Damping Factor >1000

[phase_accurate]

It is always astonishing how people believe in the magic that a high DF should be capable of doing to speaker control.

And there is ab solutely nothing gained (apart from an amp with LESS load stability) to move from an already extremely high DF of 1000 to one of 20000.
Lets do some math:
Load is 8 Ohms, we connect a cable with a total resistance of 20 mOhms (for cable AND connectors, which is a damnblodygood value).
The DF of 1000 would now be reduced to 444 and the DF of 20000 is reduced to 769 !

I don't disagree with the fact that some of these amps have indeed good LF control. But these are usually very generously dimensioned amps and I think it is the SUM of all this aspects that makes for good woofer control.

You can't have more control than the driver's TSPs allow unless you are going the NFB way or you are using an amp with negative output resistance (and don't forget to take VC heating into account when doing the latter !).
As long as your amp's output impedance is real and positive the difference in damping between an amp with a DF of 10 and a DF of 1000 is just a few % !!!

Regards

Charles

[Steven]

Originally posted by lumanauw
... I would like to ask about Steven's method. From his method, the first amp's output is treated like ground for the second amp. And this first amp output is then measured again to absolute ground. Is this right?

Correct.

Steven

[MarcelvdG]

Regarding the question of voltage versus current driving a single dynamic loudspeaker:

In the vicinity of the loudspeaker resonance frequency, the voltage across the loudspeaker terminals is largely motional voltage, that is, the back EMF of the moving voice coil. Therefore, there is a pretty direct relation between terminal voltage and cone velocity for frequencies close to the resonance frequency.

For frequencies far from the resonance frequency, the voltage is mainly determined by the product of the current and the impedance of the voice coil. Voice coil resistance changes significantly when the coil heats up and its self inductance can also vary when it moves back and forth, because the surrounding iron gets more or less influence depending on the coil position.

So, near resonance, there is a close relation between the voltage and the cone movements, and it makes sense to voltage drive the loudspeaker. Far from resonance, it makes more sense to use current drive. If you use voltage drive at a frequency far from resonance, the voltage is first converted into a distorted and compressed current by the non-linear voice coil impedance, and then the current is converted into a force driving the cone.

What all of this adds up to, is that with respect to distortion and compression in the loudspeaker, it may not be such a bad idea to equalise the low-frequency part of the loudspeaker's impedance characteristic with one or two parallel LRC tanks and then current drive the whole thing.

If the loudspeaker is designed to have a more or less flat response at high frequencies with voltage drive, but its impedance rises due to voice coil inductance, then a first-order low-pass filter in front of the amplifier will be required to prevent an increasing response at high frequencies.

[phase_accurate]

What all of this adds up to, is that with respect to distortion and compression in the loudspeaker, it may not be such a bad idea to equalise the low-frequency part of the loudspeaker's impedance characteristic with one or two parallel LRC tanks and then current drive the whole thing.

MFB is one topolgy that benefits strongly from current-drive. There would be increased stability of the loop simply due to the elimination of the voice coil inductance's effect.

OTOH it should be possible to use a mixed form of current- and voltage- drive as a compromise. Qtc would be higher, but still well defined so that it could be EQed out.
The other possibility is to make it frequency dependant: Voltage-drive around fs and current drive for any other frequencies.

Regards

Charles

[MarcelvdG]

With an LRC series circuit in parallel with the loudspeaker and a current source driving the whole thing, the loudspeaker sees a relatively low driving impedance near resonance and a high impedance everywhere else.

[MarcelvdG]

I now see that one of my previous posts is a bit confusing. When I wrote "parallel LRC tank", I actually meant an LRC series circuit connected in parallel with the loudspeaker.

[phase_accurate]

With an LRC series circuit in parallel with the loudspeaker and a current source driving the whole thing, the loudspeaker sees a relatively low driving impedance near resonance and a high impedance everywhere else.

This would work but it's neither elegant nor very efficient.

Regards

Charles

[PRR]

> Another way to measure a high DF .... take a second amplifier

There's no "correct" way, and this avoids several issues. It does require another amplifier. My main concern is that some amplifiers' output Z varies with level. For cool-running Class AB transistor, Zout is usually highest at zero voltage, so this gives a low number for DF. If you are writing the specs for the ad, you want a higher number to impress buyers. If you actually want to damp the speaker, you want a low Zout over the whole swing. If the speaker is in resonance, some variation in Zout with swing doesn't matter much; but it does increase distortion.

My way has plenty of objections too. But it needs only standard test setup (oscillator, dummy load, scope, and good AC voltmenter) plus maybe a 50Ω 1.4W resistor. It tests over typical signal swings (you can try several amplitudes to see if thy all give similar answers). It also has a conceptual simplicity, easy to understand.

