What Is the Difference Between Current and Voltage Amplifiers?

[fog37]

Hello,

Basic question: what is the difference between a current and a voltage amplifier? Naively, in principle, I would think that if either quantity is magnified, the other will be also magnified simultaneously. How can we magnify current without magnifying the voltage and vice versa?

Thanks,
fog37

 

[anorlunda]

The difference is semantic. Any device, including amplifiers, has a voltage-versus-current curve. If the curve is closer to horizontal, we call it constant-voltage. If it is closer to vertical, we call it constant-current. If it is 45 degrees, flip a coin.

Even if you call it the wrong name, the voltage-versus-current curve doesn't care.

 

[berkeman]

I would think that if either quantity is magnified, the other will be also magnified simultaneously.

One difference is in how the load impedance affects the output of the amp. If you have a CV amplifier, decreasing the load impedance just increases the output current to hold the same voltage.

For a CC amplifier, changing the load impedance changes the output voltage, since the output current is being held steady.

Both types of amplifiers have limitations. The CV output amplifier will run into overcurrent problems when the output load impedance is reduced too low, and the output regulation will suffer (the output voltage will start to drop below what the amplifier is trying to hold).

If the load impedance is too large for a CC amplifier, it will "run out of output compliance" where the output voltage starts to get too high for the CC amp to be able to regulate its output current (the circuitry that is generating the output current need some voltage drop across it in order to operate correctly).

And you can also say that CV output amps generally have a very low internal output impedance, and CC amps have a very high internal output impedance.

Does that help? Can you post some examples of the amplifiers you are looking at?

 

[LvW]

What is an "amplifier"?
To me, it is an active device that amplifies (increases) a certain input signal - that means: without changing the "nature" of the signal.
Otherwise, an "amplification factor" (gain) is hard to define.
Examples:
* Voltage-in, voltage-out (VCVS) >> operational amplifier
* Current-in, current-out (CCCS) >> Current conveyor

There are two other basic active blocks which sometimes are called "amplifier", but they really do not simply amplify - instead they transfer
* an input voltage to an output current (VCCS) >> OTA (described by its transconductance), Bipolar Junction Transistor (BJT)
* an input current to an output voltage (CCVS) >> "Transresistance" unit (described by its transresistance) .

Of course, none of the above real units can work as an ideal voltage or current source. For exact analyses we "add reality" using series or parallel resistances at the input resp. output nodes.

More than that, one should always realize that - physically spoken - an input voltage or an input current are not "amplified"; the energy of the enlarged ouput signal is taken out of the power supply. It is rather the signal content - in form of a signal voltage resp. signal current - which appears at the output as an amplified signal.

 

[anorlunda]

These curves illustrate the point. The curves are for a solar panel, not an amplifier. Amplifiers are usually designed to work in just one region, but the nomenclature is the same, and nomenclature is the subject of this thread.

 

[berkeman]

LOL "whatever region"

 

[fog37]

Thank you.

I guess I envision an amplifier as a device with an input port (two wires) and an output port (two more wires). The amplifier essentially creates a magnified version of the input signal using the energy provided by a power source (not by the input signal). An amplifier also has both an input impedance and an output impedance to consider.

I think that a voltage amplifier has very high input impedance, high voltage gain but a low output current. What about its output impedance? Why is the current output low? Because the output impedance is very high as well? So the device that we connect to the amplifier output port becomes essentially in parallel with the amplifier output impedance?
On the other hand, a current amplifier seems to have medium to high input impedance and a low impedance load can be connected to its output to provide a large current gain.

If I have a signal that is the output of a sensor, or the output of a speaker, and the signal is weak, should I amplify its voltage or its current? What type of amplifier should I use? What is the thought process?

 

[Baluncore]

What a difficult question...
Power or Signal amplifiers amplify input energy. They have an input impedance matched to the source. The output is matched to the load impedance. Impedance is the ratio of voltage to current at that port.

There are situations where a voltage or current needs to be sensed, but the source must not be disturbed by the connection. It is then necessary to have very low input power, so the product of voltage and current must be low. The identification or naming of an amplifier usually comes down to the input variable being sensed.
Voltage amplifiers sense input voltage. They have high input impedance, with very low input current.
Current amplifiers sense input current. They have low input impedance, usually with very little change in input voltage.

