# Question about class A amplifier

#### ceidas

Joined Dec 26, 2011
50
Hello everyone,
I have a question concerning class A (CC) power amplifiers and I seek your help.

So, since a common collector amplifier has a unity voltage gain and the load we want to give our signal has a constant resistance, doesnt that mean that the current throught it will be just as much as if we were connecting the source directly to the load? Because I=V/R

So, if this is correct how can someone use common collector amplifiers as class A power amplifiers? Wouldnt that be meaningless?

Thank you.

Joined Dec 26, 2010
2,148
Many of the audio amplifiers used in the world use complementary emitter-follower output, so this can hardly be the case.

The common-collector stage supplies current amplification, which allows a low impedance load like a loudspeaker to be driven. The driver stage feeding the common-collector input provides a slightly bigger voltage, but much less current.

#### ceidas

Joined Dec 26, 2011
50
I really dont get it.

Lets take an example I made in the emulator and please explain me the results.

Whats the point of this AB amplifier that doesnt even amplify the signal?

#### Jony130

Joined Feb 17, 2009
5,454
We don't see any picture in your post

#### ceidas

Joined Dec 26, 2011
50
I am sorry.

It is this picture

Is it ok now?

#### crutschow

Joined Mar 14, 2008
32,843
A common-collector amp is used to amplify current, not voltage. Thus the relatively high impedance output of an interstage amp or preamp can drive a low impedance load such as a loudspeaker.

Your simulation has an ideal source with no internal impedance so it will show no advantage in driving the load with a CC amp as compared to driving the load directly from the source. But in the real world signal generators have finite and often relatively high impedance. For example a typical op amp can supply only 20mA of output current, not enough to drive a speaker.

Joined Dec 26, 2010
2,148
The point is that an amplifier stage running at a very much lower current could be usedto supply the input of the CC stage.

The voltage generator supplying the 100 ohm resistor directly has to deliver 28mA. How much current does the generator feeding the amplifier input deliver - maybe 1mA? Try it and see.

This is very basic stuff, and there is really no doubt about the function of these things.

#### ceidas

Joined Dec 26, 2011
50
Ok, what about this simulation. High output source resistance, low load resistannce, still no good.

@Adjuster Yes, I am sure too that the current the source is giving to the amplifier is smaller than if it is connected directly to the load but so what? Isnt the current that the load is taking all we care about?

Joined Dec 26, 2010
2,148
Ok, what about this simulation. High output source resistance, low load resistannce, still no good.

@Adjuster Yes, I am sure too that the current the source is giving to the amplifier is smaller than if it is connected directly to the load but so what? Isnt the current that the load is taking all we care about?
No, the current taken by the load is not all that we care about. The current from the signal source matters too. Normally we feed amplifiers not from ideal voltage sources, but from lower power stages having restricted current capacity. In a large loudspeaker amplifier a driver stage working at a few hundred mA may control output transistors working at many amperes peak current.

#### ceidas

Joined Dec 26, 2011
50
Ok, i just changed the biased voltages of the transistors and I got what I wanted.

I really dont know what is going on here.

Joined Dec 26, 2010
2,148
Ω
Ok, i just changed the biased voltages of the transistors and I got what I wanted.

I really dont know what is going on here.

The bias voltage is quite criitcal. Had you applied 1V bias to real transistors in such a circuit, they would almost certainly been destroyed by excessive current.

With a load of only 8Ω, 1kΩ source resistance to the driver is rather too big. Try reducing it to 100Ω, and put an ammeter in series with the 1V source driving the amplifier.

#### ceidas

Joined Dec 26, 2011
50

Ok, here it is.

Joined Dec 26, 2010
2,148
The output current is almost ten times what you would get from direct connection to the 100Ω source, and over 50 times the current actually being fed to the amplifier input. This kind of output stage is very common in amplifiers. The stage driving it is usually made to provide voltage amplification, but at a much lower current level than the load will require.

This method of working is convenient for a number of reasons: among other things, the CC output stage has a low output impedance, relatively low distortion and wide bandwidth compared to a voltage amplifier stage. This is due largely to the negative feedback inherent with the stage. A wider bandwidth output stage helps to minimise phase lag, making it easier for heavy negative feedback to be applied around the whole amplifier.

