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In Reply to: RE: Becoming a believer in dc for filaments ...... posted by vinnie2 on October 27, 2016 at 05:30:25
as have been saying and see what happens. Process of elimination is what you need to do. I hope you are writing everything down you have done to cure this issue so you won't be chasing your tail after awhile. It can get confusing and hard to remember everything. The data will help steer you in the correct direction as to what is working and not.
Make sure to use shielded interconnects and internal input wiring. Also did you ever check your AC line for DC on them? That will have to do more with a physical buzz but it's still annoying. I recently put my VTA-120 on a AVA "Humdinger" and that eliminated the buzz I was starting to get on the Variac and Power transformer of the amp. I have 0.6vdc riding on the mains so the humdinger (I need to make my own now) took care of it.
hang in there!
Follow Ups:
"I have 0.6vdc riding on the mains"
Just wondering, how did you measure that?
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Might be time to call the 'lectrician.
he's gonna come out and check it.
Probably a good idea. My question though is how you measured it?
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Since the power distribution grid is transformer isolated in many layers from generator to home,there is no way to know what DC offset there may be in the various parts along the way.The only way you can eliminate it is to have your own transformer coming into your home but then you have your own issues to deal with creating standing DC on the lines.
Being it is .6v,this can be caused by running things in his house like a coffee pot,or any appliance for that matter.Notice that they have a silicon diode on the mains and it takes .6v to turn those on.There is something you can buy or build to eliminate that small amount of DC.I think there is a fix in the 1977 ARRL handbook covering this very issue.
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken
"Being it is .6v,this can be caused by running things in his house like a coffee pot,or any appliance for that matter.Notice that they have a silicon diode on the mains and it takes .6v to turn those on."
That couldn't possibly create a DC differential on the line. The impedance of the line is much too low for a diode in series with tens of KOhms to have any effect.
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My experience is that a "diode" in certain appliances does indeed put DC on the line. As an example, I have an old 1.2KW heat gun with high and low heat settings (effectively 12 ohms on high). The low setting puts a half wave rectifier in series with everything. Any energized toroid on the same circuit as the heat gun buzzes loudly when it's on low and an old school VOM set to the appropriate DC range reliably indicates a DC voltage value. In the case of my heat gun and 15A home circuit, it's about 2-3VDC.Not sure if the rectifier type range control is any longer "legal" but there are a ton of high current draw appliances using it that are still out there.
Edits: 10/28/16
"Any energized toroid on the same circuit as the heat gun buzzes loudly when it's on low and an old school VOM set to the appropriate DC range reliably indicates a DC voltage value."
I'm not sure that's evidence of DC. A distorted waveform will create those symptoms and won't necessarily be snubbed by the low DCR of the mains transformer and wiring.
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The distorted waveform is the result of a DC offset on the line. The DC resistance of the toroid primary is very low (~1ohm) so there is a significant DC current in it even with low DC voltage thus causing saturation. EI transformers on the same circuit are nowhere near as susceptible to this saturation. Plus, the main point is DC offset caused by loads with half wave rectifiers in them and the meter verifys the phenomenon to my satisfaction.
A DC meter will deflect whenever it encounters an asymmetrical (distorted sine) waveform. DC isn't required.
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An asymmetric waveform (distorted sine wave) causing an indication on an average reading DC meter contains a DC component!
No, it just fools the meter. Does a tube amp output DC from its OPT at 5% even order distortion?
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"No, it just fools the meter. Does a tube amp output DC from its OPT at 5% even order distortion?"Good point.
I suppose one could define the DC component on an AC signal as what one would measure with a DC voltmeter if a resistance R in series with a capacitor C is connected across the signal, and the voltage across C is measured, in the case where C is sufficiently large that its impedance at the AC signal frequency is negligible in comparison to R.
In you example with the signal on the secondary of the OPT in an amplifier with large second-harmonic distortion, I think one would certainly measure zero DC with this set-up.
And I would tend to agree that simply connecting an analogue DC voltmeter across the predominantly AC signal would quite likely be too susceptible to idiosyncracies of the mechanical dynamics of the meter movement when fed with a predominantly AC signal. Besides, I would certainly no want to risk my own meter on trying to pick up a 0.6V DC component "riding on the mains" AC. One would have to set the meter to a fairly low DC scale, like 10V fsd or so, while feeding 120V AC into it. Definitely something to do with someone else's meter and not one's own!
So while I don't doubt that the AC mains can certainly have a DC component on it if some asymmetrical load like a reduced-power hair-dryer running through a half-wave rectifier is connected, I would agree with you that it could well be too simplistic to claim a quantitatively accurate figure for the DC component simply by connecting an analogue DC voltmeter.
