What I am thinking is to find a way to probe whether the condition is balanced or not.
Before that, the only thing that I can rely on is the datasheet.
It is just a class B amplifier.
The opamp fixes the crossover distortion with feedback.
So high slew rate opamp helps.
Otherwise with low slew opamp. These circuits will " Crackle" and pop
Because the opamp does not like fixing the Dead zone of full class B
or absolute no bias transistor distortion.
Depending on opamp they dont do much more than 20 to 40 ma.
So the resistor would need to be far far higher than 100 ohms to limit the opamp current.
Besides the 1.2 volts of "dead zone" that the "current dumping" needs to fix.
Basically you need a clean sinewave coming from somewhere to fix the dead zone.
Which basically come down to 50 ma of clean sinewave. Needs to come from somewhere to the load for a 8 ohm load.
And about 100ma from somewhere for a 4 ohm load.
So with the 100 ohm resistor that will never ever happen.
People haven't seemed to figure out with these circuits and a Non inverting amp.
Your DC offset will be positive. So typically that slight amount of positive DC on the non biased output transistors.
Makes all the AC current go to just one output transistor the NPN
If you make DC offset negative or use inverting amplifier. Then all that AC current goes to PNP transistor.
Basically you get one transistor doing all the AC signal and you have opamp fixing crossover distortion
with differential.
Easier to just bias outputs.
Sorry to be negative, I thought this was magically circuit too when first starting audio amps.
The opamp fixes the crossover distortion with feedback.
So high slew rate opamp helps.
Otherwise with low slew opamp. These circuits will " Crackle" and pop
Because the opamp does not like fixing the Dead zone of full class B
or absolute no bias transistor distortion.
Depending on opamp they dont do much more than 20 to 40 ma.
So the resistor would need to be far far higher than 100 ohms to limit the opamp current.
Besides the 1.2 volts of "dead zone" that the "current dumping" needs to fix.
Basically you need a clean sinewave coming from somewhere to fix the dead zone.
Which basically come down to 50 ma of clean sinewave. Needs to come from somewhere to the load for a 8 ohm load.
And about 100ma from somewhere for a 4 ohm load.
So with the 100 ohm resistor that will never ever happen.
People haven't seemed to figure out with these circuits and a Non inverting amp.
Your DC offset will be positive. So typically that slight amount of positive DC on the non biased output transistors.
Makes all the AC current go to just one output transistor the NPN
If you make DC offset negative or use inverting amplifier. Then all that AC current goes to PNP transistor.
Basically you get one transistor doing all the AC signal and you have opamp fixing crossover distortion
with differential.
Easier to just bias outputs.
Sorry to be negative, I thought this was magically circuit too when first starting audio amps.
Why? 0.7 V/100 ohm < 20 mA.
You seem to make the same mistake as wahab. The op-amp only needs to supply enough current to turn on the dumpers: about 0.7 V/100 ohm plus whatever input current the dumpers require. At least that's the case at frequencies where the voltage drop across the inductor is << 0.7 V, it gets more complicated at higher frequencies.
Of course this means that, if the op-amp were class A in the first place, the class-A region would be limited to an output current range from -7 mA to +7 mA. That's about 200 uW sine wave power into 8 ohm.
I think some of the later QUAD current dumping amplifiers are non-inverting.
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Suppose you want to play Deutsche Marschmusik: Einzug der Gladiatoren at full volume. I wouldn't want to play it at all and certainly not at full volume, but others may have different tastes. The peak rate of change of this music is roughly equivalent to that of an 8 kHz sine wave.
The reactance of 4.7 uH at 8 kHz is 0.236248 ohm. Assuming a peak output voltage of 14 V and an 8 ohm load, the voltage drop across the inductor would then be 0.413434 V peak, if the full and undistorted current would flow through the inductor. At 4 ohm, it would be 0.826867 V peak.
It's clearly not negligible compared to 0.7 V. A weak point in my calculation is that it doesn't account for the effect of the distortion components that also flow through the inductor.
There are a few differences that add up. Mostly, it is the frequency response and the open-loop output impedance of the opamp that mess up the cancellation. The finite DC gain of the opamp and the non-zero DC impedance of the inductor add a bit, as do the non-ideal values of the "bridge" components.
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The 5532 model in my simulation suite is too ideal.
A real 5532 has 15 Ohm emitter resistors at the output stage. That has to be included in the calculation.
Although the really 5532 rolls off a little faster than 20dB/decade at high frequency. In general the Gain Band Product still holds true from the bode plot in the datasheet.
I would love to see someone could hook it up with a spectrum analyzer.
Look like the debate at least is going forward, so i ll add some remarks.
For one the Quad 405 low THD is due to huge amounts of NFB more than anything else, FI there s 40dB GNFB
at 10kHz and 60dB/1kHz, to wich should be added 20dB/1kHz and 40dB/10kHz through the local NFB loop
that enclose the VAS, that is, 80dB total NFB in the audio band.
Beside using as reference a non biaised OS is somewhat dishonnest as it amount to cripple the reference
and give an artificial advantage to the dumping scheme.
If we remove the weird OS and replace it with an EF2 biased at 16mA the distorsion will be comparable, and if we partly
enclose the OS in a moderate TMC loop then distorsion is lower, even at 0 bias, than with the current dumping scheme,
and this without modding at all the amp s driver characteristic, at 24mA bias it is one order of magnitude lower at low powers.
