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Full Wave Bridge Rectifier (Odd Output Voltage)

Discussion in 'General Electronics Discussion' started by spongebillybob, Aug 3, 2010.

  1. spongebillybob

    spongebillybob

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    Jul 13, 2010
    Hello Folks,

    I am teaching myself how to trouble shoot a batter charge card for a piece of equipment at work. Right now I am trying to get a baseline on the card...ie: voltages, resistances etc etc. Problem I am having is my output voltage of the rectifier. The rectifier is being fed by a 48 volt secondary signal transformer that is center tapped supplying 24 vac to the rectifier. This I have verified and it makes sence to me. Output is what is confusing me. I am getting 37 vdc. I would be expecting more or less around 14 volts (half of the ac wave form). I am not sure if I am reading/operating the meter incorrectly?
     
  2. rob_croxford

    rob_croxford

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    Aug 3, 2010
    You should be getting arround 12V DC. Where are you measuring?
     
  3. Mitchekj

    Mitchekj

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    Jan 24, 2010
    Edit: A couple more observations - There is probably a large bulk cap (or two) right after the bridge. Are you measuring this DC output voltage with no load on the rectifier? That would give you 37Vdc easily, as there is not going to be any ripple after the bulk cap if it's not asked to supply any load current. That would give you almost exactly the peak voltage of the transformer output. What is the AC voltage coming out of the transformer measuring? I'd guess it's around 26-26Vac, yeah? What is the power the load draws, btw?

    37Vdc sounds about right.

    I'm not clear how they've got the center tapped xfmr wired to the bridge. Can you sketch up the connections? I'm assuming you have 24Vac going into the bridge, yes?

    24Vac will have a peak voltage of 33.9V. 24Vac is actually an RMS reading, which is ~70,7% of the peak voltage of the sine wave. When you full-wave bridge rectify that, and supply a sufficiently large bulk capacitance to "smooth" the ripple, you end up with nearly the peak voltage, now DC, which would be somewhere in the 30Vdc region, depending on the exact AC output from the transformer, the size of the bulk capacitor, the load current, the voltage drop of the bridge diodes, etc.

    At no time should you get half the ac wave form from a full-wave bridge. It's full wave rectification, afterall. Even without a bulk capacitor you'd still end up with something like 21V avg, which would consist of all positive-going waves now at 120Hz. (or 100Hz, depending on line frequency.)
     
    Last edited: Aug 3, 2010
  4. spongebillybob

    spongebillybob

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    Jul 13, 2010
    Seems I might need to get back to the basics...

    First off, thank you both for your quick response. I appreciate the assistance. It seems I need to go back to the basics again to relearn a bit.

    I was incorrect in some of the terminology and will try to clear it up:

    120 Vac on primary of transformer => approx 24 Vac on secondary. (the transformer is not a center tap, it is wired to where you can either have one secondary winding or split it to have two seperate windings...If I am not mistaken this allows you to have either twice the voltage or twice the current capacity..correct?)

    24 -27 Vac to rectifier. Output of rectifier approx 36-37 Vdc. 220 uF Cap for Bulk. Output after bluk capacitor is then used for an unregulated power supply and also supplies an LM317T voltage regulator chip that maintains an output of 12 Vdc.

    Everything you have said is exactly what I have, now its just me trying to understand it.
    First off:

    I thought the output of a full wave bridge rectifier was half the total + and - amplitude of the input AC (I incorrectly described it as the wave form earlier). Wouldn't that mean the output of the rectifier should be somewhere near 12 Vdc since the input is at 24Vac? (here I think I am misunderstanding a basic fundamental property)

    I don't quite understand what you mean about there not being any "ripple after the bulk cap" under no load conditions. I thought once the cap was charged to max it then acts like an open to DC. Or maybe that is what you ment, once charged and there is no load on the cap and therefor no smoothing of the wave, the voltage I should read would be peak output voltage of whatever the transformer was putting to the rectfier.

    As I understand it now, when putting in a bulk capacitor to smooth out DC you also increase the voltage output from the RMS of what the transformer is putting out, to closer to peak voltage of same transformer?

    I hope I am showing that a few brain cells are starting to fire =). If not, I got a lot of restudying to do.
     
