Is there a difference in the meaning of power and current amplifier terms?

Question

I understand the difference between voltage amplifier and power or current amplifier. For example in an audio amplifier there are usually three stages: two voltage amplifiers and a power amplifier at the last stage. Okay they call the first two amplifiers here voltage amplifiers since they increase/amplify the input voltages with driving very little current from the power supply.

But at the last stage the current is amplified, hence the power. I understand that power is related with current's square but my question is about naming.

Does "power amplifier" and "current amplifier" mean the same thing? Or power amplifier term is used when an amplifier amplifies both current and voltage; but "current amplifier" is only used when there is unity voltage gain with high current gain?

Answer 1

If you ask 10 EEs this, you will probably get close to 10 answers. This is not a well-defined convention, but is roughly outlined as you guessed. A power amplifier will be able to provide more power than the previous stage. A current amplifier will be able to provide more current than the previous stage.

If you need a definitive answer, I would say ask the person that you are involved with on this project. The key is always to understand what the other person means, not necessarily what they say. Basic communication theory. :-)

Answer 2

I understand the difference between voltage amplifier and power or current amplifier. For example in an audio amplifier there are usually three stages: two voltage amplifiers and a power amplifier at the last stage. [...]
[emphasis mine, N.A]

What you have described is the difference between signal amplifier and power amplifier. The signal happens to be voltage and power happens to be current in this case.

Consider this. There is no general definition of gain, or even units of measurement for gain. The gain and its units are defined on case-by-case basis.

When you wish to drive coils in the speakers, you care about output current, while voltage is less important as long as you have enough compliance voltage. The gain¹ of the current output power stage would be defined as I_out/V_in, and units would be [A/V].

When you wish to drive piezos (for a different example), then you care about output voltage, while current takes the second seat. The gain¹ of the voltage output power stage would be defined as V_out/V_in, and units would be [V/V].

¹ I'm assuming that the signal is voltage (perhaps provided by an upstream signal amplifier).

Answer 3

The amplifier as a whole amplifies power. The output stage you're discussing amplifies current only, but you're talking about the special case of a solid-state output stage, which is almost invariably a voltage-follower. There are other topologies and other technologies. For example, a typical valve output state amplifies both voltage and current.

Answer 4

You`ve got already so many answers - here is another one (very short):

According to my experience, very often there is simply a misunderstanding concerning the term "power amplifier".

In most cases, the purpose of such a stage is not to "amplify" power, but to PROVIDE a large amount of output power to the connected load. For example, think of the popular push-pull stage (class-B operation). Due to non-linear operation of the stage, we do not speak about the amplification factor of this stage. The most important parameters are output power and efficiency.

Answer 5

I suspect the concept you are missing is "output impedance". A voltage amplifier has some voltage gain but probably a high output impedance. When it is connected to no load, it shows a nice big output signal but as soon as you attach a load, the signal disappears.

The output impedance and load impedance act together as a voltage divider.

For example, you might have a voltage amplifier producing 20V open-circuit and it has a 1k output impedance. Connected to a 100k load, the output voltage will be 20*100k/(101k) = 19.8V... but when you connect it to a 10R load, the output voltage will drop to 20*10/(1010)=0.198V.

^{simulate this circuit – Schematic created using CircuitLab}

If you want to get some power (voltage * current) into the load, you need a high voltage output with a low output impedance so that it is capable of supplying current and therefore power to the load.

In a classic audio amplifier, the output stage is NOT a "current amplifier" but a low-impedance voltage follower. "Current amplifier" has the implication that its output is some linear multiple of its input current, and that's not quite what's happening. The purpose of the final stage is to provide enough current to the load so that the load's voltage follows the open-circuit voltage of the amplifier (to grossly oversimplify).

Feedback is also important. The amplifier uses feedback to correct its internal nonlinearities and the voltage-divider effect of its output impedance interacting with the load impedance.

As the load impedance goes lower, the amplifier dissipates more power/heat. Maximum power-transfer into the load (for a given output impedance) is when the load and output impedances are equal, but that is a really bad situation for an amplifier because it will have insufficient damping factor. Normally you want to keep the output impedance well below the load impedance in order to maintain stability of the amplifier's feedback loop and to avoid the amplifier getting unnecessarily hot.

Edit for the ignorance and downvote fairies out there. Here's a classic and famous audio amplifier design, from Doug Self.

Note the three stages: an input differential stage which is used to implement feedback, a voltage-gain stage in the middle which has an output impedance of a couple hundred ohms, and finally an emitter-follower (low-output-impedance voltage follower). There is voltage negative feedback from the output back to the input differential, resulting an an overall closed-loop amplifier voltage gain of about 20.

Summary: the amplifier as a whole implements a voltage source with low output impedance. If you put an 8 ohm load on it, the load will get about twice as much current and therefore twice as much power as a 16 ohm load.