7812

Comments

• Mike Green Posts: 23,101

  2007-02-13 04:59 edited 2007-02-13 04:59

  Here's an article from EDN on this www.edn.com/filtered/pdfs/contents/images/72204di.pdf.
• Lightfoot Posts: 228

  2007-02-13 06:53 edited 2007-02-13 06:53

  How do I calculate the value of an input filter capacitor in a regulator circuit?
  
  ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
• Mike Green Posts: 23,101

  2007-02-13 07:04 edited 2007-02-13 07:04

  You don't "calculate" it. Look at the manufacturer's datasheet for their recommendation. It varies depending on the regulator design.
• Lightfoot Posts: 228

  2007-02-13 08:46 edited 2007-02-13 08:46

  I usually put the circuit on the scope and see which one clears up the most ripples. Their has to be a way to mathematically calculate the emimination of ripples with capacitors.
  
  ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
• Jim Richey Posts: 82

  2007-02-13 13:44 edited 2007-02-13 13:44

  Hey Mike
  Thanks for posting that most interesting information on utilizing multiple 78XX regulators.
  I never looked at an arrangement with diode isolation like that and will soon give it a try.
  
  Regards
  Jim Richey
  
  ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
  Thanks, Parallax!
• RDL2004 Posts: 2,554

  2007-02-13 17:50 edited 2007-02-13 17:50

  http://sound.westhost.com/power-supplies.htm
  
  See the section on input capacitor value.
  
  
  ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
  - Rick
• Mike Green Posts: 23,101

  2007-02-13 18:16 edited 2007-02-13 18:16

  RDL2004,
  This is a nice tutorial, but it may not apply. If you're designing a complete power supply taking 110VAC to regulated 12V or 5V or whatever, this is a great reference. For most microcontroller projects, a "wall-wart" is going to be used for the initial power conditioning. These can be unregulated DC supplies with the transformer, rectifier, and filter capacitor already included. In that case, you just need enough capacitance at the regulator to supply instantaneous peak demands given the inductance and resistance of the cable connecting to the "wall-wart" (and this fulfills the regulator's requirements for "stability"). Increasingly, the "wall-wart" already provides regulated power using a compact switching regulator, but maybe at a higher voltage than required by most of the project (like when there's a mix of servos and control logic) and the regulator simply dissipates the extra voltage and doesn't have to compensate for ripple. In this case, the regulator input capacitance just provides for stability (since the regulator is an amplifier and can get internal feedback) and possibly noise suppression.
• Lightfoot Posts: 228

  2007-02-13 23:24 edited 2007-02-13 23:24

  Would a good "rule of thumb" for dealing with ripple current involve using a cap with a rated ripple current value that is >= to the current output in your circuit?·
  
  ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
• PJAllen Banned Posts: 5,065

  2007-02-14 00:22 edited 2007-02-14 00:22

  Capacitors are rated in Volts.· Your "rule of thumb" is to rate twice as high as the highest voltage it'll be across.·
  If it's a 15V_pk supply voltage, then the capacitor should be rated for at least 30V.·
  If it's a nominal 15VDC, the cap should still be rated at least 30V, too.
• RDL2004 Posts: 2,554

  2007-02-14 01:53 edited 2007-02-14 01:53

  I'm sorry I missed the fact that this was a power supply for a microcontroller running off of a "wall wart". If it is, 2 amps is a lot more than the average AC adapter is going to supply anyway. You should still read the tutorial at the link I posted. Granted it's intended for people building amplifier power supplies, but it does cover the basics of linear power supplies fairly well (and it does have a section on capacitor ripple current).
  
  ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
  - Rick
• Bruce Bates Posts: 3,045

  2007-02-14 09:18 edited 2007-02-14 09:18

  Gents -
  
  Something I read recently gave me the impression that (regardless of what happens to the capacitance) if two capacitors, equal in all respects, are placed in series, the working voltage would be doubled. Is that indeed the case?
  
  In other words, two capacitors with a working voltage of 15 volts COULD be used on a 30 volt circuit. I realize that violates the 2X safety "rule of thumb" noted earlier. This is just an example.
  
  My own specific application has to do with the irksome problem of using a capacitor in a 6 volt DC application. Capacitors, as a rule, are rated at 5 volts, so one ends up using a 15 or 25 volt capacitor on a lousy 6 volt DC circuit. Sometimes one can purchase 2 x 5 volt capacitors for less than the price of a higher working voltage unit, particularly in the surplus market.
  
  Regards,
  
  Bruce Bates
  
  ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
  <!--StartFragment -->
• Jim Richey Posts: 82

  2007-02-14 13:39 edited 2007-02-14 13:39

  Hello Bruce
  
  It makes me wonder if there is any disadvantage in using a cap with a much higher
  working voltage than required.I have always assumed [noparse][[/noparse]a dangerous habit of mine] that
  as long as the rating was at least twice or higher everything would work out. But then,
  I try to keep a fire extinguisher handy when experimenting!
  
  Regards
  Jim
  
  ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
  Thanks, Parallax!
• Bruce Bates Posts: 3,045

  2007-02-14 14:01 edited 2007-02-14 14:01

  Jim -
  
  The "mitigating" factor in what I was asking was cost, nothing more. I'm sure there is no other DIS-advantage to using a cap with a higher working voltage.
  
  Regards,
  
  Bruce Bates
  
  ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
  <!--StartFragment -->
• PJAllen Banned Posts: 5,065

  2007-02-15 02:19 edited 2007-02-15 02:19

  Bruce,
  
  Yes, caps' WVDCs, in series, add [noparse][[/noparse]but, as you know though others may not, C_T = C₁C₂/(C1+C2) ].· I think that it's OK.· This assumes that the caps are the same value (same mfr., same date-code, too.)
  
  That's my 2d.
• David B Posts: 591

  2007-02-15 16:05 edited 2007-02-15 16:05

  If you connect equal capacity capacitors, having equal internal resistances, in series, they each get half the voltage and the combination gives you half the capacitance.
  
  But if one of the caps develops more internal leakage then the other then its voltage will drop, raising the voltage across the other, possibly to the point of exceeding its rating.
  
  Some people add resistors in parallel with the series caps to compensate for that case. Fairly high resistance values, like 100K, will draw negligible current for a power supply circuit, but hopefully will be so much less than the capacitors internal resistance that they will force the voltage to divide evenly. Plus for big caps, the resistors provide a discharge path to empty the capacitors when the supply is off.
  
  It's funny though. In all my years of electrical experimenting, I've never had a capacitor fail due to excessive voltage. I've had tantalums explode when I connected them with the wrong polarity, but never had a non-tantalum fail. Maybe I've been lucky, or maybe it is an industry standard to allow plenty of safety margin in the voltage ratings of capacitors.
• Beau Schwabe Posts: 6,547

  2007-02-15 17:27 edited 2007-02-15 17:27

  David B
  
  "...But if one of the caps develops more internal leakage then the other then its voltage will drop, raising the voltage across the other, possibly to the point of exceeding its rating."
  "...It's funny though. In all my years of electrical experimenting, I've never had a capacitor fail due to excessive voltage..."
  
  
  About 10 years ago I had this happen once. The result of one cap failing caused the other cap to fail in a chain reaction....BANG!! BANG!! (each one sounded like a gun shot )
  One of the cases to the electrolyte capacitor nailed me in the cheek and hurt like a {blankity blank}. Fortunately the caps were only about the size of my pinky nail. (100uF's)
  but they did pack a respectable punch!
  
  ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
  Beau Schwabe
  
  IC Layout Engineer
  Parallax, Inc.