The 10k collector resistor of the CE transistor makes a voltage divider with the 188 ohm base resistors of the power transistors.
It is the 188 ohm resistors (not the CE transistor) that turn on the output transistors but even 188 ohms does not provide enough current.

For a 10W sinewave into 4 ohms the peak current is simply calculated to be 2.25A and the peak output voltage is simply 2.25A x 4 ohms= 9V. If you buy hundreds of output transistors and measure them, then you might find some with a typical current gain of 60 so the peak base current will ne 2.25A/60= 37.5mA. Then you can add ordinary little transistors to drive each output transistor.
Use darlington output transistors instead like this old amplifier:

 

You mentioned using the expensive high current LT1886 opamp to drive the output transistors.
You did not look at its datasheet (the simulation software also does not look at maximums) to see that its max supply is only plus and minus 6.6V and its max output will be only plus and minus 4.5V.

 

You mentioned using the expensive high current LT1886 opamp to drive the output transistors.
You did not look at its datasheet (the simulation software also does not look at maximums) to see that its max supply is only plus and minus 6.6V and its max output will be only plus and minus 4.5V.

oh thanks for the warning.
i missed that part.

I tried looking for op amps which can handle a current around 150mA, LT1886 was a good choice, until the limits now.
I must find some else...

 

Hi a,
Have you checked the TIP power dissipation.?
E
Image1 is he TIP' #2 is the PSU
BTW: You have DC coupled amplifier circuit.!


Update:
Why are you using a 20mA Current source to generate a 200mV signal for the LT1886.????

 

Hi a,
Have you checked the TIP power dissipation.?
E
Image1 is he TIP' #2 is the PSU
BTW: You have DC coupled amplifier circuit.!


Update:
Why are you using a 20mA Current source to generate a 200mV signal for the LT1886.????

I changed the opamp, its LT1207 now, the Power graph issue is now very minute. May be was some issue with other opamp characteristics (especially the Limits).

* I'm using 20mA current source in input, cz, that's in my question. I've to run the 4ohm 10W from 20mA.

 

hi a,
This is what I see.
E

 

hi a,
This is what I see.
E

Actually, i'm having this on my screen

 

hi a,
This is what I see.
E

It's even better now, with 1n4007s

 

hi,
Where did you get those TIP transistor models from.??

Mine are TIP32C and TIP32C

E

 

hi,
Where did you get those TIP transistor models from.??

Mine are TIP32C and TIP32C

E

its TIP31a Tip32a.

i got them from internet library.

 

hi,
Where did you get those TIP transistor models from.??

Mine are TIP32C and TIP32C

E

+Eric,
I've a little issue.
the normal gain formula for non inverting (1+R2/R1) is not working here perfectly.
it should be 43Kohm according to formula for 8.9V output.
But it is giving 8.9V at 53k ohm.

What's the case here ?

 

The expensive LT1207 is a current-feedback opamp but you are using voltage feedback. Current-feedback opamps are never used for audio.
Since the output transistors have a minimum current gain of 12 at the peak output of 2.25A then the peak output current of the opamp must be 2.25A/12= 188mA (not 20mA) causing the opamp to burn up (too much heat) also since its power supply is way too high at plus and minus 15V.

10W into 4 ohms is 6.33V RMS which is 8.95V peak. The opamp supply must be only plus and minus 11V or 12V but it will still burn up.
The plus and minus 25V supply for the output transistors will also cause them to burn up, only plus and minus 11V or 12V is needed.
Simulation programs often have overloaded and overheated parts working wrongly.

Use power darlington or Sziklai pair transistors at the output so that an ordinary cool CE transistor or ordinary cool opamp can drive them.

 

The expensive LT1207 is a current-feedback opamp but you are using voltage feedback. Current-feedback opamps are never used for audio.
Since the output transistors have a minimum current gain of 12 at the peak output of 2.25A then the peak output current of the opamp must be 2.25A/12= 188mA (not 20mA) causing the opamp to burn up (too much heat) also since its power supply is way too high at plus and minus 15V.

10W into 4 ohms is 6.33V RMS which is 8.95V peak. The opamp supply must be only plus and minus 11V or 12V but it will still burn up.
The plus and minus 25V supply for the output transistors will also cause them to burn up, only plus and minus 11V or 12V is needed.
Simulation programs often have overloaded and overheated parts working wrongly.

Use power darlington or Sziklai pair transistors at the output so that an ordinary cool CE transistor or ordinary cool opamp can drive them.

Thank you soo much.
I tried your suggestion (using Darlington PNP NPN in AB stage) even with LM741, output is quite favourabe.


However, the input AC to AB amplifier is 8.94V,,,, while the output reduces by 0.4V to about 8.56V.

Can you please explain why does this happen ?

Because all calculations are for Power: 20W(peak), but output is 18.something Watts.
The drop is somewhere in AB circuit.

 

hi a,
Please post your calculations for the power output you are expecting.

What voltage swing would you expect across the 4 Ohm load resistor to give the required power output.?

E

 

hi a,
Please post your calculations for the power output you are expecting.

What voltage swing would you expect across the 4 Ohm load resistor to give the required power output.?

E

Hi Eric,

I need a swing of +-8.94V across 4ohm load. For power of 10Watt.

the input to AB is exactly +-8.94Vs, But it comes out +-8.5Vs at the output.

The input impedance is approximately 4.4k in this circuit.

 

hi a,
This shows a direct drive of 8.94v across 4R

E

 

hi a,
This shows a direct drive of 8.94v across 4R

E

Yes Eric,
But the issue is, i'm not getting 8.94V at Load 4ohm.
it is some what 8.5V.
I'm not sure, where that 0.44V is being reduced.

Because input to AB is 8.94, but output is 8.5V.

 

hi a,
The centre point biassing of the diode network is off the mark.
Please use F4 to place labels on the asc file circuit.

E

 

hi a,
The centre point biassing of the diode network is off the mark.
Please use F4 to place labels on the asc file circuit.

E

Thanks.

Previously i used simple AB-push pull amplifier.
And method to calculate the biasing resistances was follows:
(VCC-Vbe)/Icq=Rbias
Icq=0.04*Ic(sat)
IC(sat)=VCC/(2×RL)

so i calculated the biasing resistance. But here i'm using Darlington pair. Two transistors, so previous isn't applicable. i tried to find a way, but only think i saw was 10k being used mostly, and no calculations.
It's not deep in my mind,,, is there any way to calculate the biasing resistance for Darlington pair ?

 

hi,
This shows the NPN and PNP load drive Emitter currents, they are not being equally driven.

E