@6A3sUMMER thanks again! I wasn't aware of those:
https://en.wikipedia.org/wiki/Second-order_intercept_point
https://en.wikipedia.org/wiki/Third-order_intercept_point
So, applying the same to the point at 450V B+, 60 mA Ia, 5500 Ra, we have:
10 Wrms at 3% (-30 dBc) 2nd and 0,6% (-44 dBc) 3rd.
1 Wrms is 10 dB lower than 10 Wrms, so we can expect 2nd to be 1% (-40 dBc) and 3rd 0,006% (-64 dBc).
https://en.wikipedia.org/wiki/Second-order_intercept_point
https://en.wikipedia.org/wiki/Third-order_intercept_point
So, applying the same to the point at 450V B+, 60 mA Ia, 5500 Ra, we have:
10 Wrms at 3% (-30 dBc) 2nd and 0,6% (-44 dBc) 3rd.
1 Wrms is 10 dB lower than 10 Wrms, so we can expect 2nd to be 1% (-40 dBc) and 3rd 0,006% (-64 dBc).
zintolo,
Glad you found the writings on Intercepts.
I lived those everyday, for decades.
Now in my old age, I have to remind myself, sometimes by doing an example calculation or two, in order to make sure I am doing it right.
-64dBc is 0.06% . . . not 0.006%.
I consider 5% 2nd harmonic distortion to be at, or just outside, of the well behaved region.
With a sine wave test signal displayed on an oscilloscope, at 5% 2nd the distortion is obvious.
But at 3% 2nd harmonic distortion, it is less obvions, and surely is within the well behaved region.
If you are barely outside of the well behaved region, then when you drop the power by 10 dB, the 2nd will improve 'at least' by 10dB, and the 3rd will improve by 'at least' 20dB. (11dB and 22dB, perhaps).
For additional 10dB lower power out, we are back to the 10dB and 20dB rule, because we are comparing one power that is in the well behaved region, to another power that is also within the well behaved region.
These are always fun.
So is the listening.
Uh Oh!
I was reminded of the '1dB compression point' concept (another effect, and another measurement, but I used it in RF, but not in audio, even though it can be tested in audio too).
Just keep increasing the input to the amplifier, step by step, until the amplifier output increase lags by 1dB.
Suppose you use 10 each 1 dB steps, and the amplifier output only increases by 9dB (the 1 dB compression point).
If I remember correctly, MJ magazine from Japan had graphs showing the compression curve.
I am going to leave that one alone, at least for now.
If you do the other classical amplifier tests, you need not run the 1dB compression point test.
Glad you found the writings on Intercepts.
I lived those everyday, for decades.
Now in my old age, I have to remind myself, sometimes by doing an example calculation or two, in order to make sure I am doing it right.
-64dBc is 0.06% . . . not 0.006%.
I consider 5% 2nd harmonic distortion to be at, or just outside, of the well behaved region.
With a sine wave test signal displayed on an oscilloscope, at 5% 2nd the distortion is obvious.
But at 3% 2nd harmonic distortion, it is less obvions, and surely is within the well behaved region.
If you are barely outside of the well behaved region, then when you drop the power by 10 dB, the 2nd will improve 'at least' by 10dB, and the 3rd will improve by 'at least' 20dB. (11dB and 22dB, perhaps).
For additional 10dB lower power out, we are back to the 10dB and 20dB rule, because we are comparing one power that is in the well behaved region, to another power that is also within the well behaved region.
These are always fun.
So is the listening.
Uh Oh!
I was reminded of the '1dB compression point' concept (another effect, and another measurement, but I used it in RF, but not in audio, even though it can be tested in audio too).
Just keep increasing the input to the amplifier, step by step, until the amplifier output increase lags by 1dB.
Suppose you use 10 each 1 dB steps, and the amplifier output only increases by 9dB (the 1 dB compression point).
If I remember correctly, MJ magazine from Japan had graphs showing the compression curve.
I am going to leave that one alone, at least for now.
If you do the other classical amplifier tests, you need not run the 1dB compression point test.
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In order to do a 1dB compression test, you need two 1dB step attenuators, two 10dB step attenuators, a signal source; an output fixed attenuator to protect the output attenuators; and a power measurement device (voltage or power).
Doing the test manually with 4 attenuators is easy.
Without those, there may be a suicide in the test department.
HP step attenuators work great! (Even though I worked for Tektronix).
Doing the test manually with 4 attenuators is easy.
Without those, there may be a suicide in the test department.
HP step attenuators work great! (Even though I worked for Tektronix).
Nice to have some new things to learn!
https://www.ni.com/docs/en-US/bundle/ni-rfsa/page/nirfsa/gain-compression-point.html
Would you mind explaining the procedure more in details?
