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Here is a simple FM transmitter that I've been studying recently. What I'm wondering about currently is how the BJT amplifier in this circuit is biased. I believe that from the audio signal standpoint the BJT is working in a common emitter mode: the signal from the mic is input into the base, and the output appears at the collector and the voltage between the two changes the base-collector capacitance, modulating the signal. From RF signal standpoint it works in common base mode: unwanted frequencies are filtered from the collector to ground through the LC circuit, while the frequency of interest is blocked by it and fed as feedback through C2 into the emitter. The base is grounded for the RF signal through C1. Am I correct so far?

Back to biasing: looking at the circuit, it seems to be biased using "base bias". Let's calculate the quiescent current through the base:

\$I_{b} = \frac{9V - 0.7}{4700\Omega + 470\Omega}\approx1.6mA\$

It's impossible to know exactly what the beta is for the specific transistor, but let's say \$\beta=100\$. Then:

\$I_{c}=\beta I_{b} \approx 160mA\$

Therefore the quiescent voltage across R1 is approximately \$0.160 A * 470\Omega = 75.2V!\$ Clearly this is not possible, since the battery has only a voltage of 9V. So what is my error in understanding the biasing here?

  • \$\begingroup\$ Beware this kind of circuit is rather horrible; study it if you really want to as a curiosity, but you'll likely learn more of practical use studying designs where stages have clearly distinct functions. \$\endgroup\$ Commented Aug 19, 2019 at 15:05
  • 1
    \$\begingroup\$ What is MIC1? Looks like it could be a FET-type Electret.... it would consume considerable current through R2 (more than base current of Q1). \$\endgroup\$
    – glen_geek
    Commented Aug 19, 2019 at 15:55
  • \$\begingroup\$ @glen_geek I assumed it's just a simple condenser mic, but I could be wrong - it's not mentioned where I found this circuit.. \$\endgroup\$
    – S. Rotos
    Commented Aug 19, 2019 at 16:02
  • \$\begingroup\$ It had better be an electret microphone. A dynamic microphone would not much like the current from R2. \$\endgroup\$
    – JRE
    Commented Aug 19, 2019 at 16:09
  • \$\begingroup\$ Cbc parasitic capacitance is the main cause for FM modulation in this circuit. And Cbc changes his value due to amplitude change thanks to the signal from the microphone. And Cbc for the AC signal is in parallel with VC1 capacitor and this is why we have FM modulation together with unwanted AM modulation. \$\endgroup\$
    – G36
    Commented Aug 19, 2019 at 17:14

1 Answer 1


You used the wrong equation for the base current. The correct one is:

\$ I_B = \frac{V_{CC} - V_{BE}}{(\beta +1)R_1 + R_2} \approx 160\mu A \$

But in your circuit, the situation is more complicated because the MIC will sink current (around of \$ I_{MIC} \approx 500 \mu A\$).


$$I_B = \frac{(V_{CC} - V_{BE}) - I_{MIC}\cdot R_2}{(\beta +1)R_1 + R_2} \approx 114 \mu A $$

For \$\beta = 100\$ and \$V_{BE} = 0.7V\$

Because the base current is lower than the electret microphone current. We can neglect the base current and solve for the emitter current directly.

$$I_E = \frac{V_{CC} - I_{MIC}\cdot R_2 - V_{BE}}{R_1} \approx 12.7mA$$

  • \$\begingroup\$ Oh, I should have realized this simple mistake! Thank you very much. \$\endgroup\$
    – S. Rotos
    Commented Aug 19, 2019 at 17:55

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