Tuned amplifier basic design question

Thread Starter

georgefrenk

Joined Dec 24, 2023
54
Usual amplifier has two bias resistors, collector resistor and emittor resistor among other elements.
I have question about collector resistor (that will be replaced later with LC tank), that is connected between DC voltage source and collector pin of transistor.

Collector's resistor voltage is usually set as DC_VOLTAGE_SOURCE / 2 . And I choose Ic (collector current) from transistor datasheet - here I look for maximum Ic. By equation R = U / (Ic_max / 2), I get it's resistor value. And quiescent collector current is calculated by Ic_max / 2

But for tuned amplifier, how to calculate quiescent Ic - for purpose of DC biasing? Because the inductor, which is part of LC tank, has almost 0 ohms and that means also almost zero value of voltage on it.

I would need to get proper quiescent Ic,so I can later get quiescent Ib by equation Ib = Ic / beta and then calculate bias resistors as final step
 

crutschow

Joined Mar 14, 2008
34,455
You simply choose the minimum transistor collector bias current that will give you the frequency response you need, without concern for the collector voltage, which will stay at the DC collector supply voltage.
 

Thread Starter

georgefrenk

Joined Dec 24, 2023
54
I am planning to use Hybrid PI high-frequency transistor model equations.
Should I simply compute quiescent current Ic from equation Ic = U / Z , where U is DC_VOLTAGE_SOURCE / 2, and Z is impedance of LC tank at resonant frequency? Because in Multisim, when source signal frequency equals the resonant frequency of LC tank, the measured AC voltage on LC tank equals to the AC voltage on transistor CE resistor (internal transistor resistor between C and E pins). And if the DC_VOLTAGE_SOURCE is 9V, then LC tank has 4.5 V on it and transistor CE internal resistor has 4.5 V on it.
 
Last edited:

BobTPH

Joined Jun 5, 2013
8,989
First of all, you understanding of the collector resistor is totally off. You do not want 1/2 the max current the transistor can handle.

And secondly, when you replace the collector resistor with the a tank circuit, it will drop near zero volts with no input since the inductor has almost no resistance.
 

Thread Starter

georgefrenk

Joined Dec 24, 2023
54
I am willing to pay if someone would construct tuned amplifier from scratch with all equations and explanations, with HYBRID PI model and parameters taken from 2N2222 datasheet and if results of SPICE simulation would not deviate much from calculated parameters, especially voltage gain and power gain.
 

BobTPH

Joined Jun 5, 2013
8,989
Sorry, I misunderstood, this sounded like a homework problem.

So what is the purpose of this amplifier? What is the input? What does the output need to drive? What frequency? What bandwidth? How much gain?
 

Bordodynov

Joined May 20, 2015
3,180
I'll disappoint you. The 2N2222 transistor model in Multisim is very bad! And throw the calculations with this model into the trash! There are a lot of bad models in Multisim.
 

Thread Starter

georgefrenk

Joined Dec 24, 2023
54
1.) So, I choose for example quiescent current at 1mA .
2.) Then I calculate R of emittor resistor by Remittor = (1/10 * Vsupply) / 1mA
3.) Then I calculate Ib by equation Ib = Ic / beta
4.) Now I have beta current and I can now calculate bias resistors

And I am not using here the parallel resistor to LC tank.
Am I doing right? This is for DC equivalent circuit as first step.
 

Thread Starter

georgefrenk

Joined Dec 24, 2023
54
I wonder what beta value you are going to substitute into the formula? beta depends on current and voltage!
I use beta from transistor SPICE model parameter, named BF (Ideal maximum forward beta).
I go in reverse way. First I have my wanted quiescent current, then I calculate Re from it, then I calculate base current from it with help of BETA parameter from SPICE transistor model.

Correction: base current, not beta current
 
Last edited:

Bordodynov

Joined May 20, 2015
3,180
You are not right. I set the emitter current to 1 mA and varied the collector-base voltage from 0V to 10V. I built a graph of the static gain of the transistor versus voltage.

BF2n2222.png
 
Last edited:

BobTPH

Joined Jun 5, 2013
8,989
With an emitter resistor, the beta has very little effect in the on the bias point. You want the bias resistors to carry about 10 times the needed base current. The emitter voltage then rises to the voltage divider voltage - 0.6V.

When I have built tuned RF amps, I have simply used two 10K resistors and a 1K emitter resistor. This will give a reasonable bias point.
 

MrAl

Joined Jun 17, 2014
11,487
Usual amplifier has two bias resistors, collector resistor and emittor resistor among other elements.
I have question about collector resistor (that will be replaced later with LC tank), that is connected between DC voltage source and collector pin of transistor.

Collector's resistor voltage is usually set as DC_VOLTAGE_SOURCE / 2 . And I choose Ic (collector current) from transistor datasheet - here I look for maximum Ic. By equation R = U / (Ic_max / 2), I get it's resistor value. And quiescent collector current is calculated by Ic_max / 2

But for tuned amplifier, how to calculate quiescent Ic - for purpose of DC biasing? Because the inductor, which is part of LC tank, has almost 0 ohms and that means also almost zero value of voltage on it.

I would need to get proper quiescent Ic,so I can later get quiescent Ib by equation Ib = Ic / beta and then calculate bias resistors as final step
Hi,

Can you show your circuit? That gives us something definite to work with
 

Thread Starter

georgefrenk

Joined Dec 24, 2023
54
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
 

Thread Starter

georgefrenk

Joined Dec 24, 2023
54
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.
 
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