So it's the point where the power is 79% and the signal 89% of what would have been with a linear amp.
https://www.ni.com/docs/en-US/bundle/ni-rfsa/page/nirfsa/gain-compression-point.html
Would you mind explaining the procedure more in details?
So it's the point where the power is 79% and the signal 89% of what would have been with a linear amp.
zintolo,
Suppose an amplifier is rated to have 20dB gain, and 10 Watts output.
And suppose that coincidentally the 1dB compression point is at a little less power than 10 Watts output (1 dB less than 10 Watts).
Send a 1kHz sine wave into the amplifier input, and adjust the level until there is 1 Watt output; verify that it is putting out 1 Watt (2.828Vrms into an 8 Ohm non-inductive resistor.
Increase the voltage to the input by 1 dB (1.122 x the original signal voltage). The amplifier output should now be 1.122 x 2.828V)
Increase the voltage to the input by another 1 dB (1.122 x the second voltage you had). The amplifier output should now be increased by another 1.122 times.
Repeat these signal increases of 1.122 x, until the amp output is falling off by 1dB, versus the input increases of 1dB.
Suppose that the output did not go to 10 Watts, it went to 1dB less (2.828Vrms x Root (10)Vrms would be 8.943 Vrms);
But instead you only get 1 dB less than 10Watts (8.943/1.122 = 7.97Vrms).
As you can see, this is very complex, unless you have 1dB step attenuators (you increase the input by 1dB, and increase the output 1dB attenuator by the same 1dB step, and the voltage remains constant.
When the constant voltage output is 1 dB less (1/1.122), you have found the 1dB compression point.
This test can only easily be taught by showing it to somebody, with a complete set of two 10dB step attenuators, and two 1dB step attenuators.
Now you know why my earlier Post said that I did not want to illustrate the test.
At work it was easy, I had a complete set of two HP RF 10dB step attenuators, and two HP RF 1dB step attenuators (50 Ohms in and out).
Oh . . . is there any other world than a 50 Ohm world? Of course, but the 50 Ohm world was the most common one to use the 1dB compression test.
I know, some would like to claim that there is only a 75 Ohm world (video), but I tell them I can not "see" it.
Boy, the battles of different electronic disciplines that think they are the only ones out there.
Over 40 years of electronics including radar, TV, LF, HF, VHF, UHF communications, sonar, test and measurement, ultrasound dopplers, sub-millimeter waveguide, EMI, EMC, Cell Phone, optical fiber and OTDRs, electronic countermeasures, ship degaussing, electronic navigation, satellites, etc. told me there was more than one electronic discipline.
Suppose an amplifier is rated to have 20dB gain, and 10 Watts output.
And suppose that coincidentally the 1dB compression point is at a little less power than 10 Watts output (1 dB less than 10 Watts).
Send a 1kHz sine wave into the amplifier input, and adjust the level until there is 1 Watt output; verify that it is putting out 1 Watt (2.828Vrms into an 8 Ohm non-inductive resistor.
Increase the voltage to the input by 1 dB (1.122 x the original signal voltage). The amplifier output should now be 1.122 x 2.828V)
Increase the voltage to the input by another 1 dB (1.122 x the second voltage you had). The amplifier output should now be increased by another 1.122 times.
Repeat these signal increases of 1.122 x, until the amp output is falling off by 1dB, versus the input increases of 1dB.
Suppose that the output did not go to 10 Watts, it went to 1dB less (2.828Vrms x Root (10)Vrms would be 8.943 Vrms);
But instead you only get 1 dB less than 10Watts (8.943/1.122 = 7.97Vrms).
As you can see, this is very complex, unless you have 1dB step attenuators (you increase the input by 1dB, and increase the output 1dB attenuator by the same 1dB step, and the voltage remains constant.
When the constant voltage output is 1 dB less (1/1.122), you have found the 1dB compression point.
This test can only easily be taught by showing it to somebody, with a complete set of two 10dB step attenuators, and two 1dB step attenuators.
Now you know why my earlier Post said that I did not want to illustrate the test.
At work it was easy, I had a complete set of two HP RF 10dB step attenuators, and two HP RF 1dB step attenuators (50 Ohms in and out).
Oh . . . is there any other world than a 50 Ohm world? Of course, but the 50 Ohm world was the most common one to use the 1dB compression test.
I know, some would like to claim that there is only a 75 Ohm world (video), but I tell them I can not "see" it.
Boy, the battles of different electronic disciplines that think they are the only ones out there.
Over 40 years of electronics including radar, TV, LF, HF, VHF, UHF communications, sonar, test and measurement, ultrasound dopplers, sub-millimeter waveguide, EMI, EMC, Cell Phone, optical fiber and OTDRs, electronic countermeasures, ship degaussing, electronic navigation, satellites, etc. told me there was more than one electronic discipline.
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