Calculating Power consumption by a wireless node

[dsk2858]

Hi ,
Recently i developed a wireless sensor node to which i three analog sensors and one solid state relay.
Here i attached the figure of it

Power Requirements of MCU
Supply Voltage 2.7 - 3.6 V
Transmit Current (Typ@3.3V) 80mA
Idle/Receive Current (Typ@3.3V) 20mA
Power-down Current (Typ@3.3V) 1.6uA
Sensors power requirements
1.Soil moisture sensor 5VDC and 7mA current
2.Soil temperature sensor 5VDC and 3mA current
3. LM-35 5VDC and 60uA current
4.SOLID STATE RELAY 5Vdc and 15mA current

I used two LDO ic for 3.3v and 5.5vDC
3.3VDC and max 800mA is used for powering th MCU.
And 5VDC and max 800mA is used for powering sensors connected at the connectors and there supply is controlled by BC109 Transistor with 3.3v base voltage through MCU port pins.

Battery specifications
Using 6 in number 1.5vDC Duracell AA battery for hen added all the six is 9VDC but according to the model number on it MN1500 i found that it is capable of 2.850AH

Here i even attached the circuit diagram of it

could any one please help to calculate them and eve the procedure

 

[BradtheRad]

On the chance this is what you're asking...

Power Requirements of MCU
Supply Voltage 2.7 - 3.6 V
Transmit Current (Typ@3.3V) 80mA
Idle/Receive Current (Typ@3.3V) 20mA
Power-down Current (Typ@3.3V) 1.6uA
Sensors power requirements
1.Soil moisture sensor 5VDC and 7mA current
2.Soil temperature sensor 5VDC and 3mA current
3. LM-35 5VDC and 60uA current
4.SOLID STATE RELAY 5Vdc and 15mA current

This amounts to about 130 mA.

Battery specifications
Using 6 in number 1.5vDC Duracell AA battery for hen added all the six is 9VDC but according to the model number on it MN1500 i found that it is capable of 2.850AH

You can expect six new AA cells to provide this amount of current for 2.850 divided by .130, or 21.9 hours.

I used two LDO ic for 3.3v and 5.5vDC
3.3VDC and max 800mA is used for powering th MCU.
And 5VDC and max 800mA is used for powering sensors

The batteries will last only a few hours if these devices drain current at anywhere near these rates.

 

[dsk2858]

Thanks you for your explanation.
If most of the time the MCU would be in sleep mode then it consumes only 1.6uA .
If i would write a program in this manner that is

After every one hour the MCU wakes up from the sleep mode and it would be awake for one minute so that it would measure all the sensor readings and will transmit those back to the gateway and again goes back to sleep mode after that one minute and again wakes up after one hour duration to measure the sensor values then how do i calculate the power consumption and the battery backup time.

 

[BradtheRad]

Using the two 800 mA max figures...

for 1 minute per hour...

works out to 1.6 amps x 1/60, or .027 amp-hours. (The 1.6 uA used in sleep mode is negligible.)

Divide .027 into 2.850 battery capacity, and you obtain 107 hours battery backup time.

4.45 days.

 

[dsk2858]

Sir,
I could not understood how did you calculate .027amp-hour could you please elaborate it.

 

[BradtheRad]

When the 2 devices are on, they each draw 800 mA maximum (a total of 1.6 amps)...

and they draw that amount for 1 minute per hour...

then the average net HOURLY drain will be 1/60th of whatever is their instantaneous current draw.

(Notice if they were on for 2 minutes per hour, then the net hourly amount would be 2/60ths, etc.)

We want to calculate the drain in amp-hours (the unit of interest here) because the batteries are rated in amp-hours.

At 1.6 Amps drain, 1/60th of that is .027 Amp-hours.