Attenuators with transformer preamps, a good idea ??

Boswell

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Yes, that's true, but 6.4K across the input will be scarely noticed. It's a small difference, but is generally better than switching between 6.4K and 2K. This is all assuming that the input impedance of your receiving device is of order 10K or more.
 

pcrecord

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Great info...
Since the Focusrite ISA impedance settings goes like this : The low impedance setting is 600 ohms. The setting labeled ISA 110, which is modeled on another piece of equipment in the ISA range, is 1400 ohms. Medium impedance is 2400 ohms. High impedance is 6800 ohms.

So I should only avoid 600 ohms, I guess...
 

kmetal

Kyle P. Gushue
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Boston, Massachusetts
Nice work! I was grooving to the music in the background. Some circuits seem more complicated on paper than in actuality to my untrained eyes.

What do you think a circuit board would improve, and could heat shrink tubing be used instead of double sided tape? Also, how much more complex would it be to have used a variable, stepped gain knob, or rotary switch, instead of a toggle switch?

Looking forward to the sequel!
 

Boswell

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Also, how much more complex would it be to have used a variable, stepped gain knob, or rotary switch, instead of a toggle switch?
The essence of an attenuator box of this sort is that the attenuations are known and repeatable, both over time and over channels. Switches rather than infinitely variable controls are preferred for this.

Marco's box (and also the box that I posted a photo of) had two attenuation settings, so a simple DPDT toggle switch suffices. With some fudging, 3 settings would be possible using a centre-off toggle switch, but 4 or more postions requires a rotary switch. You need at least two resistors per additional setting, so the switch gets a bit bristly fairly quickly.
 

pcrecord

Quality recording seeker !
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Nice work! I was grooving to the music in the background. Some circuits seem more complicated on paper than in actuality to my untrained eyes.

What do you think a circuit board would improve, and could heat shrink tubing be used instead of double sided tape? Also, how much more complex would it be to have used a variable, stepped gain knob, or rotary switch, instead of a toggle switch?

Looking forward to the sequel!
Having a circuit board would eliminate any short-circuit risk. Heat shrink is a good idea but the U shape of the resistors placement make it hard to well isolate it from the input connector.. I would need to put small shrink cable on every resistor seperatly.
Frankly, I was a bit too exited to finish it to wait an buy more parts..

I think @Boswell should answer about using a stepped variable gain knob. Not sure how it would interact with the design.. EDIT : he just did in the previous post !

I have a few recording to do for customers.. but I'm also exited about the next step, listening to the difference.. ;)
 

kmetal

Kyle P. Gushue
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My other (basic) question is how do you figure out which resistor(s) are appropriate for a given gain reduction. How do you know what to use for say -10db like marco did vs -15, -20, -25db, ect, ect?

Also does it vary based on input? Like is box was receiving mic level, line, or a guitar?
 

pcrecord

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Bos, is better placed to explain since he is the one who supplied those numbers... but the values are based on line level and the values result of some basic calculations I didn't learn, yet...
I remember touching this at school 35 years ago and never thought that some day I could be interested in this :)
 

Boswell

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The table of resistor values I gave Marco were for a U-pad balanced line-level attenuator. It's made up of three resistors: two identical values in series with the positive and negative input feeds respectively and a shunt resistor across the output. It's not difficult to calculate the ratio of shunt to series resistance needed for a given attenuation if you assume the source impedance is zero and the destination input impedance is infinite. Note that the circuit is symmetrical through the centre of the shunt resistor, with this centre point being at zero volts (a virtual ground). You simply calculate the values for either the upper section or the lower section, giving the series resistor and half the shunt resistor.

If the ratio of series to half the shunt is K, then the formula for an attenuation of dB is K = Alog(dB/20) -1, where the antilog is base 10. In practical terms, round the logs with care, e.g. take log(2) as 0.3 rather than 0.301023

As an example, for a 12dB pad, K = Alog(12/20) - 1 = Alog(0.6) - 1 ~= 4 - 1 = 3, so the series resistor is 3 times half the shunt, or 1.5 times the shunt value.

However, there are other factors to consider and adjustments to be made.

