NFB Lowers Output Impedance?

[tbryanh]

It has been said that NFB lowers the output impedance of an amplifier, and the more NFB there is, the more the speaker is damped (from the lower output impedance) and the less effect the resonance of the speaker has on the sound (tighter bass).

It appears correct that the more NFB there is, the less effect the resonance of the speaker has on the sound (tighter bass), but not because NFB reduces the output impedance of the amplifier. It is because NFB reduces the output voltage of the amplifier at the resonant frequency of the speaker.

What is the output impedance of an amplifier rated at 8 ohms for example?

Suppose the primary inductance of the OT is 5 H. (5 H might be in the ballpark. See http://www.hammondmfg.com/125a.htm)

The inductive reactance of the primary at 100 Hz is about 3.15 Kohm. (2 * pi * 100 * 5).

Suppose the plate resistance of the pentode output stage is 1 Mohm. (1 Mohm might be in the ballpark. See http://www.geocities.com/tube_theory/Pentode.htm)

Suppose the impedance ratio of the OT is 400 to 1. (400 to 1 might be in the ballpark. See http://www.transcendentsound.com/amp..._impedance.htm)

The output impedance of the amplifier at 100 Hz is about 8 ohms. (3.15 Kohm / 400) (Vector sum of 1 Mohm and 3.15 Kohm is approximately 3.15 Kohm.). (The output impedance increases as frequency increases.)

Suppose the resonant frequency of the speaker is 100 Hz. (100 Hz might be in the ballpark. See http://www.churchsoundcheck.com/imp1.html)

Suppose 1000 Hz sine wave is input to the amp. The output impedance of the amplifier is about 31.4 Kohm, and the impedance of the speaker is about 6 ohms. Since the speaker impedance is less than the amplifier output impedance, the speaker is loading the amp pretty well.

Suppose there is a certain amount of NFB. The signal across the speaker at 1000 Hz will be a certain amount.

Suppose the frequency to the amp is reduced to 100 Hz while its input level is held constant. (Assume the preamp and splitter frequency response is flat.)

The impedance of the speaker is now about 18 ohm while the output impedance of the amplifier is 8 ohms.

The speaker is no longer loading the amplifier as much as it was before, and the output voltage of the amplifier starts to increase.

The increasing output voltage causes more negative feedback through the NFB loop which tends to reduce the output voltage.

The effect of the resonance has been reduced because the output voltage of the amplifier has been reduced by the NFB. The more NFB, the more the effect of the resonance is reduced, but the output impedance of the amplifier does not change when NFB changes, and the damping of the speaker does not change when NFB changes.

[Steve Conner]

You say that the output voltage of the amplifier is reduced at resonance by NFB, compared to what it would have been without NFB.

Well, that is the meaning of "damping", and also the meaning of "a reduction of output impedance". The "output impedance" can be understood by saying that we have an ideal power amp whose voltage doesn't change at all with loading, and a fictitious resistor in series with this, that forms the top arm of a potential divider, whose bottom arm is the speaker. The smaller this fictitious resistor, the less the amp's output voltage will be affected by speaker loading. This resistor is what we call output impedance. (or the real part of it, at least.)

Here's another experiment you can try: Tap the woofer cone with the power amp on standby, then tap it again with the amp live. Can you hear a difference? If the resistor we imagined above is small compared with the voice coil resistance, then we have a good "damping factor". This means that if the cone velocity is different to what the amp's output voltage demands (and this includes disturbances like tapping the cone) then the difference in velocity gets damped out by what's essentially eddy current braking. Without NFB, the damping factor is almost zero, and the cone velocity can be whatever the mechanical side wants.

Qms and Qes in the Thiele-Small parameters ("Mechanical and Electrical Q") quantify this. With the voice coil open-circuited (or driven by pentodes without NFB) the speaker resonance has a Q of Qms. With the voice coil shorted (or connected to an amp with high damping factor) the resonant Q is (Qms*Qes)/(Qms+Qes).

[tbryanh]

Thanks for the reply. I'm a little stubborn I guess, but the way it appears to me is:

Damping absorbs energy out of a resonant system causing the resonance to have less effect.

NFB reduces the amount of energy put into a resonant system causing the resonance to have less effect.

NFB controls voltage. Frequency controls impedance.

Above the resonant frequency of the speaker, the impedance of the speaker is less than the output impedance of the amplifier, so the speaker loads the amplifier well.

At the resonant frequency of the speaker, the impedance of the speaker is greater than the impedance of the output of the amplifier, so the speaker does not load the amplifier well, and the output voltage of the amplifier increases. The NFB loop senses the increase in voltage and reduces it. The result is, at the resonant frequency of the speaker, less energy is put into the resonant system by the amplifier than would be put in to it if there were no NFB loop.

