Electronics > Beginners

What does this Jfet do?


Could anyone explain what Q401 does in this circuit and is the part anything special as the original is long out of production?

David Hess:
JFET Q401 is operating as an amplifier with current feedback to the source.  Using a JFET has the advantage of lower input bias current and higher input impedance.  It looks to me like it was selected for very low gate cutoff voltage because otherwise it was intended for RF applications.

If it helps, the big-picture role of q401 is exactly the same as it would be if it was an npn BJT.   I am not saying that you can replace it with a BJT without changing anything else about the circuit, but the high-level logic of the design is the same whether it is a JFET or a BJT.   

I’m sure there are multiple ways to think of it.  In my mind, I find it helpful to think of it as a common source (inverting) amplifier for the input signal entering the gate, and a common gate (non-inverting) amplifier for the feedback signal entering the source.   So the signal at the drain will be proportional to the difference of those signals. 

I have built headphone drivers with similar circuits, except using an npn BJT for q401, a Vbe multiplier instead of D401 to bias the push-pull output stage, and a current source instead of R407.   This can be a very low distortion topology. 

By the way, if you build this, you might find that you need a capacitor from base-to-collector on q402 for stability/phase margin reasons.  Or maybe not…


Terry Bites:
The jfet a common gate stage.  As pointed out by JasonRF, they could have used a bipolar transistor here with some minor fiddling.
May be it was fun day at Sony. It's a fossil circuit but I think it was probably very cheap to produce. There are easier ways of making an amp with a 100Hz peakiness.
Its a good illustration how to use feedback to get the inverse of the filter function.


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