The amplifier output can be a voltage, a current, or a signal, that will be some proportional function of the input variable. Voltage or current outputs reduce sensitivity to variation of load.
If the output is a voltage it will have a low output impedance, driving a higher impedance load.
If the output is a current it will have a high output impedance, driving a lower impedance load.
If the output signal is power it should have a matched output impedance.

The sign of the amplifier gain is critically important when the amplifier is used in a feedback loop. But when signal gain is specified in dB, the polarity of the amplification is unspecified and will be dependent on phase shift.

An amplifier will contain some device with gain. But the device does not have to increase the signal, it may just change impedance. A signal amplifier may have a power gain of +15dB, or maybe –15dB like a passive attenuator. This can lead to "double negative" confusion where –dB attenuation is equivalent to amplification.

I think that a voltage amplifier has very high input impedance, high voltage gain but a low output current.

High voltage gain is not necessary. Output current is decided by the load.

 

[sophiecentaur]

Power or Signal amplifiers amplify input energy.

That's it in a nutshell. The impedance of the input signal need not be the impedance of the output signal for you to have gain. The voltage follower circuit is a great example; it takes in a varying voltage signal from a photo diode and dishes out exactly the same voltage signal but can feed it into a 1Ω load.
I am (over the top, no doubt) picky about the use of dBs to describe how the volts are amplified or reduced when the dB change is specifically a ratio pf Powers. Transformers always have less than 0dB gain, whatever happens to the Volts. That simple point can often missed by aficionados of the dBV idea. (Edit - I refer to the less well informed aficionados )

 

[fog37]

Thank you Baluncore. Let me summarize if possible:

Any type of amplifier has two ports (input and output).

Ideal Voltage amplifier: Infinite input resistance, i.e. $R_{input} =\infty$ , so NO current from the circuitry connected to the input port flows into the amplifier. However, the amplifier's output impedance must be $R_{output} =0$, so all the magnified voltage across the output port is applied and found across the load resistance $R_{load}$ and there is no voltage drop of the across the the amplifier's output resistance being zero. A voltage amplifier essentially behaves like an ideal constant voltage source (zero internal resistance) applying all its voltage across the load irrespective of the load's resistance $R_{load}$. The current coming out of the amplifier's port will depend on the load's resistance.

Ideal Current amplifier: Zero input resistance, i.e. $R_{input} =0$, so ALL the current from the circuitry connected to the input port flows into the amplifier. However, the amplifier's output impedance must be $R_{output} =\infty$, so NO current will go through the output resistance $R_{output}$ and will all flow through the load resistance $R_{load}$ regardless of its value. At the output port, the amplifier behave like an ideal current source. The voltage drop across the load depends on $R_{load}$.

In terms of real life examples, when we use $8 \Omega$ speakers, what kind of amplifier do we need to use? What are we trying to amplify? Voltage or current? Or Power? How is a power amplifier different from the ideal current and voltage amplifiers?

I am familiar with the concept of impedance matching to transfer power from one portion of the circuit to another without losses. But there is no amplification involved in the case of impedance matching. Impedance matching requires the impedance of the input circuit to be equal to the input impedance of the amplifier and the the impedance of the load circuit to be equal to the output impedance of the amplifier. In the case of voltage or current amplifiers, what are the requirements for the impedance of the source circuit and load circuit?

Thanks!

 

[LvW]

"A voltage amplifier essentially behaves like an ideal constant voltage source"

Just one small correction: Not "like an ideal constant..." but like an ideal controlled voltage source (VCVS, voltage-controlled voltage source)

 

[fog37]

Thanks LvW. I see.

Therefore, a current amplifier behaves like a CCCS, current-controlled current source. I guess I am clear on voltage and current amplifiers. For example, a photo-resistor produces a small current proportional to the incident light. This current signal must first be amplified (increased) with a current amplifier and then possibly be converted using a current-voltage converter to a voltage signal that is presented to a microcontroller.

So, where does a power amplifier fit? Power is $P=IV$. In order to amplify the power of an input signal, we need to amplify both $V$ and $I$ or either one while the other is maintained constant. For example, a transformer increases/decreases either $V$ or $I$ so its gain is unity (or less) so it is not a power amplifier. But the most power that can be passed to a circuit from another circuit is determined under the conditions of impedance matching, isn't it?

I guess a current amplifier could work as a power amplifier: the larger the load impedance the larger the power delivered to the load. But also a voltage amplifier could do something similar: the lower the load impedance the higher the power delivered to the load.

