shutdown if you really want it.
 

ebeowulf17

Joined Aug 12, 2014
3,274
As the others have said, you can probably ignore the slow shutdown and/or reduce the cap size quite a bit.

Having said that, if you really want a quick power down, you could do something draconian like put a MOSFET in as a switch between the filter caps and the load.

I've uploaded a very rough sketch. Forgive the crude quality - I draw circuits almost exclusively on computers, but I'm not near one now.

The idea is that when 24V supply is present, current fled through R1 and NPN base, activating NPN and pulling MOSFET gate low, turning it on... so when the power is on, the MOSFET is almost transparent, assuming it has a low enough Rds-on spec compared to the current draw.

Then, when 24V supply voltage is cut the NPN stops pulling the MOSFET gate low, so R2 pulls the gate up, turning off the MOSFET almost instantly. Charge stored in the filter caps will eventually disappear through various leakage paths, but won't power the load for indeterminate durations like it currently does.
View attachment 155739
I haven't simulated or otherwise tested this circuit, and I'm still in many ways a newbie, so don't just take my word for it, but l *think* this would give you an instant shutdown if you really want it.
I see the problem with my earlier suggestion now - the voltage regulator may leak a lot of current when reverse biased (when the 24V supply drops out, but there's still 3.3V on the caps,) providing current to keep my whole shutdown circuit from working as I originally expected. I knew it was too good to be true. I'll keep brainstorming, cause there's gotta be a reasonably simple way to do this, but what I posted above doesn't appear to be it.
 

ebeowulf17

Joined Aug 12, 2014
3,274
Ok, I found a way to defeat my leakage current problem using a Zener diode and a pull-down resistor, although this solution now uses more components than I originally hoped (there's pretty much always a better, simpler way than whatever I come up with, but this is the best I could do):

1 P-MOSFET
1 NPN transistor
1 Zener diode (I simmed with 8.6, but I think other values from ~5-10 would work equally well)
3 resistors​

This circuit will cut off power to the load in small fractions of a millisecond after the 24V supply is removed. The 24V supply, regulator, caps, and load in the image below are meant to represent the existing circuit being discussed, and the new shutoff control parts are contained in a rectangle. The other parts floating off to the top left are for simulation only.

MOSFET-stop-caps_04.png
 

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Thread Starter

Goxeman

Joined Feb 28, 2017
77
If you are drawing a low current such that it takes a long time for those capacitors to discharge then you may be able to use smaller capacitors.
A bleeder resistor will work but it will be drawing current all the time and needlessly drawing power from the battery.
Indeed my proposed solution would be drawing current all the time, but I dont think it would be too much. I would have the posibility to connect and disconnect the power to the PCB from a switch and while being used this will be connected directly to the battery of my car that is being charged by the car alternator

What is the problemwith thie ---what kind of situation are you trying to avoid? Is there anything on the PCB that does not tolerate this situation well?
In general terms, as far as I know, no circuit should be discharged too slow, there would be brown-out issues if the MCU or other IC with internal registers stay in brown-out condition for a long time. It could lock up of the ICs until a full power cycle takes it all to the ground

Ok, I found a way to defeat my leakage current problem using a Zener diode and a pull-down resistor, although this solution now uses more components than I originally hoped (there's pretty much always a better, simpler way than whatever I come up with, but this is the best I could do):

1 P-MOSFET
1 NPN transistor
1 Zener diode (I simmed with 8.6, but I think other values from ~5-10 would work equally well)
3 resistors​

This circuit will cut off power to the load in small fractions of a millisecond after the 24V supply is removed. The 24V supply, regulator, caps, and load in the image below are meant to represent the existing circuit being discussed, and the new shutoff control parts are contained in a rectangle. The other parts floating off to the top left are for simulation only.

View attachment 155745
Thank you very much ebeowulf for your special attention and interest in this matter, I see that this second circuit of yours uses more components and I have to say I woulnt have come up with this idea. So I have to evaluate the advantages/disadvanges of your proposal, why do you think is better than just a bleeder resistor?