Chris
Edits: 10/29/16 10/29/16
Chris: With due respect, unless you've tried this, you're engaging in conjecture here regarding the use of a conventional meter to detect the presence of DC on the mains. I've performed the "experiment" numerous times with the same result every time: an unmistakable unidirectional deflection of the meter pointer when the half wave rectifier is on the line and no deflection whatsoever when it's not. I make no claim as to accuracy of the magnitude of the DC but it's presence is reliably and repeatably detectable using the method described. Under these conditions I can think of no other cause of the pointer deflection other than DC on the line.
Note that I used a high voltage DC range appropriate to my mains voltage and made no attempt to measure the DC on say a 3VDC range for improved accuracy as this would surely damage the meter. Thus deflection was small and well outside the meter's accuracy capabilities. But there was always an unmistakable deflection in the range of 1-2VDC. This far, the meter has shown no ill effects of having been subjected to AC on a DC voltage range.
Also note that the main point of this exercise is to determine whether or not a specific type of "load" can cause DC to appear of the line. The details of detecting the DC are secondary.
"Chris: With due respect, unless you've tried this, you're engaging in conjecture here regarding the use of a conventional meter to detect the presence of DC on the mains. I've performed the "experiment" numerous times with the same result every time: an unmistakable unidirectional deflection of the meter pointer when the half wave rectifier is on the line and no deflection whatsoever when it's not. I make no claim as to accuracy of the magnitude of the DC"I wasn't suggesting that there would be no deflection of the meter needle resulting from the presence of DC superimposed on the line; I was questioning the accuracy of the measurement. As I said, "it could well be too simplistic to claim a quantitatively accurate figure for the DC component simply by connecting an analogue DC voltmeter."
The OP had reported a measurement of rather high precision, namely 0.6V DC superimposed on the 120V AC mains. [Edit: Sorry, Cougar, not OP.] It does seem to me that TK's original question of how that was measured is a very pertinent one.
I think the suggestion I made, of connecting a resistor in series with a suitable capacitor, and measuring the DC across the capacitor, would give a much more quantitatively reliable measurement. The idea being that the capacitor would filter out almost all the 60Hz AC, so that one is not trying to measure the rather small DC voltage against a backdrop of a superimposed AC signal whose amplitude is about 280 times larger (in the OP's example).
By using an RC filter as I was proposing, one no longer has to speculate about how a mechanical system like an analogue meter, or an electronic system like a DVM, might be handling the display of a small DC component superimposed on a vastly greater AC background. Instead, one has a simple system that filters out the AC background to any desired, and precisely calculable, degree of accuracy.
Chris
Edits: 10/30/16
I appreciate your thoughtful response. All I was attempting was verification that a household appliance could put DC on the line. At this point we have no idea how the OP arrived at the value reported. OTOH, obtaining an accurate measure of magnitude of DC offset is routine and easy with the right equipment. A friend has a Drantz circuit analyzer. I'll see if I can borrow it. If so I'll report back.EDIT: Out of curiosity I tried your RC filter method. I used a 100uF/370VAC oil filled motor run for the cap and a 6K5/5W WW resistor. With this setup, ~0.5VAC and 0.0VDC was meas at the cap terminals under standby conditions with both a VOM and DVM. With the heat gun on the circuit ~2.5VDC was meas at the cap terminals with both VOM and DVM. This value is of the same order of magnitude as the meas made using the VOM alone on the appropriate DC range.
Edits: 10/30/16
"EDIT: Out of curiosity I tried your RC filter method. I used a 100uF/370VAC oil filled motor run for the cap and a 6K5/5W WW resistor. With this setup, ~0.5VAC and 0.0VDC was meas at the cap terminals under standby conditions with both a VOM and DVM. With the heat gun on the circuit ~2.5VDC was meas at the cap terminals with both VOM and DVM."I was just trying similar experiments. I used a 10K resistor and a 220uF capacitor. I got no DC under normal conditions. Plugging in a 1200W hairdryer in its "half" power mode, where there is a rectifier in series, caused about 0.14V DC on the capacitor, if I plugged the dryer into the other outlet at the same wall socket. The sign of the DC voltage reversed, as expected, if I reversed the polarity of the hair-dryer plug. The 0.14V dropped to about 0.07V if I plugged the hair-dryer into a socket located in a different part of the house. The 0.14V increased to about 0.8V if I plugged the hair-dryer into an outlet on the same extension cord as the one where I was taking the measurements. All results in line with reasonable expectations.
I also looked at the voltage across the capacitor on a scope. The residual 140mV or so AC on the capacitor indeed displaced by exactly the same DC amount as indicated by the meter, when turning on the hair-dryer
I also tried with two different DVMs set to measure DC, across the mains supply. One of them gave a reasonable approximation to the truth; namely 0V going to the various voltages described above. The other DVM (Tenma) was completely inaccurate. It read about -2.6V DC across the mains just in normal conditions (no hair dryer). And it read -2.6V regardless of which way around I connected it to the mains. Obviously that particular DVM would be of no use at all for directly measuring DC on the line.
