Soft Start Circuit for a capacitor bank

[larryg]

I need to design a very simple soft start circuit for a super capacitor bank (12 4.7F caps in series). The bank will be charged to 24-28VDC.

I'm thinking of using a power resistor, a relay (to shunt out the resistor when the caps are charged) and some simple control circuit. My first inclination for the control is to use a 22V Zener at the output of the bank. This zener would bias the gate of a FET or BJT (which would turn on the relay). Is this a reasonable method for controlling the relay? Or is there some easier method I'm not thinking about? BTW, I do not want to use any soft-start IC's. Simplicity is crucial.

Thanks
Larry G

 

[electricuwe]

If you charge a capacitor with a resistor in series from constant voltage you will always have to dissipate an amount of energy equal to the energy stroed in the capacitor after the charging is complete.

If you are aware of that fact and your resistor is designed according to that energy your approach seems quite reasonable. You should include some positive feedback to achieve stable operation, e.g Schmitt-Trigger circuit.

 

[Lewish]

I just did a quite calculation based on what I think you said; i.e. 12 caps of 4.7F in series charging to 24volts.

J=0.5*C*V*V or for your case 16,243 joules approximately.

That is going to be one big mother of a resistor.

 

[larryg]

Allow me to fill in some more details. Basically, this cap bank is being used as a PSU hold-up on power loss. I will need 3.5A for 0.5s from 28V to 18V (which is the min. voltage). This equates to around 0.2F in capacitance. This bank will be connected between the existing off-the-shelf PSU and my controllers.

My concern involves the intial turn-on when the caps will charge up. My thought was to put a large resistor in between the PSU and the bank and force a slow charge of the cap bank, therby getting rid of the immense in-rush. Once the caps are charged, I want to shunt out the resistor via relay contacts.

Lewish, I'm not sure I follow your logic of needing a huge resistor. If I use a 120 Ohm, 5W resistor, I can limit the current to 0.2A. The main drawback is that my charge time will be long (5RC= 5*120*.2 = 144s)I may need to put a couple resistors in parallel to cut down my charge time to something more reasonable.

Does this make sense or am I offbase here?

Thanks
LG

 

[Lewish]

Ooops, I missed something. You said the caps were in series, and I repeated it, but my brain was thinking they were in parallel. Sorry about that.
You are right, about the 120 ohm resistor.

OK, now another approach. A company called Ametherm makes "Circuit Protection Thermistors". www.ametherm.com.
I use their devices in power supplies to limit inrush current. The device starts out looking like some larger resistance which reduces as the part heats due to current.
They may or may not be a good choice for your application. Give them a call and see what they recommend.

 

[larryg]

Lewish,

Thanks for the suggestion. I actually use some of Ametherm's thermistors in another application. I don't know why I didn't think about using one for this application. It would be a lot simpler than incorporating a resistor, relay, and control circuitry.

Thanks to electricuwe for posting as well.

 

[jwerthman]

Why not just connect the capacitor to the PSU through a parallel combination of your resistor and a diode. Connect the cathode end of the diode to the (+) output of the PSU and the anode side to the capacitor. When the PSU is energized, the diode will be reverse biased and the capacitor must charge slowly through the resistor. If the PSU voltage drops below the capacitor voltage (plus the diode junction voltage), the diode will be forward biased and will discharge to the DC load as desired. Seems pretty simple - no switching. One drawback is that if your PSU load terminals are every short circuited, the capacitor will discharge through the diode into the short and probably fry the diode.

 

[Bung]

Dunno, haven't thought this through fully, but can't you use an inductor instead of a series resistor? I haven't done the maths on it, so it might be a big 'un, and you might also have a few resonance problems in some situations, but it could be less lossy than a resistor, and possibly does away with relays etc.



Bung
Life is non-linear...

 

[melone]

What happens when you turn it off...

 

[peebee]

Do you have any way to slowly ramp up the voltage? Or can you use a 48v transformer source that will self-limit the output?

 

[peebee]

Your need for current limiting doesn't sound all that different from what a lamp ballast does. Maybe studying some ballast schematics would give you some ideas.

 

[peebee]

Check thread240-32406 for some nice ideas on current limiting, including one similiar to your original idea, but using a small resistance to turn off a transistor.

 

[larryg]

Hi All,

Thanks a lot for all the responses. I finally got my prototype together and found that a hefty thermistor does wonders for snubbing out my inrush. And it makes the circuit quite simple. I also added an inline diode to prevent the capacitor bank from discharging thru the PSU when power is lost.

Anyway, thanks again.

LarryG