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I think I see the problem

Posted by Commuteman (A) on August 25, 2004 at 11:00:58

In Reply to: I am sorry to say posted by Soundmind on August 25, 2004 at 04:54:19:

I'm trying to bridge the understanding gap. Hopefully you can respond appropriately.

You said:

"While it is true that the conversion of electrical energy into heat is equal to the square of the current times the resistance..."

That's true for a resistor when a current is applied, but it does not apply here. Let me explain:

Current I through resistor R results in a voltage V across the resistor. V = I x R

The power dissipated P = V x I (think of it as Joules per Coulomb x Coulombs per Second = Joules per Second = watts)

Since V = I x R, we can also write this as P = I² x R

So, if you triple the current through a resistor, you get nine times the power, and therefore heat. Right?

Let's look at the output transistor example:

The voltage across the output device V_D equals the supply voltage V_S minus the output voltage V_OUT. This is independent of the transistor characteristics (close enough for this example), since it is ultimately determined by the amplifier's feedback structure and the supply voltage.

The current through the output device I_OUT = V_OUT / R_LOAD

The power dissipated by the output device P_D = V_D x I_OUT.

As you reduce the load impedance, note that only the current changes (not the voltage across the device).

You have P_NEW = V_D(SAME) x I_OUT(x3) = P_D x 3

Since we don't have the current causing a voltage change proportional to current, we don't have the I² relationship; it's linear.

In other words, when you decrease the load impedance by a factor of 3, the current through the output device increases by a factor of 3, and the power dissipated by the output device increase by a factor of 3 (not 9).

Is there something wrong with this analysis?

I await a rational response.

Peter

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