I think first I must choose quiescent collector current . Then I must calculate Re resistance, I have already choosen quiescent collector current, so now is only question, what voltage should I apply to it (to emittor resistor) via equation, so at end I get calculated Re.

I tried this way, but it doesn't work. I choose 3 volts for Re and quiescent collector current as 0.5 mA. I get then resistance of Re by R = 3V / 0.5 mA and it is 6 Kohm . Then I run simulation for DC equivalent circuit and Remittor has voltage of 14 volts, which is half of the DC voltage supply.

 

You need another 10K from the base to the positive supply. Your circuit biases the transistor to fully off.

 

Also, 1pF is way too small for your
tank capacitor. Stray capacitance and the parasitic capacitance of your inductor will swamp that. I would go with 100pF and calculate the inductor.

 

 

What should be the Re (emitor resistor) value equation for basic - DC equivalent circuit?
Inductor as 0 ohms, and rest of voltage is shared across Re and transistor Rce internal resistor

Hi,

Bob has given you some good advice I'll just add a little here.

For DC, inductors are short, and capacitors are open.
For AC, inductors are open and capacitors are short.

That's the simplified version for AC though, but for DC that is true. Only exception is if the inductor has significant ESR and then you have to include that in the DC bias setup. Here significant means it is large enough when compared to the other parts of the circuit. Some inductors can have an ESR of 10 Ohms or more for example, and that may or may not be enough to have to include it in the DC bias calculations.

 

I think first I must choose quiescent collector current . Then I must calculate Re resistance, I have already choosen quiescent collector current, so now is only question, what voltage should I apply to it (to emittor resistor) via equation, so at end I get calculated Re.

One way to do it is to assume a particular Beta, such as 50. Once you do the calculation and get a value for say Re, then change the Beta to maybe 100 and see if the design still looks like it will work. If the transistor can have a Beta of 200, then check with that too. A decent design will be able to work with a range of values for Beta.

 

For the DC bias point beta has almost no effect on the circuit I have posted.

I calculated the resistor based in 1 mA emitter current, ignoring the Vbe. Beta does not enter into that. Beta will affect the collector current since the the base current will be 1/beta times the collector current and the emitter current is collector current + base current.

Since I posted the circuit, I realized that the gain would be higher with an emitter resistor bypass capacitor. Z found experimentally that 100pF works well, giving me 8.5V peak output for 1mV input with no load.

Larger bypass capacitors resulted in lower gain, which I do not understand.

 

For the DC bias point beta has almost no effect on the circuit I have posted.

I calculated the resistor based in 1 mA emitter current, ignoring the Vbe. Beta does not enter into that. Beta will affect the collector current since the the base current will be 1/beta times the collector current and the emitter current is collector current + base current.

Since I posted the circuit, I realized that the gain would be higher with an emitter resistor bypass capacitor. Z found experimentally that 100pF works well, giving me 8.5V peak output for 1mV input with no load.

Larger bypass capacitors resulted in lower gain, which I do not understand.

Only beginner circuit designers do not take into account the impedance of the signal source. At a certain signal impedance, your amplifier may begin to oscillate. You also need to take into account the inductance resistance, which includes the skin effect.

 

For the DC bias point beta has almost no effect on the circuit I have posted.

Yes - exactly THIS is the goal for each good design.
For the circuit in post#24, the current through the voltage divider is roughly (ignoring the unknown base current)
I1=15V/20k=0.75mA allowing a base voltage of Vb=15V/2=7.5V.

Assuming Vbe=0.7V the emitter voltage is Ve=Vb-0.7=6.8V and the emitter current is
Ie=6.8V/7.5k=0.9mA

That means: Even for a pretty large base current (neglected in the above calculation) of Ib=0.9mA/50=18µA (assumed beta=50) the calculated base voltage will practically not depend on this small current (18µA<<0.75mA).
Hence, the calculation above was exact enough assuming Ib<<I1.

Fazit: For a good design with a base voltage divider allowing a divider current which is much larger than the base current (rule of thumb: At least factor of 10) the large tolerances of beta plays practically no role.
The base current ib does exist, but its uncertainty has practically no influence on the design.

 

Only beginner circuit designers do not take into account the impedance of the signal source. At a certain signal impedance, your amplifier may begin to oscillate.

Please, can you explain a little this statement? For oscillation, a positive feedback loop is required. Where can you see such a loop?
Do you refer to the Miller effect?

 

The Miller effect means something slightly different. I mean the collector capacity. In the presence of inductive impedance of the signal source, oscillation is possible. Also, in reality, the transistor has parasitic body parameters, including inter-terminal capacitances. Let me emphasize once again that for proper calculation it is necessary to take into account the source impedance. Not taking this into account is a typical schoolboy mistake.

 

For the DC bias point beta has almost no effect on the circuit I have posted.

I calculated the resistor based in 1 mA emitter current, ignoring the Vbe. Beta does not enter into that. Beta will affect the collector current since the the base current will be 1/beta times the collector current and the emitter current is collector current + base current.

Since I posted the circuit, I realized that the gain would be higher with an emitter resistor bypass capacitor. Z found experimentally that 100pF works well, giving me 8.5V peak output for 1mV input with no load.

Larger bypass capacitors resulted in lower gain, which I do not understand.

Yes that's great. My comments were of a more general nature. I asked for his circuit and you can see in post #18 that he has a single base resistor bias.
I usually recommend going through a Beta spread test anyway though just to make sure the circuit stays biased correctly and provides enough gain even with lower Beta.

 

I already warned that the 2N2222 transistor in Multisim has a very bad model. Here's a demo:

 

Generally the very first step in designing an amplifier of any sort is deciding what the output should be, either voltage, power, or current. That will allow reasonably designing the L-C tuned circuit to satisfy the output requirements. After that will come selecting a transistor able to produce the desired output. At this point linearity and distortion requirements must also be considered with the transistor selection. After that comes choosing the operating point of the transistor, between saturation and cutoff..

 

I see for 2N2222 there are many SPICE models on web. And different SPICE models for this transistor do affect very much the measured Voltage AC gain in Multisim. I found on web the one, that has measured Voltage AC gain very much near my calculated value.

How should I choose the proper one, that I should trust? Should I rely on LTSPICE models?
Should I test every model for Voltage AC gains with different source frequencies?