Trouble understanding common-emitter amplifiers

[OldWorldBlues]

Hi there! I've been tinkering with some homemade speakers, and wanted to make a simple amplifier circuit to step up the some 3 volts that comes out of an aux jack to 9 or 10 volts. I decided that the common-emitter model looked the easiest to understand. I was mistaken. This is the simplest diagram I found while poking around the internet:

Assuming that I change the component values, it becomes immediately apparent that the input & output voltages are being crossed over on the same line. How does this work? What does the ground symbol mean at the bottom? I appreciate any advice the community can give - thanks :)

 

[berkeman]

Hi there! I've been tinkering with some homemade speakers, and wanted to make a simple amplifier circuit to step up the some 3 volts that comes out of an aux jack to 9 or 10 volts. I decided that the common-emitter model looked the easiest to understand. I was mistaken. This is the simplest diagram I found while poking around the internet:
View attachment 216001
Assuming that I change the component values, it becomes immediately apparent that the input & output voltages are being crossed over on the same line. How does this work? What does the ground symbol mean at the bottom? I appreciate any advice the community can give - thanks :)

The schematic looks to be a reasonable starting point for a prototype. What do you mean by voltages "crossing over"?

 

[NascentOxygen]

What does the ground symbol mean at the bottom?

The ground symbol indicates that the negative terminal of this "voltage boosting" circuit's battery is to be connected by a wire to the circuit ground of the device that is supplying you the driving signal. The circuit ground of the device is generally connected to a pin or shield of the socket you are plugging in to, anyway.

Some circuits show this in a clearer way, because if the input to your booster circuit uses shielded cable then this ground wire connection is done for you by the conductive shielding mesh on that cable when you plug into the AUX jack.

 

[sophiecentaur]

it becomes immediately apparent that the input & output voltages are being crossed over on the same line.

Could you be alluding to the fact that a common emitter amplifier is a Voltage Inverter? As the input volts go more positive, the collector current increases so the collector volts (set by the voltage drop across the LS) go less positive. As the collector load is somewhat inductive, the voltage swing can be more than if it were just resistive.

 

[Svein]

There are several issues with your circuit, but I will just state the main one: Your loudspeaker is biased DC-wise by the collector current. This means that the speaker diaphragm will be constantly displaced from its normal position and since the collector current can never change direction, less than half of the effective speaker volume will be available.That is why loudspeakers are always AC-connected, either via a transformer or a capacitor. See below for an example.

Even this circuit has a problem. The output transistor can pull the output low very fast end very effective, but it cannot push the output high. The only thing that can make the output go high is the 1.2k resistor. Therefore you will have a very asymmetrical drive of the loudspeaker. The solution is to use two output transistors, one to pull the output low, and another to pull it high.

 

[sophiecentaur]

but it cannot push the output high.

With a collector load of only 8Ω, there's a lot of "pull up".

 

[Svein]

With a collector load of only 8Ω, there's a lot of "pull up".

Well, the 8Ω is a fiction. The impedance of a loudspeaker is very variable:

The real trouble lies in the DC biasing which forces the loudspeaker cone away from its natural "zero" position.

 

[Rive]

Assuming that I change the component values, it becomes immediately apparent that the input & output voltages are being crossed over on the same line. How does this work? What does the ground symbol mean at the bottom?

The ground symbol at this level of understanding means an absolute '0V' reference which can sink infinite current (and will still remain a '0V').
What confuses you is that the opposite part of the ground symbol is not used in this schematic. The upper part of the battery (at this level of understanding) should be taken as an absolute 15V which can provide infinite current (and will still remain 15V).

According to this the input and the output is not crossed over, but: the input has a resistor to the power rail (this is required to provide adequate bias for the resistor) and the output is using the power rail as a source of power.

All the other issues mentioned in the topic coming only after this point.

 

[sophiecentaur]

According to this the input and the output is not crossed over,

This quotation from the OP is confusing for a start. The 'Input' signal is AC coupled and could have any mean voltage and its peak to peak swing is not mentioned. The volts on the input end of the 1k resistor could easily be more positive than the V_ce of the transistor when it's saturated. We have been discussing this in more detail than the original diagram needs. Other than it definitely being an inverting amp, there's nothing more to be said about signal values.
An emitter feedback resistor could improve matters greatly as it would define the circuit mainly in terms of the component values rather than the (unknown) transistor characteristics.

 

[Rive]

This quotation from the OP is confusing for a start.

