×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Soft Start Circuit for a capacitor bank

Soft Start Circuit for a capacitor bank

Soft Start Circuit for a capacitor bank

(OP)
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

RE: Soft Start Circuit for a capacitor bank

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.

RE: Soft Start Circuit for a capacitor bank

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.

RE: Soft Start Circuit for a capacitor bank

(OP)
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

RE: Soft Start Circuit for a capacitor bank

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.

RE: Soft Start Circuit for a capacitor bank

(OP)
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.

RE: Soft Start Circuit for a capacitor bank

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.

RE: Soft Start Circuit for a capacitor bank

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...

RE: Soft Start Circuit for a capacitor bank

What happens when you turn it off...

RE: Soft Start Circuit for a capacitor bank

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?

RE: Soft Start Circuit for a capacitor bank

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.

RE: Soft Start Circuit for a capacitor bank

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.

RE: Soft Start Circuit for a capacitor bank

(OP)
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

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources