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Charging supercapacitors in series?

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eem2am

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hi,

i have a stack of six series super-capacitors (650 Farad each and 2.5V max).

I wish to charge them up with a 1 Amp charging current.

I wish to charge them up to 2.2V each (8.8V for the stack)

How do i ensure that during charging , and subsequent trickle-charging, no single capacitor suffers a voltage of greater than 2.5V?

I am thinking that there must be standard circuits for this?
 

You just connect a similar resistor across one to share the voltage. With a 1A charging current, I would allow .1A to flow through the resistors, so 2.2V drop at .1 A = 22 ohms (X 4!!), the resistors should be accurate in value as the final voltage on each cap is dependent on their values. +-10% are not really accurate enough, go for 1 %.
Frank
 
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    eem2am

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See an Epcos datasheet excerpt dedicated to supercap cell voltage balacing. It derives a criterion, in which cases passive (parallel resistor) balancing is sufficient. You'll find active balancing with larger supercap batteries from Epcos that are e.g. used in automotive.
 

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it does not work..The problem is that I don't know the leakage current of the supercapacitors..
Click to expand...
It does not work or you don't know how make the circuit?

How do you determine that the scheme "does not work"? Which circuit are you referring to?
 

To be honest I am not sure about the resistor values that would fit my circuit.
One example that it doesnt work .Let's say one capacitor has an initial voltage of 0.6 and the other one of 1.8.
If I start charging them they will charge at the same rate but but the second one will continue charging beyond 2.5..
 

Again: Which circuit are you referring to?

The linked publication uses active balancing, thus the circuit doesn't depend on resistor values (within reasonable limits).
 

AD623 is an instrumentation amplifier rather than an OP, thus I wonder how you put it into the Fig. 5 circuit? In Fig. 5, the balancing current is only limited by R14 and the OP output characteristic.

One example that it doesnt work .Let's say one capacitor has an initial voltage of 0.6 and the other one of 1.8.
If I start charging them they will charge at the same rate but but the second one will continue charging beyond 2.5.
Click to expand...
The problem is that you apparently didn't understand the circuit's operation principle. It's expected to balance the capacitor voltages before you start to charge it.

There are of course some restrictions related to minimal OP voltage and initial operation procedures.

MAX4470 has a minimal operation voltage of 1.8 V. So in your example, in the unbalanced charging state the circuit starts to work until e.g. equal voltages of 1.1 - 1.2 V are achieved. Equal capacitance of both super-caps within certain tolerances is still required.
 
I am working on a breadboard that's why I chose a rail-to-rail amplifier (like MAX4470) but with pins.
Ok thank you for the explanation.
Do you think it will be much of a problem if I leave AD623 instead?
 

Supply voltage range and missing rail-to-rail input are a problem, I assume.
 

I changed the opamp to the one mentioned in the datasheet however there is always a slight difference of about(50mV-70mV)
between the supercapacitors?Any ideas?Do you think I should try different limiting resistor values?
 

50 to 70 mV should be tolerable. It may be caused by OP input currents and respective divider voltage drops.
 
milvapp, there is an error on the circuit diagram. The battery should be connected to the negative input side of the 470 ohm resistor NOT to the output terminal side. As it is the positive input is help at 1/2 V, so the output is held at 1/2V any difference goes to the negative input and corrects the output, which then feeds more or less current into the battery junction. What is should do, is to compare the battery junction to the 1/2 V set up by the two resistors the difference being amplified which then feeds more or less current via the 470 ohms into the battery junction.
Frank
 
milvapp, there is an error on the circuit diagram.
Click to expand...
The balance generator has an impedance of 470 ohm and thus will have a 500 sec time constant and possibly a permanent offset with higher leakage currents. But the circuit behaviour may be on purpose.
What is should do, is to compare the battery junction to the 1/2 V set up by the two resistors the difference being amplified which then feeds more or less current via the 470 ohms into the battery junction.
Click to expand...
Without additional compensation means, the suggested modification will most likely create an unstable feedback loop and force the circuit into permanent oscillations. This won't hurt the capacitors but cause an unwanted current consumption.

With a parallel feedback high-pass, it can work.
 
Thank you Frank for your correction.It makes more sense now.However if I connect the battery junction to the negative input side of the resistor (i.e. the negative input of the amplifier) where the output of the 470 will be connected?
 

I have this problem .For example if the voltage across the supercapacitors is 3.56 V
The voltage at SC1=2.1V and SC2=1.46 V where as VR1=2.138 V and VR2=VC =0.724 V .
Shouldnt VC/2 be near VR/2?
 

FvM said:
With a parallel feedback high-pass, it can work.
Click to expand...

I designed this circuit in LTspice with a HPF in parallel.
However ,I tried many different values of compensation capacitor Cc and I didnt see any difference.
What am I doing wrong?
 

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