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 22nd April 2013, 01:41 AM #1 diyAudio Member     Join Date: Mar 2013 Location: Lynden, WA USA bias spreader bypass cap considering a simple Vbe multiplier type bias spreader, how does one determine the size of the bypass cap. Cordell's book shows many examples with 10uf, but I can't find how that number was reached. I assume it needs to be a pretty good sized cap, but I want to know how to calculate it. Surely it can't be as simple as 2pi*the divider resistances*the freq I want to pass. Intuitively I feel the transistor matters in the calculation too, but I can't see how to do this.
 22nd April 2013, 01:43 AM #2 diyAudio Member     Join Date: Mar 2013 Location: Lynden, WA USA Oh, obviously in my example of what I'm pretty sure is NOT the relevant formula, I would divide one by my answer.
 22nd April 2013, 02:00 AM #3 diyAudio Member     Join Date: Mar 2009 Location: Buenos Aires - Argentina Sure does. And the value is much lower than the resistive divider suggests, because the transistor both has current gain and passes much higher current. On first approximation, consider an "equivalent resistance" of: Req=V/I and define a capacitor which has that impedance at, say, 16 Hz or something, so the transistor is bypassed in all of the Audio range.
 22nd April 2013, 02:17 AM #4 diyAudio Member     Join Date: Mar 2013 Location: Lynden, WA USA so I would imagine a resistor who's value is the collector voltage/collector current of the transistor, put it in parallel with the divider resistors and use the result as the "R" variable in the standard equation I mentioned earlier? Since the base current portion of the total current is probably on the order of 1/100th of the total, it can be ignored I assume? Wouldn't 16 hertz be kinda high? I presume what we don't want is the signal modulating the bias voltages. Wouldn't using 16 hertz as my -3dB point cause the cap to attenuate noticeably still at 20 hertz where it is audible? I'm not disagreeing. I am very much a novice here. Just trying to understand is all.
 22nd April 2013, 02:23 AM #5 diyAudio Member     Join Date: Mar 2009 Location: Buenos Aires - Argentina Yes that's the idea. This is just an estimation, to have some idea about what we are talking. The 16 Hz cutoff was just a personal estimation, personally I don't lose sleep about what happens at 20 Hz, given that most speakers don't reproduce them, there's no music program material that low and I doubt I can hear it anyway, but that's a personal/practical choice.
 22nd April 2013, 02:46 AM #6 diyAudio Member     Join Date: Mar 2013 Location: Lynden, WA USA I realize this is just an estimation, but can I assume such an estimation is sufficient in this case? I guess my thinking 16htz was high stems from the fact that I have been doing guitar amps. With tube guitar amps, many bad things come from attempting to make the lower bandwidth limit of a stage too low. Blocking distortion and such. Now that I'm learning hi-fi, I may be chomping at the bit to have, at least on paper, the crazy bandwidths that were impossible with previous projects. Don't look to hard for any real logical rational for this line of reasoning .
 22nd April 2013, 03:52 AM #7 diyAudio Member   Join Date: Nov 2010 Location: The City, SanFrancisco As JMFahey pointed out the multiplier is typically a single stage feedback regulator. The designs can vary but even the minimilast will have loop gain above 40, or a low output impedance from dc up to near a MHz. This impedance should be less than 20 ohms, thus there is really no benefit adding a large capacitor (it would take hundreds of uf's). Once the multiplier nears its crossover frequency its impedance will rise so here bypassing is critical (predrivers shouldnt be fed from an inductive source). But values near .01u are sufficient to keep the impedance near its dc value. The charge necessary to remove from the predriver bases during crossover can be minimized to the point where these 0.01u or less are sufficient. Hope this helps. -Antonio Last edited by magnoman; 22nd April 2013 at 03:53 AM. Reason: typo

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