AC component on DC bias

Discussion in 'General Electronics Chat' started by Buster, Feb 16, 2005.

  1. Buster

    Thread Starter New Member

    Feb 16, 2005
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    Why is there an AC component present in the DC portion of a full wave rectified signal?

    Whats the best way to measure AC "ripple" and why?

    I have 4 DMM's. If I measure the AC ripple (leads across the output), 3 of the meters read around 6VAC and one reads around 25VAC. Why would the one read 25VAC?
     
  2. n9xv

    Senior Member

    Jan 18, 2005
    329
    1
    The AC component is present because the DC voltage following a full wave rectifier is pulsating or rippling DC. The 60-Hz AC sinusoidal waveform entering the bridge rectifier rises from 0-volts to maximum +volts then back to 0-volts and then to maximum -volts then back to 0-volts thus completing one cycle. The voltage leaving the bridge rectifier is 120-Hz "pulsating" DC. 120-Hz because the negative going cycle is converted to a positive going cycle. Both cycles now rise from 0-volts to a maximum +volts and then back to 0-volts. This is called rippling or unfiltered DC. The AC component is this "rippling" of both cycles rising from 0 to +max to 0 to +max etc. Capacitive filtering is used to null out the AC component by maintaining the DC voltage at the maximum +value of the origional AC waveform. The only way to accurately measure AC ripple is with an oscilloscope. The DMM is measuring an AC component riding on a DC component and will give false readings.
     
  3. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Hi,

    Your meter that reads 25 VAC probably has an absolute value circuit instead of a "true RMS" circuit doing the AC measurement. The absolute value circuit is an active rectifier and smoothing folter that responds as well to DC as AC. That meter probably reads about the same if you measure the unrectified output of your transformer.
     
  4. Buster

    Thread Starter New Member

    Feb 16, 2005
    7
    0
    I hear you and understand for the most part but something is not clicking. If you look at a full wave rectified signal on an oscilloscope is a portion of the waveform going negative (a very small portion) and because a portion goes negative you now have both positive and negative swing hence the AC component? If so this is caused by "leakage" in the diode? How is it that an oscilloscope is more accurate than a DMM that reads true rms? By the way, thank you all very much for helping me with this. I sorta of stumbled onto this site by accident but am glad I did.
     
  5. David Bridgen

    Senior Member

    Feb 10, 2005
    278
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    A positively rectified a.c. should have no negative-going portion. If you observe any then it could, as you surmise, be due to diode leakage.

    An analogue 'scope can be read to an accuracy of no more than 2% - several orders of magnitude worse than even a cheap and nasty d.m.m.

    Any discrepancy between 'scope and d.m.m. readings in this instance is due to the d.c. component, as explained by n9xv.

    To obtain a more meaningful reading of ripple with a d.m.m. insert a capacitor in series with one of the test leads. 1uF, or even 0.1uF, in series with a 10M meter will introduce very little error.
    n.b. the capacitor should be a non-polarised one.
     
  6. Buster

    Thread Starter New Member

    Feb 16, 2005
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    OK, so the AC component that shows up on the rectified side is a phenomenon of the pulsating DC. In other words, the pulsating DC causes it to be there...nothing else? The DMM situation..... The one meter is reading 25V because it is looking at both the AC & DC components together....the other 3 meters condition the signal differently somehow and or block the DC portion better so that all it is reading is the AC component? So your saying the 'scope is not as acurate as a DMM??
     
  7. torpedopudding

    Member

    Feb 6, 2005
    28
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    I think verbal descriptions alone are a risky way to describe waveforms! I detect some confusion in the dialog!

    1) First, is the AC input from normal AC powerline voltage, sinewave 50-60 Hz?

    2) Secondly, one will usually AC couple an oscilloscope when measuring ripple. Assuming there is nothing wrong with the oscilloscope, this is a fine way to do it. Although 2% accuracy with an oscilloscope sounds fair, an oscilloscope is the only good way to measure ripple, unless you want to have RMS measurements and have a true RMS meter, AC RMS only. If the meter is not true RMS, the measurement of most ripple waveforms with only a meter will be a poor one because the waveform is not a sinewave. RMS ripple measurements are useful mainly for calculating heat generation in the filter capacitor. If the filter capacitor is large enough to get moderately low ripple, heating of the capacitor will not usually be a problem.