It seems to me that if the 0.6V DC reported by Cougar was happening all the time, as opposed to just exceptionally when someone was using a hair dryer or something like that, then it would be something crying out for further investigation, since it would suggest a rather substantial and continuous asymmetric loading on the mains. But given my experience with the grossly erroneous reading given by the Tenma DVM, I think speculation on his 0.6V DC offset is premature until one knows how his measurement was performed.
Edit: A quick and simple first check that Cougar could perform, assuming he just connected a meter set to read DC voltage across the mains, would be to reverse the polarity of the meter connection to the mains. If the apparent DC offset indicated by the meter were in fact purely an artifact of the meter rather than a true DC offset, then the indicated sign of the offset would not change after reversing the meter leads. (Just like with my Tenma meter.) On the other hand, if the sign of the indicated offset did reverse, then it would suggest that at least qualitatively, if not quantitatively, the existence of the DC offset is genuine.
Chris
Edits: 10/30/16
"I used a 10K resistor and a 220uF capacitor. "
That's not valid. An integrator will modify the zero-crossing point to coincide with the asymmetrical average. In effect, that's what a mechanical meter does too. Just to be clear, I'm not saying that a high current diode-fed load won't produce some offset as a function of the current and mains impedance. However, 0.6V DC is much more than I would expect, so I'm questioning the method of measurement. It doesn't seem to me that this can be done with a meter. Only a DC-coupled scope would allow one to observe the AC and DC deltas between loaded and unloaded conditions.
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"That's not valid. An integrator will modify the zero-crossing point to coincide with the asymmetrical average. In effect, that's what a mechanical meter does too."
Suppose we assume, at least at first, that the task is to measure the DC offset voltage of a single-frequency sinusoidal wave (the mains frequency). So we could suppose the waveform coming out of the mains socket is
V(t) = V0 + V1 Sin[omega t]
V1 is the amplitude of the mains voltage, so it is about 170 V.
V0 is the constant DC offset that we wish to measure, and which was alleged to be 0.6V in the earlier posting.
Setting up the differential equation for the system with the RC filter, one then finds that the solution is that the voltage U(t) across the capacitor is given by
U(t) = V0 + U1 Sin[omega t + alpha] + k exp[-t/(RC)],
with U1= V1/Sqrt[1+ omega^2 R^2 C^2) and Tan[alpha]= -omega R C.
k is a constant of integration, but it is irrelevant in the large-t limit once the system has settled down.
The key point is that the DC offset across C is exactly the same voltage V0 as the DC offset in the original signal V(t).
The merit of the RC filter is that on the mains itself the AC "ripple" amplitude V1 is colossal compared with the small DC offset V0. By contrast, across the capacitor the AC ripple amplitude U1 can be made as small as desired by making the product omega*R*C suitably large in comparison to 1, while the DC offset voltage is the same quantity V0, and it can now be easily and reliably measured with a DC voltmeter.
One could have a separate and more complicated discussion about asymmetric (distorted) waveforms, but for the purposes of the measurement of a DC offset on the mains, I think the above is sufficient.
Chris
OK, here's the SPICE simulation...
A sinewave:
The same sinewave, only now a portion of the positive-going half has been clipped:
And here are the two traces overlaid to demonstrate that they are identical, other than the asymmetry introduced into the second trace. This is how they would look on a scope set for DC coupling. Note that there is no DC offset introduced into the clipped waveform. The unclipped, negative-going halves coincide precisely with those of the unclipped waveform:
Finally, here is the result of feeding the clipped waveform into the integrator described earlier consisting of a 10K resistor and 220 uF cap. Energy contained in the asymmetrical waveform has been averaged, creating a DC offset where none previously existed. This is how the DC voltmeter is fooled:
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If you do a .four command for the waveform it will call out a DC component. This is also referred to as the Zeroth harmonic. It bothers me that even harmonics have a DC component and I still haven't gotten my head around what it really means but I trust the math doesn't lie.
dave
"If you do a .four command for the waveform it will call out a DC component. This is also referred to as the Zeroth harmonic."
Yes, that would fit with what I was saying, I think. All the non-zero Fourier modes time-average to zero, while the zeroth harmonic (the pure constant) time-averages to itself.
Chris
"Energy contained in the asymmetrical waveform has been averaged, creating a DC offset where none previously existed."Here, I think we are reaching the nub of the argument. Before discussing how to measure the DC offset, we need to reach an agreement about how it is defined, in the case of an asymmetric waveform.
The clipped waveform you have considered in your SPICE simulation is more predominantly negative than positive. To me, that means that indeed it does have a negative DC offset, and the way to make that more precise is to consider the time average of the voltage. As you note, the time average comes out to be the steady negative voltage displayed in your fourth plot. That, to me, is by definition the DC offset voltage for this case.
Let us contrast it with a different situation. Suppose you had a waveform of exactly that same clipped *shape*, but that it was coming out of the secondary of an output transformer. The shape may be the same, but the waveform would necessarily be shifted upwards somewhat, in comparison to the one in your plot. This upward shift would be precisely such that the time average of the positive portion of the wave would be exactly balanced by the time average of the negative portion of the wave. That is to say, the time average of the entire waveform would be exactly zero.