 

[sophiecentaur]

I guess a current amplifier could work as a power amplifier:

To be classed as an "amplifier" it would need to amplify power.
And, whilst we are talking about the extreme forms of amplification (current and volts) how do we stand with an amplifier that takes Current from a transducer and apples a Voltage to a 300 Ohm transmission line. This thread is in danger of turning into 'the classification game' again.

 

[CWatters]

In terms of real life examples, when we use $8 \Omega$ speakers, what kind of amplifier do we need to use? What are we trying to amplify? Voltage or current? Or Power? How is a power amplifier different from the ideal current and voltage amplifiers?

Depends what your source is. A typical "line output" from a hifi might generate up to say 0.5pk into 10kOhm.

If you wanted to build a hifi amplifier that could deliver 200w output into 8Ohms it would need +/- 40V power supply rails... 40*40/8=200w...

So you need a voltage gain of 40/0.5=80

The source current is around 0.5V/10,000= 50uA
The output current would be 40/8=5A
So the current gain needed is 5/50uA=100,000

You can work out the power gain.

 

[Baluncore]

Thank you Baluncore. Let me summarize if possible:
Any type of amplifier has two ports (input and output).

You then make unnecessary generalisations to ideals, which rapidly restrict the definition and make it inapplicable to reality.

You carelessly confuse the terms resistance and reactance. Impedance allows for quadrature currents due to reactance.
Impedance = Resistance + Reactance; Z = R + jX.

What are we trying to amplify? Voltage or current? Or Power? How is a power amplifier different from the ideal current and voltage amplifiers?

Amplifiers always source power. Ideal voltage input or current input amplifiers take no energy from the signal, so all output power must come from the power supply, all available input energy is reflected, or wasted in a resistor.

When an amplifier input is matched to the source, it can use all the available input power to drive the active element of the amplifier, which reduces complexity and cost per watt.

 

[sophiecentaur]

I guess a current amplifier could work as a power amplifier: the larger the load impedance the larger the power delivered to the load. But also a voltage amplifier could do something similar: the lower the load impedance the higher the power delivered to the load.

I'm afraid you are just going round and round the houses at this stage of the thread. As Juliette was heard to remark: "What's in a name?"
How is it that such a fascinating subject as Electronic Engineering is beset with so many pointless classification exercises? How it works is of great interest, what we choose to call it is of much less interest.
There will always be exceptions to classification rules but any description that's written about a design or process can (should) resolve problems arising from this.

 

[LvW]

How is it that such a fascinating subject as Electronic Engineering is beset with so many pointless classification exercises? How it works is of great interest, what we choose to call it is of much less interest.

Yes - I fully support this view.
Another example: Some people try to classify harmonic oscillators. However, this is really impossible because in some (many?) cases we can explain the working principle of one particular oscillator circuit in different ways (positive feedback, negative input impedance, undamping of a tank circuit,...).

 

[f95toli]

So, where does a power amplifier fit? Power is $P=IV$. In order to amplify the power of an input signal, we need to amplify both $V$ and $I$ or either one while the other is maintained constant. For example, a transformer increases/decreases either $V$ or $I$ so its gain is unity (or less) so it is not a power amplifier. But the most power that can be passed to a circuit from another circuit is determined under the conditions of impedance matching, isn't it?

The difference between power amplifiers and regular amplifiers is mainly in the purpose of the amplifier and choices that have been made when designing them.
The most important specifications for a regular amplifier tends to be low-noise, well-behaved input impedance and low distortion. This means that they are typically made using a class A configuration using low-noise transistors.
Power amplifiers are (AFAIK) nearly always voltage amplifiers but they are designed to -as the name implies- deliver a lot of power into (sometimes) difficult loads This means that they are designed in a class AB, B or "D" (switch mode amplifiers) configuration using power transistors. Power amplifiers are therefore typically verynoisy and in some cases there is also a fair amount of distortion (class B guitar amp being the extreme example).

Hence, in a typical configuration you would two stages of amplification, a low-noise pre-amplifier. with a lot of gain followed by a power amplifier with less gain (remember that power goes as V^2, a gain of x10 will give you x100 in power). As long as the gain is high enough in the first stage the noise of the power amplifier does not matter (see Friis formula)

I agree that it is pointless to try to find a "universal" classification scheme, it is almost always clear what is meant once you know the context.