I think it's confusing only because it is just about problems which are too basic for the community.
Hope the OP will soon clarify the type of the problem.

 

[sophiecentaur]

I think it's confusing only because it is just about problems which are too basic for the community.
Hope the OP will soon clarify the type of the problem.

There are many possible semi random approaches to electronics. I started off that way but never got anywhere until I started some formal learning of the subject.

 

[Borek]

There are several issues with your circuit, but I will just state the main one: Your loudspeaker is biased DC-wise by the collector current.

The solution is to use two output transistors, one to pull the output low, and another to pull it high.

Correct me if I am wrong - that's more or less the difference between class A and class B amplifiers?

 

[Svein]

Correct me if I am wrong - that's more or less the difference between class A and class B amplifiers?

No. Class A, AB etc. refers to the idle current of the amplifier.

• In a class A amplifier the idle current is greater than zero. For "single-ended" amplifiers the idle current is (at least) equal to half the peak current.
• In a class B amplifier the idle current is zero. Thus you can only have a class B amplifier with a "push-pull" output stage.
• Class AB is a sort of compromise between class A and class B. You need a "push-pull" output stage, but you allow some idle current in order to minimize "cross-over" distortion.
• Class C amplifiers are only relevant for transmitters. It is an extension of the class B in that it has zero idle current, but it does not draw any current until the input amplitude is greater than a certain level. Thus: Lots of distortion.
• Class D amplifiers are very different from the amplifiers above. It is a digital amplifier, it converts the input signal to a digital bit stream (usually Pulse Width Modulation), feeds the bit stream to the loudspeaker and assumes that the loudspeaker is able to convert the bit stream back to analog sound (which it usually is).

Check out http://www.electronics-tutorials.ws/amplifier/amplifier-classes.html for a more detailed explanation and figures.

 

[Borek]

I thought in terms of a connection between these two statements:

In a class A amplifier the idle current is greater than zero. For "single-ended" amplifiers the idle current is (at least) equal to half the peak current.

Your loudspeaker is biased DC-wise by the collector current. This means that the speaker diaphragm will be constantly displaced from its normal position

Constantly displaced diaphragm seems to me to be effect of the non-zero idle current. (So in a way these things are "equivalent"). And the second circuit you posted - with two transistors used to avoid the non-zero idle current - looks to me like a class B amplifier. I thought it makes sense to use these terms (class A and B) in the context of the original question and discussion.

Chances are I am misusing/misunderstanding the nomenclature, it is quite new to me.

 

[Svein]

Constantly displaced diaphragm seems to me to be effect of the non-zero idle current. (So in a way these things are "equivalent"). And the second circuit you posted - with two transistors used to avoid the non-zero idle current - looks to me like a class B amplifier. I thought it makes sense to use these terms (class A and B) in the context of the original question and discussion.

As I said:

• Idle current through the loudspeaker displaces the diaphragm
• Idle current through the amplifier output stage determines the amplifier class

The second circuit I posted shows a "push-pull" output stage. Whether the output stage is a class A, B or AB is not possible to determine from the schematic. Observe that the amplifier idle current does not pass through the loudspeaker. The use of R1 has nothing to do with idle current, it is part of a positive feedback from the output to the top of the bias network (there are several ways to create the same effect using more components). The figure below shows how to do it without using feedback through the loudspeaker. Observe that this circuit incorporates several advanced details like constant current generators and current mirrors.

 

[Borek]

Thanks. The only thing that is clear to me is that it is definitely out of my league. Hardly surprising

 

[sophiecentaur]

Thanks. The only thing that is clear to me is that it is definitely out of my league. Hardly surprising

The 'business end' of the circuit isn't too hard to follow. Currents into the bases of Q9 one Q10 are arranged to produce a Potential Divider and the point where the loudspeaker is (AC) coupled is driven in a nicely linear way. The Diodes D1 and D2 are there to keep the base voltages separated by the right amount to make up for the V_BE drops of both output transistors. The two transistors 'cross over' the 'mid point' between the power rails in a well behaved fashion. One turns of at the same rate as the other turns on. The Feedback in this sort of circuit transforms it from a simple amplifier into a good, linear amplifier. (Probably not 'good enough' for some applications, of course.)
There are a few tried and tested sub-circuit elements in that amplifier that you will see in pretty well every sophisticated analog circuit.

 

[Svein]

For an actual HI-Fi amplifier check out this page: https://www.eleccircuit.com/40w-amp-ocl-2n3055mj2955/. It is really good, and there is a lot of circuit explanation for those who are interested.