Regards,
 

ebeowulf17

Joined Aug 12, 2014
3,274
Thank you very much ebeowulf for your special attention and interest in this matter, I see that this second circuit of yours uses more components and I have to say I woulnt have come up with this idea. So I have to evaluate the advantages/disadvanges of your proposal, why do you think is better than just a bleeder resistor?
To be honest, I'm not totally sure either way. However, it seems like if you want a quick discharge, the bleed resistor will have to be a relatively low value, drawing a lot of current. In addition to seeming wasteful, I'm not sure (because of my lack of experience and the fact that I haven't looked up any references on this yet) whether or not this will impact the intended filter performance of the caps. It would be silly to use these giant caps for protection, just to have them partially defeated by a bleed resistor. I don't think that's the case here, but it was a possibility that crossed my mind.

In any case, the idea of bringing your circuit to a full stop without a bleed resistor intrigued me, so I decided to see if I could figure out how to do it. I like puzzles!

My circuit stops power to your load much faster, with less wasted power, and with negligible impact on the rest of the circuit. I have no idea whether it's necessary, but I just wanted to see what was possible. I was kind of hoping some of the more experienced folks here would chime in, although the impression I'm getting so far is that they think the slow brownout isn't an issue, in which case both of our solutions would be unnecessary.

Just for reference, here's a plot of voltage discharge times and current consumption with various bleed resistor sizes:
cap-discharge-timing.png
 

Thread Starter

Goxeman

Joined Feb 28, 2017
77
Many MCUs include a brownout circuit which can hold it in reset while the supply voltage is too low for correct operation.
Indeed AlbertHall, many MCUs include a brownout circuit, like the one I am using has the LPBOR option but not the bluetooth module I would like to include. In your opinion it doesnt matter to leave fully charged capacitors always?

To be honest, I'm not totally sure either way. However, it seems like if you want a quick discharge, the bleed resistor will have to be a relatively low value, drawing a lot of current. In addition to seeming wasteful, I'm not sure (because of my lack of experience and the fact that I haven't looked up any references on this yet) whether or not this will impact the intended filter performance of the caps. It would be silly to use these giant caps for protection, just to have them partially defeated by a bleed resistor. I don't think that's the case here, but it was a possibility that crossed my mind.

In any case, the idea of bringing your circuit to a full stop without a bleed resistor intrigued me, so I decided to see if I could figure out how to do it. I like puzzles!

My circuit stops power to your load much faster, with less wasted power, and with negligible impact on the rest of the circuit. I have no idea whether it's necessary, but I just wanted to see what was possible. I was kind of hoping some of the more experienced folks here would chime in, although the impression I'm getting so far is that they think the slow brownout isn't an issue, in which case both of our solutions would be unnecessary.

Just for reference, here's a plot of voltage discharge times and current consumption with various bleed resistor sizes:
View attachment 155763
Indeed the bleeder resistor I calculated should to be 47ohms giving me a discharge speed of 99ms until 0.79V. As you say, I didnt think about that, it is not useful using such a capacitor using that bleeder resistor because then the purpoise of that giant capacitor is wasted.

Indeed, it would be great if as you said others could give their opinion of how to give a quick discharge and shutting down the circuit after disconnecting the power to the whole circuit.

I learned many things being curious, asking and wondering how to do things, that is why electronics always intrigue me.
 

DickCappels

Joined Aug 21, 2008
6,384
I use brownout protection on the AVR controllers that I use. Sure makes life a lot simpler than in the old days when you had to make your own. Before that series had brownout protection, external brownout circuits were the norm.

The bleeder resistor approach just was not a reasonable solution when the controller is executing many instructions each microsecond. A lot of damage could, and sometimes was, done with only a few instructions.
 

ebeowulf17

Joined Aug 12, 2014
3,274
I use brownout protection on the AVR controllers that I use. Sure makes life a lot simpler than in the old days when you had to make your own. Before that series had brownout protection, external brownout circuits were the norm.