    3) When you AC couple the oscilloscope, you indeed see a negative and positive portion of the waveform. If the oscilloscope display is set to 0 divisions when AC coupled and input is grounded, you can easily add and subtract the DC (meter) and AC (oscilloscope) readings. For example:

    4) Lets say your meter says you have 14 volts DC on the filtered waveform. When you measure the ripple waveform with the oscilloscope AC coupled, you see that the waveform goes (assume) 0.5 volts above 0 and 0.75 volts below. Then:

    4a) If you could have a highly accurate scope that could easily measure 0.1V out of 12V (DC coupled) you would see that the peak voltage is 14+ 0.5 = 14.5.

    4b) Likewise, you would see that you have a valley (minimum) voltage of 14-0.75=13.25.

    4c) So if your capacitor is rated 16V it would be operating within ratings. And if you had a 12V linear regulator after the capacitor, and it has a 1.0V dropout requirement, it would also be within ratings as it has 14 - 0.75 - 12 = 1.25 V available to dropout.

    5) An important point in the above is that any simple averaging circuit will give you the same average. While you want to be careful about AC measurements, any properly working meter should give you the same DC measurement. That DC measurement can be mentally placed at the "0 volt" line of the AC coupled oscilloscope readout assuming that the scope has been zeroed with no input.
     
  8. Buster

    Thread Starter New Member

    Feb 16, 2005
    7
    0
    Confusion.....yeah I think your right. I am still confused. I appreciate everyones patience. Lets forget the meters for a second and back up to my first question. First off, lets assume the AC input is 60hz. That 60hz signal gets fed into a full wave bridge. The output of the bridge is a full wave DC output or a pulsating DC output what ever you want to call it. For now, lets assume there is no filtering of this signal. This is where I am stuck.....The waveform is now 120hz because the converted AC waveform is going from 0 to max back to 0 to max and back to 0 and so on and so on. I get this part of it.

    In order for a waveform to be considered to be AC or alternating, the waveform has to be above 0 (positive) and then cross 0 (goes negative) correct? OK, on the output side of the bridge I have pulsating DC which does not cross 0....once it reaches 0 it goes positive again hence the term unfiltered DC. So, where does this AC portion come in?

    I know this AC component is always there but just don't understand how it is present in the first place if the output is pulsating DC that does not cross 0??

    How does this AC component "ride" on the DC component? Again, I appreciate your help and paitence !!
     
  9. n9xv

    Senior Member

    Jan 18, 2005
    329
    1
    The output of the bridge is rippling DC @ 120-Hz. This 120-Hz waveform is NOT AC but DC. It is periodic in nature because it starts at 0 then rises to a maximum + value then back to 0. This "0" to "0" point represents a period. A period of time that last for 1/120-Hz or 8.3-mS. A waveform does not have to go negative to be periodic. Periodic simply implies that the waveform (through time) comes back to the point where it origionally started, thus replicating itself. Period and time are of course the same thing. The term AC component is derived from the fact that the "ripple" content is rippling because it was once AC. Rippling DC (from the bridge rectifier) is not AC by any means. It no longer "alternates" from - to + to - to + etc. It is "periodic" because it rises from 0 to a +maximum then back to 0 etc. Remember that the typical DMM is calbrated to 60-Hz for AC voltage/current readings which is another reason why you should not rely on it for "AC ripple" measurements. A better approach would be to measure the peak-to-peak ripple on the O'scope and do the mathematical conversions to RMS etc.

    Hope this helps and did'nt cause you to blow a head gasket! :D
     
  10. Buster

    Thread Starter New Member

    Feb 16, 2005
    7
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    n9xv,

    Blown gasket? Lets try total melt down! Again, lets back up and try to forget I asked anything about how to measure the AC component. Let me try to explain what I know.

    I know your thinking this should be good ! haha

    With a bridge rectifier circuit, four diodes are used. One output terminal of the bridge network is a common ground for the return of the load current. The other output terminal is connected to the load.2 of the diodes conduct during the positive period of the AC sine wave (input) and 2 diodes conduct during the negative period of the AC sine wave (input). In other words...the pulsating DC has positive half-cycles on both alternations of the AC input voltage. I understand that this output contains an average DC value and an AC portion called "ripple".

    I also understand that this "ripple" is an unwanted portion of the DC output and typically it is filtered using capacitors, etc. Unwanted in that it can cause problems with circuits that have digital inputs, etc.

