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Battery Calculations for the Microcontrollers

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Bjtpower

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Hello Guys,

i am Developing the the Project where i have to use the Battery of any capacity.

Application is Like:
Microcontroller will Send signal in every 100ms (Data transmission time would be only 100uS in 100mS)

Transmitting current will be 22mA MAX

Now i want to calculate required battery Capacity for 300 days (Daily circuit will be on for 8 hours)

Regards
Marx
 

100uS / 100ms = 0.001
So, basically, your device will operate 0.001 duty cycle in normal mode and 0.999 in idle mode.
Let's imagine, that in idle mode it takes 1uA, and in normal mode 1mA. So, 0.999 * 0.001 + 0.001 * 1 = ~2uA average current. For 300 days = 300d * 24h * 2uA = 15mAH capacity needed.
 
Thank You..!!

I am using an multiplexer 74HC4067

I want to understand the Current of This Multiplexer when Mux is 100 % Loaded.

What parameter in the datasheet mentioned that current..??

So that i can also calculate the Battery Consumpion for the Multiplexer
 

Bjtpower said:
Microcontroller will Send signal in every 100ms (Data transmission time would be only 100uS in 100mS)

Transmitting current will be 22mA MAX

Now i want to calculate required battery Capacity for 300 days (Daily circuit will be on for 8 hours)
Click to expand...

You have not specified the idle current. The duty cycle is 0.1%; Avg current for 8 hrs will be 22uA. Add 5uA for the idle state and will be having total avg load 22/3+5=12uA cont for 24 hrs.

365*24*12=105mAH for an year or 86mAH for 300 days. You can safely use a battery with double the capacity with low leakage (self-discharge current should be low).
 
Hi,

I am using an multiplexer 74HC4067
Click to expand...

If not otherwise noted: Expect the 4067 supply current to be independent of load current.

It is a switch, so the power for the load will be supplied by the "transmitter" ... at the other endo fo the switch.

Klaus
 

c_mitra said:
You have not specified the idle current. The duty cycle is 0.1%; Avg current for 8 hrs will be 22uA. Add 5uA for the idle state and will be having total avg load 22/3+5=12uA cont for 24 hrs.

365*24*12=105mAH for an year or 86mAH for 300 days. You can safely use a battery with double the capacity with low leakage (self-discharge current should be low).
Click to expand...

Hi Mitra
I don't know how 86mAH you calculated..?

Here I have calculated like this
22mA*(100uS/100mS)=22*0.001=0.022*8(hrs)*300=52.8mAH
 

Easyrider83 said:
100uS / 100ms = 0.001
So, basically, your device will operate 0.001 duty cycle in normal mode and 0.999 in idle mode.
Let's imagine, that in idle mode it takes 1uA, and in normal mode 1mA. So, 0.999 * 0.001 + 0.001 * 1 = ~2uA average current. For 300 days = 300d * 24h * 2uA = 15mAH capacity needed.
Click to expand...

Hi..
It's fine that we got the average current..
Let me understand that that average current consumed in 1 hour.. if consumed then where is the multiplier of 1 hour..?
Like later multiply it by 24 hours and 300 days
 

Easyrider83 said:
100uS / 100ms = 0.001
So, basically, your device will operate 0.001 duty cycle in normal mode and 0.999 in idle mode.
Let's imagine, that in idle mode it takes 1uA, and in normal mode 1mA. So, 0.999 * 0.001 + 0.001 * 1 = ~2uA average current. For 300 days = 300d * 24h * 2uA = 15mAH capacity needed.
Click to expand...

Does it means that in 1hour it takes only 2uA of current..??
How..??
or do we need to multiply it by the (second to hour)=60x60=3600..??
Formal Doubt
 

Bjtpower said:
Let me understand that that average current consumed in 1 hour.. if consumed then where is the multiplier of 1 hour..?
Like later multiply it by 24 hours and 300 days
Click to expand...

Sorry for not being clear enough.

First thing you need to understand that the AH specs does not bother about the voltage. However, we assume that the battery supplies the specified current for the specified duration while the voltage stays within usable limits.

If the av current taken by the circuit is 7mA, you will need a battery of 7mAH capacity to run the same circuit for one hour.

