Capacitor between emitter and collector in fm modulator

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The circuit is a FM transmitter and that capacitor is used for Oscillator Feedback..
 

could you tell me how it works?
if we use ac analys it we know in small signal it should be shorted.how does it work?
 

baby_1 said:
could you tell me how it works?
if we use ac analys it we know in small signal it should be shorted.how does it work?
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You can not do AC analysis with this circuit because it's an oscillator that works autonom.
You can make a transient analysis by providing start-up conditions.

Note:AC analysis is done in "steady state occured" circuits.
 
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Thanks alot dear bigboss.could you tell me a reference about this circuit(analyse them and explain more)?
 

The base of Q2 is grounded to RF. If a little bit of noise gets into Q2's emitter such as to go positive, this reduces the forward bias on its base-emitter junction, so its collector current falls and the collector voltage rises. So the voltages at Q2's emitter and collector are in phase. The tuned circuit L1 & C6 only provide gain at their resonant frequency, so as the gain is peaked at this frequency the circuit oscillates at this frequency. C5 is to provide the positive feedback and stops the low input impedance of the emitter damping the Q of the tuned circuit by having a high impedance.
Frank
 
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Thanks a lot dear chuckey .
could you tell me how can we calculate the C5 feedback capacitor value accroding the circuit & frequecny?
Thanks
 

The input to the "amplifier" is the emitter impedance of the transistor. I can't remember the formula, buts something like Hfe/Ic(in mA), anyway its low, say 25 ohms. So the voltage across R6, sends some current down through R6 to earth and some into the emitter (the importent bit!). Because of the action of the transistor this current (or at least 99% of it), flows through the tuned circuit, causing a voltage to appear across it, there is also the output resistance of the transistor shunting the tuned circuit (high resistance 20K?). Its is this voltage that sends the current through C5 into the emitter and causes the volt drop across R6.
Frank
 
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Hi Baby,

some value must be wrong in the schematic.
As is, Q1 goes in saturation.
C7 should be 22 pf or 0.022 nF.
Regards

Z
 

Well spotted Zorro! . If Q1 takes 1mA, then Ve = 1V, Vb= 1.6V, so Vcc -1.6V = 7.4V, so R2 : R3 = ~5:1, so I would make R2= 100K. As for C7, ever tried buying a .022 Pf capacitor?, I reckon 22Pf is about right.
Frank
 

Thanks dear frank.
didn't remeber where you read this anaylise or book that you have been read before?
 

I can't remember, I first came across transistors in about 1962. Wow!, that was almost fifty years ago. I have been learning about them ever since. Perhaps one day I'll understand them. ;-)
Regards
Frank
 
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