    I also know that if I measure across the output of this bridge rectifier with an oscilloscope (AC COUPLED) I will "see" an alternating waveform. It is at a frequency of approximately 120hz and has a peak-topeak voltage of approximately 25 volts, i.e. the waveform starts at 0...goes to max +...come back to 0...goes to max - and comes back to 0.

    So, if I have rectified the input sine wave (input to bridge) and have no leakage in my diodes (lets assume) I end up with a full wave rectified output (pulsating DC)with no filtering and an AC Component riding on this DC output. I just can't get it in this thick skull how there can be an AC signal when I rectified the original signal. You say the AC component is not AC but I can see the alternating waveform on the scope???

    Oh my head hurts. hahaha This is all going to hit me like a ton of bricks sooner or later.
     
  11. n9xv

    Senior Member

    Jan 18, 2005
    329
    1
    Try DC coupling with the scope. The AC coupling (which is actually capacitive coupling) is whats causing you all the greif!

    The output of the bridge is DC not AC. Remember that after the bridge rectifier it is rectified. The rippling DC waveform does not go negative, just positive. Your being "fooled" by the AC coupling method. Try it again with DC coupling.

    AC bad :eek: , DC good :D

    Dont go postal yet! :lol:
     
  12. David Bridgen

    Senior Member

    Feb 10, 2005
    278
    0
    Hi Buster,

    Let's try to do it with pictures.

    The first waveform is that of a 20Vpp a.c. sinewave.

    The second is the raw d.c. after full-wave rectification (which doesn't necessarily, by the way, have to be a bridge.)

    The third part is both the d.c. and a.c. components which are added together to produce the second waveform.


    If you short a 'scope's input (switch it to "ground") the vertical position of the trace represents zero volts. Now switch it to a.c. and look at a raw d.c. such as that in the second waveform. It will be displayed on the 'scope as I have indicated in the third part of the diagram, with both positive and negative-going portions.

    Does that make a bit more sense?
     
  13. Buster

    Thread Starter New Member

    Feb 16, 2005
    7
    0
    Didn't help at all ?! It just simply brings me back to the question I asked in the begining. The 3rd waveform you show is exactlly what I have been looking at all along and is what prompted the question in the first place. It appears to be an AC waveform, although it is not a pure sine wave, in that it goes above and below 0. Where does it come from? I see it because I AC couple the full wave rectified signal and whala...there it is. Yet there is no AC because it has been rectified right n9xv? So why do they say there is an AC component present? Is clear as mud. My head doesn't hurt anymore though guys. It's simply numb !! :blink: :D
     
  14. n9xv

    Senior Member

    Jan 18, 2005
    329
    1
    By switching the scopes input to AC to look at a raw DC waveform you are introducing an error directly related to capacitive coupling. The second waveform in your example is indeed correct in showing the fullwave rectified waveform as pulsating DC with peaks reaching +10-volts. The DC value of this "unfiltered" pulsating DC is equal to .636 X Peak voltage or .636 X 10-volts = 6.36 volts. The "filtered" value of DC voltage is equal to the "Peak" value of the DC voltage or 10-volts. Thats what filtering does - it filters out the rippling or pulsating of the "raw" DC from the rectifier and attempts to maintain it at that Peak level.

    The third waveform is in error due to the capacitive effects of AC coupling. There is absolutely no AC present in a fullwave rectified waveform following the bridge rectifier. There is the rippling or pulsating component before filtering. But this is NOT an AC component. It is a periodic DC component. If you look (with a scope) at the waveform on the output of a simple capacitively filtered DC power supply you will see nothing but a flat line at the rated DC voltage of the power supply. If you then try to draw an excessive amount of current from this supply, the output voltage will drop somewhat and you will begin to see the ripple or pulsating component of the DC voltage. You only see the ripple because the demand for current is too great and the capacitive filter can not keep up with this high current demand and therefore can not maintain the rated DC voltage at the Peak level as provided by the fullwave rectifier. Keep in mind that we are talking about a simple supply without all the fancy regulation & current managing circuits etc. My understanding is that buster is dealing with a basic power supply circuit.

    >>>>>>>> Try it again with DC or direct coupling guys! <<<<<<<<

    Buster, take 4 ibuprofen and call me in the morning :lol:
     
  15. David Bridgen

    Senior Member

    Feb 10, 2005
    278
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    No. It is the r.m.s. value, 0.707 x peak.


