# Large ac input class AB biasing

I'm trying to bias the following circuit to power an 8 ohm speaker with an 8Vptp output. The following are the calculations I've done to get this current setup but the output is only good up to 1Vptp before it starts to lose gain and get crossover distortion. What am I overlooking here?

Taking power output as 1W, V=sqrt(P*R)=sqrt(1*8)=2.82. sqrt(2)*2.82=3.99V so this should give me an 8Vptp output.

I=sqrt(P/R)=0.35A, sqrt(2)*0.35=0.5A

Taking hfe of 100 then Ib should be 5mA

12V-1.4 for the diode drop =10.6 and dividing by 2 to bias equally =5.3V

5.3V/5mA = 1060 ohms

Fixed design with transistor at correct orientation:

Actual output:

What can be done to eliminate this clipping in the practical circuit?

• Did you get the BJTs wired up correctly in that diagram? – jonk May 9 '19 at 20:22
• You do know that the Diode characteristics are important and you may need to increase their current to get sufficient Q1-Q2 bias and then they need to be thermally connected. – Tony Stewart Sunnyskyguy EE75 May 9 '19 at 20:29
• Your Q2 is shown as an NPN device, but annotated as "BD140", which is a PNP device. Can you fix the schematic to make clear what the design is? – The Photon May 9 '19 at 20:34
• Matt, I suppose while you are mulling over the other comments I'll add still another. (I'll assume you fix up the diagram, someday.) If you imagine pulling up on the signal input line and indirectly through the capacitors to the bases of the two BJTs, then this should move the diode pair upward. If Q1 is sourcing into the load there is less current in R9 exactly when you need more current in R9 for the base drive of Q1. It will work after a fashion, but it will very, very much limit your useful output span. – jonk May 9 '19 at 20:41
• @ThePhoton Q2 does have the correct symbol, it is just "upside down". Emitter and collector need to be reversed to make this work properly. OK, it can work like this as well but Q2 will be in "reversed active" mode and then you'll get an extremely low beta. – Bimpelrekkie May 9 '19 at 20:41

## 1 Answer

Taking hfe of 100 then Ib should be 5mA

That assumes that hFE remains constant at 100 and this is by no means the case when the transistors start to work harder and are delivering current to the load with a much smaller volt drop to the rail.

Take a look at the BD139/140 data sheet; with a current delivered of 0.5 amps and a volt drop of 2 volts between collector and emitter, the hFE can be as low as 25.

Also remember that the current of 0.35 amps you have calculated is only the load current - there could be another 100 mA flowing due to class AB biasing.

Fix the orientation of Q2 in the top diagram of your question please.

• Fixed the orientation in the next photo, do I have enough current flow with this new setup? – Matt May 10 '19 at 9:52
• Q2 in your top diagram is still wrong. – Andy aka May 10 '19 at 10:26
• Your new design is clipping on the top rail because the common emitter feeding the signal to the darlington bases can't drive higher than the positive rail AND, the output transistors being darlington means anything up to a couple of volts lost top and bottom. Please don't evolve this question any more. This site is not a forum. – Andy aka May 10 '19 at 10:29
• This is why BJT Op Amps are not Rail to Rail. final comment. i.stack.imgur.com/EATnV.png but improved by CC regulation for linearity and short circuit protected. – Tony Stewart Sunnyskyguy EE75 May 10 '19 at 17:50