Controlling the Low power mode of Buck Boost converter using a Microcontroller

[newbie_hs]

I am using TPS63000 in my design and I need to control PS/SYNC pin using a micro-controller.

The input to the buck boost converter is 3V to 4.3V which is coming from a battery. The output of the converter is 3.3V

The microcontroller is working at 3.3V. This 3.3V is coming from TPS6300

I am using a BJT switch to control the PS/SYNC pin.

When micro-controller output is high the BJT will turn on and PS/SYNC will be connected to GND and TPS63000 will go to low power mode.

My circuit is given below .May I know is it fine or not.

 

[KlausST]

Hi,

At first: there is the TPS63000 ... and the TPS63000-Q1 (Automotive version).
They have different datasheets.
All your text and schematic refers to the nonAutomotive one, while your datasheet is for the Automotive one.

--> please clarify

Functionally it should be simple:
The PS pin has two states according it´s applied voltage: H and L. Both thresholds are given in the datsheet.
Also the datasheet specifies the input current.
So it´s easy to check on your own if the correct levels are met.

***
Now the microcontroller can decide which mode to chose
* mic_port = HIGH --> PS = LOW --> power save enabled

****
But for me your circuit/ function does not make much sense.

There is this rather high value resistor with 1MOhm ... allowing a max of 4.3uA to flow. This means humidity, dirt, may cause stray currents ... and your high ohmic resistor may be not sufficient to drive this additional current.
I see no problem for a clean, new circuit in normal humidity ...

On the other hand you have this 1kOhms resistor. It draws about 2700uA. And it wastes this rather huge current exactly when you tell the converter to go into power save mode. This is a huge contradiction for me.

A BJT is considered a "amplifier" ... but you drive the input with more than 600 time the expectable output current.
If input current is higher than output current ... then I don´t call it amplifier. Then something is wrong.
For sure one may say in this case the BJT is not used as a true amplifier, it rather is used as inverter or as openCollector driver .... but still i would go with I_C > I_B.

You are familiar with Ohm´s law --> so use it. (If you are not familiar: learn it.)

Klaus