How Does Energy Compensation Occur in Amplifier Systems?

[japam]

well i have this simple question, its about the energy in a amplifier system, whatever class is it, you have at first a tiny electromagnetical signal in the space with an amplifier system with its resistances, condensers, transitors,coils, calculated to work on certain constant voltages and currents with stability.
and then after the signal passes through the amplifier its boosted, so you have the same energy in the amplifier, because its stability of polarization and aggregate the power boost of the signal, so there's an increasing of the energy in the total system signal-amplifier

so my question is: this grow in energy has to be compensated by a decay of energy in the amplifier? , where exactly, in the transistor?

 

[berkeman]

You are using energy from the power supply to boost the energy in the signal. The transistors are the amplifying elements, but the power gain comes from the power supply.

 

[japam]

...

then any amplifier must have a limited power for load , because the transistor is designed to just work on certain voltage-current range,

and that means also that the current-voltage changes in the transistor according to load or not load in the output , with the same power source ,right?

 

[chroot]

A transistor will indeed behave quite differently when different loads are presented to it. Most audio power amplifiers, for example, are designed to drive speaker loads, which are typically in the range of 2 to 4 ohms.

- Warren

 

[Cliff_J]

Minor correction Warren, most speakers for home use would be a nominal 8 ohms and 4 ohms for car use. Although in some car subwoofers, the 2 and 1 ohm impedance exists for the loudly motivated. In the pro audio world, 16 ohm and 24 ohm drivers exist too, and then there are public address speakers that sometimes have an autoformer (transformer) to work on 70.7V systems.

Japam - think of the amplifier as a voltage increasing device. If a 2V signal is fed into it, and it has a gain of 20 then it will output a voltage of 40V.

Since we know that E^2/R = W then 400/8 = 50W, 400/4 = 100W and so on. At some point, the current will exceed what the amplifiers power supply can handle and at that point the power will drop off because it is not able to handle that load very well (if at all). Or if the impedance were to go too low, its possible the transistors will attempt to pass more current than they can handle and burn out.

Boost isn't really the most handy way to think of it in my opinion, its more a frame of reference.

Maybe a more handy way to think of it would be that you have a big power reserve waiting to do something and a small signal voltage that controls the power delivery. Kind of like how a car engine makes a lot more power than your leg could, but your leg controls it with the accelerator pedal.