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|12th February 2017, 02:04 AM||#1|
Join Date: Mar 2016
Output Impedance & Planars
So I've read the following.
Musings on Headphone Amplifier Output Impedance | InnerFidelity
NwAvGuy: Headphone & Amp Impedance
So Ill explain my understanding. Under the assumption that the amp sends out an impulse in a really small time frame, the signal hits the driver. The driver will extend and make sound. But from the driver extending to its peak, and it returning, a back current will be generated (as it is an electro mechanical device ). Output impedance of the amplifier ideally is close to zero such that this current can be sank quickly, and minimize further oscillations. (If I short a generator, will it require more force due to back emf? How does a non zero load alleviate this?)
Im still struggling with the following concepts.
In the case of the dynamic driver, how do we get a different frequency response where impedance peaks? Aka if I have a impedance peak at 100hz, by increasing Zout there is a few db change up in frequency response at 100hz. Oscillation in the circuit?
Would I be correct in saying that planars (since they are purely resistive) will experience next to no change? If there is change would it be to impulse response only since their impedance is flat across the frequency range?
Where does nwavguy get the following numbers " For the output impedance to create a -1 dB change, you have antilog(-1/20) = 0.89." I realize dB's are logarithmic, but doing some math I get 20 log (.89). What is 20 and .89?
Last edited by Cata1yst; 12th February 2017 at 02:07 AM.
|12th February 2017, 04:27 AM||#2|
Join Date: Apr 2011
Location: Upper midwest
Since Rout is constant, if Zspeaker increases, the voltage across the speaker increases.
You're right, a constant Zspeaker results in a constant voltage across the speaker, since Rout is constant.
The formula is: -1dB = 20 x log (0.89). The 20 is actually two scale factors multiplied together.
We need a x2 factor to account for using voltage instead of power, since power is Vsquared/R.
We also need a x10 factor to account for using dB (decibel) instead of B (bel). So, 2x10 accounts for the 20.
The 0.89 is the factor of reduction of the amplifier output voltage due to the drop across the Rout.
If the amp puts out 1V open circuit, then the voltage across the speaker is 0.89V, because 0.11V
is dropped across the amplifier's Rout. Then the load response is -1dB down.
Last edited by rayma; 12th February 2017 at 04:46 AM.
|12th February 2017, 02:16 PM||#3|
Join Date: Oct 2010
Take a (perfect) voltage source and two resistors in series.
The first resistor Zout is our output impedance, the second Zl our load impedance (the headphones).
Set the source to 10V. Zout=0. What voltage drops across Zl? The full 10V, regardless of the resistance of Zl.
Now we assume Zl=100 ohm and set Zout to 100 ohm as well.
Now 5V drops across Zout, 5V across Zl. That is -6 dB through the headphones.
In a dynamic headphone system, Zl varies with frequency. Usually, you see a peak in impedance around 100 Hz and a minimum around 1 kHz.
So let's assume Zout=32 ohm, our headphone has 32 ohm at 1 kHz and 64 ohm at 100 Hz.
That's -6 dB at 1 kHz, -3.5 dB at 100 Hz. Effectively a +2.5 dB boost at 100 Hz.
The math: 64/(32+64) = 0.6666
20*log10(0.6666) = -3.5 dB ... that is how a deci bel is defined for an amplitude ratio.
So what is a spike in the time domain would be a flat line in the frequency domain.
From above example we get a 100 Hz boost however. Any change in the frequency response leads to ringing in the time domain.
(In a minimum phase system simple equalization back to a flat frequency response will perfectly undo that ringing however.)
So far we've only talked about voltage.
With the perfect voltage source (with Zout=0) the output voltage will perfectly follow the input signal. The current will be whatever the load draws with the simple I=V/R relationship.
As you increase the output impedance you'll get closer to a perfect current source, that is that the output current will perfectly follow the input voltage and the output voltage will be whatever V=I*R dictates.
So in the first case, around 100 Hz the load impedance is higher so less current will be output. Another way to look at this is that the efficiency near 100 Hz increases for the dynamic driver, needing less current as a result.
This also means that the "ringing" mentioned earlier will happen in both cases. In the first case you'll see it in the output current, in the second case you'll see it in the output voltage.
Last edited by xnor; 12th February 2017 at 02:26 PM.
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