Hello,

I am in the process of designing a circuit with ultra low consumption, which consist of a battery with 3.6V nominal voltage, a step down converter to regulate to 3.3V, a MCU, a reed switch and a few other components.

For the step-down converter I found a suitable one for my purpose, a ISL9123 from Renesas that has a bypass feature, it has an enable pin.
The enable pin allows me to activate it for the first time with a classic momentary reed switch, once the reed switch closes for a brief period it should keep the step down converter on until end of battery life.
For this I have used a diode (D2) that connects the outputs back to the enable pin, theoretically once Vout is present the enable pin will stay on.

Diode D1 is present so I can read the state of the reed switch for another purpose with my MCU when the Step down is on.

The issue I'm facing is that the Load Switch needs a logic HIGH or LOW at the enable pin, no floating.
A simple pull-down resistor would resolve this issue but this means current loss through it when the enable pin logic high, even with big resistor values such as 1Mohm would consume 3.3uA which is way too much, also I assume with big resistor value it would cause issues?

I come here for your help, maybe someone have an idea I could avoid using a pull-down resistor?



Thank you

Regards

 

I suggest reading the specifications for the MCU device to see what voltage range it can accept for operating from. It does not seem reasonable that 3.6 volts will be above the allowable voltage supply range. Certainly it is worth investigating.
An alternative would be to abandon the reed switch and use a toggle switch for OFF/ON control Then there will be no extra power consumption.

 

I suggest reading the specifications for the MCU device to see what voltage range it can accept for operating from. It does not seem reasonable that 3.6 volts will be above the allowable voltage supply range. Certainly it is worth investigating.
An alternative would be to abandon the reed switch and use a toggle switch for OFF/ON control Then there will be no extra power consumption.

Thanks for your reply.
Indeed this was planned however I have a few other components such as sensors and transceivers that cannot accept a voltage higher than 3.5V.
Battery voltage also changes with temperature.

As for the the ON/OFF switch this isn't possible, I need to use a magnetic switch.
I know that latch reed switch exist but this is still not possible for my application for numerous reasons.

 

Replace D2 by the resistor.
Don't forget there could be as much as 300nA leakage on the EN pin, so that limits the upper value of the resistor

 

Replace D2 by the resistor.
Don't forget there could be as much as 300nA leakage on the EN pin, so that limits the upper value of the resistor

Hello, thanks for your reply.
By replacing D2 with a resistor EN will still be floating since Vout will be floating if the step down converter is disabled?

 

Hello, thanks for your reply.
By replacing D2 with a resistor EN will still be floating since Vout will be floating if the step down converter is disabled?

Won't it be pulled down to 0V by whatever is connected to Vout?

 

Won't it be pulled down to 0V by whatever is connected to Vout?

Vout is connected to some sensors and and a load switch in parallel, how would it interpret it as 0V? leakage?

 

Vout is connected to some sensors and and a load switch in parallel, how would it interpret it as 0V? leakage?

Presumably, but you didn't show what was connected to Vout.
As long as the leakage is >300nA, it will work.

 

Presumably, but you didn't show what was connected to Vout.
As long as the leakage is >300nA, it will work.

I see.
Also I was thinking about this earlier on:

Adding a diode in series with a pull-down, this way when EN turns HIGH no current will pass through the resistor.
no idea if this would work, what do you think?

 

Adding a diode in series with a pull-down

It seems like a good idea - the reverse leakage of a diode should be a few nA, and the EN logic high is only 1.6 volts, so forward drop isn't an issue.

 

1Mohm would consume 3.3uA which is way too much,

That's not much current.
Divide the Ah rating of your battery by that current to calculate the time it would take to drain your battery.
I think you will find it's a very long time.

 

For the "ENABLE input, use a double throw (form C) reed switch, between Vbat and common (ground). No floating and no extra resistors.

IT would be very informative to tell us more about the mysterious project/load.

 

That's not much current.
Divide the Ah rating of your battery by that current to calculate the time it would take to drain your battery.
I think you will find it's a very long time.

Indeed it might be not much but with the rest of my circuit it adds up really fast.

For the "ENABLE input, use a double throw (form C) reed switch, between Vbat and common (ground). No floating and no extra resistors.

IT would be very informative to tell us more about the mysterious project/load.

With a double throw reed switch once the magnet is removed the EN pin will go back to low which will disable the step down which is not what I'm trying to do.
Once the reed switch closes for a brief period it should keep the step down converter on until end of battery life.

As for the load the step down output is connected to multiple sensors followed by a TPS22916 load switch that is activated by the sensors to be able to power the an ESP32.

 

Indeed it might be not much but with the rest of my circuit it adds up really fast.

Tells me nothing about that actual current draw of the rest of the circuit.
That pull-down current is still likely not a significant factor in the battery life.

 

After 6 posts, still missing some information -

1. Battery part number / amp-hour rating

2. Max tolerable standby current before activation

3. Maximum possible wait time before circuit activation

4. Minimum required system run time

5. Long-term average load current when circuit is operating

ak