It must be like that; if you put an RC filter on the output of a transformer, you must get zero volts across the capacitor (neglecting the small residual ripple, which can be made arbitrarily small by making R*C big enough). If you were going to say that the clipped waveform as you have displayed in your plot had zero DC offset, then you would have to claim that the shifted-up one that came out of the transformer secondary does have a DC offset!
To me, the very definition of the DC offset in the waveform is the extent to which its time-averaged voltage is displaced from zero.
In all of our discussion I have been taking this as my definition of a DC offset. Thus the RC filter I proposed comes very close to allowing a DC measurement of precisely that time average. (The extent to which it remains imperfect is the extent to which an AC ripple remains across C, but this can be made arbitrarily small by making R*C large enough.)
My reason for distrusting the reading of a DC voltmeter connected directly across the mains is that one now has to make a model (mechanical, in the case of an analogue meter; electrical, in the case of a DVM) for how it will handle a waveform that is predominantly AC, with a very tiny DC component superimposed.
For example, take the case of an analogue meter, and suppose, for the sake of argument, it were set to a 170V fsd range. If the 60Hz mains frequency were turned down low enough, the needle would swing between full-scale deflection at the positive peak, and would be compressed hard against the left-hand end-stop at the negative peak. As we crank the frequency up to 60Hz, the moment of inertia in the coil plus needle will prevent the needle from swinging that far to the right before it starts moving back to the left. If the mechanical system were ideal, then the wiggling needle will indeed hover around the time-averaged voltage, But there are actually asymmetries in the mechanics of the meter. The RC filter I proposed is a much more reliable and precise integrator than the coil and needle of the meter. Plus, also, with the RC filter one can safely turn the sensitivity of the meter much higher, and get a much more precise measurement of the offset.
DVMs are even more of an unknown quantity than mechanical meters. When set to DC, the DVM is expecting to see essentially the same voltage across its terminals every time it samples. If instead it is connected to 120V AC, it will instead be seeing any voltage between +170V and -170V when it samples. Who knows what it will do with that. I tried two DVM's and one of them gave a reasonably sensible answer, but the other one showed -2.5V (regardless of which way round it was connected).
But these discussions of the problems with using meters set to measure DC when looking for small DC offsets are secondary to the main point. First, one has to give a precise definition of what one means by a DC offset in a waveform. I have given my definition (the time-averaged voltage of the waveform). What is your definition?
Chris
Edits: 10/31/16
"To me, the very definition of the DC offset in the waveform is the extent to which its time-averaged voltage is displaced from zero."
By this definition, the mere introduction of even-order distortion into a sine wave creates a DC offset. That's not the same as offsetting an otherwise undistorted sine wave above or below the zero crossing. The latter is composed of a fundamental frequency plus a DC component. Its composition as such (AC plus DC) can be ascertained at any single point along the waveform. The distorted sine wave, on the other hand, exhibits no offset, and as evidenced by the real-time oscilloscope display, all points along the waveform other than those modified by harmonic energy coincide with the unmodified waveform. Thus, while additional processing of the wave through averaging extracts a DC component, no offset exists in the wave itself. This offset, or lack thereof, is my definition of DC for purposes of this discussion.
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"By this definition, the mere introduction of even-order distortion into a sine wave creates a DC offset."One has to be careful about the precise statements here. Suppose we do what Steve O was just talking about, and do a more extreme cutting off of the entire positive half of the sinewave. For simplicity, suppose the original sinewave had amplitude 1 V, and the angular frequency omega was unity, to save having to type omegas all over the place. So the original sinewave is simply
Vsine(t) = Sin[t]
Now we chop off the top halves completely. Working out the Fourier series for this half-wave rectified waveform, we get
V(t) = -1/Pi + 1/2 Sin[t] + 2/Pi Sum_{n> 0} 1/(4n^2 - 1) * Cos[2 n t] ..........(1)
So we have a fundamental frequency sinewave 1/2 Sin[t] , plus an infinite number of even-harmonic oscillatory cosine terms, and also a pure constant, -1/Pi.
You would be quite right in saying that if you merely added the oscillatory even-harmonic cosine terms to the 1/2 Sin[t], the waveform would continue to have zero DC offset. But it would be a half-wave rectified waveform in which the flat tops were sitting at a positive voltage (in fact, 1/Pi). The time average of the whole waveform would be zero.
By contrast, the waveform that is literally just the original sinewave but with the positive halves chopped off has the Fourier expansion given in equation (1) above; it includes the -1/Pi term. This term represents a constant downward displacement of the half-wave rectified waveform, so that the flat tops now sit at exactly zero volts. This term, and only this term, survives when we perform the time average. Thus, as Steve O said, the sinewave with the top halves chopped off has a net (negative) DC offset, -1/Pi.
[Edit: I originally had a factor of 2 error overall in the Fourier expansion, which I have now corrected.]
