The simulation program does not know that beta is a range of numbers. The sim program simply uses the "typical" beta number then when a transistor with low beta or high beta might not work when the circuit is built.

A simulation program is also stupid enough to allow a low power transistor to be severely overloaded but not show any problem.

 

The simulation program does not know that beta is a range of numbers. The sim program simply uses the "typical" beta number then when a transistor with low beta or high beta might not work when the circuit is built.

A simulation program is also stupid enough to allow a low power transistor to be severely overloaded but not show any problem.

The part .model does provide a specific value for beta, Bf, but the value is only good under very specific conditions. Other variables such as Vce and actual Ic will push that value one way or the other in the .model.

 

I think a simulation program should be smart enough to "sweep" the range of beta to show you if a transistor will be cutoff or saturated if it has passing low or high beta.

 

I think a simulation program should be smart enough to "sweep" the range of beta to show you if a transistor will be cutoff or saturated if it has passing low or high beta.

Can be done.

 

Transistor '2ntest', Ib set to 1 mA. Beta (x axis) stepped from 25 to 1000 every 25, measured Ic (y axis).

 

Hi all
Out of curiosity I tried a simpler output stage.
I thought I biased it correctly.
There is a lot of distortion.... A lot...
Since I have to use those two output transistor do you think I can improve this circuit behaviour?
Why the Sziklai version performs better?
Cheers

 

Your new circuit has the parts vertically so far apart that I could barely see them. I put the parts vertically closer together and extended the timebase so see fewer sinewaves. It looks fine.
EDIT: your current source block was only 25mA and it works much better at 100mA.

 

Regarding the simple output stage (not the sziklai one)
I thought that the current source just needed to be greater than the maximum Ib at Ie_peak=1.4A
The gain of that transistor is 66 (from LTSpice parameters) so I did Ie_peak/66=Ib_peak=21mA
I left 4mA for the Vbe multiplier and I thought I was ok.
I don't understand how I can choose that value. Just trying to increasing it and looking at the result?

 

Where on earth can you buy a 2N3055 with a beta of exactly 66 at a collector current of 1.4A??
A simulator program is just guessing that the model says a beta of 66.

Your parts are so far apart that your text cannot be seen.

 

I'm confused.
The current that you mentioned are quiescent currents right? Is there a way to choose them? I think the answer to my question is related to that. Which Ic is right for the output transistor? The answer is related to the distortion that I can accept, right?
The current I mentioned are peak currents and I thought I dimensioned them right.
Maybe I was wrong since beta can vary...

 

Since you cannot select the beta of a transistor, if you want every circuit to work properly then you design with the minimum beta printed in the datasheet for the current you need. The curves in a datasheet are usually only for a "typical" device but many devices have lesser spec's.
Or you could buy many extra transistors, test them all and throw away or sell the minimum ones on ebay.

 

I don't think AG believes in sims. How did you determine that the 2n3055 .model had a beta of 66 at IC=1.4 amps? I see Bf=73 in the .model, but that value usually falls at higher IC.

Simple test circuit attahed

 

I don't think AG believes in sims.

What happens if you build a 100W audio amplifier and turn it up to produce 100W and use a little pair of 2N3904 and 2N3906 transistors as the output pair?
They instantly glow like an LED then blow up.

But they work fine in a simulation.

 

Simple test circuit attahed

Great simple test circuit.
I haven't thought about it.
I've done the same for the other BJT.
So, at the end of the day I have to look at the greater Ib current of those two transistor at peak Ic (considering beta variation as in the image explained) and that should be the minum current (plus the current for the Vbe multiplier) that my current source should give to the circuit. Right?
Cheers

 

Hi all
I've compared a Darlington version of the circuit with a Sziklai one.
I can't understand why I cannot have the full +-14V swing at the output with the Darlington one.
Have I done something wrong? How can I improve the situation?
I don't understand if that's due to something wrong I did with the Darlington version of the circuit or it's due to the fact that Sziklai have half tension across Vbe multiplier.
Happy new year

 

Are you going to build this amplifier in real life or only in the virtual world of simulation?

 

Are you going to build this amplifier in real life or only in the virtual world of simulation?

This is just an electronics college course exercise. Maybe in the future I will try to build it using proper transistors and so on but for now it's just a theoretical project.

 

Either way, you should simulate the whole amplifier with negative feedback.

https://forum.allaboutcircuits.com/...l-maximum-output-voltage.156364/#post-1351368
https://forum.allaboutcircuits.com/...-current-drain-quiescent.156216/#post-1349535

 

Amazing... In just a few minutes... I thought that if I couldn't get the full swing at that stage I could not proceed. May I ask why did you set the input impedence at 22k? At least I think that's the purpose of R11.
Are R7 and R8 speedup resistors? Is it better to connect them like that than connecting base emitter of Q1 and Q2?
Thanks

 

May I ask why did you set the input impedence at 22k? At least I think that's the purpose of R11.

Why not? Also, you wrote: "amplifier input resistance >= 10kohm" and for sure 22kΩ is larger than 10kΩ.

Are R7 and R8 speedup resistors?

Yes, they help to discharge the Cbe capacitance. And they help improve the thermal stability of an output stage.

Is it better to connect them like that than connecting base-emitter of Q1 and Q2?

What will be the difference if R1 = R2 = 0.22Ω