I tried to measure the Zout of my EL84 SE amp with the calculator on this page: http://www.sengpielaudio.com/calculator-InputOutputImpedance.htm
I had it connected to a 8.7 Ohm dummyload (output 15.9V), and then put a 10.6 Ohm resistor in series with the dummy. Output 19.84V (all peak to peak)
The outcome was negative?! -2.11 Ohm.
I've used this calculator for preamps, but this is my first power amp, I'm probably missing a lot but I don't know what.
Thanks in advance!
I had it connected to a 8.7 Ohm dummyload (output 15.9V), and then put a 10.6 Ohm resistor in series with the dummy. Output 19.84V (all peak to peak)
The outcome was negative?! -2.11 Ohm.
I've used this calculator for preamps, but this is my first power amp, I'm probably missing a lot but I don't know what.
Thanks in advance!
If output voltage rises with rising load, impedance is not negative.
Tell us WHICH values you used, and WHERE
A screen capture would be fine.
Tell us WHICH values you used, and WHERE
A screen capture would be fine.
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That 15.9 V at the calculator (V1) is not open circuit voltage. You told you loaded it with 8.7 ohms.
The calculator uses the open-circuit voltage you don't have.
What you have is V1 = 19,84V and a load of 8,7Ω giving a current I1 = 1,028A
and V2 = 15,9V with 10,6 + 8,7 = 19,3Ω for I2 = 1,828A current
Zsource = (V1 - V2) / (I2 - I1) = 4,93Ω
But that's not the end of it, with a different load the gain of the output stage change and so does the open-circuit voltage. To get a more reliable value repeat the exercise with two allmost equal loads.
Mona
What you have is V1 = 19,84V and a load of 8,7Ω giving a current I1 = 1,028A
and V2 = 15,9V with 10,6 + 8,7 = 19,3Ω for I2 = 1,828A current
Zsource = (V1 - V2) / (I2 - I1) = 4,93Ω
But that's not the end of it, with a different load the gain of the output stage change and so does the open-circuit voltage. To get a more reliable value repeat the exercise with two allmost equal loads.
Mona
Use software?
Use logic, and understand what output impedance and damping factor mean . . .
I set the amplifier output voltage to 1.0V, with the output unloaded.
I connect an 8 Ohm load and measure the loaded voltage (for example, 0.8V loaded).
When the amplifier is loaded, the voltage across the amplifier's output impedance, X Ohms, is 0.2V.
The amplifier output impedance and the 8 Ohm load are a series circuit.
Now do a proportion equation:
8 Ohms / 0.8V = X Ohms / 0.2V
Cross Multiply: (8 x 0.2) / 0.8 = 2 Ohms, the output impedance.
8 Ohm load / 2 Ohm output impedance = 4.0.
The damping factor is 4.0
I hate software (except I love the software that is in my HP model 11 calculator).
The numbers of my measurements are at least to 3 places, so I use my HP 11 calculator, and its internal software to get an accurate damping factor.
Use logic, and understand what output impedance and damping factor mean . . .
I set the amplifier output voltage to 1.0V, with the output unloaded.
I connect an 8 Ohm load and measure the loaded voltage (for example, 0.8V loaded).
When the amplifier is loaded, the voltage across the amplifier's output impedance, X Ohms, is 0.2V.
The amplifier output impedance and the 8 Ohm load are a series circuit.
Now do a proportion equation:
8 Ohms / 0.8V = X Ohms / 0.2V
Cross Multiply: (8 x 0.2) / 0.8 = 2 Ohms, the output impedance.
8 Ohm load / 2 Ohm output impedance = 4.0.
The damping factor is 4.0
I hate software (except I love the software that is in my HP model 11 calculator).
The numbers of my measurements are at least to 3 places, so I use my HP 11 calculator, and its internal software to get an accurate damping factor.
I know nothing about measures, but if the number you got was inverted (x -1) it should be close to what it should be.
dave
dave
The loading measurement method is invalid/inaccurate for a number of reasons (impedance is a vector, it is non-linear, etc.).
The only valid way to do it (and in a sim, you have no excuse for not using it) is to drive the output (muted) with an external source/amplifier, delivering 1A AC for example.
If you measure the residual voltage appearing at the output, you obtain directly the magnitude of the impedance, and it will be the true one.
If you make a vector measurement, you can even extract the reactive part (the output impedance of practically all amplifiers is inductive)
The only valid way to do it (and in a sim, you have no excuse for not using it) is to drive the output (muted) with an external source/amplifier, delivering 1A AC for example.
If you measure the residual voltage appearing at the output, you obtain directly the magnitude of the impedance, and it will be the true one.
If you make a vector measurement, you can even extract the reactive part (the output impedance of practically all amplifiers is inductive)
THAT.
FIRST understand the Basic Principles , THEN calculate.
Pencil and paper, in your mind, using a simulator, pick one, but useless without the first part.
Just throwing numbers at simulators and taking results at face value, without analysis, is a dead end.
well, yes,and that is part of the measured generator impedance value.
Think "black box" .... why/how it outputs such and such voltage depending on load is moot in this case.
That does not change the measurement principle, but as long as they are different .... fine.
Sorry but no.
Using the proper voltages asked by the simulator the result will be correct, and impedance will be positive.
Sort of nitpicking, you suggest at a WAY more detailed level which is neither asked nor merited here, for a basic "damping factor" measurement and calculation.
You are free to calculate and post graphs or tables as detailed as you want, at all frequencies, following speaker impedance curves, etc. if you wish, but that was not asked for and far exceeds the question.
For one, no speaker brand and model was stated, no speaker impedance curves were shown, nor transformer curves.