Chris
Edits: 10/31/16 10/31/16 10/31/16 11/01/16
...It seems TK is having trouble with the concept of decomposing an asymmetric waveform into independent AC and DC components. The details of this as applied to electrical apps were worked out and well understood decades ago in the power conversion field. If one begins with TK's 2HD waveform and takes it to a limit where the positive side is entirely cliped to zero, the net result is a half wave rectified waveform. This waveform has been thoroughly analyzed and shown to contain a DC component =(1/pi)*Vpeak which incidentally is exactly what a decent DCVM will indicate.
It will be interesting if TK decides to take Dave Slagle's ".four" spice operator challenge and what his comments are regarding the results.
I'd like to see this one flushed out in its simplest terms since there are a few things at odds here and I suspect it may just come down to proper definitions.
Pushing the example of even order distortions causing a DC component and the concept that transformers will block DC we have a pretty sincere conflict unless we accept that a transformer can block even order distortion.
plus... if even harmonics do indeed have a DC component, why don't they saturate minimally gapped magnetics?
dave
"Pushing the example of even order distortions causing a DC component and the concept that transformers will block DC we have a pretty sincere conflict unless we accept that a transformer can block even order distortion."Here are a couple of plots, which perhaps are helpful:
Figure 1 shows the Fourier series expansion for a pure half-wave rectified sine wave of amplitude 1 V. The top halves of the sine waves are simply chopped off, and the resulting flat tops are at zero volts. Note that Fourier series has the fundamental 1/2 sin t, an infinite sum of oscillatory even-harmonic cosine waves, and also a constant term -1/Pi. The DC offset is therefore -1/Pi.
Figure 2 shows the Fourier series expansion when the -1/Pi constant is omitted. The waveform is then a pure sum of oscillatory waves; the fundamental and the infinite sum of even-harmonics. There is no DC offset in this case. Note that the flat tops of the half-wave rectified waveform are now at +1/Pi volts.
In general, if we only ever add oscillatory sine and/or cosine wave harmonics, they will not produce a DC offset. The DC offset arises only if there is an actual constant term in the Fourier expansion of the waveform, as in the example in Figure 1.
Chris
Edits: 11/01/16 11/01/16
Pushing the example of even order distortions causing a DC component and the
concept that transformers will block DC we have a pretty sincere conflict unless we accept that a transformer can block even order distortion.If one considers even order distortion of a sine wave (infinitely repetitive, not a short term transient situation) to be composed of a DC component and an independent AC component, the apparent dilemmas you pose disappear. In the long term, transformers "block" only the DC component of the composite waveform: the AC component remains and its long term average value is zero.
As you probably know, any repetitive waveform can be represented (or duplicated) by an infinite series of sine waves of specific harmonics, amplitude and phase. An example is the classic unipolar
square wave = K+sin(1*B)/1-sin(3*B)/3+sin(5*B)/5......... We also know that a good transformer can pass a pretty decent representation of a square wave. At first glance this would seem impossible because the "flat" portions look like "DC". What's really happening is that the transformer strips out the (constant)DC component represented by the "K" at the beginning of the series and attempts to pass the remaining (AC) "frequencies" represented by sin(n*B)/n...portion of the series. If the transformer is good, the resulting waveform will "look" just like the original but will have a longterm average value of zero because the DC component, the "K" was not passed from primary to secondary. IOW, only the AC component (that transformers readily "pass") define the envelope of the waveform we see. Note that the "K" component does exist in the primary where it could cause issues.In a similar fashion, a continous sine wave with even order distortion is "passed" by the transformer because the AC component can be decomposed into an infinite series of sine waves in a manner similar to the square wave example. The DC component is again not passed and the AC waveform will "appear" unchanged but will have a long term average value of zero. Thus it can be seen that a transformer doesn't "block" even order distortion, just the zeroth harmonic as you referred to it. All of the important higher order harmonics that define its shape remain...mostly.
plus... if even harmonics do indeed have a DC component, why don't they saturate minimally gapped magnetics?
I suspect that if the even harmonics are generated in the stage driving a transformer, saturation is indeed a possible issue. Isn't this one of the reasons true toroids are not often found in audio apps?
Edits: 10/31/16 10/31/16 10/31/16
I think chris makes a good point about the transformer being able to support the DC but I still see a problem with the "subtraction" of the DC / zeroth harmonic.
If one considers even order distortion of a sine wave (infinitely repetitive, not a short term transient situation) to be composed of a DC component and an independent AC component, the apparent dilemmas you pose disappear.
take a sine wave through a diode and then feed a transformer. The fourier analysis shows the same DC component on the secondary and the primary.
In the long term, transformers "block" only the DC component of the composite waveform: the AC component remains and its long term average value is zero.
again take the above diode creating a "whole bunch" of cyclical even order distortion and no amount of time will ever average the value to zero but if you replace the diode with a 9V battery and the transformer will indeed block that 9V DC component.
