Maker Pro
Maker Pro

Automatic switching between parallel and series connection

NK808

May 17, 2017
16
Joined
May 17, 2017
Messages
16
Hello all !
I have 4 supercapasitors connected in parallel to harvest energy from very weak power source ( indoor
PV cell or similar ). As voltage across capacitors reach some level ( for example 0.2-0.3 volt ) I swap wires to
connect supercapacitors in series to get 0.8-1.2 volts and unload energy to supercapasitor # 5. Duration of
equalization is about 60 seconds, current is 3A and drops to 0A.
As of now I do it simply by using dpdt toggle switches. Please advise circuit to automate this process.
Thank you!Basic circuit.JPG
 

Harald Kapp

Moderator
Moderator
Nov 17, 2011
12,613
Joined
Nov 17, 2011
Messages
12,613
Welcome to EP.

To switch with as little loss as possible between the two configurations you'd need low-threshold MOSFETs plus a controller circuit.
Unfortunately the controller circuit would need to be able to run from 0.5 V ..1 V otherwise the circuit would not be able to start-up.
This is kind of a hen an egg problem. I'm not aware of a suitable solution.

It is probably a better solution to use a spezialized energy harvester chip to directly up-convert the input voltage to a higher value and store the energy on only one supercapacitor. Many companies produce suitable chips. Here's an example of a selector page where you can enter your parameters (source of energy, power available etc.) to find the right chip.
If you think you don't find what you're looking for, try other chip manufacturers like TI, ONsemi etc.
 

NK808

May 17, 2017
16
Joined
May 17, 2017
Messages
16
Hi Herold!
Thank you for your reply.
Yes, I tried to use energy harvester based on LT3108 chip. Great device , starts at just 0.02V and can boost up to 2.3V, 3V, 4.1 or 5V. But losses are huge. For example: I use 10 TEC ( TEG ) connected in parallel sandwiched between two heat sinks with delta ~5 degree Celsius. Power output is about 0.2V at 0.08A. If to connect it to supercapacitor 2.7V , 350F for 5 min, then energy stored enough to run tiny motor from the phone for about 10 min. If to connect those TECs to LT3108 upconverter wired for 2.3V ( for best efficiency ) and from LT3108 to the same supercap, after 5 min, it stores energy to run tiny motor for 30-35 seconds. I repeated this process many times. Outcome is always the same. Difference is huge.
So far I can think of solid state relays controlled by Arduino or similar, but as you pointed out energy harvested is far less then needed to run control unit. Probably, best way to go would be to use 3000F capacitors and and do manual switching once a day. My knowledge of electronics is next to none and I was hoping that someone has an answer already.
Thanks!
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Joined
Jan 21, 2010
Messages
25,505
I've not used any of the energy harvesting chips, but I suspect that if you're getting the losses you observe then your doing something wrong.

Can you provide a link to the datasheet for the energy harvesting board you have and for the chip and I'll do a bit of reading.
 

NK808

May 17, 2017
16
Joined
May 17, 2017
Messages
16
I repeated this experiment many times and with different configuration of TECs.
Conclusion: energy "upload" directly from weak source to supercapacitor has almost no losses compare to usage of DC to DC upconverter.
Disadvantage : very low and useless voltage.
Hence my initial question. Perhaps, just "go" with series connection of TECs lowering amperage?
In any case it would be interesting to come up with automation of series to parallel connection circuit.
Thanks!
 

Harald Kapp

Moderator
Moderator
Nov 17, 2011
12,613
Joined
Nov 17, 2011
Messages
12,613
Perhaps, just "go" with series connection of TECs lowering amperage?
This thought came to my mind when seeing yout TEC setup. The total energy will be the same (E=V*I*t), but you need only one supercap and no switches. And no additional controller to step-up the low voltage.
 

NK808

May 17, 2017
16
Joined
May 17, 2017
Messages
16
Hi Harold!
You are very right. Total sum of energy available is the same. Task is to collect it as efficiently as possible.
Observation shows that from weak source to capacitor "upload" is not linear, but more like logarithmic.
Assuming 0.5V source, voltage in capacitor from 0 to 0.1 takes 4-5 min, from 0.1 to 0.2 takes 8-9 min and from 0.2 to 0.3 ~20min etc.
It is somewhat similar to two buckets with water connected with hose, one is full, one is empty. Longer we keep valve open slower equalization becomes.
Another experimental observation shows that by connecting TECs in series does increase voltage but decrease time of reaching 0.1 volts level in a ratio .
Observation 1 = 14 TEC in series gives 1.3V and takes 8 min to get S-cap to get from 0 to 0.1
Observation 2 = 14 TEC in 7series by 2 parallel gives 0.7V and takes 7 min to get S-cap to get from 0 to 0.1
Observation 3 = 14 TEC in 2series by 7 parallel gives 0.2V and takes 5 min to get S-cap to get from 0 to 0.1
I'm not sure what it is, may be wire gauge or faradness of capasitor or the way TEC converts delta T into electrons flow? Many more experiments needed or a bit more brains on my behalf, most likely both.
My goal in this case is to automate process and minimize " manual intervention" to the best present technology offers while reducing energy losses on such automation.
Thank you!
 