These things have been around for a while, I think that the original idea for the complementary symmetry follower was thought bup in the 1950s, but widespread application had to wait until suitable transistors were easily available. They have certainly been in use for decades, so there is not much point imagining that they are not useful. Manufacturers would have not been making these things for the last forty or fifty years if there were an obviously better (or more importantly cheaper) way.

#### Audioguru

Joined Dec 20, 2007
11,248
With an emitter current of 64mA, the base current of a typical 2N3904 or 2N3906 is 0.36mA but could be as high as about 1.3mA.

But such a low current into an 8 ohm speaker makes a very low level of sound.
The amplifier might have a peak output voltage of 3.2V like an LM386 amplifier that has a 10V supply so the 200mA max rating of the 2N3904 and 2N2906 is not enough and the base current will be much higher.

Joined Dec 26, 2010
2,148
A valid point, but the OPs problem appears to be more fundamental. He seems to be doubting the usefulness of any stage which does not provide voltage amplification.

I hope that by now he may have conceded this point - it might be good to get an update from him about that before going much further.

#### ceidas

Joined Dec 26, 2011
50
Believe it or not I've been studying electronics for a while but there are some moments when I am trying to practically apply some things in my own constructions and it is like hitting a wall. This is because a lot of things that I have studied in theory are not so simple in practice.

This problem of mine is one of them. The book is telling everything about the three known amplification topologies but never mentioned anything about real applications when the source has output resistance.

The book I have studied the most is the classical Microelectronic circuits by Sedra and Smith (i am sure most of you have heard of it). Its output stage (power amplifiers) chapter is not so clear about how we achieve power amplification and what the difference between power and small signal amplifiers is.

@Audioguru If I am not wrong, what you want to say is that 2N3904 and 2N3906 are not the proper ones when it comes to the much higher current the 8ohm loudspeaker requires and I have to buy myself other (power) transistors. Right? By the way, I own a few LM386 but I want to make my own audio power amplifier.

Last edited:

Joined Dec 26, 2010
2,148
Power amplifiers have to deliver power, and generally with a reasonably high efficiency. This makes the resistive internal loads often used in lower level amplifiers unattractive, as they waste power.

Sometimes a transformer can be used as the load, but this is generally not preferred for wideband amplifiers, and is inherently unsuitable if the response needs to go down to DC. Earlier audio amplifiers used transformers a lot: valve (electron tube) amplifiers still do.

Generally, a circuit using two devices forming a potential divider across the power supply is the most usual method nowadays. The CC complementary pair is one of the favoured ways of doing this.

Apart from reasons mentioned already, with a common-emitter complementary stage, the choreography of the bias currents can get rather fraught! With CC, the transistors are reasonably easy to bias so that they hand over current appropriately, whereas the CE equivalent would be harder to manage. The penalties for getting this wrong are crossover distiortion, which sounds really lousy, or excessive current drain - the latter can easily burn things out.

Finally, what do you suppose is inside an LM386? Look at the simplified schematic in the datasheet, and there is a modified CC output stage. The lower transistor is however a composite called a Sziklai pair instead of a straight PNP.

http://en.wikipedia.org/wiki/Sziklai_pair

#### ceidas

Joined Dec 26, 2011
50
Ok, I think most of my questions have been answered, thank you for your response. So, tell me, in my device there is going to be a single 5V supply from batteries so what is best for the 8ohm 1 Watt speaker? My own AB or A amplifier or the LM386 amplifier?

Joined Dec 26, 2010
2,148
LM386 will just about go down to 5V, but it will only give barely 250mW into an 8Ω speaker, not quite double that into 4Ω

To get a watt into 8Ω from a 5V supply, something more involved like a bridged amplifier would be needed, as otherwise there is simply not going to be enough voltage to get that much power in an 8Ω speaker

Nowadays, perhaps one of the specialised amplifiers made for the job would be the way to go. Here is an example - but not a recommendation: I don't even know if it is currently available.

#### ceidas

Joined Dec 26, 2011
50
Well, I have seen an other device using a tda2822 with 4,5V (3x1.5V batteries) but I just dont have any of these. So, there is no other solution?