Supply Cut Circuit Idea Needed

Thread Starter

mishra87

Joined Jan 17, 2016
906
Hello Guys,

As per shown in below circuit my LDO is enabled by switch event for the 1st time. Once 3V3 makes microcontroller wakes up and it takes control of EN pin so 3V3 always appear at outputs.

So the problem here is if switch is pressed unintentionally and being pressed for longer duration so it will drain the battery because always remain in ON condition.

Could anyone suggest any idea to rectify this problem ?

1614591268140.png

Thanks in Advance !

Regards,
 

ericgibbs

Joined Jan 29, 2010
12,548
hi mishra,
Your circuit shows a momentary push button, what is the drain on the battery via the p/b and the enable pin.? must be quite low.?
Use a timer in the MCU to switch Off the Enable pin after a preset time.
E
 

Thread Starter

mishra87

Joined Jan 17, 2016
906
hi mishra,
Your circuit shows a momentary push button, what is the drain on the battery via the p/b and the enable pin.? must be quite low.?
Use a timer in the MCU to switch Off the Enable pin after a preset time.
E
hi mishra,
Your circuit shows a momentary push button, what is the drain on the battery via the p/b and the enable pin.? must be quite low.?
Use a timer in the MCU to switch Off the Enable pin after a preset time.
E
Thanks for reply !
3V3 is having some load e.g. microcontroller.
if EN is high 3V3 at out pin.

Still i did not understand your suggestion !

Momentary press will give high to EN pin and 3V3 at out pin. microcontroller wakes up and the now EN will driven high as long as microcontroller wants.

Now now you are saying microcontroller pull low to EN pin so that 3V3 become 0V. In this case microcontroller will be off after pulling EN pin low.

Since switch is pressed continuously again microcontroller wakes up and so ?

Could you please elaborate a bit more ?

Thanks
 

Thread Starter

mishra87

Joined Jan 17, 2016
906
.... or you could use the switch to apply an enable pulse via a series capacitor.
What is use of series capacitor with switch ?
Does this series capacitor will full fill the requirement if switch is falsely pressed.

May be these are silly question but i wanted to understand.

Regards,
 

ericgibbs

Joined Jan 29, 2010
12,548
hi m,
You said.
So the problem here is if switch is pressed unintentionally and being pressed for longer duration so it will drain the battery because always remain in ON condition.

Pressing the p/b will only pass a small current into the LDO Enable pin , so someone pressing the button for a long time would not drain the battery.

Do you want the LDO to stay On until the MCU switches it Off.??

E
 

Thread Starter

mishra87

Joined Jan 17, 2016
906
hi m,
You said.
So the problem here is if switch is pressed unintentionally and being pressed for longer duration so it will drain the battery because always remain in ON condition.

Pressing the p/b will only pass a small current into the LDO Enable pin , so someone pressing the button for a long time would not drain the battery.

Do you want the LDO to stay On until the MCU switches it Off.??

E
Thanks for your quick reply !

I do not want LDO to stay ON until MCU switches it OFF.

Pressing the p/b will only pass a small current into the LDO Enable pin , so someone pressing the button for a long time would not drain the battery.
So in this case there will always a voltage at LDO Vout : am i correct ? and MCU will stay off
Since no wakeup event MCU will remain in sleep ?


blue bypass line in SW(p/b) indicates SW(p/b) is shorted if some how it is triggered false. This is not actual condition. This is false condition.
1614595102835.png
 

Thread Starter

mishra87

Joined Jan 17, 2016
906
Thanks for your quick reply !

I do not want LDO to stay ON until MCU switches it OFF.

Pressing the p/b will only pass a small current into the LDO Enable pin , so someone pressing the button for a long time would not drain the battery.
So in this case there will always a voltage at LDO Vout : am i correct ? and MCU will stay off
Since no wakeup event MCU will remain in sleep ?


blue bypass line in SW(p/b) indicates SW(p/b) is shorted if some how it is triggered false. This is not actual condition. This is false condition.
View attachment 231755
My understanding says once MCU pull DO pin to GND the LDO will switch off and MCU will switch off and once MCU leave the control of EN pin LDO will wakeup again since EN pin see again High Voltage.

Could you please confirm my perception if i am correct ?
As long as LDO see voltage at EN pin it will remain ON.

Regards,
 
Last edited:

sghioto

Joined Dec 31, 2017
2,336
my understanding says once MCU pull DO pin to GND the LDO will switch off and MCU will switch off
That is correct
and once MCU leave the control of EN pin LDO will wakeup again since EN pin see again High Voltage.
Not correct.
If there is no voltage to the MCU then the Enable pin will stay low until the button is pressed and the LDO is reactivated.
What is the part number of the LDO?
Some additional components may be required depending on the LDO as shown below including a manual OFF switch.
1614644178819.png
 
Last edited:

Thread Starter

mishra87

Joined Jan 17, 2016
906
hi mishra,
This clip shows the point I was making about the Enable input current.
Added the full datasheet.
EView attachment 231910
Hi,
Thanks for reply !
That means EN pin will draw only 0.01 uA once MCU switches OFF or disable the LDO even if EN pin see 4.2V. So there should not be any issue.

I think i have understood the above point.

Still having one doubts, Once MCU switches OFF or disable the LDO, will LDO wakeup again since EN pin will have again battery voltage.
 

Thread Starter

mishra87

Joined Jan 17, 2016
906
hi mishra,
Look at this option.

E
Hi E
Sorry for a bit late reply !
Thanks for beautiful circuit.
Since we have tactile switch and once the switch is pressed LDO will make supply momentarily and it goes off. Since it does not have control with microcontroller.

And the second point is if the switched is pressed accidently for longer duration how LDO will switch off the supply

Regards,
 

ericgibbs

Joined Jan 29, 2010
12,548
hi mishra
Ref that circuit.
The user presses and holds the switch until the MCU powers Up and applies a 5v signal to GP0, then he releases the push switch.
The C1 value can be selected to hold the Enb high long enough to ensure power up.

To power down the MCU pulls GP0 low, which disable the TLV.

E
 

Attachments

Thread Starter

mishra87

Joined Jan 17, 2016
906
hi mishra
Ref that circuit.
The user presses and holds the switch until the MCU powers Up and applies a 5v signal to GP0, then he releases the push switch.
The C1 value can be selected to hold the Enb high long enough to ensure power up.

To power down the MCU pulls GP0 low, which disable the TLV.

E
Hi E,

Thank you very much for your reply !
Now i am more interested and have more questions -

1. C1 value should high enough unless and until MCU wakes up and give signal to GP0 pin. Now GP0 pin holds control of LDO EN pin. So this is for me and i understand.

2. Lets come to the second topic if the SW is accidently pressed for longer duration. How the MCU disable the LDO because LDO will always see divider voltage at its EN pin ?

I hope you will understand my concern.

Will MCU be able to switch off the LDO power supply if it see the SW is pressed for longer duration ?

I will ask more question once you reply this.

Regards,
 
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