Better approach for powering up sensor nodes ?

Thread Starter

ronit991

Joined Aug 6, 2021
2
I have a sensor node containing an MCU & radio running at 3.3 V and a sensor running at 5 V.

What is the better approach for powering up the system (in terms of efficiency, stability, etc.) :
1. High Voltage Battery + Buck Converter
or
2. Low Voltage Batttery + Boost Converter?
 

Deleted member 115935

Joined Dec 31, 1969
0
Yes,
with what you have given us, wither will do.

How about a bit more details.

How far apart the sensors / MCU
how much power do each part need
where is th eorriginla power source ?

All very open, draw us a picture of what you have already planed before we get in to this details.
 

Ian0

Joined Aug 7, 2020
3,463
If you have long cables to the sensor then High Voltage Battery & Buck Converter.
If your radio might not like the interference from the buck converter then about 7V-9V and a linear regulator.
 

Thread Starter

ronit991

Joined Aug 6, 2021
2
Yes,
with what you have given us, wither will do.

How about a bit more details.

How far apart the sensors / MCU
how much power do each part need
where is th eorriginla power source ?

All very open, draw us a picture of what you have already planed before we get in to this details.
Distance between sensor & MCU : ¸ 1 meter
Sensor current consumption: 35mA (peak)
MCU + radio current consumption: 120 mA (peak), <1mA (average)

Original power source: This is what the question is. I want to know whether I should use a 6-7 V battery + buck converter or a 3.5 V battery + boost converter
 

Ian0

Joined Aug 7, 2020
3,463
In that case - I'd choose buck.
A buck regulator makes the most switching noise on its input, a boost regulator makes the most switching noise on its output.
The battery* won't give a damn about a bit of noise - the sensors and radio might.

For what percentage of the time is it at PEAK? If (Average/Peak) < 1/120 as your figures suggest, then I'd still recommend linear - the standby current of a switcher is much greater than that of a linear regulator, and may be greater than the overall standby current. In other words, the energy lost by a switching regulator during standby is more than the energy lost by a linear regulator during peak.

*Unless it's got one of those BMSs which shut everything down when a fly sneezes in the next room.
 

Deleted member 115935

Joined Dec 31, 1969
0
Distance between sensor & MCU : ¸ 1 meter
Sensor current consumption: 35mA (peak)
MCU + radio current consumption: 120 mA (peak), <1mA (average)

Original power source: This is what the question is. I want to know whether I should use a 6-7 V battery + buck converter or a 3.5 V battery + boost converter
So you have chosen battery as original power source,

To reinforce @Ian0 reply

Looking at total power over time,
A switcher at 1mA is going to likely be very in efficient.

A LDO is going to be in efficient at 120 Ma,
but if thats a true Peak, then providing a super cap or such like after the LDO could handle that ,

My first go would be to send the "raw" volts to all units, and regulate locally with LDO's ,.

As for switch up or down,
I'd probably look at down
but its a heads you loose, tails you don't win situation,
 

Yaakov

Joined Jan 27, 2019
3,565
One thing is that a single prismatic Lithium cell at 3.7V might be simpler to implement and maintain than a more complex multi-cell battery. Since you have losses in one place or another, that simplicity might be a vote in favor of lower voltage, depending on design goals.
 
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