FWIW: in simulation I use both techniques: unloaded/loaded and external source.

> I think damping factor is also have to do with the power supply. The more bulk and stiff we make, offcourse the dip will be smaller for the same load.

No. Or, not with any significant amount of feedback, and assuming the amplifier (including the power supply) is not overloading and distortiong the signal. You can put a 5534 chip on two 9V batteries and demonstrate a damping factor over 1,000 in 8Ω. You will need a very good meter, because the peak output of a 5534 into 8Ω is less than 0.3 volts, possibly o.1 volts if the 9V batteries are not very fresh. But it will damp an 8Ω load. Power supplies are for power, not control.

> Again from PRR's answer, we can make high damping factor by making high feedback.

I had assumed that was obvious to anybody designing an amplifier. You can always reduce Zout with more feedback.

Without feedback: a pentode's Zout is always higher than a good-power load impedance. A triode Zout is usually about hald the Z of a good-power load, though as an extreme you can make it 1/10th. A transistor collector is always very-high impedance compared to a good-power load. A transistor emitter impedance is usually low compared to a good-power load.

The most popular output transistor affair is the emitter follower. What is the impedance of an emitter? Remember Shockley? The dynamic resistance of the emitter at various currents is:

30Ω at 1 milliAmp
3Ω at 10 milliAmp
0.3Ω at 100 milliAmp
0.03Ω at 1 Amp
0.003Ω at 10 Amp

This is true for ANY Silicon bipolar transistor.

Note that a simple Class A emitter follower working at 1 Amp has a damping factor in 8Ω of over 200, without any added feedback (and ignoring some very practical details of bias).

However that runs hot. We more often run Class AB, bias about 50mA, Zout about 0.6Ω, DF is about 10 for small signals changing to about 200 for large signal peaks. That is sure to distort. Also we have ignored thermal stability bias resistors. You can use bias resistors to also stabilize output impedance, but Zout stabilizes around 0.1Ω-0.5Ω, DF around 20.

> the gain of a single transistor is like collector resistance/emitor resistance, so if I eliminate all emitor resistance, all the gain will be infinite

No. And never trust any thought that says gain can be infinite.

There is always emitter resistance. If you don't have an actual resistor, you have Shockley's junction relation above: the emitter always has a dynamic impedance.

The collector has an impedance too. It is usually 300 to 3000 times larger than the emitter impedance.

And you have a LOAD impedance. In power amps, we never think about the collector impedance because the load impedance is much lower than collector impedance.

Go back to a Class A amplifier, biased at 1 Amp, but make it a common emitter (collector follower) instead of the usual emitter follower. The emitter impedance is 0.03Ω plus zero external resistance. The collector impedance is maybe 30Ω. The load impedance is 8Ω. So the collector load is 30 in parallel with 8. 30||8= 6.3Ω. The emitter is 0.03Ω. 6.3/0.03= 210 voltage gain.

But: look at the input impedance of this transistor. It is Beta times emitter impedance. Say 50*0.03= 1.5Ω! How is your previous stage going to drive 1.5Ω? It can be done, but you won't get any voltage gain.

It -may- be possible to get DF=1000 into 8Ω with three stages. I think you have to sacrifice a lot of other important audio goals to get there.

> I still waiting for your answers in "Very low voltage preamp"

It seems to me you know what you want, and I have nothing to add.

[PRR]

> with respect to distortion and compression in the loudspeaker

Distortion will be unchanged. Voltage, current, same thing just factored by impedance.

You should never be running speakers into significant thermal compression in non-commercial work. All the parameters shift, performance droops. Get bigger speakers.

In commercial work, some now consider it "necessary" to run in thermal compression.

With constant voltage drive, as the speaker gets hot, it takes less power. Burn-out is delayed.

With constant current drive, as the speaker gets hot, it takes more power. It goes into thermal runaway (except in practice, with practical amplifiers, it will first clip like hell).

> one or two parallel LRC tanks

An expensive and slightly wasteful way to do what constant voltage drive does naturally.

You can drive at any impedance, and then EQ-out the response. It won't even take more power, if you just EQ to the constant voltage response. You will get in trouble if you run in thermal compression.

The synergy between a dynamic small cone loudspeaker and a constant voltage amplifier is a very practical and beautiful thing. People have tried every other thing, but for 85+ years low-Z drive has held its ground (except in millions of cheap pentode radios). Any other plan requires the amp to be mated to the speaker, adds a lot of complication, and almost never gives better results.

[phase_accurate]

According to a paper I own the voltage-transfer function (motion vs voltage) has an expression of (B*l)^2 in the denominator where you don't have this for current drive.
This would give less distortion for current drive.

Regards

Charles