I guess this is just another one of those "grey areas" in audio where all the good stuff hides.
dave
Actually, it's not grey at all.Regarding the rectifier scenario, as Chris' plots show, the input to the transformer contains the entire infinite series inc the constant 1/pi term that represents the DC component. If you analyze the transformer output you'll find that everything is the same EXCEPT that the 1/pi term is missing. The 1/pi term plays no role in defining the "shape" of the half wave waveform. A DC coupled scope connected to the secondary will show no DC offset or component. The flat top portion of the waveform will be displaced above the zero line by 1/pi and the average value of the portion of the waveform above the zero line exactly equals the average value of the portion below the zero line so the net value is exactly zero within real world limits of signal generation and measurement.
Bottom line is that there is no magic or mystery in how seemingly DC containing waveforms can be "passed" by a transformer (or capacitor for that matter). All of this was worked out and fully understood decades ago.
Edits: 11/01/16
"take a sine wave through a diode and then feed a transformer. The fourier analysis shows the same DC component on the secondary and the primary."
Hmm...I'm not sure I would agree about the DC component. Why do you say that?
The waveform on the secondary might have the same *shape* as the waveform on the primary (neglecting effects resulting from imperfect behaviour of the transformer, etc.). To get the same shape, it means that the oscillatory sine and cosine waves in the two Fourier series for the primary and the secondary waveforms are the same. But the constant term in one Fourier series can be different from the constant term in the other Fourier series. That would not affect the shape; it would just represent an upward or downward shift of the whole waveform.
It would be pretty much like those two plots I showed in my previous posting. They differ only in that one is shifted vertically relatively to the other. The plot in Figure 1 would be a bit like the situation on the primary of the transformer, with a DC offset component. The plot in Figure 2, with no DC offset, would be like the situation on the transformer secondary.
There can be no DC offset on the secondary side of a transformer, but there certainly can be on the primary.
Chris
Hmm...I'm not sure I would agree about the DC component. Why do you say that?
I don't say that... spice does with its .four analysis. I know spice doesn't make mistakes but I also know the person behind the keyboard often tells it to do the wrong thing.
I am clear conceptually on what you are saying but still on shakey ground in bringing that concept to reality.
dave
"I don't say that... spice does with its .four analysis. "
On the face of it, if .four in SPICE is saying that the constant mode in the Fourier series gets transferred through from the primary to the secondary of the transformer, then it sounds like maybe there is a bug there.
An extreme example would be to try it when one turns off all the oscillatory Fourier modes, i.e. if one just connects a DC source to the transformer primary. By the same token, if SPICE says the constant Fourier mode (which is all there now would be) gets transferred to the secondary, then it would seem that it would be claiming that there would be a steady DC appearing on the transformer secondary, which would obviously be wrong.
A slightly less extreme example would be to apply DC plus a very small superimposed sine wave, and see what .four shows on the secondary.
I'm not familiar with the .four command in SPICE. Can you recommend a good place to read up about it?
Chris
I think I essentially did what you suggest by placing a 9V battery in series with the primary of a transformer and no DC component showed up on the secondary and I highly doubt there is a bug in spice.
The DC offset does seem to be the difference of the RMS positive and negative going waveforms so we are talking about ever changing signals that have a net difference over time and the difference over time suggests an AC behavior. Given enough inductance a transformer will pass a 1hz or lower signal so that goes a long way to connecting the dots for me.
Maybe a future test would be to put a 1hz sine in series with a 60 and sample for a few 60hz cycles at different points in the 1sec wave and see what happens on the secondary of the transformer.
dave
"The DC offset does seem to be the difference of the RMS positive and negative going waveforms so we are talking about ever changing signals that have a net difference over time and the difference over time suggests an AC behavior."
Ah, OK. I'm not familiar with using .four. I guess it is performing an analysis on a transient signal, not on a steady state system in the limit t goes to infinity? That could probably explain the DC component on the secondary. I was thinking of a steady state Fourier analysis, once things had settled down. There couldn't be any DC component showing up on the secondary in such a situation.
Chris
Indeed the t=infinity throws a wrinkle into it. I was simming for about 10 cycles after 10 seconds and I increased the time to 100 seconds and the DC does approach 0.... the really odd thing is the DC on both the primary and the secondary both go to zero even with the diode in place and the clipped sine at startup becomes a near perfect sine at the end.
in the pic below I dropped the frequency to 1hz to get a more meaningful visual example of what spice says. Adding more inductance seems to slow down the rate of increase. Its odd since it almost looks like a cap charging.
I bumped the frequency back up to 1Khz and here are the last few cycles. Even more interesting is the output (secondary) now has 10X the DC of the primary?
now my head really hurts!
dave
Your findings using a half wave waveform are indeed odd. How "perfect" is your simmed transformer? If the primary DC resistance is zero or very low, I would expect a large DC current in it and near zero DCV across it over the long term. Is there a load across the secondary?
