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#9281 |
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diyAudio Member
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You can start with just one BiB set for 35V ... I suggest a minimum 300mA current.
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RC |
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#9282 |
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diyAudio Member
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Hi Salas
A learning tech question for you. Imagine I would like to lower input noise and use 4 x k170 there. This would double Id current so I would need to increase Vdd to around 60V in order to have enough curren over R4 right ? This is only a suposition, but using the same R13 (1500r) I would also increase gain to around 47dB right ?
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#9283 | |
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diyAudio Member
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Quote:
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#9284 |
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diyAudio Chief Moderator
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The demands are monopole +35V @ 30-35mA per channel. We usually set the SSLV regulator much higher for it to work its MOSFETS better. Usually at 150-300mA CCS per channel depending on how much sinking you are prepared to use.
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#9285 | |
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diyAudio Chief Moderator
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Quote:
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#9286 |
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diyAudio Member
Join Date: Dec 2002
Location: Milan, Italy
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#9287 | |
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diyAudio Member
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Quote:
I found your post here: Simplistic NJFET RIAA Just do not know how you determined the 12r for k170 Rs
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#9288 |
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diyAudio Chief Moderator
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4 Rs in parallel amount to 1/4 the value, its just we use one each for better current sharing, auto matching by local feedback in other words. So we get a total current in a single virtual "super transconductance'" JFET device with lower noise, more gain, more gate capacitance. Then the CCS resistor (1K in that example) steals that current from the PSU node just shy of what we need left for the PNP cascoding BJT to run through its resistive load folded to ground (2K2) for centering it voltage wise so we get a symmetrically clipping sinewave at max input drive.
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#9289 |
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diyAudio Member
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So if you used only one common Rs for the four k170, it would have to be 3r to generate the same Vgs right ?
Now I did not think of the 1k resistor as a CCS.... that explains a lot. That resistor must be correctly dimensioned to let go only the right amount of current so if it runs on the fets and also on the folded load resistor. If we decrease the CCS resistor, more current will flow and as the current through the fets is fixed by Rs, there is more current through the folded load resistor and so more Vdrop there. That is how the trimmer works ![]() ![]() ![]() Thank you again for your cristal clear explanation.
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RC |
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#9290 |
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diyAudio Member
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Now in the double K369 schematic you used the same 2r2 as in the single K369 schematic (Not 4r4).... The difference in this case is in the CCS resistor that is smaller (1k2 instead of 2k2)
So your choice of Rs does not follow a strict trend... It must come from your simulation and experience.
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RC |
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