Using the same circuit, you can use for 300 days and 8 hours per day (2400 total hours) if you use a battery of capacity 2400X7 mAH=16800mAH or about 16 AH capacity battery. You can add 10-20% margin and round up to a round figure!!

AH says about capacity (basically it is 3600 Coulombs)- but the specs are not easy to interpret.

All batteries have leakage current OR self-discharge current that is being used EVEN IF you are not connecting it to anything. That is the reason battery will not last longer than the shelf life specified.
 
c_mitra said:
Sorry for not being clear enough.

First thing you need to understand that the AH specs does not bother about the voltage. However, we assume that the battery supplies the specified current for the specified duration while the voltage stays within usable limits.

If the av current taken by the circuit is 7mA, you will need a battery of 7mAH capacity to run the same circuit for one hour.

Using the same circuit, you can use for 300 days and 8 hours per day (2400 total hours) if you use a battery of capacity 2400X7 mAH=16800mAH or about 16 AH capacity battery. You can add 10-20% margin and round up to a round figure!!

AH says about capacity (basically it is 3600 Coulombs)- but the specs are not easy to interpret.

All batteries have leakage current OR self-discharge current that is being used EVEN IF you are not connecting it to anything. That is the reason battery will not last longer than the shelf life specified.
Click to expand...

Not Cleared enough.

What my doubt is that,
by datasheet IC is taking 22 mA (What is this..?? Average current..??)
Now,
by going the straight way calculations,
22mA in 1 second then how much current in 1 uS..??
by using cross multiplication
22mA X1uS=1S x Required current
Therefore, Required current=(22x10^-3)x(10^-6)/1=220x(10^-9)=220nA

Now converting it into the 1 hour
1 hour=60 Min=60x60second=3600 Second

In 1 hour Current required will be,
220nAX3600=792000X10^-9=0.79mA

For 8 hours=0.79mAx8=1896mA
For 300 days=1896mAX300=568800mAH

Pls Elaborate your calculation step by step.
 

Hi,

I can't follow your calculations:
(I didn't check if your initial values are correct)

22mA in 1 second then how much current in 1 uS..??
Click to expand...
Maybe it's easier with a LED.
A LED consumes 20mA. In one second, in one hour in one year...and in one us. 20mA when it is ON.
But in your case it is ON only 100us in 100ms. This is 0.001. (No unit, because it is "second/second")
So the average current is 20mA x 0.001 = 0.020mA. (Average ON time current)
If now the LED cosumes 1mA (dim) during OFF time then
The average OFF time current is 1mAx 0.999 = 0.999mA.
The total average current is 0.999mA + 0.020mA = 1.019mA

22mA X1uS=1S x Required current
Therefore, Required current=(22x10^-3)x(10^-6)/1=220x(10^-9)=220nA
Click to expand...
The idea is wrong. And the calculation is wrong, too.
(You could calculate charge per us: 22mA x 1us = 22nAs)

220nAX3600=792000X10^-9=0.79mA
Click to expand...
22nAs x 3600s/h = 792nAh = 0.792uAh per hour,

For 8 hours=0.79mAx8=1896mA
Click to expand...
The calculation is wrong.
0.792uAh x 8 = 6.3336uAh is about 6.3uAh (per 8 hours)

For 300 days=1896mAX300=568800mAH
Click to expand...
0.792uAh x 24 × 300 = 5790uAh = 5.7mAh

Klaus
 
Bjtpower said:
by datasheet IC is taking 22 mA (What is this..?? Average current..??)
Now,
by going the straight way calculations,
22mA in 1 second then how much current in 1 uS..??
by using cross multiplication
22mA X1uS=1S x Required current
Therefore, Required current=(22x10^-3)x(10^-6)/1=220x(10^-9)=220nA
Click to expand...

22mA is 22mA current; rate of flow of charge; it may be for 1 sec or 1 hour (it does not matter).

22mA in 1 sec = 22 mC (ampere X sec = Coulombs)

22mA in 1 us = 22 nC (m X u = n)

If this current is only for 1 us and only once per sec, then we have 22 nC /s or 22 nA (this is avg current)

Over one hour, we shall be having 22 nAH battery capacity is needed; 1 AH = 3600 C
 
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