Transistor audio amplifier load impact

[Uros11]

Hello,

I am building an audio amplifier using transistors, and I'm using circuit from "Art of Electronics" as a reference. In the book, circuit is without load and everything is working correctly, and I understand how that circuit works. But in my case, I when I put 8 Ohm load resistor at the output, the characteristic is cut at the bottom and I don't understand why. I would like if someone can explain or give some hints for calculations of resistors, and explain the impact of load resistor on this circuit. I changed circuit a little bit, and when I experiment with values I can get some better looking sine on the output, but I don't feel good about myself if I get it experimentally, and don't know how to get those values.

 

[AnalogKid]

Q4 pulls up the output stage and can saturate, so the base of Q3 is at (V3 - 0.1 V). R6 pulls down the output stage, and at some point (as the voltage at the bottom of the diode stack decreases with the signal) cannot sink enough current out of the Q5 base to support enough collector current to make the negative signal peaks into (out of) the load.

Start with the negative peak current into the load, divide that by the transistor gain. then calculate the Q5 base voltage at the negative peaks. Now you can use Ohm's Law to calculate R6. Hint - you might not like the answer.

ak

 

[Ylli]

This is a common issue. The negative output is from an emitter follower, and the base of that emitter follower is connected to the - rail through a 1K resistor. When the output of the amplifier slews toward the negative rail the voltage across that 1K resistor becomes less and less, and eventually will not be able to provide sufficient base current to the output transistor.

The trick is to use 'bootstrapping'. Coupling the output back to provide a negative voltage greater than the rail so that the output transistor is not deprived of base current.

Using your existing circuit, change that negative side 1K resistor into two 250 ohm resistors in series. Now add a 1000uF cap between the output and the junction of the two 250 ohm resistors. See if that doesn't improve you output range before clipping.

Also, it is good practice to use small value resistors in the emitters of each of the output transistors. Add a 1 ohm resistor between each output transistor emitter and the 'out' point.

 

[Uros11]

Q4 pulls up the output stage and can saturate, so the base of Q3 is at (V3 - 0.1 V). R6 pulls down the output stage, and at some point (as the voltage at the bottom of the diode stack decreases with the signal) cannot sink enough current out of the Q5 base to support enough collector current to make the negative signal peaks into (out of) the load.

Start with the negative peak current into the load, divide that by the transistor gain. then calculate the Q5 base voltage at the negative peaks. Now you can use Ohm's Law to calculate R6. Hint - you might not like the answer.

ak

You were right, I don't like the answer, but I found out that my output transistors were small signal BJTs, with maximum collector current of 0.2 A. I changed them and it is working fine.

This is a common issue. The negative output is from an emitter follower, and the base of that emitter follower is connected to the - rail through a 1K resistor. When the output of the amplifier slews toward the negative rail the voltage across that 1K resistor becomes less and less, and eventually will not be able to provide sufficient base current to the output transistor.

The trick is to use 'bootstrapping'. Coupling the output back to provide a negative voltage greater than the rail so that the output transistor is not deprived of base current.

Using your existing circuit, change that negative side 1K resistor into two 250 ohm resistors in series. Now add a 1000uF cap between the output and the junction of the two 250 ohm resistors. See if that doesn't improve you output range before clipping.

Also, it is good practice to use small value resistors in the emitters of each of the output transistors. Add a 1 ohm resistor between each output transistor emitter and the 'out' point.

In addition of transistor changes, I put two 500 Ohm resistors and 100uF cap. I can see change at the output, but it is not that big. I think the main problem was small-signal BJTs.

Thank you both for your answers, I appreciate it.

 

[AnalogKid]

What part numbers did you change them to?

 

[Uros11]

I changed all NPNs to FZT849 and PNPs to 2SAR542P, but on the output my positive and negative amplitude is not equal, it varies a little bit, but it shouldn't be a problem.

 

[Ylli]

Here is how I'd tweak it. Different selection of transistors (used ones available in LTspice), modified bias to reduce idling current, added the bootstrap and emitter resistors, changed input resistor and first stage load to better balance the input long-tailed pair.

 

[Uros11]

Here is how I'd tweak it. Different selection of transistors (used ones available in LTspice), modified bias to reduce idling current, added the bootstrap and emitter resistors, changed input resistor and first stage load to better balance the input long-tailed pair.
View attachment 118579

Thank you for the help, it looks very good. One more question, when I plot output power, the amplitude is not constant. So is there a way to calculate exact power of this amplifier, or I can just take the average value of amplitude?

 

[Ylli]

Without doing a more thorough analysis, it looks like I can input about 800 mV before I see obvious limiting.

Right click on the graph and select "Add Trace". On the form that opens, in the "expressions to add" box enter the output node voltage * the current in the load resistor. Using my circuit, that translates to V(n008)*I(R10). That trace will be instantaneous power. After it plots, put the mouse on "V(n008)*I(R10)" and Control-Click. A box will pop up and give you average power. With the 800 mV input, I see 10 watts out.

 

[Nigel Goodwin]

You need to be aware that this isn't a practical working circuit, it's only a cut-back theoretical partial circuit (to demonstrate the principle), and has a number of serious issues.

 

[Ylli]

What Nigel said.