How many caps do I need to compensate for ripple current

Discussion in 'The Projects Forum' started by emc2k99, Jun 25, 2011.

  1. emc2k99

    Thread Starter New Member

    Mar 12, 2011
    Let's say I have a transformer (24V 10A CT). I found this capacitor that I want to use to buffer the source current:

    It's Ripple current is rated at 4.17A. Based on my research, if I only use 1 cap on my circuit, the current should cause the cap to overheat and be ruined, right? Shouldn't I use 3 in parallel in order to compensate for the 10A output?
  2. #12


    Nov 30, 2010
    Not that simple.

    The ripple voltage at the capacitor is divided by (the square root of) (the Xc of the cap^2 + ESR^2) at the rectifying frequency to find ripple current.

    First you have to find the ripple voltage which might be:
    .707 C Er F = I (Help me out guys. Confirm this or deny it.)

    .707 Capacitance Ripple voltage Frequency = current

    But before you do that, you have to learn that you can not get 10A DC out of a 10 A transformer!

    Lets start with what form of rectifier you want to use, look up how many amps you can really get without melting the transformer, and calculate from there.
  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    Let I = C * delta V / delta T be your guide to picking a cap. The first thing it will show you is you need lots more cap there to support that current!

    The cap must supply the full 10A in between pulses from the bridge, and if we assume this current to be constant we can rearrange to get

    dV = I * dT / C

    where dT = 1/(2 * 60) = .0083s (for a 60 Hz FW bridge )

    so dV = 10A * .0083s / 3,900 uF = 21.3 volts

    That is the ripple with the cap you picked, so first off you need lots more cap there!

    If we assume you can live with 1 volt of ripple:

    C = I * dT / dV = 10A *.0083s / 1V = 83,333 uF

    The ratio will hold so if you can live with 2 volts then you only need half the cap.

    Now someone else (who remembers how <grin>) can help you get the ripple current of that.
  4. emc2k99

    Thread Starter New Member

    Mar 12, 2011
    I'm sorry for not adding more info #12, even though the transformer is advertised I had suspected (and hoped) that I wouldn't get a full 10A out of it. But without a datasheet I have to wait until it arrives in the mail to do any testing... I'm using 2 schottky diodes to rectify the CT Trans however with even 5+ amps wouldn't that hurt the capacitor?
    Thanks for the help. I'll have to do some more research into this. Sadly I have to go to my PT Job, So I'll probably have to log back in tomorrow to reply
  5. #12


    Nov 30, 2010
    @ Ernie

    How come you don't tawk wit a Noo Yawk aksent wen anserin' questions?
  6. SgtWookie


    Jul 17, 2007
    The diode you linked to is only rated for 35v, which isn't enough.

    The peak output voltage of your transformer under its' rated load will be 24v/.707107 = 33.94v; and with no load it will go significantly higher.

    Also, at 10A the rectifier will have a ~600mV drop across itself, which means 6 Watts of power dissipation. You will have to have a heat sink to mount your rectifiers on; TO-220 packages won't handle that much power dissipation without a heat sink.
  7. #12


    Nov 30, 2010
    You can get 8.333 amps DC out of that configuration. The diode RMS current will be 1.35Amps per amp delivered. Better make them at least 6 amp rated.

    You don't need a datasheet to find these formulas. I used these books:

    National Voltage Regulator Handbook, 1982
    National Audio Handbook, 1977

    Sadly, I have to go do some math and I will post this before the site thinks I'm taking too long and forgets my logon.

    edit: The configuration you posted while wookie was typing indicates a lower voltage than wookie expected. However, the actual power line voltage is high where I live and it makes transformer outputs higher than expected. I will include that in my math.
    Last edited: Jun 25, 2011
  8. #12


    Nov 30, 2010
    I'm gonna show you some math. It won't get the "right" answer. It will equip you to get the right answer.

    For 1 volt peak to valley ripple, radical2 C Er F = I
    1.414 Capacitance 1Volt 120Hz = 8.33333amps
    49105 uf

    That will require 13 of the capacitors you presented.
    13 x 3900 uf = 50,700 uf
    Xc = 1/2PiFC
    Xc = 1 over ( 2 x 3.14159 x 120Hz x .0507 F)
    Xc = .02616 ohms

    The ESR of each cap is .015 ohms
    there are 13 of them in parallel
    .015/13 = .0011538 ohms

    The square root of (.0011538 ^2 x .02616^2) = .02618 ohms

    I think .707 times the peak ripple voltage will work here for RMS ripple voltage
    I = .707 / .02618 = 27 amps
    27amps /13 capacitors = 2.076 amps each

    This will work with 13 capacitors or you can pick a larger capacitor.

    anyone that can dispute the line that starts with "I think" is welcome to correct me.

