Hello all,
I'm looking for some help determining if the N rank 2sk68a (IDSS ~ 6 mA) will be an ok substitute for the L and M rank parts originally supplied in the Kenwood KA-9100
I will attach photos of the circuits with one being the phono input and the other being the control amp input.
I am also hoping that this issue will help me understand JFET's and JFET circuits better so please understand if I make assumptions in what I write here that are totally wrong.
first the phono circuit that specifies L rank 2sk68a's that are currently 2-2.5 mA IDSS parts.
I can only guess how the circuit would react to the higher IDSS parts. But it would seem that with the given values on the schematic, the current flow through the JFET would increase roughly in proportion to the difference in IDSS between parts. However, since the circuit is biased with fixed resistors, the operating points of the circuit would shift. Can anyone explain or model what would happen here, using the higher IDSS parts?
Same here for the control amp input: where they used different ranks within the differential pair although based on the circuit it makes sense why that might be required, I can't say exactly why.
I'm looking for some help determining if the N rank 2sk68a (IDSS ~ 6 mA) will be an ok substitute for the L and M rank parts originally supplied in the Kenwood KA-9100
I will attach photos of the circuits with one being the phono input and the other being the control amp input.
I am also hoping that this issue will help me understand JFET's and JFET circuits better so please understand if I make assumptions in what I write here that are totally wrong.
first the phono circuit that specifies L rank 2sk68a's that are currently 2-2.5 mA IDSS parts.
I can only guess how the circuit would react to the higher IDSS parts. But it would seem that with the given values on the schematic, the current flow through the JFET would increase roughly in proportion to the difference in IDSS between parts. However, since the circuit is biased with fixed resistors, the operating points of the circuit would shift. Can anyone explain or model what would happen here, using the higher IDSS parts?
Same here for the control amp input: where they used different ranks within the differential pair although based on the circuit it makes sense why that might be required, I can't say exactly why.
Thank you, do you mind explaining the math? In the phono section, it looks like the current through the tail resistor / drain should be around 2 mA which is close to the IDSS of the L rank 2sk68a. It seems like the jfet would thus be operating around pinch off which would be ideal, correct? With a much higher IDSS, the JFET would be operating below the pinch off and in a non linear part of the curve. Wouldn't the downstream bipolar pair be impacted as well concerning bias / operating points? With this assumption, shouldn't we say that it would actually be a poor substitution? Again, im totally shooting in the dark here so my assumptions may be very far off.
Your "math" is correct: the standing current is half of that of the tail resistor, provided the latter is big enough. Of course, a real CCS will have much better static performances.
Re. the operating point, your reasoning, if I understand it correctly, is also valid, but I'd say it's immaterial within the context of your amp, where others things could be better optimized if cost were no object. Sorry for hurting your feelings.
Re. the operating point, your reasoning, if I understand it correctly, is also valid, but I'd say it's immaterial within the context of your amp, where others things could be better optimized if cost were no object. Sorry for hurting your feelings.