DC Offset Blocker/Killer - where to buy in the USA
FWIW, my reading of the thread leaves me with essentially zero doubt that the OP is being entirely truthful. And I agree with him that it is very likely that investing the time, physical effort, and money that would be involved in sending the amp to McIntosh would result in a finding that it is functioning as designed. Regards, -- Al |
That all seems perfectly consistent with the explanation I proposed in my previous post, Jim. The hairdryer draws considerably different amounts of current in the positive half-cycle than in the negative half-cycle. The resulting DC offset corresponds to that difference in current x the resistance of the wiring that current is drawn through, **for wiring that is in the path between the outdoor AC wiring and the measurement location.** For example: I then plugged the Low pass filter into the first outlet. (Outlet closest to electrical panel. First outlet on home run feed.) At that measurement location the DC offset resulting from the current differential between the two half-cycles will not reflect the resistance of the wiring between the first outlet and the last outlet of that branch. Regarding the 1.577 vs. 0.734 difference, note that 12-2 has about 60% more resistance per unit length than 10-2, and the length of the 12-2 was about 20% greater than the length of the 10-2. 0.734 VDC x 1.6 x 1.2 = 1.41 VDC. The additional difference of 1.577 - 1.41 = 0.167 VDC is probably accounted for by a combination of the resistances of the connections to the several intervening outlets in the room with the 1.577, and imprecision in the estimates of the run lengths. Best, -- Al |
If Al is still following this thread I am sure he will have a better technical answer. Not necessarily, Jim :-) Your finding that having the hairdryer plugged into the outlet where the measurement was taken resulted in a DC offset of around 730 mv, while measuring at the same outlet but with the hairdryer plugged into various other circuit branches on the same leg resulted in around 115 mv, is indeed a bit of a headscratcher. But not being familiar with the design of hairdryers I did a little research and found this interesting paper: http://www.idc-online.com/technical_references/pdfs/electrical_engineering/MEASURING_ACOUSTIC_NOISE_EMITTED_BY_POWER.pdf As stated on pages 4 and 5 and as depicted in Figure 13, at least when hairdryers are operated "at lower power" they place a half-wave rectifier diode in series with the load they place on the AC, or at least a substantial part of the load. So DC offset results from the difference between the amount of current that is drawn during the positive half-cycle of the AC waveform and the amount of current that is drawn during the negative half-cycle, and the differing voltage drops that occur in the resistance of the AC wiring between the two half-cycles as a result of that current difference. In tests 2, 3, and 4 essentially zero current was being drawn through the dedicated wiring between the service panel and the outlet where the measurement was taken, so no voltage drop would have been occurring in that wiring. And the DC offset that was measured would have resulted essentially from the voltage drop differential between half-cycles that was occurring in the panel and in the outside wiring, since the additional voltage drop in the wiring between the panel and the outlets where the hairdryer was plugged in would not have been in the path to the outlet where the measurement was being taken. And presumably the outside wiring is considerably heavier gauge than the 10 gauge Romex used for the dedicated line, and therefore it would present a lower source resistance (per unit length, at least) for DC offset to develop across as a result of the asymmetrical current draw. Whereas in test 1 that asymmetrical current was being drawn through an additional resistance in the path to where the measurement was taken, corresponding to the sum of the resistances of the dedicated line’s two 75 foot 10 gauge conductors, which amounts to about 0.15 ohms. Looking at it quantitatively, if my theory is correct the difference in the amount of current drawn by the hairdryer between the positive and negative half-cycles would be: (730 mv - 115 mv)/0.15 ohms = 4.1 amps. In the context of the large current draw of an 1875 watt hairdryer, and one that is placing a diode in series with much of the load it presents, I suppose that is consistent with your findings. Best, -- Al |
P.S. to my previous post: I didn’t mention the units that should be used for R and C in the various calculations. If R is expressed in ohms and C is expressed in Farads, the calculation described in my last paragraph will provide an answer expressed in seconds, and the calculation of bandwidth that is described earlier in the post will provide an answer expressed in Hz. Best, -- Al |
Hi Jim, They both appear to be reasonable values for this purpose. The 10 uF/100K combination will charge up to the final value about twice as fast as the 22 uF/100K combination, which isn't important, but it won't be down quite as far at 60 Hz as the latter. That difference probably isn't important either, although I can't say that with certainty as I'm not familiar with the internal design of modern digital multimeters. Specifically, the 3 db bandwidth of the low pass filter (i.e., the frequency at which 3 db of rolloff will have occurred) = 1/(2 x pi x R x C). Let's refer to that as "bw" (bandwidth). For 10 uF/100K, bw = 0.16 Hz For 22 uF/100K, bw = 0.07 Hz The rolloff at 60 Hz will be 1/(square root(1 + ((60/bw)squared))), converted into db based on 20 x logarithm of the resulting numerical value. (That looks more complicated than it really is; it could be illustrated more simply if it didn't have to be shown as text. Some people depict "squared" as "^2" and "square root" as "^(1/2)", but I'm not sure if for most people that would make the equation more clear or less clear). For 10 uF/100K that calculates to -52 db at 60 Hz, which would reduce 120 volts to about 0.3 volts. For 22 uF/100K that calculates to -59 db at 60 Hz, which would reduce 120 volts to about 0.14 volts. Also, the time to charge to very close to the full value of the DC that is present will be approximately RC x 5, which is how I derived the 10 or 15 second figure for the 22 uF/100K combination. Best regards, -- Al |
Jim (Jea48), good posts as always. My only comment on your low pass filter project is that with the 22 uf and 100K values it will take around 10 or 15 seconds after the AC is applied for the cap to fully charge up to whatever amount of DC may be present. Which is fine, of course, but it’s just something to be aware of when the measurement is made. Best regards, -- Al |
