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I have been reading the following article about Class-A and Class-B amplifers. http://www.antonine-education.co.uk/Pages/Electronics_2/Amplifiers/Power_amps/further_page_12.htm#Question 1.

I am not sure I understand the point of having an Op-AMP in the class-B amplifier ! Why do we use a set up pull-push in the Op-AMP's feeback loop? Why don't we just use OP-AMP by itself? And, how does OP-AMP helps to improve cross-over distrotion?

Also, in the same article, right above question 3, it states the following:

"The op-amp holds the base-emitter voltage of the npn transistor at 0.7 V. When there is a negative signal, the op-amp changes over, dropping 1.4 V, so that it turns on the pnp transistor. The voltage gain is 1, but the power gain is almost infinity."

Could someone please explain the above in relatively easier way?

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Let's start our with a single op amp and work out way toward a class AB design. Consider the voltage follower.

schematic

simulate this circuit – Schematic created using CircuitLab

In this case \$V_{in}=V_{out}\$. Why? Negative feedback. Negative feedback forces the inverting pin's voltage to match the non-inverting pin. In other words, the op amp will do whatever it takes with its output to make \$V_{NI}=V_{INV}\$.

Let's take that a step further. We need more power to drive a low impedance speaker. Well, the average op amp will only have a few tens of milliamps of drive capability. That is where we add the power stage.

schematic

simulate this circuit

By tying the inverting pin to the output of the power stage, we have created a voltage follower. \$V_{in}=V_{out}\$, but now the circuit has the ability to deliver much more current than the op amp output ever could. Because the op amp has negative feedback, \$V_{NI}=V_{INV}\$. The cross over distortion is eliminated by the op amp doing whatever it takes to satisfy that relationship. As an exercise, build this up on a bread board, put a sine wave into \$V_{in}\$, and observe the output of the power stage and output of the op amp. The two will not look anything alike!

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  • \$\begingroup\$ This makes a lot of sense, thanks. Can we use the regular negative feedback to acmplish the same thing? Something like this? upload.wikimedia.org/wikipedia/commons/thumb/4/41/… \$\endgroup\$ – Rudy01 Feb 12 '14 at 0:04
  • \$\begingroup\$ @Rudy01 Sure can, as long as the power stage is inside the negative feedback loop, and the signal is within the power rails after having the voltage gain applied. \$\endgroup\$ – Matt Young Feb 12 '14 at 0:19
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Without biasing, a class B output stage is going to produce cross over distortion. On the other hand an op amp is almost perfect at producing a clean output.

The trouble is, an opamp can't supply any serious power to a load but, in conjunction with a class B amplifier it can.

Because, theoretically, an op amp has infinite open loop gain, when the loop is closed around a class B push pull stage it can compensate for the class B crossover inadequacies.

In reality, as frequency and amplitude demand rises, a normal op amp struggles to overcome cross over distortion but there are plenty of amps out there that do a reasonable job in this configuration.

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The funny thing is was looking for the answer myself, here is my understanding adding on to Matt's answer.

  • The Class B and Class AB circuits are both in common-collector configuration.
  • The voltage gain, Av = roughly 1
  • It has very high input impedance
  • As a result:Sorry that should read V(in)

    enter image description here

  • Where Ai is the current gain,, In is the input signal, Vin is the input voltage, Rin is the total input resistance for the AC analysis.
  • So the current gain increases and the input resistance is increased.

And:

  • Ap is:

enter image description here

So having a Class B or AB increases the current gain which is what is needed to operate the speakers. And, in increasing the current gain we are also increasing the Ap, which is the power gain. And, we can go one step further and say that the power gain is increased as the input (Rin(tot)) is increased.

I hope any answer helped. It certainly cleared up my misunderstanding.

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