The bleeder resistor approach just was not a reasonable solution when the controller is executing many instructions each microsecond. A lot of damage could, and sometimes was, done with only a few instructions.
Do you have any thoughts on the circuit I proposed above? I haven't tried it, except in the simulation above. It seems to me like it makes sense, but I'm not totally confident.

***EDIT: Better yet, what sort of circuit did you use as external brownout protection? Would the sort of brownout protection you were using in the past be better than what I suggested above?

Mine doesn't do anything about voltage variations in the incoming (12/24V) power supply, it just shuts down the low voltage to the micro immediately when the 12/24V supply goes below the Zener voltage (more specifically the Zener voltage plus the transistor's Vbe.)
 
Last edited:

DickCappels

Joined Aug 21, 2008
6,384
I did not use external brownout protection back them and as a result suffered a lot of problems (that was nearly 20 years go, a lot of progress had been made since then!
 

ebeowulf17

Joined Aug 12, 2014
3,274
I found out a component that in my opinion is doing all I was looking for, it is a load switch, all in one component

I am reading a component from TI that is really good a explains the difference with a power MOSFET
http://www.ti.com/lit/an/slva716/slva716.pdf
That's an interesting app note and a good read. It doesn't discuss what happens when Vin is lost; l don't see any indication that it helps in slow power down/brownout conditions.

How were you planning to incorporate it? If you were thinking of something similar to my circuit earlier, this part could probably replace 5 of the 6 new components, but you'd still need the Zener in front of it. If you were thinking of a totally different layout, I'd be curious to see what you have in mind.
 

Thread Starter

Goxeman

Joined Feb 28, 2017
77
That's an interesting app note and a good read. It doesn't discuss what happens when Vin is lost; l don't see any indication that it helps in slow power down/brownout conditions.

How were you planning to incorporate it? If you were thinking of something similar to my circuit earlier, this part could probably replace 5 of the 6 new components, but you'd still need the Zener in front of it. If you were thinking of a totally different layout, I'd be curious to see what you have in mind.
Read point 6.3, second paragraph
Quick output discharge is a load switch feature which discharges VOUT through an internal path to ground when the switch is disabled. This provides a known state on the output and ensures that all loads have been discharged and are turned off.
 

ebeowulf17

Joined Aug 12, 2014
3,274
Read point 6.3, second paragraph
Quick output discharge is a load switch feature which discharges VOUT through an internal path to ground when the switch is disabled. This provides a known state on the output and ensures that all loads have been discharged and are turned off.
I'd already read that. I believe that's talking about when your power supply is still running at full voltage, but you switch a GPIO pin low to signal the chip to turn off power to your downstream device.

So far, you've not been describing a situation in which you had a control signal to feed this chip - you don't have (or at least haven't discussed) a controller on the 12/24V side which could tell this new chip when to shut off your device.

I don't think this new chip has any special properties regarding a slow ramp down of the 3.3V supply like you want to address. It certainly looks like a good compact switch that would replace many of the components I suggested earlier, but if you want it to cut power to your microcontroller and Bluetooth device in a sharp, crisp way instead of slowly ramping down, you'll have to provide a control signal to it that tells it when to cut power.

A connection from the 24V side, through a Zener, might be one way to provide that signal (although even then, depending on input characteristics of that chip, leakage through the Zener might keep it turned on when it shouldn't be, and you might need a pull down resistor between the Zener and the chip, in which case you've saved 4 components instead of 5.)
 

Thread Starter

Goxeman

Joined Feb 28, 2017
77
I'd already read that. I believe that's talking about when your power supply is still running at full voltage, but you switch a GPIO pin low to signal the chip to turn off power to your downstream device.

So far, you've not been describing a situation in which you had a control signal to feed this chip - you don't have (or at least haven't discussed) a controller on the 12/24V side which could tell this new chip when to shut off your device.

I don't think this new chip has any special properties regarding a slow ramp down of the 3.3V supply like you want to address. It certainly looks like a good compact switch that would replace many of the components I suggested earlier, but if you want it to cut power to your microcontroller and Bluetooth device in a sharp, crisp way instead of slowly ramping down, you'll have to provide a control signal to it that tells it when to cut power.

