Yes. I initially modeled it with LSK489A with standing current below the lowest possible Idss, and 3k9 was the optimal value of load resistors for it. I did not update them after replacing the input device with JFE2140 and increasing the current to ~4mA per side.
Thank you for your detailed analysis, Marcel.
So it would seem that changing the way I connect them ... is not really going to affect the resulting capacitance load on the cart much?
Not much, just a little.
BTW, if by "each JFET is biased at about 2 mA of drain current", you mean the current which is flowing through each jfet, from Drain to Source - I have this at about 4.8ma (75% of Idss).
If they draw 9.6 mA together, then the voltage drop across the 46 ohm source resistor must be about 0.44 V.
At -0.44 V gate-source voltage, the current is far below 4.8 mA according to this Toshiba graph, at least at 10 V drain-source voltage.
If they draw 9.6 mA together, then the voltage drop across the 46 ohm source resistor must be about 0.44 V.
At -0.44 V gate-source voltage, the current is far below 4.8 mA according to this Toshiba graph, at least at 10 V drain-source voltage.
The value of 46 ohm is what I need for R4 in the LTspice sim.
In 'real life' - ie. actually measuring the built circuit - the value of R4 is 4R4 and the Source voltage I get ... leads to a combined current of about 9.6ma. Drain-Source voltage is about 8v.
With 4.4 ohm and about 30 mS per FET (transconductance at 4.8 mA from a Toshiba datasheet graph), only 20.886 % of the input AC voltage will drop across the resistor, so you can only get about 20 % of capacitance reduction. The remaining capacitance is then 48 pF when you connect C3 to the resistor, 50.4 pF when you don't.
This seems to work.
I had to introduce compensation to make it stable.
AC analysis shows a 2dB peak at 3.2MHz. It's a good thing that there will be a passive low-pass filter implementing the 75us RIAA pole after this stage.
I had to introduce compensation to make it stable.
AC analysis shows a 2dB peak at 3.2MHz. It's a good thing that there will be a passive low-pass filter implementing the 75us RIAA pole after this stage.
With 4.4 ohm and about 30 mS per FET (transconductance at 4.8 mA from a Toshiba datasheet graph), only 20.886 % of the input AC voltage will drop across the resistor, so you can only get about 20 % of capacitance reduction. The remaining capacitance is then 48 pF when you connect C3 to the resistor, 50.4 pF when you don't.
Thanks, Marcel. 👍
So ... not really worth doing. 🙁
But I'm confused by sk's post here:
But if you tie the base (or the gate) of the upper device to the source of the lower device, you’re preventing any voltage variation from developing between the drain and the source of the input device(s), which not only mitigates the Miller effect, but also effectively eliminates the capacitance of the input device altogether.
How can I do what he says ... ie. "eliminate the capacitance of the input device altogether"? What change in my circuit do I need to make, to achieve this?
There are two in parallel, so you should draw a 92 ohm loadline to find the current per JFET. Hence my estimate of 2 mA for an average IDSS.
Thanks, Marcel. 👍
So ... not really worth doing. 🙁
But I'm confused by sk's post here:
How can I do what he says ... ie. "eliminate the capacitance of the input device altogether"? What change in my circuit do I need to make, to achieve this?
stratokaster83 assumes perfect voltage follower behaviour from gate to source. You cannot completely achieve that, but you could get a lot closer by replacing the FETs with some multistage negative feedback circuit (for example a Sziklai pair with a JFET as its first stage and a current source or resistor for biasing it) or by using a much larger source resistor - which would reduce the transconductance, which is unacceptable according to one of your earlier posts.
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