Harald Kapp

Moderator
Moderator
Nov 17, 2011
12,613
Joined
Nov 17, 2011
Messages
12,613
The energy stored oln a capacitor of a given capacity is E=1/2 * C * V^2. Therefore when you double the voltage Vm the energy E becomes 4 times a high. This may be the rason for your observed 'more like logarithmic' characteristic (which isn't logarithimic at all, but square).
Assuming your original circuit of N capacitors with capacity C in parallel such that N equals the number of TECs (this is not necessary but simplifies the following consideration) and a max. voltage of the TECs of V0 (again the exact values are unimportant for the following consideration), the energy of the system at full charge state is
E0 = 1/2 * N * C * V0^2
Now connect all N TECs in series and use only a single capacitor (N=1), then the energy becomes E1 = 1/2 * C * (N*V0)^2 = 1/2 * N * C * V0^2 * N
The energy is now N times as high as in the parallel connection although you used less capacitors (1/N) and the same number of TECs.


Another experimental observation shows that by connecting TECs in series does increase voltage but decrease time of reaching 0.1 volts level in a ratio .
Small wonder:
When charging a capacitor wit a current I, the voltage increases as V(C)= C*integral(I) dt. Assuming (for simplicity) each TEC gives off a small fixed constant current I0, then this equation becomes V(C) = C* I0 * t.
Therefore for a single TEC (or stack of TECs) time to reach 0.1 V is t0 = 0.1V/(C * I0).
When you connect the TECs in parallel, the currents add up.
Therefore for a parallel connection of N TECs (or stack of TECs) time to reach 0.1 V is t(N) = 0.1V/(C * N * I0) = 1/N * t0.
 

NK808

May 17, 2017
16
Joined
May 17, 2017
Messages
16
Awesome points you are making, Harald!
Just now I realized that y=x^2 is identical to x=y^2! You are right it is square still only lays at 90 cw or ccw.
Though, more gear's turning will take for me to sort out and understand mathematics of it, but somehow it make sense already. I'm going to re-do and re-verify all 1x14, 2x7,7x2 and 14x1 again.
Thank you!
 

AnalogKid

Jun 10, 2015
2,665
Joined
Jun 10, 2015
Messages
2,665
Observation 1 = 14 TEC in series gives 1.3V and takes 8 min to get S-cap to get from 0 to 0.1
Observation 2 = 14 TEC in 7series by 2 parallel gives 0.7V and takes 7 min to get S-cap to get from 0 to 0.1
Observation 3 = 14 TEC in 2series by 7 parallel gives 0.2V and takes 5 min to get S-cap to get from 0 to 0.1
I'm not sure what it is,
A TEC has a relatively high output impedance, effectively a resistance in series with the voltage source. A discharged capacitor is presents a very low equivalent resistance to a DC voltage source, so the cap is basically "shorting out" the TEC output, and charging is based on the current available. As you have more and more TECs in parallel, the output impedance goes down and the output current goes up. Both of these contribute to faster capacitor charging.

ak
 

NK808

May 17, 2017
16
Joined
May 17, 2017
Messages
16
Thank you AnalogKid! Your explanation supports experimental data.
Voltage as a speed of water in a pipe I understand, Amperage as a diameter of the pipe I understand, thus W=V*I make sense for me.
V=I*R is difficult for me to understand. What is Resistance and why at times it is Impedance. Is it friction of water in the pipe? Is it sudden change in diameter of pipe? Or water analogy completely irrelevant and lead to totally incorrect understanding?
So, do I hear you correctly: more TEC ( TEG ) I connect in parallel to a given capacitor - less losses I will get or improve TEC generator efficiency overall?
Thanks!
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Joined
Jan 21, 2010
Messages
25,505
In that analogy, voltage would be pressure, current would be the rate at which water is flowing, and resistance is the difficulty the water has flowing through the pipe. Resistance could be inversely related to the pipe diameter or roughness of the pipe or, well, resistance to flow...
 
Top