![]() The design has an ugly group delay and other issues. ![]() However, despite the cabinet size, you are not getting the best performance from that woofer. My earlier calculations were for a cylindrical vent. If am correctly understanding your stated box size, that vent is tuned to 20 Hz. Ps Missed this further on in your thread: Some box/driver combos do use protective high pass filters below tuning.some don't need it.and you may not have content that low to worry about it.yet? While needing eq to get to subwoofer territory, the box would be much easier to get right." This would be a difficult box to do well as it needs really thick walls and significant bracing.Īs stated earlier, a more optimal box for this subwoofer is a 2.5 sealed one. These results are BassBox Pro simulation.īassBox Pro set for extended bass requires a 7.4 cu ft box tuned to 20 Hz. "Right now, your box is tuned to about 13 Hz. But solving a mechanical problem in the mechanical domain is still the best route, if possible.Looking at your other thread did some calculating for you and arrived at the conclusion your box was tuned to 13hz for your driver: A decent starting point might be a common value like. If we don't believe the assumptions, we'd need to either measure accurately or prepare to experiment. If we believe all of these WAG assumptions, we calculate C=1/2pi(25Hz)(1M Ohm) = 6n36F give or take A LOT. We don't have any design frequency in mind anyway, other than 25Hz seemed to work OK. No good reason to choose this number other than convenience. So, again, how do we guess the load resistance without knowing exactly (because details matter here) the fed-back stage(s) gain? Let's take a WAG and call the total (sum of source and load) R 1M Ohm. Our desired pole is set by the product of the coupling cap with the sum of source (maybe 160K) plus load "resistance" (2M grid leak in parallel with summing junction loading due to feedback). Without feedback, the loading side of the coupling cap sees only the 2M grid leak. But it's not even close to infinite rather it's somewhere much nearer unity, and very dependent on the gain of the particular valve in use. If the open loop gain were infinite (and therefore there would be a resistor across the 330pF cap to set the 50Hz pole) the impedance at the second stage grid would be zero, and the RC time constant / high pass pole would be determined simply by the coupling cap and the source resistance of maybe 160K Ohms. How to interpret this? The designer apparently believed that the fed-back stage(s) would run out of open loop gain at 50 HZ, because they included no resistor across the 330pF cap to set the 50Hz RIAA pole (and have also pushed the 500Hz and 2122Hz pole components to compensate). ![]() On the other side of the coupling cap is the 2M grid leak resistor in parallel with a complex impedance formed by the amplified current loading from the 330pF RIAA feedback cap. Let's call the resulting source resistance 160K. So, how could we guesstimate? The first stage's output resistance could be sloppily guessed to be the anode load of 330K in parallel with valve anode "resistance" of 80K plus mu of 100 times cathode resistor of 2K2, so 300K (very rough estimates, but probably good enough for what we're doing. If all ECC83 valves were entirely predictable forever it would only be a little easier. And, there's no well defined 50Hz pole, to make it even more nebulous. You're right that the first cap's rolloff frequency is hard to define (without an exact LTspice model of the whole project) because it depends, as you say, critically on two somewhat "soft" resistances, the source impedance from the first stage (resistively degenerated by un-bypassed cathode resistor) and the summing junction's impedance (poorly defined because it depends critically on open loop gain within the feedback loop which is, at best, only slightly higher than closed loop gain). ![]()
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