The schematic in post 18 doesnt work, or rather any distorsion reduction doesnt come from the alleged
current dumping but from the huge GNFB provided by the 5532 wich is set at very low CLG, that s roughly
80dB NFB at 1kHz enclosing the OS, hence the +-0.6V OS threshold will produce no more than 60uV
switching distorsion by the virtue of those 80dB GNFB.
The 100R "current dumping" resistance only influence here is to force the NE5532 to output +-6mA
to reach the OS input threshold, beside any current dumping circuitry is supposed to make the work within
this +-0.6V dead zone, and this require 0.6/8 = 75mA pk wich the 5532 is perfectly unable to supply,
let alone with a 4R load and a 150mA pk requirement.
current dumping but from the huge GNFB provided by the 5532 wich is set at very low CLG, that s roughly
80dB NFB at 1kHz enclosing the OS, hence the +-0.6V OS threshold will produce no more than 60uV
switching distorsion by the virtue of those 80dB GNFB.
The 100R "current dumping" resistance only influence here is to force the NE5532 to output +-6mA
to reach the OS input threshold, beside any current dumping circuitry is supposed to make the work within
this +-0.6V dead zone, and this require 0.6/8 = 75mA pk wich the 5532 is perfectly unable to supply,
let alone with a 4R load and a 150mA pk requirement.
If you say so, so be it, i stand my case that in this schematic the only thing that is at work to correct the OS
non linearity is the 5532 providing 80dB GFNB, this amount is a fact and the 100R resistance change nothing.
Edit : As you reduce the 5532 gain from 20dB to 6dB the crossover distorsion is reduced accordingly by 14dB,
thing is that i did some sims before stating so, while you are just talking out of random estimations.
non linearity is the 5532 providing 80dB GFNB, this amount is a fact and the 100R resistance change nothing.
Edit : As you reduce the 5532 gain from 20dB to 6dB the crossover distorsion is reduced accordingly by 14dB,
thing is that i did some sims before stating so, while you are just talking out of random estimations.
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The output stage is ordinary, and so is its wingspread. The original Quad used an underbiased quasi-complementary stage (its low bias and no need to keep it constant were selling points I believe). I remember it was pulled away from the zero bias point by the current of the preceding common-emitter stage. Complementary output stages (as in the Q17) and/or higher bias values are totally possible and help with performance. However, once you get thermal runaway under control, there is no real need for current dumping anymore, as increasing loop gain provides better and more consistent results than Walker-style cancellation, at least on measurement.
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But its completely underbiased, so not ordinary surely? The resistor is filling the gap and this ought to have a distinctive shape in the wingspread?
There cant be a smooth transition from the driver, actualy the VAS, wich supply 42mA, to the driven power xtor
when feeding the load.
The driver is supposed to feed the load up to 84mA pk, (0.65V pk) while being in class A, notice that at idle those 42mA are enough
to bias the other side of the OS, 25mA goes to the other driver and the remaining 17mA to the complementary power device
while the upper power device can be left switched off.
when feeding the load.
The driver is supposed to feed the load up to 84mA pk, (0.65V pk) while being in class A, notice that at idle those 42mA are enough
to bias the other side of the OS, 25mA goes to the other driver and the remaining 17mA to the complementary power device
while the upper power device can be left switched off.
Attachments
Ordinary for completely underbiased, then:
The model is for the Quad 909 OPS, which runs open loop here. Although the model includes the above mentioned resistor, it cannot show any distortion cancellation - you need to close the feedback loop and carefully balance the resistor and the inductor's series resistance for that.
Yes - but the distortion doesn't change appreciably, whether you connect the feed-forward resistor or not. I did not build it, or course, just simulated with your values:
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If you implement current dumping properly, the crossover distortion will not show up from the output. It doesn’t matter if you bias it or not. The dominant factor is the class A amp(the opamp in my case). That is why it is not biased in Quad 405.
You really need to read through the links from the OP to understand how current dumping and feedforward resistor work. Yes, the feedforward resistor is required.
In theory, it won't. (Trust me, I know how it should work). But for that to happen, that distortion must show up at the opamp input:
The amplitude and the bandwidth of the distortion signal far exceed those of the signal being amplified, and it doesn't help the opamp's linearity. A full cancellation of the crossover distortion is tricky to achieve, but before you get there, the linearity of the opamp becomes a problem - in particular, the linearity of its input stage. Because of that, current dumping works much better wrapped in another, outer feedback loop. It also helps to have a more linear output stage than a "current dumper".
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I had the same doubt that the opamp model is too ideal in the simulation. However, after I simulated with a "discrete opamp", right now, I am pretty sure it is going to work. Below, the BD139/BD140 is biased at 16ma.
PS: I think the inductor play a big role to filter out a big portion of the distortion.
I see nothing like your 26dB difference in THD in my simulation - more like 1dB difference - but if it works for you, I guess it works.
It would be interesting to find the source of the difference in these simulations. Alex, did you use the same component values as those posted above, particularly around the bridge?
Here is a good paper about current dumping. It explains current dumping without visualize the "Bridge". I like this approach, because in this approach, there is no need to add the capacitor to complete the bridge. For most opamps, you don't have the access to the internal Cdom. Please check the file attached to read the full paper.
Here is the most important part in the paper.
The balance condition "6" mentioned in the the paper attached.
Note, the k is the feedback ratio: k = Ra / (Rb + Ra)
"A" is the finite voltage gain of the amplifier.
For a single-pole-compensated opamp, the voltage gain rolls off proportionally with the frequency. Fortunately, the "Gain Bandwidth Product" stays same(**See note). You can find Gain Bandwidth Product (GBP) from the datasheet.
Thus:
A = (GBP) / freq
Then the condition "6" above becomes.
R3 / R4 = k * (GBP) / freq
R4 is an inductor. R4 = 2 * pi * freq * L
Then you can caculate R3:
R3 = k * (GBP) / freq * 2 * pi * freq * L
The "freq" is cancelled out. Thus the condition can be balanced regardless the frequency.
R3 = 2 * pi * k * (GBP) * L
By given L, you can calculate R3. Conventionally, the output inductor ranges from 1uH to 5uH. To maximize R3 (to minimize the current), I'll pick 4.7uH for the inductor.
**NOTE: If you really look into NE5532, the compensation looks more like not one, not two, but a 1.3-pole compensation, with 120 degree phase shift. That might throw off the balance condition. This might be worth farther investigation.
Here is the most important part in the paper.
The balance condition "6" mentioned in the the paper attached.
Note, the k is the feedback ratio: k = Ra / (Rb + Ra)
"A" is the finite voltage gain of the amplifier.
For a single-pole-compensated opamp, the voltage gain rolls off proportionally with the frequency. Fortunately, the "Gain Bandwidth Product" stays same(**See note). You can find Gain Bandwidth Product (GBP) from the datasheet.
Thus:
A = (GBP) / freq
Then the condition "6" above becomes.
R3 / R4 = k * (GBP) / freq
R4 is an inductor. R4 = 2 * pi * freq * L
Then you can caculate R3:
R3 = k * (GBP) / freq * 2 * pi * freq * L
The "freq" is cancelled out. Thus the condition can be balanced regardless the frequency.
R3 = 2 * pi * k * (GBP) * L
By given L, you can calculate R3. Conventionally, the output inductor ranges from 1uH to 5uH. To maximize R3 (to minimize the current), I'll pick 4.7uH for the inductor.
**NOTE: If you really look into NE5532, the compensation looks more like not one, not two, but a 1.3-pole compensation, with 120 degree phase shift. That might throw off the balance condition. This might be worth farther investigation.
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