UCD180 questions

[ewildgoose]

Quick question to Bruno on volume attenuation. We covered this a little way back when the UCD180 was just out, but I wanted to double check before I built anything

Basically, with the UCD we have a differential input, but as you point out, this is differential and we don't need a mirror image signal on each leg.

So assuming I want to attenuate a balanced source component, can I just wire up a standard log pot, pretty much as normal, but using the -ve input leg wired to the ground on the pot (rather than wiring the log ground to audio ground). This way the -ve signal to the amp will be the -ve signal from the input, and the +ve signal to the amp will be an attenuated version of the +ve input signal.

OK, we reduce the amount of common mode rejection that we achieve, and the amp will see a signal that is effectively a DC offset from the original signal (ie not symmetrical about ground level), and the +ve amp input may well be below ground potential at low listening levels. However, is this design viable for a passive attenuator for the UCD?

I would prefer to use a shunt design with two resistors on each input leg, and the variable resistor tied between them (which would give a symmetric input signal), but using a log pot as the variable part means that it is no longer linear in volume, but gets louder much too quickly (obvious when you think about it). However, it's hard to get anything of reasonable quality in anything other than log pot style (goldpt.com do a few).

...and I also just bought a big DACT attenuator, so I would like to make something work using that..!!

Confirmation or suggestions for a new design appreciated

[Bruno Putzeys]

You've opened a can of worms there. The use of a single pot to attenuate a balanced signal is not trivial.

When the pot is at zero volume, the common mode is not DC, it's half the audio signal (the half which is riding on the negative side). Secondly, the impedance balance is gone which indeed degrades CMRR. So this option would only make sense if the source ties cold to gnd (asymmetrical source).

The second option (2 fixed resistors, one variable) works fine for a symmetrical source, but... if the source is asymmetric, the common mode voltage is half the audio signal, even if the volume is at zero.

So far the best method is two ground-referred precision attenuators per channel (and I mean Precision, because the matching will be what determines CMRR).

I'm still looking hard for something more elegant, but the above is the best I've been able to come up with so far.

[ghemink]

Originally posted by Bruno Putzeys

No, just biasing one of the output transistors off to remove power rail modulation on the rail that has worst PSRR.
Placing a follower after an op amp (inside or outside the loop) is not a panacea. I've often heard the sound degrade, so it's not something you'd "just do".

Hi Bruno,

I searched the net a bit and found the j511. So it is a current source that forces a constant current between output and negative rail. So you turn off one of the output transistors (the one to the negative rail) and the other one is now biased in CLass A, so you have a single ended Class A output stage.

How much bias makes sense, the J511 gives you about 4.7mA. You want to stay always in the class A bias state, so worst case, when we see about 1kOhm input impedance at one of the UcD inputs (was at the negative one?) (when biased symmetrically), we could go to max output voltage of -4.7V, if more, then the current source can not deliver enough current and the transistor in the opamp to negative rail has to contribute to the current. So with 4.7V on the + input and 4.7V on the - input (of the UcD of course, not on the opamps), we have 9.4V input voltage and with a gain of about 4.5 for the UcD module itself, we would get 42.3V (peak). For my UcD180 that would just be OK, however, for UcD400, you may want to have a bit more current (two of them in parallel?) to make sure you always stay in Class A.

Does the above all make sense or am I mistaken somewhere?

Best regards

Gertjan

[classd4sure]

Hi Gertjan,

Here is one article on it for you

http://tangentsoft.net/audio/opamp-bias.html

You probably found that one already.

It seems the key is to have the bias current just greater than the max possible load current in order to always keep it properly biased in order for this to work, like you were saying.

I think you have it figured out anyway, I have seen some say they've used up to 10mA and others 2mA, as long as it's more than the load current and within the SOA of the op amp it should work right?

Yet another edit:

Every single reference to this I've found online always say "tie it to the negative rail"

Bruno say's go with the rail that has the worst PSRR though, I'd trust him. So if I recall correctly opa134 and opa627 both would tie to the positive rail as they both have around 20db worse PSRR on them.

[Bruno Putzeys]

Ideally you'd like to bias it in class A all the time, that much is true. It's not something I'd bother to do because we're looking at momentary excursions only. I cannot think of a musical signal that would spend sufficient time outside that range in order for any effect to be noticeable.

[peranders]

I have heard, don't know if it's true or not, is that the NPN transistor is much better than the PNP, therefor it's better to only use the NPN and connect the current source down to negative rail.

