What happens inside an amplifier?

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Well the one thing I must add here is from transistor theory during USAF tech school. We were all, as a class tracing signals through drawings.
We had changed over from Tubes to Transistors in our circuits and the big difference that was mantra'd into our brains when we went thru a transistor is
"what does a transistor do to a signal? (all together) "Invert & Multiply". Of course transistors being digital the square wave was emitted larger and upside down. Which brings up Boolean algebra & flip flops. After all the transistor is no more than a amp.
 
We were all, as a class tracing signals through drawings.

Doing something like that would really help me, but I majored in English, unfortunately. I seem to learn the most from actually doing stuff and making mistakes, to the great chagrin of whatever amp I'm "fixing."
 
Love the foot on the accelerator analogy. Another one I've heard is a hand controlling a water spigot. A small turn one way or the other increases or decreases a stream of water from a trickle to a fire hose. Imagine the hand turning rapidly back and forth in time to the music.

Thank you. This all helps me.
 
The current needed by the transistor varies, and the power supply must be able to deliver high current very quickly and smoothly.

So when Kenwood put out a "high speed" receiver that was what they were talking about? Great summary; thank you!
 
Well the one thing I must add here is from transistor theory during USAF tech school. We were all, as a class tracing signals through drawings.
We had changed over from Tubes to Transistors in our circuits and the big difference that was mantra'd into our brains when we went thru a transistor is
"what does a transistor do to a signal? (all together) "Invert & Multiply". Of course transistors being digital the square wave was emitted larger and upside down. Which brings up Boolean algebra & flip flops. After all the transistor is no more than a amp.

Oh good ole' Holly Hall at Keesler!?
 
This is actually a pretty challenging question. If you know what a relay does - using a small amount of power to control a large amount of power - then an amplifier operates in a kinda similar way. Except the controlling (input) signal is continuously varying (like a sound wave) - and the controlled (output) signal is likewise continuously varying. And it has the same shape (inverted or otherwise) - but more power (more current, or voltage, or both).
 
So when Kenwood put out a "high speed" receiver that was what they were talking about? Great summary; thank you!
Typically, "high speed" or "fast" in reference to an amplifier is little more than marketese; it isn't a useful term in electronics, at least without some reference to a particular metric like bandwidth or slew rate. So it may refer to an amplifier with a relatively high bandwidth, but that's really only meaningful in comparison to something else. Otherwise, how fast is "fast", how "high" is high?
 
a simple amplifier here .
amp-1.png
 
Oh, the other thing you should know, if you don't, is Ohm's Law.

E = I * R

E = voltage (volts)
I = current (amps)
R = resistance (ohms)

Power (P, in watts) = I^2 * R = E * I

Those interrelationships enable pretty much everything in electricity and in electronic circuits.

ohms_law.gif
Not quite right,
More normally V =I*R where V is in volts. R specifically is resistance which is entirely real- that is, no capacitance or inductance involved. Z is the more general form of Impedance which can involve reactive elements (caps or inductors).
Energy (E or W is the SI unit symbol) is defined as the amount of charge moved through a voltage, V, in a time t which is the same as the integral of power over a time t, and is in units of Joules (J), or often in the US kilowatt-hours.
People often get energy and power confused so that they will say energy when they really mean power which can muddy discussions.
 
Not quite right,

Since mhardy defined E to be voltage, E = I * R is correct. I imagine the E came from electromotive force or electric potential, aka voltage. Since 'voltage' comes from the unit name (the volt), rather than the quantity, E is in fact the better, although far less common letter to use.

Energy (E or W is the SI unit symbol) is defined as the amount of charge moved through a voltage, V, in a time t

Neither E nor W is the SI unit symbol for energy; that is J, for joule. We are discussing quantities (energy, electric potential, charge, time, etc.), not units (joule, volt, coulomb, second). I don't believe SI specifies the symbols to be used when representing quantities in equations; that is up to the user to define, just as mhardy did.

https://physics.nist.gov/cuu/Units/units.html

In terms of charge and voltage, energy = charge * voltage. Time does not come into it.
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Since mhardy defined E to be voltage, E = I * R is correct. I imagine the E came from electromotive force or electric potential, aka voltage. Since 'voltage' comes from the unit name (the volt), rather than the quantity, E is in fact the better, although far less common letter to use.



