A CE amplifier I built successfully

Thread Starter

alphacat

Joined Jun 6, 2009
186
I designed and built the following CE amplifier.

DC operating point:
Measured VCE = 2.382Vdc.
Measured VBE = 0.6Vdc.

Small signal data:
Measured input: vin = 10.2mVpeak
Measured output: vout = 0.955Vpeak.
=> voltage amplification of -93.6 (there was a 180 phase between input and output, as expected).

I have several questions pleaes:

1. Could you please tell me your opinion about my design? how would you improve it?
I'm a bit worried that i had to use such a large RBASE (it took the capacitor long time to charge).
I chose such large a resistor since for that value, Vout was around 2.5V, which was my goal in order to receive a large output voltage swing.

2. Is there any way to derive β from my measurements?

3. Is it possible to connect a speaker to the vout node (from the Collector to Ground) and hear the sound which the signal generator (input) creates?
Would it work?


Thank you very much!

 

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thatoneguy

Joined Feb 19, 2009
6,359
Missing an emitter resistor, which stabilizes gain even for different transistor temperatures.

The output impedance is too high to drive a speaker efficiently, not enough current could go through the 10k Rc to drive an 8Ω speaker.

In addition to an emitter resistor, a capacitor should be placed in parallel with the emitter resistor so the signal doesn't change the calculated gain (Rc/Re)

The input with decoupling capacitor should be to the left of the base resistor, with one end of the decoupling cap at the junction of R1 and R2 biasing resistors, the other end of the cap connected only to the signal generator.

That's off the top of my head.

Here is an auto-designed CE audio amp from multisim:

 

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Thread Starter

alphacat

Joined Jun 6, 2009
186
Thank you very much.
How can you know what is the base current in the configuration you offered?

Moreover, you said that there isnt enough current to drive the speaker.
You mean that not enough AC current would pass through the speaker if i connect it in prallel with the CE junction?
 

hobbyist

Joined Aug 10, 2008
892
The amp schem. "thatoneguy" shows is the classic all around cleass A amp,

But for this being your first attempt at a class A amp design,
According to your stats, looks like it was a successful design,

you got the bias voltage at the output, close to half supply, and you got a voltage gain, at the output, so even though it is a rough design, you still got it to work.

Now study what "thatoneguy" said in his post and youll pick up real quick and you should have a good reliable amp designed sometime soon.
 

Jony130

Joined Feb 17, 2009
5,488
.
How can you know what is the base current in the configuration you offered?
Ib=[ (RB1*Vcc)-Ube*(RB1+RB2) ] / [ Re*(β+1)*(RB1+RB2)+(RB1*RB2) ]
So for β=150; Vbe=0.68 we get
Ib≈14.76uA
Ic
2.2mA
Voltage gain Au=Rc||RL/re≈40*Ic*(Rc||RL)≈2.8V/V

And both circuit are horrible and nobody besides amateurs use this simple circuit.
 

electr

Joined May 23, 2009
49
@Jony
You used beta in order to determine IB(dc), but you can never know the exact beta since beta varies for different operating points.


About the speaker, you said that there isnt enough current to drive the speaker.
You mean that not enough AC current would pass through the speaker if i connect it in prallel with the CE junction? (using a coupling capacitor to block DC current).

If so, how would adding an Emitter resistor RE help?
That would only decrease the current gain of the amplifier.
 

thatoneguy

Joined Feb 19, 2009
6,359
If you have it built, hook a speaker up to it with a signal generator input. It won't be terribly loud, but it will be audible. The Re is for long term stability.

To get power that will produce good volume, one ends up working with push-pull amplifiers, but for a 1st shot at a Class A, you did very well!
 

electr

Joined May 23, 2009
49
Thank you very much, i'll try it out.

Sorry for writing from a different user name, its the user name there's here at work.
 

Jony130

Joined Feb 17, 2009
5,488
@Jony
You used beta in order to determine IB(dc), but you can never know the exact beta since beta varies for different operating points.
If you proper design the the circuit Ib<<Idz (current that is flow through R1 and R2) then Ic is almost independent form β
For example RB1=4.7K; RB2=3.3K; Re=1K; Vcc=5V; Vbe=0.67 and for β=100 we get
Ib_= 22uA and Ic=2.2mA
for β=400
Ib= 5.6uA ---> Ic=2.225mA
As you can see Ic is almost independent of β despite that Ib changes in a wide range.

If so, how would adding an Emitter resistor RE help?
That would only decrease the current gain of the amplifier.
RE don't change the current gain.
RE reduces the voltage gain of a amplifier.
Of course, this reduction of a voltage gain occurs only if we don't use Ce capacitor.
Amplifier with RE but without Ce has a smaller voltage gain but has more input impedance and less harmonic distortion. Because RE apply a negative feedback to the circuit and that helps a lot to reduce the distortion.
 

Thread Starter

alphacat

Joined Jun 6, 2009
186
If you proper design the the circuit Ib<<Idz (current that is flow through R1 and R2) then Ic is almost independent form β
For example RB1=4.7K; RB2=3.3K; Re=1K; Vcc=5V; Vbe=0.67 and for β=100 we get
Ib_= 22uA and Ic=2.2mA
for β=400
Ib= 5.6uA ---> Ic=2.225mA
As you can see Ic is almost independent of β despite that Ib changes in a wide range.
Thank you very much for this explanation.

RE don't change the current gain.
RE reduces the voltage gain of a amplifier.
Of course, this reduction of a voltage gain occurs only if we don't use Ce capacitor.
The AC current gain is equal to:
hfe / (1 + hfe*RE)
So surely the current gain is degraded when using an Emitter resistor.

Amplifier with RE but without Ce has a smaller voltage gain but has more input impedance and less harmonic distortion. Because RE apply a negative feedback to the circuit and that helps a lot to reduce the distortion.
Agree.
 
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