1
\$\begingroup\$

I know that you adjust the positive feedback just below oscillation. But how can this result in a higher gain? The feedback would not just compensate the resistive loss in the tank circuit and thus increasing the tank's Q?

\$\endgroup\$
  • \$\begingroup\$ just as negative feedback reduces gain, while positive feedback increases gain. When too much you get saturation with hysteresis, but when added just below unity gain, you get ringing.. That's regenerative feedback. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 16 '16 at 6:32
2
\$\begingroup\$

In a regenerative receiver the signal is amplified, then fed back into the input where it is amplified again (and again, and again, and...). Just before the point of oscillation the total gain is almost infinite. If the input has a tuned circuit then frequencies further way from resonance will be amplified correspondingly less, so selectivity is improved.

The problem with a regenerative receiver is that when you add feedback it becomes more sensitive to any change in circuit parameters, increasing the risk of breaking into oscillation. At very high gain, just moving your hand closer to certain parts could be enough to set it off.

A related design that got around that problem was the Reflex Receiver. It amplified the RF signal just once, then sent the demodulated audio signal back through the same amplifier. Because the signal being fed back was at a lower and unrelated frequency, it didn't reinforce the original RF signal and so wasn't prone to oscillation. Gain was only doubled, but this was worth doing when the amplifying device (valve or transistor) was much more expensive than all the other components needed to complete the circuit.

\$\endgroup\$
  • \$\begingroup\$ Is it correct to think a analogy to this process as a infinite series like 1+1/2+1/4+1/8..., so the value in the detector converges to some specific value intead going infinity or oscillating? \$\endgroup\$ – Lucas Sep 16 '16 at 11:41
  • \$\begingroup\$ Yes, but only when the 'loop gain' (amplifier gain x feedback ratio) is less then 1. Consider 1 + 1/1 + 1/1 + 1/1... \$\endgroup\$ – Bruce Abbott Sep 16 '16 at 16:14
  • \$\begingroup\$ Just to clarify. A loop gain less then means that the amplifier is actually decreasing the signal, right? \$\endgroup\$ – Lucas Sep 16 '16 at 17:06
  • \$\begingroup\$ sorry, i wanted to write : a loop gain less than 1 \$\endgroup\$ – Lucas Sep 16 '16 at 17:29
  • \$\begingroup\$ No, a loop gain (not input to output gain) of less than 1 means it is an amplifier rather than an oscillator. en.wikipedia.org/wiki/Positive_feedback#Basic \$\endgroup\$ – Bruce Abbott Sep 16 '16 at 18:13
1
\$\begingroup\$

The regenerative feedback antenna/coil bends the incoming EM wave. Any resonant device has this capability, but due to loss this phenomena can be neglected in ordinary antennas. With regenerative feedback, the indicent wave is bent and directed to the receiving coil, in a region of half wavelength. A laic answer would be, that regenerative receiver "sucks" the EM wave.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.