[ewildgoose]

Hi Bruno, I don't quite understand a few of the points you are making (yeah, I get the big picture, just not a few details)

Firstly, assuming I have a source which is providing a symmetric output (It's a pro audio soundcard, an RME9632). The approach of using a standard log pot only has nearly full CMR at full volume. The CMR rejection reduces as the volume is lowered, but arguably this is at least still a benefit and not a hinderance? It is surely still better than just using one leg of the source component and going single ended?

Secondly though I am not sure what you mean by "impedance balance is gone"? Are you referring to the fact that we have different impedence on each input line?

However, my problem with using my *log* pot in a double-shunt style circuit is that of course it is only "log" when you compare the resistence of the output relative to the resistance of the other part of the potential divider. If I "cheat" and just use the variable part of the device, then it actually changes faster than "log" (if you like). Now there is a "law faking" idea for using linear pots as a kind of log type circuit, but I guess I need the opposite here!

(Yeah, I could use a linear pot, but I have already ordered this log thing and it's quite expensive, so I just need to do the best thing possible with it really)

We are somewhat off-topic here I know, but I think there are others who are probably thinking about the same kinds of thing

Any thoughts on the best circuit I can build for my asymmetric output using this log stepped attenuator device?

Thanks, Ed

[ghemink]

Originally posted by classd4sure
Hi Gertjan,

Here is one article on it for you

http://tangentsoft.net/audio/opamp-bias.html

You probably found that one already.

It seems the key is to have the bias current just greater than the max possible load current in order to always keep it properly biased in order for this to work, like you were saying.

I think you have it figured out anyway, I have seen some say they've used up to 10mA and others 2mA, as long as it's more than the load current and within the SOA of the op amp it should work right?

Yet another edit:

Every single reference to this I've found online always say "tie it to the negative rail"

Bruno say's go with the rail that has the worst PSRR though, I'd trust him. So if I recall correctly opa134 and opa627 both would tie to the positive rail as they both have around 20db worse PSRR on them.

Thanks for the link, looks very interesting, will raed it later at home or during lunch break, had not seen it yet.

Gertjan

[Bruno Putzeys]

Originally posted by peranders
I have heard, don't know if it's true or not, is that the NPN transistor is much better than the PNP, therefor it's better to only use the NPN and connect the current source down to negative rail.

Whoa. That's what happens when correct observation (=listening test) is incorrectly interpreted (simply looking at the output stage).

The hidden truth revolves around PSRR. An op amp always uses one rail as an implicit reference. The PSRR of this rail is zero, with only loop gain standing in for rejection. Thus, great improvement is to be had if you can make sure that rail only sees a constant current (thus making it cleaner). Tying a current source across the output transistor that ties to the "bad rail" turns it off and replaces it by the constant current source.
This has nothing to do with "class A" operation, even though for reasons of simplicity we call it that.
Most bipolar IC processes have p-jfets but no n-jfets. For this reason, the majority of IC jfet op amps has the VAS on the negative rail.

[Bruno Putzeys]

Originally posted by ewildgoose
Firstly, assuming I have a source which is providing a symmetric output (It's a pro audio soundcard, an RME9632). The approach of using a standard log pot only has nearly full CMR at full volume. The CMR rejection reduces as the volume is lowered, but arguably this is at least still a benefit and not a hinderance? It is surely still better than just using one leg of the source component and going single ended?

Secondly though I am not sure what you mean by "impedance balance is gone"? Are you referring to the fact that we have different impedence on each input line?

Any thoughts on the best circuit I can build for my asymmetric output using this log stepped attenuator device?

When the pot is fully ccw the amplifier sees the audio signal on both inputs common-mode. Common-mode rejection is not frequency-independent though - CMRR always degrades at higher frequencies. This means you'll get a weak signal that no longer has a flat frequency response.
UcD's CMRR is limited by resistor tolerances at low frequencies and by capacitor tolerances at high frequencies. One module may have 70dB CMRR up to a few kHz, then drop to 55dB at 20kHz. Another may have 60dB throughout. The phase of the small residual signal is can be 0 or 180 degrees.
A CMRR of typ 60dB is good for insuring hum and other interconnection problems are eliminated (this is why the amps have a balanced input), but it is not intended or sufficient to correctly extract a differential-mode signal riding on a common mode signal that is much larger.

I'll keep you posted on how I fare with my own balanced preamp project.