The transformer is pretty ideal but changing the coupling factor and DCR has little effect on the sim. (no saturation effects are included in the model) The load is a 1K resistor.of further interest is how the inductance value changes the rate at which the DC offset disappears. (again on both the primary and secondary at the same time) Here is the sim with 10hy and 20hy of pri / sec inductance and the distinct difference is "rise times" can be seen.
here is a shorter sample from 0 of 10 and 100hy's to show the "expected" diode clipping only happening the first few cycles and the slower rate of change to the sine.
following this pattern (whatever it says) i would expect infinite inductance to take an infinite time to show this behavior.
As for the effect of frequency which is directly related to inductance in most of these cases a similar pattern does not show up in the same manner as frequency is changed. Here is a 100hz and 1Khz sine for the first 1/10th of a second. Where a 10X change of inductance gave a notable change in the "ramp up" the 10X change in frequency doesn't appear to have anywhere near the same effect other than the "slope" of the top of the "clipped" waveform.
as a final bit of interest, the load seems to have a similar effect to the inductance in that a 1K load slows slows the reduction of the DC offset over a 10K load much in the way increasing the inductance does.
for completeness here is a screen grab of the sim.
Edits: 11/02/16
Interesting! I'm going to wire up an actual circuit in the next day or so and see how real-world components react.
I just wired up a circuit to determine how real world components react to the half wave rectified waveform. Sketch of circuit to be edited in later.Bottom line: A 1:1 transformer fed a half wave rectified waveform behaves almost exactly as predicted by theory. The AC+DC waveform at the primary was faithfully reproduced at the secondary minus the DC component. Appropriate transformer loading was required to accomplish this. Everything was performed with 60Hz from the mains but stepped down and isolated for safety.
Edit: The test circuit
Edits: 11/02/16
Interesting... as you have it drawn It sims exactly like it should with no delay. Why the need for the 45 ohm and 100 ohm resistors on the primary? If I remove the 45 ohm from the sim nothing changes but that 100 ohm below the transformer makes all the difference in the world for the sim. What is its purpose?
dave
The 45 ohm resistor is a stable "load" for the rectifier system. Probably not absolutely necessary but it does assist with raw voltage regulation. The 100 ohm resistor at the "bottom" primarily serves as a current limiter for the DC portion of the waveform because the T3 primary (and secondary) has a DCR <1R0: essentially a dead short for the DC component. Without it, T2 and the rectifier protested hotly and loudly. The resistor also serves as a defined source impedance and as a current sense point. Interestingly, there was almost no DC voltage component across the primary of T3, just the AC component but the current thru the sense resistor contained both components in approx the predicted magnitudes.
interesting. The lack of DC across the primary doesn't surprise me sing you have a 100:1 voltage divider with the DCR vs. the current limiting resistor,
why the whole thing goes crazy without the resistor is odd though... could it be dumping so much DC through the primary that it saturates it causing the impedance to approach zero causing no load for the AC component?
dave
I don't know spice but I suspect you're approaching an undefined condition or "singularity" similar to 1/0. Kinda like sin(x)/x. You can't evaluate this @ x=0 using traditional math but other math techniques do arrive at a value.
Does spice allow t=infinity?
"Interesting... as you have it drawn It sims exactly like it should with no delay. Why the need for the 45 ohm and 100 ohm resistors on the primary? If I remove the 45 ohm from the sim nothing changes but that 100 ohm below the transformer makes all the difference in the world for the sim. What is its purpose?"
If I am interpreting the SPICE schematic you posted previously correctly, it seems to me that you have a very low DCR for the transformer primary (1 ohm, for a 100H inductance?), and thus the DC component on the waveform would mean that the current could get really quite (unrealistically) large. It might perhaps account for the strange behaviour in the simulation? Using 100 ohms instead might well indeed bring things better under control.
Chris
I was using reasonable numbers for the DCR.... the odd thing is that adding the 100 ohms in series with the coil has a different effect than using a coil with a 100 ohm DCR.
dave
"I was using reasonable numbers for the DCR.... the odd thing is that adding the 100 ohms in series with the coil has a different effect than using a coil with a 100 ohm DCR."
Do you mean that the composite component, comprising an ideal inductor with 100 ohm resistor in series, behaves differently from an inductor with 100 ohm DCR specified? That would seem to be rather odd. Of course, if you compare the voltages measured across just the inductor itself in these two cases, then they will clearly differ. But if the voltages differ when measured across the composite R+L component, versus the L with specified DCR, then that would be strange.
By the way, I found that the SPICE sim can be neatened up quite a lot by introducing a second diode, across the primary of the transformer, such that it prevents the voltage from going negative. One now gets a rather clean looking half-wave rectified voltage being fed to the transformer primary. It looks pretty much the same as if one had an ideal half-wave rectified signal from a low output impedance source driving the transformer.
The half-wave rectified signal on the secondary settles down to the expected one with zero DC offset, when t gets large enough.
