rjb5191
Well-known member
Hello all, I hope this is the correct forum for this!
I'm working on rebuilding a Kenwood KA-9100 integrated amp (relevant schematic below) and the need has come to replace the input differential pair JFETS (Qi 21 and 23) in the phono section. The original were 2sk68a L rank parts. The replacement parts I received are N rank parts with different Vgs on / off and IDSS parameters. I thought it would still be possible to use these new parts with some reconfiguration of the source and drain resistors in the circuit. I will present my math and reasoning below and I'm hoping that someone more knowledgeable than I would be willing to check my math and reasoning.
The original Jfet had the parameters Vgs OFF: - .45V Vgs ON: 0.22V IDSS: 2.2 mA
Current through the drain resistors (RI 105, 107) is 16.5V / 22K = 0.75 mA each
voltage drop across source resistor (Ri 109) is 1.5 mA * 10K = 15 V
Current through Ri 113 is 14.1 V / 3.3k =4.27 mA
Current through Ri 111 is approximately 4.27 mA - 1.5 mA = 2.77 mA
The new JFET parameters are Vgs OFF: -.8V Vgs ON: -0.1V IDSS 5.5 mA
For the new circuit, I will maintain the drain voltage of 11.5 V for proper biasing of the downstream stage. I am planning on using a source voltage of 350 mV (about 50% of the difference between Vgs on and off). I am planning to use a drain current of 2.5 mA (nominally around 50% of the IDSS).
New drain resistor values for RI 105 and 107 is 16.5V / 2.5 mA = 6.6 Kohm each
For calculating the source resistor i need to take into account the increased voltage drop through Ri 113 due to the increased current as this will lower the available voltage at the junction of Ri 109, 111, and 113. I used a bit of a guess and check method here instead of setting up a system of linear equations because I am not fresh out of school at this point. I know I need 5 mA through the source resistor Ri 109 so I know that the current through Ri 113 must be greater than that to allow current to flow to ground through Ri 113 (at least I think that is the correct assumption). In this case I assumed a current of 6.5 mA through Ri 113
Thus the voltage at the Ri 109, 111, 113 junction is 6.5 mA * 3.3k = 21.45 V, 29V-21.45 = 7.55 V
Thus the new source resistor voltage drop is 7.55 V + .35V (source voltage) = 7.9 V and thus the new source resistor value is 7.9 V / 5 mA = 1580 ohms
I'm hoping that the voltage divider with ri 111 isn't critical for some reason that Im unaware of and is ok with the reduced current through it compared to the original
design. The math is simple enough I hope but I really need help with checking the assumptions I used. Hopefully the new values I came up with would work better than leaving the stock values!
I'm working on rebuilding a Kenwood KA-9100 integrated amp (relevant schematic below) and the need has come to replace the input differential pair JFETS (Qi 21 and 23) in the phono section. The original were 2sk68a L rank parts. The replacement parts I received are N rank parts with different Vgs on / off and IDSS parameters. I thought it would still be possible to use these new parts with some reconfiguration of the source and drain resistors in the circuit. I will present my math and reasoning below and I'm hoping that someone more knowledgeable than I would be willing to check my math and reasoning.
The original Jfet had the parameters Vgs OFF: - .45V Vgs ON: 0.22V IDSS: 2.2 mA
Current through the drain resistors (RI 105, 107) is 16.5V / 22K = 0.75 mA each
voltage drop across source resistor (Ri 109) is 1.5 mA * 10K = 15 V
Current through Ri 113 is 14.1 V / 3.3k =4.27 mA
Current through Ri 111 is approximately 4.27 mA - 1.5 mA = 2.77 mA
The new JFET parameters are Vgs OFF: -.8V Vgs ON: -0.1V IDSS 5.5 mA
For the new circuit, I will maintain the drain voltage of 11.5 V for proper biasing of the downstream stage. I am planning on using a source voltage of 350 mV (about 50% of the difference between Vgs on and off). I am planning to use a drain current of 2.5 mA (nominally around 50% of the IDSS).
New drain resistor values for RI 105 and 107 is 16.5V / 2.5 mA = 6.6 Kohm each
For calculating the source resistor i need to take into account the increased voltage drop through Ri 113 due to the increased current as this will lower the available voltage at the junction of Ri 109, 111, and 113. I used a bit of a guess and check method here instead of setting up a system of linear equations because I am not fresh out of school at this point. I know I need 5 mA through the source resistor Ri 109 so I know that the current through Ri 113 must be greater than that to allow current to flow to ground through Ri 113 (at least I think that is the correct assumption). In this case I assumed a current of 6.5 mA through Ri 113
Thus the voltage at the Ri 109, 111, 113 junction is 6.5 mA * 3.3k = 21.45 V, 29V-21.45 = 7.55 V
Thus the new source resistor voltage drop is 7.55 V + .35V (source voltage) = 7.9 V and thus the new source resistor value is 7.9 V / 5 mA = 1580 ohms
I'm hoping that the voltage divider with ri 111 isn't critical for some reason that Im unaware of and is ok with the reduced current through it compared to the original
design. The math is simple enough I hope but I really need help with checking the assumptions I used. Hopefully the new values I came up with would work better than leaving the stock values!