1. The reference resistance of (say) the shunt: e.g. Ohms, KOhms or MOhms?
2. Resistor noise if it's for low-level signals
3. Correcting for non-zero source impedance
4. Correcting for non-infinite load impedance
5. Mapping the corrected calculated values on to the standard set of available (preferred) values in the E24 or E48 ranges
6. Checking the power dissipation for high-level signals, including the change of resistance with temperature (from heating effects)
7. Checking the resistance-voltage characteristics of the chosen type of resistor (often overlooked, but usually unimportant for leaded parts)
8. Repeating all of the above when your supplier is out of stock of the values you have chosen

Ahh, the life of a design engineer...
 

kmetal

Kyle P. Gushue
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Jul 21, 2009
Location
Boston, Massachusetts
The table of resistor values I gave Marco were for a U-pad balanced line-level attenuator. It's made up of three resistors: two identical values in series with the positive and negative input feeds respectively and a shunt resistor across the output. It's not difficult to calculate the ratio of shunt to series resistance needed for a given attenuation if you assume the source impedance is zero and the destination input impedance is infinite. Note that the circuit is symmetrical through the centre of the shunt resistor, with this centre point being at zero volts (a virtual ground). You simply calculate the values for either the upper section or the lower section, giving the series resistor and half the shunt resistor.

If the ratio of series to half the shunt is K, then the formula for an attenuation of dB is K = Alog(dB/20) -1, where the antilog is base 10. In practical terms, round the logs with care, e.g. take log(2) as 0.3 rather than 0.301023

As an example, for a 12dB pad, K = Alog(12/20) - 1 = Alog(0.6) - 1 ~= 4 - 1 = 3, so the series resistor is 3 times half the shunt, or 1.5 times the shunt value.

However, there are other factors to consider and adjustments to be made.

1. The reference resistance of (say) the shunt: e.g. Ohms, KOhms or MOhms?
2. Resistor noise if it's for low-level signals
3. Correcting for non-zero source impedance
4. Correcting for non-infinite load impedance
5. Mapping the corrected calculated values on to the standard set of available (preferred) values in the E24 or E48 ranges
6. Checking the power dissipation for high-level signals, including the change of resistance with temperature (from heating effects)
7. Checking the resistance-voltage characteristics of the chosen type of resistor (often overlooked, but usually unimportant for leaded parts)
8. Repeating all of the above when your supplier is out of stock of the values you have chosen

Ahh, the life of a design engineer...

Awsome, thanks! Cant wait to give this a try.
 

Boswell

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Looks like a lot of work went into that video, Marco - well done!

One thing you showed is that the ISA can deliver clean signal at more than 10dB greater than the FF800 will handle at its line input, which corresponds with the specifications of the two boxes. The ISA output levels before clipping are very difficult to find in the literature, and, as far as I can find, vary in the range +24 to +26dBu between the different ISA models.

If you want to find out what your ISA output clipping sounds like, you could connect those attenuators in series to get enough dB reduction so that the FF800 line input does not overload. Because of impedances, you won't get exact multiples of 10dB reduction, but that doesn't matter in this case.

That was a test I carried out when I first got my API 3124+ pre-amp some 12 years ago, with the effect I was looking for being output transformer saturation. In the Focusrite ISA range, the models vary as to whether they are a dual-transformer design or have only an input transformer, so you would have to check the architecture of your particular model.
 

pcrecord

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Location
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If you want to find out what your ISA output clipping sounds like, you could connect those attenuators in series to get enough dB reduction so that the FF800 line input does not overload. Because of impedances, you won't get exact multiples of 10dB reduction, but that doesn't matter in this case.
I thought about this, I will certainly try it. If I finally get more mojo out of it, I'll do one more video about it...

What surprised me was that even with the clipping led on the ISA is still clean. It just reinforced my love for them ! ;)
 

kmetal

Kyle P. Gushue
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Location
Boston, Massachusetts
Well done. I wish more pres came with an output attenuation knob. That's one very useful thing Warm audio added to their api clone. With digital level to disk being somewhat standard, vs tape, the output knob is really crititcal imho.

Boz brings up interesing design differences among the ISA models.

I have to say this video and conversation has me leaning a bit more towards the RND 511 pre vs the ISA for my next pre.

The attenuation box seems like a very useful peice.
 
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