1
\$\begingroup\$

As I understand, class-C amplifier can only generate signal of constant frequency and amplitude. So the question is, how can it be useful and what are its applications?

enter image description here

\$\endgroup\$
6
  • 1
    \$\begingroup\$ Amplitude doesn't have to be constant, and frequency can vary within the passband of that LC filter. So, AM radio and FM radio transmitters. \$\endgroup\$ – user_1818839 Feb 9 at 0:18
  • \$\begingroup\$ @BrianDrummond How can amplitude and frequency vary? \$\endgroup\$ – Rodion Degtyar Feb 9 at 0:26
  • \$\begingroup\$ They both come from the input signal so of course they can vary. \$\endgroup\$ – user_1818839 Feb 9 at 0:45
  • \$\begingroup\$ @BrianDrummond: Class C. The output amplitude is largely determined by the supply voltage. But -- still good for FM, or on-off keying, or PSK. \$\endgroup\$ – TimWescott Feb 9 at 0:52
  • 2
    \$\begingroup\$ The output is driven by the transistor -- the tank circuit is passive, it can only be driven. Typically, the tank circuit has some bandwidth over which it works; as long as the signal that's being amplified is within the tank circuit's bandwidth, the amplifier will work. \$\endgroup\$ – TimWescott Feb 9 at 1:06
1
\$\begingroup\$

Narrow band RF applications are where the class C amplifeir is usefull .FM ,PM,FSK,PSK schemes dont rely on amplitude linearity so the good efficiency and simplicity of class C is an advantage .AM can be implemented by modulating the power supply voltage which is also simple.

\$\endgroup\$
1
  • \$\begingroup\$ @Rodion Degtyar Classic AM modulator for C-class amps adds AC (for. example an audio signal) to the power supply DC. The secondary winding of a transformer is in series with the DC supply. The method has a drawback: As much AC power is needed from the transformer as there's needed DC power to generate the AM carrier. I the power of the AM carrier is say 5W in a walkie talkie, a 5W audio amp is needed to feed the modulation transformer. \$\endgroup\$ – user287001 Feb 9 at 1:11

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.