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Old 22nd November 2020, 11:45 AM   #11
mjf is offline mjf  Austria
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yes, it is inverting.........."minus gain " means inverting.
post 8:
The gain from the terminal voltage at the input to the output voltage is ideally -Rfb/R1
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Old 22nd November 2020, 11:51 AM   #12
SMABB is offline SMABB  Denmark
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Thanks a lot for clarifying Marcels comments. mjf

To be honest I hadn't really noticed the "-" before the formula, so really great of you to address this, and thus also answering my question.
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Old 22nd November 2020, 03:49 PM   #13
MarcelvdG is offline MarcelvdG  Netherlands
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It's also mentioned explicitly in the manual you posted in post #3.

When you type op-amp inverting mode in duckduckgo.com or any other search engine, you will find dozens of websites that explain how this feedback configuration works. In your case the op-amp is not an IC but a discrete amplifier optimized for use as MC headamp, but the principle stays the same.
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Old 22nd November 2020, 05:54 PM   #14
SMABB is offline SMABB  Denmark
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It is - and I fully agree; however I had doubts whether this was applied to the MC prepre in particular or the preamplifier itself.

I follow you on the search pattern, but I had focused my searches around the MC-2 unit itself.
Simply because I am not an electronic wizard, but excels more in the acoustical area of music reproduction. Hence, I did not know what to search for. In addition it was only this morning that I stroke luck and found the preamplifier manual while extending my search for information.
I highly appreciate what you have put into light, Marcel, and fully understand the principle of inverted configuration. It all adds up nicely with the theory.

Now - since the sound of the unit in basic configuration is so amazing, I wonder whether it would make sense to build a new power supply instead of that mediocre type that was supplied. As circuit diagram notes voltage is 24 V DC and current is 35 mA on the lousy plug in. I'm pretty sure that there is gains to be had.

What do you all think? Would a shunt regulator following the principles of Borborly or Salas be sufficient?
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Old 23rd November 2020, 01:55 PM   #15
Hans Polak is offline Hans Polak  Netherlands
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The simplest way of finding out whether it pays to change your current power supply is to temporarily replace it with six 9Volt alkaline batteries.
With 580mAh they will last long enough to find out whether any acoustical benefit can be noticed.
Alkaline batteries are amongst the cleanest sort of supply available.
If nothing changes, just keep things as they are now.

Hans
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Old 23rd November 2020, 03:20 PM   #16
MarcelvdG is offline MarcelvdG  Netherlands
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The amplifier seems to need only one floating 24 V source, the 10 V Zener (or actually avalanche) diodes then generate +/- 10 V from it. 27 V may already overheat the diodes, so I guess you would need three 9 V batteries in series and a 47 ohm resistor to limit the current. Just three batteries might do if their internal resistance is high enough.
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Old 23rd November 2020, 07:55 PM   #17
PRR is offline PRR  United States
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Help needed understanding this prepreamplifier
Quote:
Originally Posted by MarcelvdG View Post
...27 V may already overheat the diodes...
Put a pencil on it. If we trust the numbers, it already has 606mW in the 400mW diodes.

But do we understand how much power-crap reduction is possible here? The R-Z well over 20dB. The cap-multipliers are not pulling their weight at about 40dB. All paths from a rail to the output are via Collectors and typically >>60dB. Maybe 70dB each but three paths so pencil 60dB. Plus 40 plus 20 is 120dB rail rejection relative to output. To input another 20dB-30dB.

There's things I don't like but power cleanup does not look necessary. The designer is not as dumb as we think.
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Old 23rd November 2020, 08:00 PM   #18
MarcelvdG is offline MarcelvdG  Netherlands
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606 mW is assuming that all current flows through the Zeners. The rest of the circuit draws about 20 mA (rough estimate), so it is very much on the edge: it needs to draw 20.6 mA to reduce the power dissipation of the Zeners to 400 mW.

Last edited by MarcelvdG; 23rd November 2020 at 08:04 PM.
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Old 23rd November 2020, 09:52 PM   #19
Hans Polak is offline Hans Polak  Netherlands
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Quote:
Originally Posted by MarcelvdG View Post
The amplifier seems to need only one floating 24 V source, the 10 V Zener (or actually avalanche) diodes then generate +/- 10 V from it. 27 V may already overheat the diodes, so I guess you would need three 9 V batteries in series and a 47 ohm resistor to limit the current. Just three batteries might do if their internal resistance is high enough.
Right, only one single 24 volt supply is all thatís needed.
In that case two small 1.3Ah 12Volt lead acid batteries could be a perfect choice.
They can be found for ca Ä7,- each, giving you many hours of time to compare.

Hans
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Old 24th November 2020, 05:07 PM   #20
SMABB is offline SMABB  Denmark
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Thanks members

So - is the conclusion then, that I do not need either to think or build a new power supply with sufficient power, since the unit draws only the tiniest amounts of current?
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