Electrical – High Power Capacitor Charger

chargingsupercapacitor

I am trying to put together a circuit that can charge a 3000F Maxwell supercapacitor in about 10 mins.
I am looking to use a 150W laptop adapter and a Buck converter.
Problems that I am encountering:

  1. No High power Buck converters with 2V-3V output to charge supercapacitor safely
  2. I do not have enough knowledge on the concept of having several Buck Converters in parallel to increase power output.
  3. Cost

Best Answer

Let's do a power calculation first: The charge \$Q\$ on a capacitor with capacity \$C\$ under voltage \$U\$ is:

$$ \begin{align} Q&= CU\\ &= 3000\,F \cdot 3\,V\\ &= 3000\,\frac{As}{V} \cdot 3\,V\\ &= 9000\,As\,\text. \end{align} $$

That's quite a handful of Energy:

$$ \begin{align} E&= QU\\ &= 9000\,As \cdot 3\,V\\ &= 27\, kVAs\\ &= 27\, kWs\\ &= 27\, kJ\,\text. \end{align} $$

27 kJ is not a fun thing. If you wanted to charge that within 10 min = 600s, your average current and power would have to be

$$ \begin{align} I_{avg} &= \frac{9000\,As}{600\,s}\\ &= \frac{90}6\,A \\ &= 15\,A\,\text{,}\\ P_{avg} &= \frac{27\,kJ}{600\,s}\\ &= 45\,W \end{align} $$

so, assuming a typical 20V for your laptop supply, and following your claim it delivers up to 150W, it's instantaneous maximum output current is 7.5A, less than 15A – in other words, you'll really need that step-down converter if you want to do this in 10 Minutes.

Sadly, capacitors don't charge continuously – they have exponentially decreasing voltage gap == current, so at the beginning, you'd have an immense charge current.

So, you could certainly design a very beefy step-down (buck) converter that is able to source 100s of ampere for a short time, but you could also just design something that avoids the whole inductivity thing – after all, you don't care about the voltage at all, until it reaches your desired final value; that will take longer, will burn a lot of energy, but it's also much easier to implement:

Basically:

schematic

simulate this circuit – Schematic created using CircuitLab

Another, better, since less inefficient, method would be using an Opamp over a smaller shunt resistance instead of R1 to sink a constant current (==max current of your supply) into the capacitor, and use the comparator/reference voltage only to turn that off.

But: you could also take this comparator-only circuit and omit R1, probably. Why? Because consumer electronic, sufficiently cooled, will probably just shut off in an overload situation, and switch back on as soon as things have cooled down. That way, you can use the step-down power supply that your laptop supply essentially is ... I wouldn't call this a good solution, but it's definitely the easiest one, and worst case, it costs you a laptop supply.