John Curl's Blowtorch preamplifier part III

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What have-you done ? Reduce the levels of the signal at the input and output of the power stage, in order to can use a comparator with +-12V rails, extract the error signal and amplify-it with discrete devices (in class A) to apply it as a local feedback at the input of the OPS ?

No, the signal taken from input and output is not attenuated.
Supply for the error correction opamps is a floating LS Output +/- 5 Volt.

Hans
 
yes, please.

-Richard

Here it is.
AEF_in comes from the RC filter after the Voltage Amp.
Control_in goes to the OPS.

Hans

Error Correction.jpg
 
Funny that you would post this, because just the other week I was doing some experiments with rather inductive 110 Ohm power resistors before soldering them into a ballast to test amps.

I thought along the same lines as you that theoretically it might happen possibly, but there is no effect. The magfields are just too far removed for the resistors to even see each other.

Not with the small surface mount resistors there is decent coupling. For my RF test chamber load, I use thin film resistors paralleled, interdigitated and and zig-zagged. It works to above 200 megahertz!

To test amplifiers I have loads from 150 watts to 2.5 KW but they do require compensation to keep the impedances flat. Fortunately that is not really needed, as I have yet to need high power at high frequencies even in my slightly largish systems.

If you really need it I can probably make for you a couple of 8 ohm resistors that will be good to low RF frequencies, just let me know the power handling requirements.
 
Of course it is a joke Hans. 🙂
Sorry to burst your bubble but when you post a joke, it needs to be at least the level shown in the quote below plus a bold punchline (last paragraph).

"my high end audio designer friend came over one day and pulled out some XLR cables. He said to try them out and see if I notice if they sound different at all from the cables I normally was using. I didn't expect anything audible so I asked, what if I don't hear any difference? He said, then you need to get your ears checked. Haha, I thought, very funny - NOT.
...
Wow! Everything sounded better, less distorted, and the difference was easy to hear! I was basically stunned, never expected it.
...
his own wire manufactured based on a lot of trial and error research he did. He also designed and had manufactured some speaker wire. Since then I have had an opportunity to try those too and they sound better than what I was using before.
...
businesses will start selling cables designed by him. Most here will probably never hear what the cables can do because they are going to be pretty expensive, although less so than many high end audio cables. That's something I feel is unfortunate, but probably not likely to change soon.
...
EDIT: By the way, I did not write the above for any commercial purpose. My intent is only to describe the facts of what happened that I learned something from. Same as all the other posts in this thread when I was learning about how to make my modded dacs sound better. Now I have learned another way to make my system sound better, less distorted, more clean and true to the audio coming from the source device. No reason why it should be secret when I have talked about my other personal discoveries here.
"

But he's on your ignore list ...😛
Often claimed but hardly acted upon. 😉

Richard's Thailand factory already built enough to illustrate. One unstable is one too many.
Factory to build amps? Well, he must not be in audio business... :scratch2:
 
Tourny,

If you place the 20 resistors side by side (JN technique) and then have the connection essentially zig zag the the inductance will be reduced. That yields the best result. I would use the Dale metal film resistors. The power rating of the stack should be at least four times what the actual peak dissipation could be.

Another issue is that the value of the single resistor should be equal to the resistor on the source input side of a differential pair.

Some here would try the input side resistance as high as 100,000 ohms. I would use 10,000 ohms. With a normal 20/1 attenuator that would make the feedback resistor 200,000 ohms.
Even with 1/10 watt power rating of some surface mount resistors that would allow your power supply rails to be several hundred volts!

JC would probably prefer a lower resistance value to avoid the effects of stray capacitance and as he uses a DC servo matching resistances on both sides of the differential pair is not as important.

Inverting or noninverting? DC- or AC-coupled and if the latter, what sort of pull-down resistor are we talking about? What's the voltage of the input, i.e. do we really *need* much/any gain in this circuit or is it essentially a buffer? What is the front-end topology? Are we optimizing for current noise or no?

Surely there's a lot that needs consideration for sizing the feedback resistors, and that's before playing games like N-identical series resistors or what-have-you to compensate for thermal drift.
 
No, the signal taken from input and output is not attenuated.
Supply for the error correction opamps is a floating LS Output +/- 5 Volt.

Hans
Ok.
"At the end of the day" (unreasonable assumption at my age), I will try this error correction on OPS as my next project.

Output signal and signal at the OPS input both reduced at good levels for an OPA. Error correction extracted (Signal lot lot lower than the original one with all the benefits expected) and amplified, then applied, phase reversed, to the input stage of the OPS.
Gentle comments welcomed ;-)


I am thinking of using a very low distortion OPA from A.D. to gain an unconditional support from a eminent member of our forum.
 