Chris
"In a similar fashion, a continous sine wave with even order distortion is "passed" by the transformer because the AC component can be decomposed into an infinite series of sine waves in a manner similar to the square wave example. The DC component is again not passed and the AC waveform will "appear" unchanged but will have a long term average value of zero. Thus it can be seen that a transformer doesn't "block" even order distortion, just the zeroth harmonic as you referred to it. All of the important higher order harmonics that define its shape remain...mostly."Yes, I absolutely agree. A periodic function with fundamental angular frequency omega has a Fourier expansion which is a sum over terms in Sin[n omega t] and Cos[n omega t], where n ranges over the non-negative integers. The Sin terms arise just for positive integer n, and are all oscillatory. The Cos terms are also all oscillatory, except for the n=0 term which is a pure constant.
It is this pure constant term in the Fourier expansion that defines the DC offset. Since the time averages of all the oscillatory modes in the Fourier expansion give zero, the DC offset, i.e. the constant, or "zeroth harmonic" term in the Fourier expansion, is indeed exactly equal to the time average of the waveform.
One may ask why we are concerned about the DC offset term in particular. One reason, which may be quite an important one in some situations, is that a transformer fed with a waveform that is a sum over oscillatory sine and cosine functions can handle these straightforwardly (assuming they are within the frequency and voltage range it is designed for). The transformer doesn't "block" any harmonics, even or odd, as long as they are withing the frequency range it can handle.
But that humble constant term in the Fourier expansion could cause a problem; it represents a DC offset, which cannot be handled by the transformer. At best, it is "blocked." At worst, it might saturate, or significantly saturate, the core.
Chris
Edits: 10/31/16 10/31/16
" The details of detecting the DC are secondary. "If not for the deflection of the DC meter underlying the claim of DC on the mains, we wouldn't even be talking about it. It's hardly worthwhile to discuss the mechanism by which DC might be created on the line unless it has actually happened. I don't believe this can be demonstrated with a meter, analog or otherwise. That's why I was curious as to the means of measurement. I was wondering whether the claim was based on an observation of the waveform, but that's apparently not the case. Unfortunately, the OP hasn't come back to let us know whether his measurement was also made with a meter.
What do you suppose the DCR is looking back into the house wiring and the mains transformer? I have no small difficulty with the concept that a blender motor can create standing DC on the line in the face of a 200 Amp service. Of course, the OP never claimed that was the cause...
--------------------------
Buy Chinese. Bury freedom.
Edits: 10/29/16
The discussion has moved beyond the claim of the OP. The possibility of DC on the line due to a household appliance is not speculation, it can and does occur and can be measured. The situation is analogous to a high current half wave rectifier/load on the secondary of a PT. It's common knowledge that this is poor design practice because it produces DC in the secondary and greatly reduces transformer capacity.
The DCR of the mains is easily estimated. Just for starters, in my home the branch circuit from the outlet to the panel is ~ 50ft of #14 or 100ft round trip. This is ~0.26 ohm. The DC component of my heat gun on low ~3A so the DC voltage ~0.8V from just the branch circuit alone. Bottom line is that the DCR of all the wiring supplying a typical home is significantly greater than zero.
As I note elsewhere, I have access to a professional grade circuit analyzer. I'll see if I can borrow it and report back.
> > No, it just fools the meter.
Can you explain the mechanism by which a mechanical DC meter movement could be "fooled" by an asymmetric waveform that does not also contain a DC component? Would a clean sine wave "fool" the meter movement too? How would a quality DVM react?
> > Does a tube amp output DC from its OPT at 5% even order distortion?
No, the transformer removes the DC component such that the average value of the waveform = 0 assuming a long term repetitive waveform. This is similar to the action of a series cap in the long term. However, there will be increased DC in the primary if the distortion originates in the stage driving the transformer and it's not cap coupled. OTOH, if the 2HD is caused by the load on the secondary, there will indeed be a DC component in the output.
There are two issues here: 1. Can a load cause a DC current to flow in the load winding of a transformer and can this load cause a measurable DC voltage? Clearly, the answer to both is yes. 2. Can a mechanical DC meter detect this voltage and can the presence of the DC voltage create issues with other devices on the circuit? All evidence indicates yes to both of these situations too. Which of these issues are you disputing?
FWIW, math analysis of typical idealized AC waveforms and distortions is possible. There are professional instruments avail designed to detect and quantify all kinds of AC power issues. I think I have access to a decent sample from Dranitz. Also, the situation ought to be susceptible to a spice sim and you're good with spice aren't you? Care to pursue this further?
Might be coming off the cable with a ungrounded service?
I am still a ways away from the actual build. I want to hear it in stereo first, and then I will need to acquire some more parts. Also have a few other ideas I want to check out on the breadboard. Almost every time I have gone from breadboard to final build all my little problems like residual hum and buzz tend to disappear, so I am not too worried. I will have to check out the mains for dc. How does that get on there when it does?
All the tips have gone in a notebook I keep for future reference. I print out the post and file them in a three ring binder.
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