    Yes, but we're not dealing with the filtered value.


    As I mentioned, it isn't necessarily a bridge.

    The raw d.c. is, mathematically, the algebraic sum of the two parts of the third waveform.

    This is true of any d.c. which isn't a "straight line."


    The rest of it, while correct, is irrelevant to the discussion.
     
  16. n9xv

    Senior Member

    Jan 18, 2005
    329
    1
    RMS and DC are two different values. The RMS value of the 20-volt Peak-to-Peak example is 7.07-volts. But the average or DC value of this "unfiltered" pulsating DC is equal to .636 X Peak voltage or .636 X 10-volts = 6.36 volts.

    DC (Average value) = .636 X Peak voltage [UNFILTERED]
    DC (Average value) = .9 X RMS [UNFILTERED]

    RMS value = .707 X Peak voltage
    RMS value = Peak-to-Peak voltage / 2.828

    How is a bridge rectifier not necessarily a bridge. It is a bridge circuit like any other bridge. Resistance bridge, capacitance bridge etc.

    The raw "unfiltered" DC (from the output of the fullwave bridge) is simply .636 X Peak voltage or .9 X RMS AC voltage being input to the bridge.

    My statement about the "filtered" DC being equal to the Peak voltage is important in that it hopefully gives buster the complete picture from AC input to the bridge to filtered DC output of the bridge.

    >>>>>>>> DC coupling guys, I'am tellin ya! <<<<<<<<
     
  17. pebe

    AAC Fanatic!

    Oct 11, 2004
    628
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    I think you are splitting hairs. For as long as I can remember it has been the convention to refer to the ripple as the 'AC component' of a waveform. 'Component' implies it's part of another signal. You can consider the waveform as an AC signal sat on a DC pedestal and alternating about that DC level.
     
  18. David Bridgen

    Senior Member

    Feb 10, 2005
    278
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    Yes, I already established that.

    I say again. No. It is the r.m.s. value of the peak, 7.07V

    Yes, I too remember those from my college days. Well done.

    The original post mentioned full-wave rectification. It didn't say if this took place in a bridge or in two diodes across a centre-tapped transformer. You are jumping to conclusions. The blind skunk did that and look where it got him.

    No. It is the r.m.s value.
     
  19. n9xv

    Senior Member

    Jan 18, 2005
    329
    1
    DC (Average value) = .636 X Peak voltage [unfiltered]
    DC (Average value) = .9 X RMS [unfiltered]
    RMS value = .707 X Peak voltage
    RMS value = Peak-to-Peak voltage / 2.828


    RMS value = .707 X Peak voltage

    This equation is stating the fact that the calculated RMS (AC) value will produce the same heating effects as the calculated DC/average value.

    EXAMPLE: The wall outlet in you home is rated 120-VAC (RMS). This is 340-Volts Peak-to-Peak. 340-Volts Peak-to-Peak / 2 = 170-Volts Peak. The DC/average value is .636 X (Peak voltage of 170) = 110-Volts. The RMS value is an "AC" value not a DC value. But this 120-VAC RMS value will produce the same heating effects as the DC/average value.

    The DC/average value of 110-Volts (calculated) will produce the same heating effects as the 120-VAC (actual voltage).

    Take a 12.6-VAC RMS secondary of a transformer and connect it to a simple bridge rectifier and this is what you would get:

    12.6-VAC RMS = 36-Volts Peak-to-Peak.
    36-Volts Peak-to-Peak / 2 = 18-Volts Peak.

    18-Volts Peak is what you will see on the O'scope as measured on the DC output terminals of the bridge. Now, you can calculate the RMS value of .707 X 18-Volts and get 13-Volts. If however, you "measure" the DC voltage on the DC output terminals of the bridge with a DMM you will see a DC/average voltage (unfiltered) of .636 X 18-Volts = 11-Volts. The RMS value and DC/average value are two different values.


    If you remember these from your college days then why is the formula for unfiltered DC/average value such a problem for you. Could it be that all my engineering books and design references have a typo? If you can show me via printed text (with reference to author etc.) where this formula is wrong then please do so. I will then render my engineering degree useless and mail it to you with a written appology for driving you crazy on this point!
     
  20. David Bridgen

    Senior Member

    Feb 10, 2005
    278
    0
    n9xv (Kevin?)

    I apologize unreservedly. I was wrong. You are correct.

    Regards.
     
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