Powering a 16 bit ADC present in wearable device using a buck boost converter

Thread Starter

hoyyoth

Joined Mar 21, 2020
528
Basic Explanation of my System:

The proposed design is a battery-powered wearable device that integrates advanced sensing and power management features. The system operates primarily from a single-cell Li-ion battery (3.7 V nominal) and includes a USB Type-C interface for charging and external power supply.

The device uses a battery charger with power-path management ( BQ24075) to ensure seamless operation:

When USB power is connected, the system is powered directly from the USB input while simultaneously charging the battery. When USB is disconnected, the system automatically switches to battery power without interruption. A fuel gauge IC monitors battery health parameters such as voltage, state of charge (SOC), and remaining capacity, reporting these to the microcontroller via I²C.

The system includes an electrochemical front end (AD5941) with a 16-bit ADC for high-precision sensor measurements.

Sensor signals are routed through a multiplexer (MUX) to the AD5941, which interfaces with an STM32 microcontroller development board for data acquisition and processing.

Power Tree:

1765382476575.png
Power Supply Design:

I designed the below circuit using for powering the ADC and remaining circuitry using ADP2503. My simulation results and circuit diagram are given below.The ferrite bead P/N is MPZ1608S601ATA00. The simulation results are given below.

1765382546886.png

1765382660591.png


The voltage rails are respectively 3.2551427 V (AVDD) , 3.2995143 V (DVDD & IOVDD)

  1. Can I go ahead with this power supply design.Or do I need to go with Buck-Boost + LDO
  2. Could you please check my battery current consumption calculation.
  3. If I am using internal reference, will the ripple on the AVDD line is an issue?
 

panic mode

Joined Oct 10, 2011
4,930
things don't add up...
you have two load resistors of 80Ohm - why?
each of them draws more than 40mA for total draw of some 83mA. how is that comparable to 12.449mA?

also 90% efficiency is not there... read the datasheet.

90% efficiency for 3.3V output is possible if input voltage is more than 4V and load current is 0.2-0.6A.
for low current, i see less... closer to 75%:
1765388464254.png
 
Last edited:

Thread Starter

hoyyoth

Joined Mar 21, 2020
528
things don't add up...
you have two load resistors of 80Ohm - why?
each of them draws more than 40mA for total draw of some 83mA. how is that comparable to 12.449mA?
My mistake I will correct it .
Please see the new calculation below.
In the light loads efficiency is the range of 75 to 80%( from the datasheet ,efficiency vs output current curves for PSM)

Assuming 75 % efficiency,the input current taken by the IC (ADP2503) is given by

(3.3 V *18mA)/(3.7 V *.75) = 21.41mA ;

3.3V -->O/P voltage
18mA -->O/P Current
3.7V--> I/P voltage(from battery)

May I know can I use this circuit to power the ADC
 

Ian0

Joined Aug 7, 2020
13,112
"Electrochemical" suggests to me a slow-moving signal that doesn't need to be sampled very often. As the AD5941 has an idle mode that cuts is power by >90% it is worth the trouble of adding all that noise into the system?
You can reduce the current by 3.3/3.7 which saves 50uA at idle, but requires 34uA to run the converter.
 

MisterBill2

Joined Jan 23, 2018
27,314
I want to know about where the TS got an affordable 8 digit voltmeter!!
And there is a potential noise issue with that low an operating voltage. Simply this: The signal voltage peaks tend to be less than the supply voltage, while all of the noise voltage is not limited. So as the supply voltage is reduced the signal to noise ratio is also reduced. IF noise will not be a problem then this is not an issue.

BUT with a 16 bit A/D converter, it possibly will be, since each digital bit represents a very small voltage increment.

So I suggest a careful study of how much noise the system can have before the performance or accuracy suffer. It is much less expensive to see the problems early in the design process than after the system is built. And if you catch the flaw early enough you can appear to be incredibly wise, by producing systems with no problems.
 

MisterBill2

Joined Jan 23, 2018
27,314
In post#1 you stated two voltages as "The voltage rails are respectively 3.2551427 V (AVDD) , 3.2995143 V (DVDD & IOVDD) ""
Those values must have been calculated, because I am not aware of many voltmeters able to provide that great a resolution. Most of the "high end" digital voltmeters will provide four digits of resolution.
Those two numbers describe the voltage to the tenth of a microvolt. That is very unusual
 
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