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Ripple increase on higher load: at rectified output: 18Vac to 25.5Vdc

  1. Apr 24, 2008 #1
    Hello All,
    glad to be a part of this elite community,
    i have been following the forum threads for quite few days now, i am really interested to hear from Mr.Berkeman.

    here is my problem:
    we are facing a problem at the rectified output of our power supply, we have used a bridge rectifier to convert 18VAC to 24VDC. the board derives 18VAC and 30Vac, the latter one for battery charger part.
    the issue is that the ripple at the rectified output is shooting up to 1.5V as load increases.
    the 24VDC is fed onto a 12V linear regulator(LM350) and 15V linear regulator(KA7815).
    any load connected results in increase of the ripple.
    we tried LC And Phi filters for solving the issue, but of no use.
    the transformer being used can source up to 2A at each Output (18VAC and30VAC).
    the total load of the board may consume little over 2.5A, but this is supplied by the regulators.
    Please throw some light on this.
    P.S: Berkeman has been emphasising on power supply safety, and that has been taken care of.

    Thanks in advance.
     
  2. jcsd
  3. Apr 24, 2008 #2
    BTW the current RC filter at the rectified output is:
    10K and 4700uF,63V
     
  4. Apr 24, 2008 #3

    berkeman

    User Avatar

    Staff: Mentor

    The ripple is just going to come from the times that the input bridge is not conducting (because the input AC voltage waveform is below the rectified voltage). All you can do to minimize the ripple at a given Iout load, is to keep increasing the size of the storage capacitor.

    Except, there is a better way, if you are willing to do a bit more circuit design. The problem with the simple input rectifier architecture is that it has a poor "power factor". That is, current is not drawn from the AC power input waveform for the whole AC cycle -- current only flows into the storage cap through the input bridge during the peaks of the AC voltage waveform. To have a "good" power factor (around 1.0), the input current would need to track the input voltage, like you would get with a purely resistive load.

    There are ways to design a 2-stage switching power supply (a boost followed by a buck, generally) that makes the input current drawn from the line track the voltage waveform much better, and fairly high power factors can be obtained using these approaches. Here's a basic wikipedia.org article on PFC:

    http://en.wikipedia.org/wiki/Power_factor_correction

    And here's a great paper by ON Semiconductor with more detailed information:

    http://www.onsemi.com/pub_link/Collateral/HBD853-D.PDF

    If you don't do active PFC, you are stuck with a larger storage capacitor after the bridge rectifier, and you are not efficiently drawing power from the AC source....
     
  5. Apr 25, 2008 #4
    Cheers,
    I am going through the material, will get back when I get a query.
    BTW the ripple is expected to increase when the load is inductive. but the Bridge rectifier is itself a non linear device. and so the ripple increases even for a resistive load, so reactive components alone cannot solve the problem as they cant eliminate the harmonics of the reactive components. The best solution is to use a Buck-boost or a Boost buck converter to achieve the correlation in phase. Is my interpretation right?
    If so can you suggest such a device where input is 25.5V and output 24V?
    or can u suggest a Filter structure for it( meanwhile i will go thru Onsemi Paper).
    Thanks in advance
     
  6. Apr 29, 2008 #5
    I don't think the regulator supply current by itself. Current is ultimately supplied by the transformer. If the transformer can only source 2A, the regulator won't be able deliver 2.5A. Your choices are a more capable transformer and/or more capacitance. Since you are trying to source more current than the transformer is rated at, voltage will drop and capacitors will starve. Are you sure *any* load creates high ripple? How about if you connect say a 10kohm resistor.
     
  7. Apr 29, 2008 #6
    are you sure a regulator cant supply an output of 5V@2A, by drawing 12V@200mA?
    We already have the idea of replacing the transformer, but we just want to research the characteristics of the current design. the current drawn from 24V is around 1.5A. where as the load consumes 2.5A.
    and say if we replace the 2A transformer by a 5A one, can we expect a decrease in ripple at the rectifier output? is the non linearity of the bridge rectifier countered?
     
  8. Apr 30, 2008 #7

    NoTime

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    Science Advisor
    Homework Helper

    Yes, I'm sure. You can't get something for nothing.
    At 100% efficiency you could get 5V@0.5A

    12V@200mA = 2.5w
    5V@2A = 10w
     
  9. Apr 30, 2008 #8
    The simple thing would be to swap around the terminals.
    Use the 30VAC for the work and the 18VAC for the battery charging.
    You can up the voltage to match a truck battery voltage of 24VDC(is that what it is?) by using a dc to dc convertor.
    You wouldnt want to fast charge your battery regularly anyway as it shortens the life....
    The 18VAC should be more than enough to charge the battery.

    If you are reading 24VDC on the bridge terminals from the 18VAC transformer circuit, then its likely you are using a digital ohmeter which has a high input resistance and is reading the open circuit voltage which tends to be higher than the rating...
    The peak for the 18VAC should be approx 25volts....
    If you design based on that, you will have ripples that you cannot get rid of no matter how much caps you use....
    This estimator page will return your ripple values based on the factors at hand..
    http://tangentsoft.net/elec/ps-est.html#python
     
    Last edited: Apr 30, 2008
  10. May 1, 2008 #9
    thats exactly what i have been thinking. i tried that out, and ripples are less, BTW the 43VDC out of the BR for 30VAC is being converted to 24Vdc for Battery charging.
    And the output across BR for 18VAC is 25.5V.
    you said "fast Charging" the battery, what exactly does that mean?
     
  11. May 2, 2008 #10
    Thats when you just dump the full current on the battery for an extended period.
    Normaly, you would like to trickle charge your cells so that they regain their state gradually.
    Fast charging just juices up the battery and lowers its service life.
    An intelligent charger will trickle feed while watching things like temperature and current changes.
    Find a prebuilt circuit battery charger on the net and modify it to suit your charger....should be easy enough....
    dont worry about the voltage drops across the bridge as a regulator should compesate easily enough for the difference needed at dc.....especially with lower currents.

    The voltages you are measuring reflects the peak voltages of the ac...
    30 x sqrt 2 =42.4
    18 x sqrt 2 = 25.5
    to measure the voltages properly, a RMS meter setting should be used.

    all in all.....a nice little challenge...so enjoy trying!
    g'luck.
     
    Last edited: May 2, 2008
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