  5. shrtrnd

    shrtrnd

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    453
    Jan 15, 2010
    Mitchekj pretty-much gave you the scoop. If you have no load on the transformer, you're going to read higher voltage than when there IS a load. The caps after the transformer are to help smooth out the AC (ripple) from your transformer sine wave, You're confusing the 'effect' of a capacitor on an AC circuit, ('effective dc') with the DC you want out of your LM317, which IS a Voltage Regulator. The transformer output is raw AC, the large caps work with the voltage rectifier circuit to convert the voltage to a cleaner DC voltage, ... and the LM317T makes sure the dc is regulated to run your circuit.
     
  6. Mitchekj

    Mitchekj

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    Jan 24, 2010
    See photos: I threw some spare parts together to try to demonstrate how it all works...

    A 24Vac from the transformer is an RMS reading of it's peak voltages... which is about +34Vpk and -34Vpk, for a peak-to-peak of 68Vpp. That's probably where you're getting confused about half the input. It's still a 24Vac voltage, just half is positive and half is negative. At no time will you get a potential difference of both half-cycles. (In this particular example, anyhow.)

    Photo 1 - Ac input. - A 12Vac input (I used a variac to take 120Vac down to 12Vac) and this is the waveform on the scope. It is positive half the cycle, and negative the other half. RMS reading is 12Vac. Notice that the peak voltage is ~18Vpk. Peak to peak is ~35.2Vpp. The little number 1 with the arrow (on the left of the scope) is the zero voltage point. The zero reference (ground) is always in the middle of this AC wave, hence positive and negative. We never end up with a 24Vac potential from ground at any point in time.

    Photo 2 - Bridge output, no cap. - I connected a bridge recitifier to the 12Vac, and applied a load which drew ~1A. Notice that now all the negative cycles have become positive. Peak voltage is still ~18Vpk. We aren't missing any part of the waveform, all we've done is take all the negative cycles and rectified them to be positive. That's all the bridge rectifier does. We have also effectively doubled the frequency from 60Hz to 120Hz.

    Photo 3 - Bridge output, with cap - I connected a 680uF cap (what I had handy on the bench) and now you see the minimum voltage has started to come up. Instead of dropping to 0V each cycle, now we're up to ~12V or so minimum. This increases our voltmeter reading to ~15Vdc. Notice the 4.8V ripple... that's about 26% of the peak, which means I would need to use a larger cap to get a decent DC voltage w/ low ripple, (I'd probably try a 1000uF cap for this 1A load.) That can all be calculated though. The more current (power) the load draws, the lower this minimum voltage will sag (with a constant capacitance.) If we were to increase the capacitance to 1000uF, for example, we'd raise our output DC voltage, since the minimum voltage will rise. I'd expect it'd go from 15Vdc to somewhere between 16 and 17Vdc.

    Photo 4 - Bridge output, with cap, no load - Now I've disconnected the load. Now we have a DC voltage with near zero ripple. All we've done here is charge the cap up to the peak voltage level... since there is no load to discharge the cap, it just remains at 18Vdc, and each cycle will top off what little it loses through leakage. We will measure 18Vdc with a voltmeter, even though the input is 12Vac.

    If we were to use an infinitely large bulk cap, we would also get 18Vdc with almost zero ripple, no matter how much power the load was drawing.
     

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  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    But it would also require an infinite amount of time to charge ;)
     
  8. Mitchekj

    Mitchekj

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    Jan 24, 2010
    Haha, truth!
     
  9. spongebillybob

    spongebillybob

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    Jul 13, 2010
    Thank you for the effort and patience on explaining this to me. Yes, I do see that I have a few principles mixed up but this definately clears up alot. Now I just need to sit down and go over it. Again, thank you very much.
     
  10. davenn

    davenn Moderator

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    Sep 5, 2009
    and whilst talking about power supplies and DC smoothing and to follow on from what Mitchekj said earlier .... 1000uF for a 1 Amp load

    something that was taught to me many years ago as a good "Rule of Thumb"

    a minimum of 1000uF per 1A of load for good smoothing so a 10A load = at least 10,000uF of capacitance 15A 15,000uF etc etc

    Cheers
    Dave
     
  11. Sateesh kumar

    Sateesh kumar

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    Jul 16, 2012

    Hello dear, the rms output dc voltage of a bridge or centre tap rectifier is ACin x √3. Measuring 37vdc as output of fullwave rectifier is correct.
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I think you meant √2 :)
     
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