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How does this circuit perform Frequency modulation. I know the LC generates the carrier frequency. How does the transistor act as a varicap to perform FM. Which part of the circuit acts as a varicap and how ?

To people who marked this as duplicate - My question is how is the miller capacitance "effectively" in parallel with the inductor ?


1 Answer 1


All BJTs have what is known as "miller capacitance". It can be "altered" by a greater or lesser reverse bias voltage across the collector-base region. Note that C1 shunts this small capacitance to ground and that C3 returns this capacitance to the positive rail thus ensuring miller capacitance is effectively in parallel with the inductor.

This change in capacitance alters the tuned circuit in the collector and shifts (or deviates) the frequency. Given that the average DC voltage on the collector is pretty much 9V (due to the inductor), moving the base voltage up and down modulates the carrier.

Also, pretty much any reverse biased diode I've come across exhibits this effect and that is how varactor diodes work as well.

  • 1
    \$\begingroup\$ I understand how the miller capacitance acts as a varicap to provide FM. But, what I don't understand is how the Ccb(collector-base capacitance) is in parallel with the L(inductor) ? \$\endgroup\$ Commented Mar 29, 2014 at 3:35
  • \$\begingroup\$ C1 and C3 ensure it is in parallel. C1 and C3 act to short the base to 9v rail for HF signals like the carrier. \$\endgroup\$
    – Andy aka
    Commented Mar 29, 2014 at 9:57
  • \$\begingroup\$ Wouldn't a single capacitor from base to 9v suffice ? Why is C1 connected from base to gnd, and then gnd is connected to 9v through C3. \$\endgroup\$ Commented Mar 29, 2014 at 13:24
  • \$\begingroup\$ I built this circuit on the breadboard and it worked fine the first time itself. I really want to explain it to a bunch of other guys, but I can't understand the role of C1 and C3. \$\endgroup\$ Commented Mar 29, 2014 at 13:29
  • \$\begingroup\$ A single cap from base to 9v is fine dude. \$\endgroup\$
    – Andy aka
    Commented Mar 29, 2014 at 13:46

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