walid Posted August 7, 2009 Report Share Posted August 7, 2009 Hello guysThe attached photo is a crystal oscillatorit is a voltage divider config. but replacing the RC with LC tuning tankI need to know the out put impedance Zout??Is it XC// XL at the operation freq 27MHz???please help methank you in advancehttp://NOTE: sorry L= 3.48 uH not 1.2 uH as in the circuit Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 8, 2009 Report Share Posted August 8, 2009 I didn't calculate the resonant frequency of the tank but if it is 27mHz then it is an extremely high impedance. Then the circuit will not start oscillating or it will stop oscillating if it has a load. Quote Link to comment Share on other sites More sharing options...
walid Posted August 8, 2009 Author Report Share Posted August 8, 2009 thank you gurubut if it is 27mHz then it is an extremely high impedanceBe sure that the resonance freq is 27MHzis the Zout = XC//XL; the parallel equivalent of the two reactancess (coil and ca)???regards Quote Link to comment Share on other sites More sharing options...
Hero999 Posted August 8, 2009 Report Share Posted August 8, 2009 Yes, the output impedance will be equal to the impedance of the inductor and capacitor in parallel. The impedance of an ideal inductor and capacitor in parallel is infinite, in reality is won't be infinite but it'll be very high and depend on the Q of the tank. Any load connected to the output will reduce the amplitude and stability of the oscillator because it will affect the Q of the resonant circuit. Quote Link to comment Share on other sites More sharing options...
walid Posted August 9, 2009 Author Report Share Posted August 9, 2009 Any load connected to the output will reduce the amplitude and stability of the oscillator because it will affect the Q of the resonant circuitWhat the solution then?? Do I have to replace the coil with a stepdown RF transfrmer?? What is the turn ratio then if i want to match with 100K ohm load?thank you in advance. Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 9, 2009 Report Share Posted August 9, 2009 A 100k load is a high impedance to the LC parallel tuned circuit and will be fine. A 1k load would be much too low. Quote Link to comment Share on other sites More sharing options...
walid Posted August 9, 2009 Author Report Share Posted August 9, 2009 A 1k load would be much too low.what to do if the load is 1k only??Do I have to replace the coil with a stepdown RF transfrmer?? Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 9, 2009 Report Share Posted August 9, 2009 If you use an output transformer to increase the load impedance then it steps down the signal voltage.An emitter-follower transistor will increase the load impedance without the signal voltage loss. Quote Link to comment Share on other sites More sharing options...
walid Posted August 9, 2009 Author Report Share Posted August 9, 2009 Hi GuruThank you for attentionWhat do you think of this circuitfrom: http://www.talkingelectronics.com/projects/27MHz%20Transmitters/27MHzLinks-1.htmlThe author use a transformer then PI network filter for impedance matching the low Z antenna (50 ohm) to the very high (near infinity) Z tank circuit.I am a week ago, trying to understand, and I hope that you help me to understand Thanks Quote Link to comment Share on other sites More sharing options...
Hero999 Posted August 9, 2009 Report Share Posted August 9, 2009 An alternative to building a transformer is to simply put a tap in the middle of the inductor, the apparent impedance of the load will be increased by a factor of four and the voltage across it will be halved so the power will be down by a factor of four. Quote Link to comment Share on other sites More sharing options...
walid Posted August 10, 2009 Author Report Share Posted August 10, 2009 Hello guysIn the following link, http://www.electronics-tutorials.com/amplifiers/negative-feedback.htm the author speaks about how to made the necessary calculations to determine the transformer's turns ratio and also the LC needed for impedance matching a 50-ohm antenna to the preceding stageRegardless of some of the details that the writer is talking about, espicially his talking about the dc biasing and the o/p voltage swing, I will focus in the discussion with you on the method of calculating the turns ratio.Look at the fig:the author said:Here we are only mainly concerned here with the principle of a tuned circuit as a load. In this example we have a final load of 50 ohms connected to the output link coupling of our tuned circuit. Assume, just for discussion purposes, that in this example we needed a power output of 100 milliwatt from the amplifier and that we have available a power supply of 12V. Further, Now: 2 * Po = [Vcc - Ve]^2 / R or [2* 0.1] = [12 - 3.22]^2 / R or [2* 0.1] = [ 8.78 ]^2 / R and R = 77 / 0.2 = 385 ohms Therefore the load presented to the output of our amplifier needs to be 385 ohms. The impedance ratio is 385 / 50 or 7.7:1 and the turns ratio on the transformer (inductor) is the square root of that number or 2.775:1. My questions are:1)Are these calculations are true in general?2) Why '2' in 2.Po = [Vcc-Ve]^2/R, i learn that P = V^2/R3)Why [Vcc-Ve]^2, I think it must be [Vcc-Vc]^2Eagerly awaiting your reply Thanks Quote Link to comment Share on other sites More sharing options...
walid Posted August 20, 2009 Author Report Share Posted August 20, 2009 My questions are:1)Are these calculations are true in general?2) Why '2' in 2.Po = [Vcc-Ve]^2/R, i learn that P = V^2/R3)Why [Vcc-Ve]^2, I think it must be [Vcc-Vc]^2Eagerly awaiting your reply ThanksI discover the answer of the second question: the RMS value of the peak voltage = Vpeak/squre root of 2and when we squre that root we get 2 which appear in the equation.i need answer to question 3thank you very much Quote Link to comment Share on other sites More sharing options...
walid Posted August 21, 2009 Author Report Share Posted August 21, 2009 I appeal to my friend audioguru to respond to the subject I waited for his reply for more than a week Thanks Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 21, 2009 Report Share Posted August 21, 2009 Hi walid,I have not done RF work in my career and I haven't done school-work for about 42 years.The transistor is a linear amplifier so its collector swings down to about 4.5V and swings up to the same amount above the supply voltage because of the tuned circuit. So the voltage swing is doubled.Read further in the article where the author explains that the transistor is clipping and that its current should be higher. Then re-calculate the impedance of the transformer's primary. Quote Link to comment Share on other sites More sharing options...
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