ultra-capacitor based battery for data acquisition system

Thread Starter

jimit5

Joined Sep 26, 2016
17
I'm trying to makeblock diagram.JPG a project that uses ultra capacitor as an battery for a data acquisition system
the system will contain number of slave devices (nodemcu esp8266) that i have to run for almost 8-10hrs (only 1 on each battery) i know that they require 3.3 volts at 80mA current to work
Im trying to find a formula to calculate the charging/discharging time of the capacitor to compare it with normal batteries i have ultra capacitors rated at 500f 2.7v that im going to use for the same
I found this post and wanted to know if it could be used to calculate the charging/discharging time of the capacitor
http://fiziknota.blogspot.in/2010/01/relationship-between-energy-voltage.html
The basic block diagram is attached if it helps
 

wayneh

Joined Sep 9, 2010
16,399
I found this post and wanted to know if it could be used to calculate the charging/discharging time of the capacitor...
Before we start in on the math, is there a reason you're using a capacitor? A battery is superior for storing energy on a per-cost or per-volume basis. It might even allow you to eliminate the boost circuit.
 

OBW0549

Joined Mar 2, 2015
3,503
...the system will contain number of slave devices (nodemcu esp8266) that i have to run for almost 8-10hrs (only 1 on each battery) i know that they require 3.3 volts at 80mA current to work
3.3V * 0.08A = 0.264 watts. 10 hours = 36000 seconds. 0.264 watts * 36000 seconds = 9504 watt-seconds (joules) of energy. That is the amount of energy you need to run each node for 10 hours.

Im trying to find a formula to calculate the charging/discharging time of the capacitor to compare it with normal batteries i have ultra capacitors rated at 500f 2.7v that im going to use for the same
A 500F capacitor charged to 2.7 volts stores 0.5 * 500 * 2.7^2 joules of energy, or 1822 joules. That is how much energy is available from the capacitor, assuming (a) you can successfully extract ALL of the energy stored in the capacitor until it is discharged completely (you cannot), and (b) you can do so with 100% conversion efficiency (again, you cannot).

You would do much better using a rechargeable battery; ultracapacitors really aren't very "ultra."
 

Thread Starter

jimit5

Joined Sep 26, 2016
17
I know batteries are better than capacitors but it is just a proof of concept that it could be done the actual size doesn't matter.

According to what OBW0549 said I would require 9504j of energy to run my setup for 10hrs approx so by the energy equation E=0.5*c*v^2
I would need an 760f cap at 5 volts correct
So can I calculate the charging time using

E=VIT

Where E is the energy
V is the voltage
I is the charging current
T is the time
 

Papabravo

Joined Feb 24, 2006
14,189
Did you compute the voltage on the capacitor at the end of the 10 hours? Will the boost converter be able to maintain the output voltage as the input voltage drops continuously during the discharge? I don't think you will be able to make it work. Don't believe me, try simulating it in LTSpice before you run off and order parts and build one.
 

Thread Starter

jimit5

Joined Sep 26, 2016
17
For now everything is on paper only ill try to simulate the system and let you know


Did you compute the voltage on the capacitor at the end of the 10 hours? Will the boost converter be able to maintain the output voltage as the input voltage drops continuously during the discharge? I don't think you will be able to make it work. Don't believe me, try simulating it in LTSpice before you run off and order parts and build one.
 

OBW0549

Joined Mar 2, 2015
3,503
I know batteries are better than capacitors but it is just a proof of concept that it could be done the actual size doesn't matter.
You don't need any "proof of concept" to show that it can be done; of course it can. The issue isn't whether or not it is possible, but rather whether or not it makes sense. In the end, all you will end up "proving" is that an ultracapacitor-based solution is bulky, heavy and expensive.

So can I calculate the charging time using

E=VIT

Where E is the energy
V is the voltage
I is the charging current
T is the time
Wrong. That equation is not applicable because as a capacitor charges, the voltage is constantly changing.
 

Thread Starter

jimit5

Joined Sep 26, 2016
17
You don't need any "proof of concept" to show that it can be done; of course it can. The issue isn't whether or not it is possible, but rather whether or not it makes sense. In the end, all you will end up "proving" is that an ultracapacitor-based solution is bulky, heavy and expensive.


Wrong. That equation is not applicable because as a capacitor charges, the voltage is constantly changing.
Thank you for your reply i know the issue that it is bulky and heavy but we are aiming to show the features that it has such as fast charging, life and capability to work at lower temperature
i also found the time constant (RC) can be used to find the charging time for the capacitor ie:~5*RC right ??
 