    Ps, 25 volts will be a sufficient voltage rating for the capacitors in this circuit.
    12V x 1.414 = 16.968V
    16.7 x 125 power line volts / 115 rated volts on the transformer primary = 18.44V
    Add 20% for unloaded condition = 22.13V
    Subtract a bit for the diode voltage loss and it's still under 25 volts.
    Last edited: Sep 21, 2011
  9. WellGrounded


    Jun 19, 2011

    The question to ask is "What type of load are you using on the output?" A mostly resistive load doesn't need smoothing. Many electronic PCB boards have a partial ripple smoother built into them based on their voltage and current requirements. Therefore "some" ripple smoother effect would be OK on your output, but not necessarily to the extent that you wish to pursue.

  10. CDRIVE

    AAC Fanatic!

    Jul 1, 2008
    This is a very odd statement. That's like saying... "I ordered an 8Ω 200W resistor but I hope it can only handle 100 Watts".

    Are you under the impression that the transformer's current rating is the governing factor of ripple current? In other words, do you think the higher the current rating, the higher the ripple current?
    Last edited: Jun 25, 2011
  11. #12


    Nov 30, 2010
  12. MrChips


    Oct 2, 2009
    I think there are a whole bunch of things missing here.

    (1) Firstly, let's begin with the transformer:

    We don't know enough about the transformer. Not all 24V 10A CT transformers are created equal. Many are grossly overrated and rated at no load. The voltage drops under load.

    (2) Are we going to use full-wave bridge or full-wave center-tap?
    You get half the current rating with a bridge or half the voltage with center-tap.

    We will probably have to derate the supply to 10V at say 10A.

    (3) What is the load? At full power the load is 1 ohm.

    (4) The next question you have to ask is how much ripple are you willing to tolerate. From this we can calculate the capacitance required.

    (5) Next you have to calculate the peak current through the diodes.
    You may be surprised to learn the peak current can be more than ten times the load current... 100A!
  13. #12


    Nov 30, 2010
    The question is: "how many caps do I need to survive the ripple current".
    The ripple current depends on the delivered current, and that is derived from the rectifier type and the amp rating of the transformer. Those have been established.
  14. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    So the question about how many of the 3900uF capacitors (the minimum no.) are required in parallel to prevent failure hasn't been answered. Let's propose a fixed load resistance of say 1.5 ohms. Perhaps also assume a center-tapped full-wave rectifier using the non-bridge two diode configuration.
    Last edited: Jun 27, 2011
  15. #12


    Nov 30, 2010
    True. The difficulty is in deriving RMS ripple voltage.
    I have vowed to use that web page in post #11 until I get the math down, but family problems take my time for a few days.

    You are welcome...ANYBODY is welcome to present a working formula
    (pretty please with sugar on it)

    My last 2 answers for the coefficient of "I think" were .288
    That was when I invented my signature line.
    t_n_k likes this.
  16. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    OK - I'll stick my neck out and say for the proposed parameters that perhaps 3 caps in parallel give about 5V ripple and 3.7A rms ripple current per capacitor. So 3 capacitors might be sufficient to prevent failure.
  17. #12


    Nov 30, 2010
    Yebut...5 volts ripple on a 12 volt supply is huge.

    I'm trying to get a "universal" coefficient for practical use.
    A few more hours of math and research might get me an answer.
    If I can't get it from the internet, I can connect circuits and measure temperatures!
    But I have to go to work now.
  18. wayneh


    Sep 9, 2010
    Only if you're on the right track. (no offense intended) I spent a lot of time on modeling the DC output from a standard bridge rectifier + filter cap, as it depends on the (purely resistive) load current. My goal was to use DC measurement to understand the AC input, so I needed this back-calculation to estimate how much the measured DC voltage was affected by the filter cap. We all know this varies with the load. It's way more complicated than most folks would think! But it is doable.

    The problem is that the conduction phase of the rectifier is just at the tip of the sine wave and it's not symmetrical, or even a continuous function. At lower load, the voltage during conduction starts slightly after the sine peak, comes down along the sine curve until conduction stops, and then decays along the RC curve. So solution of voltage versus time requires calculating exactly where the conduction stops (which depends on the DC load and voltage) and then knowing which region you're in - sine versus exponential - throughout the cycle.

    I have an Excel spreadsheet that does all this nicely, including a graph of the voltage profile. I'll share it with anyone that's interested, and I'll even run a few examples if anyone wants to provide the details.
  19. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    Agreed it is a lot of ripple. But that wasn't the OP was asking in the original question. They asked how many caps of a given type in parallel were needed to ensure the individual cap current ratings were within spec.
  20. #12


    Nov 30, 2010
    I started testing that link above.

    They present:
    V ripple p-p = Vpeak time / RC
    Some algebra changes that to:
    C Er F = Ipk

    I made up a circuit with 1uf poly cap on a FWBR with a load of 59,368 ohms on 10.00 Vpeak in order to get 1V p-p ripple and demonstrated that the formula
    radical2 C Er F = I
    which I got from Natinal Voltage regulator Handbook, 1982, page 8-6 section 8.2
    is accurate within 1.4 percent, and I expect that 1.4% is accountable as human error (my inability to measure any closer than that.

    I believe the university page is "busted".
    I invite anyone to duplicate my experiment.
    (It's called, "peer review".)