I recently designed several variable power supplies to test tube amps, actually I designed more than one at the same time with different characteristics. They all have one component in common with the rotator. The rotator is used as an active inductance, that is, it simulates the behavior of an inductance with regard to the trend of impedance based on frequency. Note: unlike a real inductance, it does not release energy, then work on a single quadrant, which in no way affects its functioning as regards filtering. With high frequencies, and this is not the case, tends to act as a constant current generator, with 100Hz in the circuit in which it is inserted it performs the desired function, that is, by filter inductance in a CLC pi-Greek circuit. In itself it is very efficient as it is a condenser filter followed by a low pass LC
the simplified value of the simulated inductance is
L = C*R1*R3
we see the case of the circuit in which the following values are used: R1 = 100k R2 = 100k R3 = 27 Ohm C = 22uF and compare it with an inductance from 59 Henry. The image below shows us the curves of the rotator in question (in viola) and that of inductance ( yellow ) as we can see the curves are almost superimposable.
from here on, if you want to continue reading there is the discussion relating to the circuit in question both at a mathematical and engineering level, the mathematical one is an end in itself and extremely precise, engineering, tends to simplify things when the error is negligible in order to manage something difficult to manage even at the price of a small discrepancy between the theoretical and simplified calculations.
THE TURNER: MATHEMATICAL TREATMENT
the circuit of the rotator used is the following:
and its representation for small signals is as follows:
I highlighted the currents and using the methods of the tensions at the nodes and the mesh currents I can represent it with the system of four linear equations in four unknowns below.
simplifying we get
the solution is not exactly simple and it is far from short you can give it to digest at wolphram / alpha, or you can calculate it by hand, hoping not to run into one of the classic oversight errors that would make hours of calculation futile. we only carei1
we know that
but since we work on only one quadrant we can assume that
so the replacement we will make will be
and our solution will be:
as we can see the situation is far from simple and the formula unless you have a scientific calculator in which to insert it is practically unusable. If you are still reading, I will try to simplify things in the following paragraphs up to the simplified initial formula.
THE TURNER: PRACTICAL TREATMENT
Here the engineer mentality comes in handy, a bit’ common sense and simplification: assuming a capacitor of at least a dozen microfarads, a 100Hz, which is the frequency of the residual ripple ( ripple ) that we want to eliminate, this has an impedance of a hundred ohms, THEREFORE much lower than R2, and then for calculation purposes R2 can be neglected. our circuit then becomes the following:
—- NOTE!! as far as the polarization of the MosFet R2 is concerned, it is ESSENTIAL, while C should not be considered The equivalent circuit is therefore the following and the relative formulas are these:
Replacing and simplifying
we know that
and simplifying further
we are interested in a formula that resembles that of real inductance and that is dependent on C, therefore something like this
our inductance is therefore
and ignoring also 1 / Gm which is negligible compared to R3 we obtain that the simulated inductance value is
L = C*R1*R3
By far a widely usable formula that allows excellent filtering without resorting to extremely expensive capacities or even worse if you use physical inductances from the cost, not indifferent weight and dimensions in case of power supplies for valves.