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Ok.
"At the end of the day" (unreasonable assumption at my age), I will try this error correction on OPS as my next project.

Output signal and signal at the OPS input both reduced at good levels for an OPA. Error correction extracted (Signal lot lot lower than the original one with all the benefits expected) and amplified, then applied, phase reversed, to the input stage of the OPS.
Gentle comments welcomed ;-)


I am thinking of using a very low distortion OPA from A.D. to gain an unconditional support from a eminent member of our forum.
No, when the OPS produces 70V, input to the OPS will also be 70 Volt.
This should but cannot be supplied by the opamp you sketched.

Hans
 
No, when the OPS produces 70V, input to the OPS will also be 70 Volt.
This should but cannot be supplied by the opamp you sketched.
I don't understand your remark. Both the input and output of the OPS (70 V pp) are attenuated by two divider bridges to, let's say, some 7V pp.
After being "subtracted" the remaining error signal will be some mVs (let's say 0.07V for 1% of distortion). That can be amplified in the same proportion with the margin needed for the loss in the passive (or active) mixer at the input of the OPS.
Am-I more clear ?
 
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Inverting or noninverting? DC- or AC-coupled and if the latter, what sort of pull-down resistor are we talking about? What's the voltage of the input, i.e. do we really *need* much/any gain in this circuit or is it essentially a buffer? What is the front-end topology? Are we optimizing for current noise or no?

Surely there's a lot that needs consideration for sizing the feedback resistors, and that's before playing games like N-identical series resistors or what-have-you to compensate for thermal drift.

Derfy,

The basics of using an opamp is that when the differential inputs see the same resistance then the input will drift less with temperature. Ideally the resistance should be set so that the input offset current times the resistance matches the input offset voltage. That is because they often have inverse polarity.

The AC coupling comes at the input. In most audio power amplifier designs there is a capacitor breaking the feedback loop to ground to allow for DC feedback to be 1/1 and the AC to have gain. This often requires fairly large capacitors.

A better method is to isolate the AC and DC feedback paths, virtually unknown in the audio world. The DC feedback resistance goes from the output to the inverting input. This value should match the resistance from the non-inverting input to ground used to establish a 0 volt reference. These resistors can usually be a high value such as 100,000 ohms although it is useful to match the input offset current and voltage as before.

For the AC feedback a resistor divider is used as before except there is no capacitor in the ground leg. A capacitor is placed between the inverting input and the tap on the voltage divider.

Why is this better? Because the input impedance of the inverting input at AC frequencies is much higher than the resistance of the lower leg of the voltage divider. A smaller value capacitor may be used! So where a 100 uF capacitor might have been required with a 1,000 ohm lower leg resistor to get minimal phase shift and less than 1 dB of loss at 20 hertz a 1 uF capacitor might have the same effect!

So if you don't want to use a servo adding one resistor to an otherwise common circuit will get you most of the way there.
 
I don't understand your remark. Both the input and output of the OPS (70 V pp) are attenuated by two divider bridges to, let's say, some 7V pp.
After being "subtracted" the remaining error signal will be some mVs (let's say 0.07V for 1% of distortion). That can be amplified in the same proportion with the margin needed for the loss in the passive (or active) mixer at the input of the OPS.
Am-I more clear ?

His assumption is that the output stage has no voltage gain. You did show it driven by 7 volts so it would have the gain required to accept a similar correction voltage.

Most output stages of course are merely followers and do not have voltage gain but lots of current gain.
 
I don't understand your remark. Both the input and output of the OPS (70 V pp) are attenuated by two divider bridges to, let's say, some 7V pp.
After being "subtracted" the remaining error signal will be some mVs (let's say 0.07V for 1% of distortion). That can be amplified in the same proportion with the margin needed for the loss in the passive (or active) mixer at the input of the OPS.
Am-I more clear ?

When correcting a 70 Volt OPS input signal, the correction signal will effectively be in the sub Volt region and should hover around this 70 Volt to correct the input signal.
Suppose input signal is 70 Volt at a certain moment and should be error corrected to 70.5 Volt. This will not be possible with your opamp circuit.

Hans
 
His assumption is that the output stage has no voltage gain. You did show it driven by 7 volts so it would have the gain required to accept a similar correction voltage.
No, no and NO. Its a confusion between error correction and feedback.
There is NO original signal after subtraction in the LTP of the first OPA. Just an error signal (the distortion of the OPS that I hope is not 100% ;-) that can be amplified enough to be applied in phase opposition to the input stage of the OPS.
How can-I explain better ?
 
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