amplifier design using AC line

if I try to make a basic amplifier. For example using bc548. The datasheet give me that at Vce=5V and Ic=2 mA the small signal current gain is 230, and give the hie for thus value too. So it makes sense to use these parameters, 5 V to collector emitter voltage and 2 mA for collector current.

But I know that the Q point must be in center of ac line. The equation of ac line is

Ic = -(Vce/Rc+Rl) +(Vceq/Rc+Rl)+Icq.

if , Vce(cutoff)= Vceq + Icq(Rc+Rl)

So Vceq=Icq(Rc+Rl)

In this case i'am using Rc=1.5k and Rl=1k , so

Vceq= 0.002 A * 600 Ω =1.2 V

How can be posible that the Vce must be in 1.2 V? It's near to saturation.

Maybe you are asking. Why I chosse Rc=1.5k. Because I got it using dc line. Because it was the first thing I figured. And I must use 10V like source.

My question is that I don't know how i have to do this. I have to do first the dc line and then the dc line? but it is not going to change values? like resistor for example.

Summarizing, for 10 V source and Rl=1k. I do dc line for Vceq=5V and Icq=2mA , I got Rc=1.5K , Re=1k, R1 40.8k , R2=100k. Then the midle of ac line that i got it's (1.2 V, 0.002 A) . What i must to do in here?
greeting

Re: amplifiee design using ac line.

Please post your schematic.

Re: amplifiee design using ac line.

With Ic = 2mA and Re = 1K, then Ve = 2 V. As you want Vce to be 5V, then Vc = 5+2 = 7V, Then Rl = (10 - 7)/2 K = 1.5 K.
Whats the problem?
Frank

Re: amplifiee design using ac line.



that's the circuit. And it's works phenomenal Frank. But from what I've read there isn't maximum symmetrical excursion. Because the Q-point is in the midle of dc line but it isn't in the midle of ac line. making calculations (which i did before in the first post) i get that the Q-point must be with Vec=1.2 V but it change all.

So when I want to make an amplifier the first that I have to do is the ac line, And whit thus value find the resistors value. Or not?

thanks.

Re: amplifiee design using ac line.

I guessed at your schematic and simulated your circuit. I increased the input level until the output clipped.
I changed your 40.8k resistor to 51k for a symmetrical output swing.
There is no voltage gain, it is less than a piece of wire.

Re: amplifiee design using ac line.

The voltage gain is 43. I get it using hibrid model and the proteus simulator give me it too.

Re: amplifiee design using ac line.

julian403 said:

The voltage gain is 43. I get it using hibrid model and the proteus simulator give me it too.

Click to expand...

No.
The voltage gain of your transistor is only 0.6 due to the unbypassed emitter resistor providing negative feedback.
I bypassed the emitter resistor with a capacitor and the voltage gain increased to 46 times at low levels. At high levels (but below clipping) the distortion is severe and the output level cannot be measured.

you are right i forgot to put the capacitor in the schematic. So, for what it's the ac charge line? just to get the maxim Δvce? when we have already designed. Or maybe it's usefull in disign?

The impedance of the emitter capacitor is a dead short at audio frequencies so the voltage gain is (Rc//Rl)/re where re is the internal emitter resistance (25/IE in mA). With the 1k load the AC output clips symmetrically as I show when the bias point of the collector is at about 6.36V. Then the collector can swing up to 7.8V and swing down to 5.0V.