Super capacitor over voltage

Thread Starter

k1ng 1337

Joined Sep 11, 2020
413
Hi,

I have a 2.7 500F super capacitor model # mh47765. I have been performing low power experiments with it as to learn it's nature. Immediately prior to charging with a buck boost converter (specs below), I had the supercap placed in a crude boost converter circuit of my own design that had reduced the voltage to 800mV.

I (previously) set the DSN6000AUD to 2.6V and began to charge the supercap from 800mV monitoring the entire time as I've done before. After ten minutes I returned to find the supercap voltage was at 3.3V! What the heck happened here? For safety I connected a motor and reduced the voltage to 2.7V (which took nowhere near the calculated runtime) and have been monitoring the self-discharge. In about 12 hours the voltage has self discharged from 2.7V to 2.0V and seems to be holding at 2V. I do not intend to use the supercap until I get to the bottom of this as my search results were unsuccessful.

Model Specification:DSN6000AUD Automatic Buck module
Module Properties: Non-isolated boost (BOOST)
Rectification: Non-Synchronous Rectification
Input Range:3.8V ~ 32V
Output Range:1.25V ~ 35V
Input Current:3A ( max ) , no-load 18mA (5V input , 8V output , no-load is less than 18mA. Higher the voltage , the greater the load current . )
Conversion efficiency:< 94% ( greater the current , the lower the efficiency )
Switching frequency:400KHz
Output Ripple:50mV ( the higher the voltage , the greater the current , the greater the ripple )
Load Regulation:± 0.5%
Voltage Regulation:± 0.5%
Operating Temperature:-40 ℃ ~ +85 ℃
Dimensions:48mm * 25mm * 14mm ( L * W * H )
 
Last edited:

Thread Starter

k1ng 1337

Joined Sep 11, 2020
413
Switching-Regulator Noise.
.
.
.
Can you elaborate? The buck boost was only loaded by the supercap and was the only device powered by the source which is also a SMPS.

My working theory is my experiments require greater control. I knew using the buck boost to charge made calculations difficult so it's not really a surprise tolerance was exceeded. Would you consider the supercap damaged having been charged to 3.3V?
 

tsan

Joined Sep 6, 2014
123
Input Current:3A ( max ) , no-load 18mA (5V input , 8V output , no-load is less than 18mA. Higher the voltage , the greater the load current . )
There is a minimum load requirement and it's possible that the capacitor was charged to a higher voltage because there was only the capacitor as a load.
 

drjohsmith

Joined Dec 13, 2021
376
If you put 5v across the capacitor,
it will charge to that also,
but not for long,

The super capacitor voltage is a limit you need to follow else you damage the part.
 

Thread Starter

k1ng 1337

Joined Sep 11, 2020
413
By my calculation, a 500F capacitor can charge from 0.8V to 3.3V in 10 minutes at 2A. So, it is not impossible that it did so.

Bob
How did you come to this conclusion? Since the buckboost module has feedback, is it modeled as a voltage or current source in this situation since the cap voltage is constantly changing therefor the module is forced to compensate? I'm having some trouble understanding the differences when charged from a voltage vs current source and how to properly apply the formulae. I did observe transients on the buck boost when it was unloaded and then loaded though I still do not understand how it could deliver that much power for so long when set to 2.6V.

Regards,
 
Last edited:

BobTPH

Joined Jun 5, 2013
4,738
You cannot charge a super capacitor directly from from a voltage source.

The way it should be done is by a current source with a max voltage. Can your buck boost module do that? Do you know how?

Bob
 

MrChips

Joined Oct 2, 2009
25,919
This is a peak detect circuit.

1640633514915.png

Every time the diode is forward biased the voltage on the capacitor will rise. If there is no discharge path across the capacitor, the capacitor will hold the charge until the next peak arrives.
 

Thread Starter

k1ng 1337

Joined Sep 11, 2020
413
You cannot charge a super capacitor directly from from a voltage source.

The way it should be done is by a current source with a max voltage. Can your buck boost module do that? Do you know how?

Bob
I've been basing my calculations on this formula E=CV^2/2, τ=RC so a resistor was used between the cap and the voltage source. I've read that when charged from a constant current source, the voltage across the cap will rise linearly with time although I have not found formula that met with present understanding of a capacitor. An example of how to prepare this kind of simulation is spice with a graph or array output would be most appreciated.
 

Thread Starter

k1ng 1337

Joined Sep 11, 2020
413
This is a peak detect circuit.

View attachment 256199

Every time the diode is forward biased the voltage on the capacitor will rise. If there is no discharge path across the capacitor, the capacitor will hold the charge until the next peak arrives.
Therefor a diode should have been used when charging from SMPS or in general? When accounting for inductances in the circuit I can imagine voltage spikes then for both the on and off states of the buck boost while switching as the cap is discharging (or at least trying to) into the buck boost during the off state?
 

MrChips

Joined Oct 2, 2009
25,919
Therefor a diode should have been used when charging from SMPS or in general? When accounting for inductances in the circuit I can imagine voltage spikes then for both the on and off states of the buck boost while switching as the cap is discharging (or at least trying to) into the buck boost during the off state?
No. Just saying that if there is no discharge path, charge will accumulate on the capacitor and the voltage will rise.
If there are noise spikes above the nominal charging voltage then don't be surprised to see a voltage higher than what you think is the average charging voltage. The same goes for ripple on the output of an AC-to-DC rectifier circuit. The capacitor will charge to the peak voltage, not the average DC voltage.
 

BobTPH

Joined Jun 5, 2013
4,738
The equation for a capacitor is:

dV/dt = I/C

This is the only equation ever needed for a capacitor, any other equation you might find can be derived from this.

If you want to see it in LTSPICE, just connect a constant current source to a capacitor and plot the voltage.

Bob
 

Thread Starter

k1ng 1337

Joined Sep 11, 2020
413
@k1ng 1337

Don't suppose yo could post a circuit diagram as to what your doing please
I must admit, reading back through the post , it all seems very confusing,
I agree it is confusing and it's no longer on the breadboard. I've concluded that I need to learn more about caps and SMPS before experimenting with both at the same time. Additionally, I need to better control my experiments which will enable me to present better findings when asking for help / debugging.

I'll end this topic here, thanks everyone for the responses.

Happy Holidays,
Mark
 

vanderghast

Joined Jun 14, 2018
57
I've been basing my calculations on this formula E=CV^2/2, τ=RC so a resistor was used between the cap and the voltage source. I've read that when charged from a constant current source, the voltage across the cap will rise linearly with time although I have not found formula that met with present understanding of a capacitor. An example of how to prepare this kind of simulation is spice with a graph or array output would be most appreciated.
Q=CV and Q = I * LengthOfTime (if the current, I, is constant ). That assumes that C = constant too (versus the stored voltage, in the cap, ... big assumption not easily satisfied for a supercap).

Numerical example: I = 2 A, C = 500F, and V = 3.3 - 0.8 Volt. So 500 * (3.3 - 0.8) = 2 T, giving T = 10.4 minutes (or 625 s ).
 
Last edited:
Top