Neither E nor W is the SI unit symbol for energy; that is J, for joule. We are discussing quantities (energy, electric potential, charge, time, etc.), not units (joule, volt, coulomb, second). I don't believe SI specifies the symbols to be used when representing quantities in equations; that is up to the user to define, just as mhardy did.

https://physics.nist.gov/cuu/Units/units.html

In terms of charge and voltage, energy = charge * voltage. Time does not come into it.
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Power is the work done per unit time=QV/t by definition, so energy = QV (total work done) or the integral of power over time, as I said. t is necessary in both cases to relate the quantities of power and energy although it does not explicitly occur in the equation for energy that contains Q and V. Constant power implies, in this case, that either the fixed charge Q is moving through a fixed change in voltage each second, or in the case of constant voltage, that the charge is linearly increasing each second. In either case, there's a time dependence even if the actual work done is only a function of the value of the charge and the difference in the potential of the electric field between the finishing point and the starting point.
You are correct that I conflated the concept of unit and symbol- I thought it would simplify things, but that was clearly not the case.
http://physics-help.info/physicsguide/appendices/si_units.shtml
The symbol commonly used for energy is W or E. E is not preferred as it is usually used, in electromagnetic theory, to designate the Electric Field while B is the Magnetic Field.
E as voltage is archaic and no longer used (See above). V is used, and for clarity one cannot simply assign any arbitrary symbol to a quantity when common usage suggests otherwise.
 
Power is the work done per unit time=QV/t by definition, so energy = QV (total work done) or the integral of power over time, as I said.

You said energy, not power:

Energy (E or W is the SI unit symbol) is defined as the amount of charge moved through a voltage, V, in a time t

and for clarity one cannot simply assign any arbitrary symbol to a quantity when common usage suggests otherwise.

If one clearly defines the symbols being used, it is acceptable, and not incorrect.

I agree that V is far more common, but, with definitions as stated, E = I * R is not incorrect.

That's quite enough pedantry for today...
 
You said energy, not power:





If one clearly defines the symbols being used, it is acceptable, and not incorrect.

I agree that V is far more common, but, with definitions as stated, E = I * R is not incorrect.

That's quite enough pedantry for today...
This is tough without calculus...
Yes I know that Energy does not have time as a dimension- but that does not mean that it doesn't have a time dependence.
If not, how do you relate power to energy without having a dependence on t in the energy equation as power is defined as the change in energy per unit time? If the charge is moving through a fixed electric field the energy changes with time- i.e. energy is a function of time, and power is the derivative of the energy. If the charge Q transitions instantaneously from V1 to V2 then the power would be infinite but for an infinitesimal time, and the integral would still be constant. If the transition occurs linearly over a time, say, T, then the power over that interval would be a constant QV/T and the integral from 0 to T would be the same value as before. In reality, of course, Q does not instantaneously transition between the end points in the field, so the power is never infinite and T is never 0- hence my chosen words.
As you said, enough pedantry.
My attempts to add clarity and consistency clearly have failed.
 
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If not, how do you relate power to energy without having a dependence on t in the energy equation

QV is the potential, or stored energy. It has no time component. Just like a weight held at some height. It has that potential energy forever, unless that energy is used to do work. The rate at which that energy is used gives you the power.
 
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QV is the potential, or stored energy. It has no time component. Just like a weight held at some height. It has that potential energy forever, unless that energy is used to do work. The rate at which that energy is used gives you the power.

Well, let's conduct a gedanken-experiment and consider an ideal resistor- say 1 ohm - connected to an 1V "battery" through a switch. Let's also assume that the battery has zero internal resistance and contains 10 coulombs of charge and hence 10J of energy before the switch is closed, at which point a 1A current flows until the switch is opened and then 0A flows.
The power dissipated in the resistor during the time the switch is closed is thus 1W (1A*1A*1ohm) and the rate of energy transfer from the +ve terminal of the battery is 1J/sec.
Let's connect the battery for t=1 sec. The energy that has been removed from the battery is thus -1J (i.e. -1A*1V*1sec). and the integral of the power in the resistor is 1W*1sec= 1J. (conservation of energy at work)
The total charge that has been transferred is 1 coulomb and it has passed through -1v, so that energy too comes out to -1J.
The energy stored in the battery once the switch is open again is 9J, and the charge is 9 coulombs.
The energy has indeed been removed from the battery and transferred to the resistor which heats up and eventually thermalizes.
A quote from my original post:
"Energy (E or W is the SI unit symbol) is defined as the amount of charge moved through a voltage, V, in a time t which is the same as the integral of power over a time t, "
 
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