OBW0549

Joined Mar 2, 2015
3,503
i also found the time constant (RC) can be used to find the charging time for the capacitor ie:~5*RC right ??
Since we have no idea what the value of R is, you'd be much better off using the charge equation,

C * V = I * T

where C is the capacitance in farads, V is the fully-charged voltage (assuming the charge starts out at zero) in volts, I is the available charging current in amps, and T is the charging time in seconds. This can be rearranged to solve for T, given the other terms:

T = C * V / I

Simple.
 

wayneh

Joined Sep 9, 2010
16,399
i also found the time constant (RC) can be used to find the charging time for the capacitor ie:~5*RC right ??
Five time constants gives nearly full charging or discharging. For many applications, just 3 time constants is good enough.

After each time constant, ~37% of the gap (distance between current state and long run state) remains. So three time constants gives 0.37^3 = 5% remaining gap and five gives 0.37^5 = 0.7%.

You may find in your application that you never want to drop below 20% level of charge, or 50%, or whatever. That just means you'll need a still-larger capacitor to get the same amount of current.
 

AnalogKid

Joined Aug 1, 2013
8,532
i also found the time constant (RC) can be used to find the charging time for the capacitor ie:~5*RC right ??
If you are charging a capacitor through a fixed (constant) resistor connected to a constant voltage source, then the voltage across the capacitor will reach 95% of the source voltage in 3 time constants and 99% in five time constants. To calculate how this relates to the percentage of total energy at those voltages compared to the total energy at the source voltage, you calculate the two and subtract.

The total energy in a capacitor is 1/2 CV^2 -- 1/2 x the capacitance (in farads) times the square of the capacitor terminal voltage. For a 1 F cap charged to 1 V, the energy is 0.5 J (old term: watt-seconds).

For a 10 F cap charging to to 5 V, the max energy is 125 J. At 3 time constants, the energy in the cap is 112.8 J.

At 5 time constants it is 122.5 J. It all of that energy is magically converted to 3.3 v for an 80 mA load (0.264 W), the load will run for 7.7 minutes.

Note the word "magically". Capacitor energy calculations always are misleading because you cannot extract all of the energy in a useful circuit.

BTW, years ago Coleman (the camping lantern guys) introduced a cordless screwdriver based on ultracaps. It's run time on one charge was below average for other devices at the time, but it recharged from 0% to 80% in 1 minute, 100% in 90 seconds.

ak
 

Thread Starter

jimit5

Joined Sep 26, 2016
17
If you are charging a capacitor through a fixed (constant) resistor connected to a constant voltage source, then the voltage across the capacitor will reach 95% of the source voltage in 3 time constants and 99% in five time constants. To calculate how this relates to the percentage of total energy at those voltages compared to the total energy at the source voltage, you calculate the two and subtract.

The total energy in a capacitor is 1/2 CV^2 -- 1/2 x the capacitance (in farads) times the square of the capacitor terminal voltage. For a 1 F cap charged to 1 V, the energy is 0.5 J (old term: watt-seconds).

For a 10 F cap charging to to 5 V, the max energy is 125 J. At 3 time constants, the energy in the cap is 112.8 J.

At 5 time constants it is 122.5 J. It all of that energy is magically converted to 3.3 v for an 80 mA load (0.264 W), the load will run for 7.7 minutes.

Note the word "magically". Capacitor energy calculations always are misleading because you cannot extract all of the energy in a useful circuit.

BTW, years ago Coleman (the camping lantern guys) introduced a cordless screwdriver based on ultracaps. It's run time on one charge was below average for other devices at the time, but it recharged from 0% to 80% in 1 minute, 100% in 90 seconds.

ak
Thanks for your reply i know its really not worth making this when we compare it to other power sources but its just a prove of concept efficiency is not something that we are aiming for i will be using 100f or 500f caps connected in series and parallel just to get more rum time
the problem that im facing is designing an balance charger for it i think i can use comparator and mosfets to do it but a pre-build circuit would help
 

Thread Starter

jimit5

Joined Sep 26, 2016
17
can someone suggest some other circuit cant get all the parts here and ordering online is very costly
 

wayneh

Joined Sep 9, 2010
16,399
can someone suggest some other circuit cant get all the parts here and ordering online is very costly
Are there particular parts that are expensive, and you need alternatives, or is everything expensive? That circuit is not terribly elaborate and if it's too expensive, I'm not sure there's much anyone can do.
 
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