Hi Extremist
I told my questions to my instructor and then she gave me some hints;
1)In order to set the input resistance easily she suggested me to use MOS instead of BJT.Because in BJT's small signal modell the input resistance consists of R1//R2//R∏ (where R1 R2 are Base Voltage Divider resistances and R∏ is the resistance occurs from fabrication process).She told me that you can easily set the input resistance by only choosing Voltage Divider resistances in MOS.
Go with which device you are most comfortable with for sure for this assignment or which will be the greater challenge for you. Learning and taking away added knowledge is the objective. I thought I would also take it on as a refresher so I chose using BJT's.
Now the complete equation for a BJT's input impedence, not to contradict your professor, is as follows:
Zin = Ra // Rb // B(Re + 26mv/Rei) where Ra & Rb are the base resistors, as you noted, Re is the
non-bypassed emitter resistance of the emitter circuit, the 26mv is a
physical constant of the emitter/base diode at 1ma (silicon or germanium), which is linearly scaleable and also referred to as "re", Rei is the emitter current and B is the device beta. This yields a very close approximation below about 10Mhz, above which one must resort to using the "h" parameters from the datasheet of the device and all those laborious equations. Use of a spreadsheet helped me converge on the solution, which resulted a Zin of 50,012 ohms using E96 resistors (1%).
The above described configuration is what I meant by thinking out of the box. I'll post the basic schematic without values below in case I haven't been clear.
2)How can i calculate the gain? She asked this question ;
You can calculate it by dividing Output/İnput.In the project that gave you, i mentioned input , by saying "input senstivity". So our input sensitivity is our input (2,5mV RMS).By doing appropriate calculations (RMS-Peak) you can calculate your gain.
Given Av = Vout / Vin, both terms must be either RMS, Peak or Peak-Peak. Since the input was given in RMS and the output in Peak voltage, it can be written as either Av = 2V / 2.5mv*2sin 45 or Av = 2V*sin 45 / 2.5mv, which both yield the same solution of Av=565.7.
3)How Hum or THD effect our circuit design?
She said that do not think about them i gave that specifications just for extra information.The important thing is your feedback configuration.
In this project we can see that that is voltage amplifier.But how can i decide its feedback configuration shunt-voltage or series-voltage?
Consider the purpose of feedback in general. If one wishes a signal to be increased it must be in phase, which is a prime tenet of Birkoff's(sp?) Law of oscillation; regenerative feedback of the proper amplitude. On the other hand, to limit a signal's amplitude, the feedback must be degenerative in nature; of the proper amplitude but
180 deg out of phase to obtain the desired output level.
If you know what the input and output levels are at the second stage, the ratio gives you the amount of feedback required. Since the overall gain is to be 565.7, it is obvious that your overall gain
before feedback must be somewhat greater. How about an Av of 800, which is precisely 3db higher meaning a required feedback of that degree.
So after you design your amplifier and the gain is known, look at it and find the best location for the feedback that will not interfere with the I/O impedances, that does not impact DC bias levels and is the correct phase.
I hope this helps. Good Luck.