Transistor Linear Region for CC Amplifier

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
Hey Fellows,

I am trying to design a CC audio amplifier and wanted some help. My goal is to put the 2N3904 transistor into it's linear region and away from its saturation region. I want it to be as efficient as possible. In the attachment you'll find the basic schematic and so I'm trying to find the resistor values. R3 and R1 form a voltage divider and I'm wondering if it would be better to take out R1. Vcc=9V and the maximum gain I can provide is 30 for a current of 22mA. So based on the 2N3904 datasheet, how to I accomplish my goal?

I'd appreciate the help!
 

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Audioguru

Joined Dec 20, 2007
11,251
Your transistor is not a common-collector type and does not have a power supply connected anywhere.

A common collector transistor is an emitter-follower that has a voltage gain of only 1 when its load resistance is fairly high. It is not a mic preamp circuit.

You need to use a common-emitter transistor circuit. You also need to bias the electret mic.
 

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Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
Your transistor is not a common-collector type and does not have a power supply connected anywhere.

A common collector transistor is an emitter-follower that has a voltage gain of only 1 when its load resistance is fairly high. It is not a mic preamp circuit.

You need to use a common-emitter transistor circuit. You also need to bias the electret mic.
Audioguru,

I chose CC due to impedance transformation but I'll take your word for CE. How do I bias the electret mic? Also, since I am now doing the CE configuration, do I take Re out or should I have Re?
 

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Audioguru

Joined Dec 20, 2007
11,251
Re is needed to help set the DC operating point of your Ce transistor circuit. Re needs to be bypassed with a capacitor to avoid signal loss.

the circuit needs a supply bypass capacitor so that the transistor does not oscillate.

An electret mic needs about 0.5mA through a resistor (10k) fed from 3V to 9V. The voltage feeding the resistor should be decoupled with a 1k resistor and 47uF capacitor.
 

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
Re is needed to help set the DC operating point of your Ce transistor circuit. Re needs to be bypassed with a capacitor to avoid signal loss.

the circuit needs a supply bypass capacitor so that the transistor does not oscillate.

An electret mic needs about 0.5mA through a resistor (10k) fed from 3V to 9V. The voltage feeding the resistor should be decoupled with a 1k resistor and 47uF capacitor.
Thanks Audioguru,

What value should this bypass capacitor be? Is there a way to calculate it or do you just pick a standard value (such as 10uF)? Could you show me the things you just explained on a schematic please (especially with the electret mic)?

Thanks!
 

studiot

Joined Nov 9, 2007
4,998
I think you will be sadly disappointed by your circuit because the input impedance of your amplifier will be an order of magnitude too low, compared to the impedance of the microphone.

With a ceramic or crystal mike and a BJT input you need to go for some impedance raising techniques such as bootstrapping, cascode connection.
 

Jony130

Joined Feb 17, 2009
5,020
Calculation may look like this:
Au=30V/V; Vcc=9V; Vbe=0.65V
I pick Rc1=3.3K and Re1=RC/Au=110=100Ω
Ic_opt=0.5*Vcc/(Rc1+Re1)=4.5V/3.4K=1.32mA
VRe1=Ic*Re=0.13V a bit too small.
So I pick Rc1=1K and change the schematics (add re2).
Ic≈4.5V/1K=4.5mA
Re1=1V/4.5mA=220Ω
I assume that Hfe_min=100
Ib_max=Ic/hfe=45uA
So current that is flow through R1 and R2 mus be large then 5..30*Ib_max.
R2=(VRe1+Vbe)/( 5..30*Ib_max)=1.64V/220uA=7.2K=10KΩ
Idz=1.64V/10K=164uA
R1=(Vcc-Vb)/ (Idz+Ib)=7.36V/210uA=36KΩ

re+(re2||Re1)=Rc/Au=33Ω
re=26mV/Ic=5.8Ω
re2||Re1=27Ω
re1=1 /( 1/27-1/100)=36Ω
Ce=0.16/(F*re2)=220uF (F=20Hz)
C1=0.16/(F*Rin)=3.3uF
Rin=R1||R2|| [ (Hfe+1)*(re+re2||Re1)]
CL=0.16/(F*RL=)=1uF



I did not take into account source impedance.
 

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
Jony,

I'm not saying that anything is wrong with your design, but I could stick any transistor in place of the 2N3904 as long as it meets the requirements. You see, I want to take a look at the specs of a transistor (the 2N3904 in this case) and look at the graphs and be able to see where to put that sucker in its linear region. Could someone take me through the specs and take out what information I need and then use that to design a linear audio amplifier? I just want to be able to make that transistor as efficient as possible using external components.

My goal is to learn how to do this and with practice I can design with transistors not only using class A amplifiers but the others as well; and not just using one transistor but more than one transistor.
 

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
Well Jony, sounds like you know your stuff. Thanks for the help.

Could someone tell me why the common-emitter configuration is best for this? I suppose it's for impedance reasons?
 

Audioguru

Joined Dec 20, 2007
11,251
Could someone tell me why the common-emitter configuration is best for this? I suppose it's for impedance reasons?
The common-emitter transistor circuit with its emitter resistor bypassed has high voltage gain for use as a mic preamp and has a medium input impedance.
But it has high distortion at high output levels and its max gain might not be high enough. An audio opamp is much better.
 

Jony130

Joined Feb 17, 2009
5,020
You as a young designer don't need to know all this stuff.
Design your circuits by a help of a Ohms law and set the voltage on collector (emitter for CC amplifier) equal 1/2*Vcc.
And remember that is always nice to Rin>>Rs and Rout<<RL
For example for source impedance 10K it will be good to have amplifier with 100K input impedance or higher.
Similarly situation is with Rout. If our load impedance is 10K then output impedance of our amplifier should be 1K or lower.
And the size of a capacitors is selected that Xc is much smaller then resistance of a cooperating resistor at the lowest frequency of operation Xc=0.1*R.
Xc=1/(2*PI*F*C)=0.16/(F*C)
C=0.16/(F*R)
 
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Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
You as a young designer don't need to know all this stuff.
Design your circuits by a help of a Ohms law and set the voltage on collector (emitter for CC amplifier) equal 1/2*Vcc.
And remember that is always nice to Rin>>Rs and Rout<<RL
For example for source impedance 10K it will be good to have amplifier with 100K input impedance or higher.
Similarly situation is with Rout. If our load impedance is 10K then output impedance of our amplifier should be 1K or lower.
And the size of a capacitors is selected that Xc is much smaller then resistance of a cooperating resistor at the lowest frequency of operation Xc=0.1*R.
Xc=1/(2*PI*F*C)=0.16/(F*C)
C=0.16/(F*R)
Yeah, but I want to know it! :rolleyes: Thanks for your help!
 

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
Ok, so the capacitor in parallel with Re increases gain, correct? I believe you calculated this value with the input frequency from a function generator but what If I am using a microphone? Do I use the average voice frequency for man?

Thanks!
 

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
Yes. Capacitor eliminates negative feedback. So
Au=Rc/re≈ 40*Ic*Rc
Ok, thanks for that.

Yes, I I typically choose between 10Hz..20Hz
Alright, that sounds good.

I don't seem to be thinking clearly today, my mind seems a bit slow (I think it's that darn coffee! :D) Anyway, what is Au again? I know I've used it before but I can't remember it right now! Arrrgh!
 
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