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#1 |
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diyAudio Member
Join Date: Aug 2008
Location: Evanston, IL / Rochester, NY
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Hello,
I have a question about the power output the simulations are giving me for each amplifier. One amplifier is a 60 watt and the other is a 120 watt. Both have the same driver and feedback configuration. Why is the peak power dissipated by the 8 ohm output resistor roughly the same? The 60 watt amp should have around half the peak power as the 120. Is it in the transformer model or something? Everything is there. Just open it in LTSpice and it should work. |
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#2 |
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diyAudio Member
Join Date: Aug 2008
Location: Evanston, IL / Rochester, NY
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Sorry, I forgot to attach the file.
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#3 |
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diyAudio Member
Join Date: Feb 2009
Location: Greater Seattle Area
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Maybe the 120 W requires 1.4x the input voltage to reach 120 W output power.
~Tom |
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#4 |
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diyAudio Member
Join Date: Aug 2008
Location: Evanston, IL / Rochester, NY
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Is that the case in real life, you need about a volt more to produce the greater power?
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#5 |
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diyAudio Member
Join Date: Jan 2009
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For your two designs Tomchr is correct because both designs have about the same voltage gain, because the feedback networks are the same.
In the 120W amp, you can either change R36 to 10K or change R37 to 330R. That will increase the voltage gain of the 120W amp by approx. sqrt2 and make the input sensitivity about the same for both amps at their rated outputs. Simon |
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#6 |
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diyAudio Member
Join Date: Feb 2009
Location: Greater Seattle Area
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Output power, P = E^2/R, where E is the output voltage, R is the load resistance. Hence, for double the power, you need sqrt(2) = 1.414 times higher voltage for the same load resistance.
P1 = E1^2/R; P2 = E2^2/R; P2 = 2*P1 --> E2^2/R = 2*(E1^2/R) <--> E2^2 = 2*E1^2 <--> E2 = sqrt(2)*E1 As you and EssB point out, the feedback networks of the two amps are the same. Thus, under the assumption that the two amplifiers have large loop gain, their closed-loop gains, i.e. the gain from input to output with feedback applied, will be identical for the two amps. If they have the same gain, but one is spec'ed to provide twice the output power of the other, then it must mean that it requires sqrt(2) times the input voltage to reach that power. It also means that the amp with the higher max output power will be able to produce an output voltage with higher amplitude. ~Tom |
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