Applying the inductive reactance formula to the inductance of the primary of the OT and dividing the results by the impedance ratio of the OT is how the output impedance of the amplifier is found. NFB is not part of the inductive reactance formula.

[Steve Conner]

What you say is correct, but it is also possible for NFB to absorb energy out of the system, as well as just reducing the amount put in. If you try the cone tapping experiment, you'll see this for yourself.

NFB makes the real part of the impedance smaller, by causing the power tubes to work against any error in the output voltage. In doing so they damp the error and dissipate the energy associated with it as heat in their plates. Or as you said, in the case of the speaker resonance, they damp it by just not supplying as much power as they would in the absence of NFB.

The inductive reactance of the OPT is a pure red herring. Inductance is imaginary impedance, but only the real part of the impedance can perform damping, because imaginary impedances can't dissipate energy.

Connecting an inductor across speaker terminals will not damp it, just raise the resonant frequency, as if you'd stiffened the suspension. This is because the suspension compliance looks like inductive reactance once it's been transformed into the electrical domain, and putting another inductor in parallel with it makes it smaller.

Note: the transformation done by the motor is current -> force, velocity -> voltage. Compliance is displacement/force, which transforms to (time integral of voltage)/current. (since displacement is the time integral of velocity)

Differentiating both sides with respect to time, we have voltage proportional to rate of change of current, which is the equation for an inductor: V=L di/dt. This proves that inductance and compliance are analogous.

By a similar argument we can say that cone mass transforms to capacitance, and mechanical damping to resistance. This is why you see a RLC circuit used in some power soaks to model the speaker resonance.

[tbryanh]

Thanks for the good response. I think it’s getting clearer to me. Of course I am still being stubborn. As it appears to me now:

Damping is the result of potential difference and impedance.

Looking at one half cycle, as the voltage of the amp increases from zero volts, the amp puts energy into the resonant system, but later in the half cycle, the speaker generates electricity, and the amp absorbs energy from the resonant system.

As was previously mentioned, NFB causes less energy to be put into the resonant system, but NFB also causes more energy to be removed from the resonant system during the time that the speaker generates electricity.

As the speaker generates electricity, a potential difference exists between the plates of the output tubes and the voltage reflected back to the primary of the OT from the voice coil, and current flows through the impedance into the amplifier from the speaker.

NFB increases the potential difference during the time that the speaker generates electricity, so more current flows into the amp from the speaker. The speaker is damped more than it would be with no NFB.

The impedance, however, remains fixed at resonance regardless if NFB exists or not.

[Steve Conner]

Well, what you said above seems quite reasonable, so I guess we're just quibbling over details now. However, the way that I use the concept of impedance is the way that it's traditionally used in engineering (where do you think I learnt it?)

I've posted explanations in the past of exactly what the "output impedance" of an amplifier is. Here is a practical example:

I take a push-pull power amp wth an ultralinear or triode output stage, and no overall NFB. This may have an output impedance roughly the same as the speaker impedance. For convenience, let's say it has an output impedance of 8 ohms.

I set it up on the bench with a signal generator and an 8 ohm dummy load, and adjust it so that it produces 10 volts (RMS, peak, whatever, doesn't matter as long as we're consistent) Now I remove the dummy load so the amp operates into an open circuit. The output voltage will increase to 20 volts.

Now I modify the amp to have overall NFB with a feedback factor (sometimes called AB) of 10x. According to the theory, this will reduce the output impedance by a factor of (1+AB), which is to say, it will reduce it from 8 ohms to 8/11 of an ohm.

I now test it again as above, and I find that when I remove the dummy load, the output voltage doesn't increase to 20 volts. It only increases to 10((8/11)+8)/8 = 10(1+1/11) = 10.9-whatever volts.

These are the physical facts you'll observe in a lab, and the industry standard model is that the amp is composed of an ideal voltage source with an impedance in series. This is called the Thevenin equivalent model. There is also the Norton equivalent: a current source with a resistor in parallel, which is probably more relevant to tube amps, but the two models can be shown to be equivalent.

In order to explain the results above, the value of the Thevenin or Norton equivalent resistance must be decreased by NFB; and this is why we say that "NFB lowers output impedance".

A similar effect happens when you bridge a PA amp. If you connect an 8 ohm load to a bridged amp, the ends of the load are driven with equal and opposite voltages. This implies that the center of the load is at zero volts, so we could ground it without affecting the circuit in any way. Therefore, each side of the amp sees a 4 ohm load. Nothing has changed about the load: it's still physically 8 ohms, but a form of "feedback" makes it look like 4. This is similar to the way in which NFB "lowers" the resistances of the physical parts an amp is made from.