For one the Quad 405 low THD is due to huge amounts of NFB more than anything else, FI there s 40dB GNFB
at 10kHz and 60dB/1kHz, to wich should be added 20dB/1kHz and 40dB/10kHz through the local NFB loop
that enclose the VAS, that is, 80dB total NFB in the audio band.
Beside using as reference a non biaised OS is somewhat dishonnest as it amount to cripple the reference
and give an artificial advantage to the dumping scheme.
If we remove the weird OS and replace it with an EF2 biased at 16mA the distorsion will be comparable, and if we partly
enclose the OS in a moderate TMC loop then distorsion is lower, even at 0 bias, than with the current dumping scheme,
and this without modding at all the amp s driver characteristic, at 24mA bias it is one order of magnitude lower at low powers.
The problem with current dumping is that the linear amplifier has to slew extremely quickly to traverse the crossover notch in the class B amplifier. The less feedforward is applied, the faster the slew rate needs to be. This is an inherent flaw.
Ed
Ed
You don’t need slew rate to make it work. The slew rate is still limited by the miller cap in quad 405 or with opamp.
Sorry, your math is wrong, again
0.7v dead zone on the class A stage, but on the output side, the dead zone is only 0.7/25 =0.028v. Exceeding that range, the dumper would be engaged. Thus there is not extra slew rate required.
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OK, I admit that my analogy is not quite right either. The beauty of the current dumping is you don’t need perfect signal at the dump. Extra slew rate to reconstruct a perfect signal is not required. If you play it a little bit, you will find it doesn’t worth the effects to minimize the dead zone or bias the dump. You get diminishing of returns.
Of course. Current dumping amplifiers use a clever combination of negative feedback and error feedforward.
If you want to establish whether the error feedforward in a current dumping amplifier reduces distortion, you obviously have to keep the output stage and the amount of feedback the same. Otherwise all you prove is that class C output stage distorts more than class AB, which is not contested by anyone.
Peter Walker and the others at QUAD were very rational people. When they developped the QUAD 405, they just wanted to design an amplifier that is good enough for all practical purposes and is very suitable for mass production. The distortion is negligible compared to the distortion of a vinyl record and is so low you still don't hear it when you much reduce the desired signal in a subtractive test, the built-in high-pass filter keeps subsonics out of the woofer and the absence of trimming points was very convenient for production.
I'm not sure what effect the inductor has, but without it, it is easy to see that @EdGr is right. Suppose you play an 8 kHz, 10 V peak sine wave. The rate of change around the zero crossing is then about 0.5 V/us. The transition from -28 mV to +28 mV then takes about 112 ns.
In this same 112 ns, the op-amp output has to go all the way from -0.7 V to +0.7 V. That's about 12.6 V/us.
The slew rate is actually very complex topic. First, lets say we have an opamp with infinity slew rate. Will it fix all the distortion? The answer is no, Here is reason.
The dynamic slew rate the amp produced is always proportional to the error voltage on the positive pin and negative pin. Let's say we need 25V/us to fix all the distortion on the dumper. In order to get 25V/us slew rate, 0.1v error voltage has to be presented between the positive pin and the negative pin. You can't fixed all the distortion while it has 0.1v error at the same time.
A good low THD amp, actually doesn't need much slew rate.
Back to the dumper, with 100 Ohm feedforward resistor, 4 Ohm load, Even if you have 25x max slew rate, you still can't fix all the error. If the dumper is used without the "bridge" balanced, no matter how fast the slew rate the amp has, you still get big distortion. As said, the actually slew rate is proportional to the error voltage. You can't have slew rate without distortion.
I played a little bit with the dumper. With 100 Ohm feedforward resistor, 8 Ohm load. Same circuit in the OP. 20KHz 1.6Vp output. The orange trace is the opamp output, while the red trace is at the emitter of the transistors. The slew rate of 20KHz 1.6Vp sine wave is about 0.2V/us. The slew rate in front of the dumper is measured about 1.6V/us. Note: The opamp has 9V/us max slew rate configured. Thus, it does not reach the limit. Yet, it won't go beyond 1.6V/us, because for example, the 5mV error voltage detected only can produce 1.6V/us.
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Double the amplitude, and the required slew rate doubles. Increase amplitude ten times to 16Vp, and your opamp will be slewing. Since you have your model, why not try it out?
The output is about 10Vp, 20KHz. The opamp is almost running out of current with single pair of transistors, so I didn't go any higher.
The slew rate is about 2.6V/us measured. As said, if the THD is low, you should not observe high slew rate. The formula for the slew rate is: (Vin_pos - Vin_neg)*gm/Cdom. You have less error to start with, you don't need big slew rate.
Let put this cleverness under test and see what is the break down of this combination.
In the Quad 405 i remove the 47R feedfoward resistance and bias the upper transistor at 5.4mA,
this will keep the lower power transistor switched off since all this current is drained by its driver that sink it
through the power device 22R base/emitter resistance.
The result is slightly lower crossover distorsion, wich is indicative that the FFWD scheme actualy worsened
things in respect of a classical implementation with the upper device at just 5.4mA bias and the lower power
device totaly switched off, notice that the output stage is left unchanged as the same quasi complementary
and the upper power device directly driven by the VAS.
So the combination is nothing else than 99.99% NFB and eventualy 0.01% error margin.