Error-correction power amp: the basics….
Here is the basic concept, for clarity applied to an amplifier stage with gain of A=1, say a power amp output stage.
We have the main amp with gain ‘A’ which is almost, but not quite, one. The amp’s gain of course varies in time and with the signal and with frequency, which leads to distortion. Error correction looks at the difference between the amp output and the amp input. Any difference is returned to the input and added to or subtracted from the input, in precisely the right amount to make the output identical to the input.
Let us assume that the output stage A is ideal and has a perfect gain of 1. In that case, the correction signal at Vc is 0, because there is no difference between Ve and Vout. We could just throw away the whole error correction part. But if the gain A is just a bit less than 1, Vc is a negative signal which is subtracted from Vin and the effect is that Ve increases. This works as a positive feedback loop. In the case where the gain A would be larger than 1, the signal at Vc is a positive signal, which will be subtracted from Vin to decrease Ve. In this case, the system works as a negative feedback loop. So, what you see is a combined feedback loop that can act either as a positive or negative feedback loop, or not at all, depending on the error of the main amplifier block. It looks as if the phase and level of the loop adjusts itself as needed.
The interesting thing is of course to find an elegant way to implement this principle in a real amplifier.
Error-correction power amp: the concept….
After a lot of searching, I decided to use a current conveyor
to develop the error signal and to add it to the input signal. Current conveyors work as follows: The
output Z generates a current identical to the current in terminal Y, but
of opposite direction. Y has almost zero input impedance and will follow
the voltage at X. (If you think that this sounds just like an ideal
transistor, you're right). So, with X (and Y) the same, we see that the
current Ierr into Y must be (Vout-Ve)/R.
Suppose that at a certain point A is too high by an error Verr. That causes a current Iec=Verr/R into Y, and, because of the CCII working, out of Z. This generates a voltage across the input R of Iec*R so Ve becomes Vin-(Iec*R). If we work that out we get Ve=Vi-Verr. Since Vout was just Verr too high, subtracting Verr will neatly bring Vout back again exactly to one…
This concept also works with amplifiers with a gain other than one. The
calculations and the circuit are a bit more involved, but it really
works in practice.