A connection from the 24V side, through a Zener, might be one way to provide that signal (although even then, depending on input characteristics of that chip, leakage through the Zener might keep it turned on when it shouldn't be, and you might need a pull down resistor between the Zener and the chip, in which case you've saved 4 components instead of 5.)
After checking the concrete datasheet of one component from TI: http://www.ti.com/lit/ds/symlink/tps22918-q1.pdf

I read point "8.3.2.2 Internal QOD Considerations" and it says "quickly discharge a load after the switch has been disabled". So I understand that this chip is discharging the load after switch is off, it can even discharge with internal resistor up to 200uF

Related to how to control the "ON" pin, I was thinking about two options:
1. Using a transistor directly from the main supply
2. Connecting directly Vin and ON pins with a pull down resistor (10K) to enssure that when the DC converter is off is giving me a 0
 

ebeowulf17

Joined Aug 12, 2014
3,274
After checking the concrete datasheet of one component from TI: http://www.ti.com/lit/ds/symlink/tps22918-q1.pdf

I read point "8.3.2.2 Internal QOD Considerations" and it says "quickly discharge a load after the switch has been disabled". So I understand that this chip is discharging the load after switch is off, it can even discharge with internal resistor up to 200uF

Related to how to control the "ON" pin, I was thinking about two options:
1. Using a transistor directly from the main supply
2. Connecting directly Vin and ON pins with a pull down resistor (10K) to enssure that when the DC converter is off is giving me a 0
Option 1 might work, depending on the specifics of how you set it up. In addition to your transistor and associated resistors, I'm pretty sure you'll still need a Zener to prevent reverse leakage through the regulator from flowing through your new transistor and holding the switch on.

Wait! I may have been thinking about your 3.3V supply wrong - are you using a linear regulator, or some sort of switching supply? Depending on what you're using to step down the voltage, all my harping about reverse leakage may be moot. Every linear regulator I've looked into acts just like a simple diode when reverse biased, providing a huge leakage path with only a small drop in voltage. But I don't think switching supplies behave the same way. I'd have to look at specific datasheets to say with any confidence, but the reverse leakage path may not be an issue after all.

Option 2 doesn't sound right to me. The caps on the DC converter output will hold the Vin and ON pins both high as the caps slowly discharge - the exact same issue you currently have. I think this will be no different than having just that pull down resistor, acting as a bleed resistor.
 

Thread Starter

Goxeman

Joined Feb 28, 2017
77
Option 1 might work, depending on the specifics of how you set it up. In addition to your transistor and associated resistors, I'm pretty sure you'll still need a Zener to prevent reverse leakage through the regulator from flowing through your new transistor and holding the switch on.

Wait! I may have been thinking about your 3.3V supply wrong - are you using a linear regulator, or some sort of switching supply? Depending on what you're using to step down the voltage, all my harping about reverse leakage may be moot. Every linear regulator I've looked into acts just like a simple diode when reverse biased, providing a huge leakage path with only a small drop in voltage. But I don't think switching supplies behave the same way. I'd have to look at specific datasheets to say with any confidence, but the reverse leakage path may not be an issue after all.

Option 2 doesn't sound right to me. The caps on the DC converter output will hold the Vin and ON pins both high as the caps slowly discharge - the exact same issue you currently have. I think this will be no different than having just that pull down resistor, acting as a bleed resistor.
Now I getting confused... hahahaha

I did an schematic of my power design, then is easier

The idea is to introduce the load switch after C8, right before the load, so Vin of the load switch would be directly connected to C8. Once the switch is opened the small load after the load switch will discharge through the internal resistor, the only "problem" is that the capacitors would remain charged before the load switch but then is not damaging any device

As you said the option of using a high pull down resistor would keep the "ON" pin high for really long time. But I dont see how I could fix this issue just by using the transistor as far as I can not connect the "ON" to more than 5V
 

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