FM Tansmitter - AC Equivalent Circuit - Capacitor Analysis

This is a follow on question from the following two questions:

From link 1, an AC equivalent circuit is derived as shown below:

My question: If C1 is shorting all AC signals directly to ground - how do they propagate through the circuit? Usually the AC signal is applied across the transistor. Why is that not the case here?

EDIT: I had a further look at common base BJT configurations and re-arranged the circuits to match the fundamental example (see below).

For the signal generator lets assume we are going to use a microphone. Would it be correct to assume that it would be wired as shown in the AC equivalent circuit?

After re-arranging the circuits, I realised on the AC circuit everything from the collector side has now been shifted onto the emitter side. Is this correct? If so..why?

• Simple: for the RF frequencies it works as a common base. There does not need to be an RF signal at the base, just the (Low frequency) modulation signal. This modulates the $C_{be}$ capacitance of the NPN resulting in FM modulation. – Bimpelrekkie Dec 5 '17 at 14:59
• I'm just having trouble visualizing the 'path'. I automatically assume since its connected to ground...the signal dissipates like a lightning strike hitting ground (probably not the best analogy, but gets my point across). I also appear to have my terminology mixed up. I assumed an 'RF signal' was any AC signal...even the low frequency ones to be modulated. Is my assumption incorrect? – SheerKahn Dec 5 '17 at 15:10
• @SheerKahn Hi I have a couple of questions; What is the frequency range you are expecting as input to the base of BJT? – rsg1710 Dec 5 '17 at 15:42
• @rsg1710 I don't know exactly what this circuit was intended for but I imagine 85-255Hz (Human voice) – SheerKahn Dec 6 '17 at 10:44

C1 shorts 100MHz. The reactance is about 1.6 ohms. At audio frequencies, say 1 kHz, its reactance is 100000 times higher and the effect accordingly smaller.

At audio frequencies (below 20 kHz) C1 causes a little attenuation, but the effect is so small that the audio signal well makes the operating point of the transistor to swing. Thus the internal capacitance of the transistor changes and causes FM.

Between 20kHz and 100MHz the effect of C1 increases gradually as frequency increases. But there's nothing happening between the audio band and the transmitting frequency in this circuit. At the transmitting frequency C1 has so low reactance (about 1,6 ohms at 100MHz) that you can think the base being grounded and that's needed to make the oscillator to work.

It's very common in electronics that the same circuit work totally differently in different frequencies at the same time. Mostly this is caused by frequency dependent reactances. The effect can be wanted as just in this case. Unfortunately many parts also have limited usable frequency range due unwanted reactances.

ADD2: DC source is a short circuit in the equivalent AC circuit. You must have separate equivalent AC circuits for 100MHz and audio due their big difference. In Audio AC circuit the 100MHz tank circuit coil is a short circuit and those few pF capacitors are open (=left out). In 100MHz AC circuit few pF capacitors and L1 are essential, the 1nF capacitor is a short circuit.

BTW. Your common base amplifie principle diagram belongs to the trashcan. Its taken from 1940's. BJT's of today work a little differently than early transistors. Unfortunately the tradition to use the same teaching image still continues. You can see the same in many explanations of radio waves. The sign of shit there is a pendulum, resonant circuit and gradually opening gap of the capacitor. Shit = nice story, unfortunately not physics.

• What is the benefit of that? Just to clarify, are you saying it shorts anything below 100Mhz? – SheerKahn Dec 5 '17 at 15:15
• @SheerKahn The answer is updated – user287001 Dec 5 '17 at 15:24
• So in this case..would that input signal be amplified and then passed into the tank circuit for modulation? Or simply passed through? On another note, is R1 and C2 forming a High Pass Filter? – SheerKahn Dec 5 '17 at 15:35
• @SheerKahn Audio signal does not alffect in Tank Circuit L1VC1. Audio signal affects to transistor's internal capacitance which still is between C and B. B happens at 100Mhz be grounded by C1 and that makes the capacitace to be effectively in parallel with VC1. C2 is the feedback route in the oscillator. R1 is needed for DC current, taransistor operating point stabilization and making some resistance at 100MHz for AC, too to make the feedback from collector (=output) to emitter (=input) effective. The feedback does not work at audio frequencies, so R1C2 is a high-pass filter. – user287001 Dec 5 '17 at 18:18
• Question updated – SheerKahn Dec 6 '17 at 10:52