UPDATE: see post 2 for new details!
Last year I released a 4-channel gain stage with PS circuit (gerber files can be found HERE) that was intended for bringing up the level of 1Vrms output DACs so that they could drive amplifiers with higher input voltage sensitivity, or provide extra voltage headroom.
This year I am in need of a BALANCED line driver circuit that accepts an unbalanced signal as input. As I started to look around for options, and not wanting to use an integrated solution (e.g. from THAT), I realized that I could simply add a couple of passive components to the outputs of the 4-channel gain stage and get a balanced signal. I thought I would share that in this new thread to stimulate discussion and get some feedback on the concept.
The ah-ha moment came when I was looking at a circuit that Rod Elliot put up on his website. See the circuit described in Figure 4, in the section called "Hey, that's cheating!":
Balanced Transmitter and Receiver II
The take away is that a balanced circuit does not necessarily need to have both hot (+) and cold (-) signal lines actively driven. Instead, "balanced" refers to matched line impedances as seen by the receiver. The closer the impedances are matched, the better balanced they are and the better the cancellation of any noise picked up on the lines in between TX and RX. In fact from what I understand it is quite common in lower priced pro gear to use exactly this kind of approach because it reduces the parts count and therefore cost.
The only thing to consider is that, compared to a circuit that drives both conductors, the gain will be lower (6dB lower). But if we are driving our "cheater" balanced output with a gain stage we can just use 6dB higher gain on the driven conductor to compensate for this, or we can choose whatever gain is needed to accommodate the rest of the gain structure in the system given the maximum voltage output level of upstream components.
So how can the unbalanced 4-ch gain stage be used, exactly? I refer you to Figure 4 at the link, above. There is already a series resistance on each gain stage output, so it is as simple as adding another (matched) resistor in series with the ground return plus an RC zobel between driven and ground output after the series resistors. The two series resistors should be matched to a high tolerance (e.g. 1% or better). Finally, choose a gain that provides +6dB or whatever is needed to properly driver your amplifier, as before. Now you can use the balanced inputs on your amplifier and get the benefits of hum and noise cancellation.
This seems like a great way to re-use an existing circuit, adapted for a balanced line, and with the ability to tailor the gain to suit my needs. I'd be interested to hear thoughts about this approach, especially if there are any downsides to balanced circuits in which only one conductor is actively driven, for instance from a DC offset voltage, etc.
Last year I released a 4-channel gain stage with PS circuit (gerber files can be found HERE) that was intended for bringing up the level of 1Vrms output DACs so that they could drive amplifiers with higher input voltage sensitivity, or provide extra voltage headroom.
This year I am in need of a BALANCED line driver circuit that accepts an unbalanced signal as input. As I started to look around for options, and not wanting to use an integrated solution (e.g. from THAT), I realized that I could simply add a couple of passive components to the outputs of the 4-channel gain stage and get a balanced signal. I thought I would share that in this new thread to stimulate discussion and get some feedback on the concept.
The ah-ha moment came when I was looking at a circuit that Rod Elliot put up on his website. See the circuit described in Figure 4, in the section called "Hey, that's cheating!":
Balanced Transmitter and Receiver II
The take away is that a balanced circuit does not necessarily need to have both hot (+) and cold (-) signal lines actively driven. Instead, "balanced" refers to matched line impedances as seen by the receiver. The closer the impedances are matched, the better balanced they are and the better the cancellation of any noise picked up on the lines in between TX and RX. In fact from what I understand it is quite common in lower priced pro gear to use exactly this kind of approach because it reduces the parts count and therefore cost.
The only thing to consider is that, compared to a circuit that drives both conductors, the gain will be lower (6dB lower). But if we are driving our "cheater" balanced output with a gain stage we can just use 6dB higher gain on the driven conductor to compensate for this, or we can choose whatever gain is needed to accommodate the rest of the gain structure in the system given the maximum voltage output level of upstream components.
So how can the unbalanced 4-ch gain stage be used, exactly? I refer you to Figure 4 at the link, above. There is already a series resistance on each gain stage output, so it is as simple as adding another (matched) resistor in series with the ground return plus an RC zobel between driven and ground output after the series resistors. The two series resistors should be matched to a high tolerance (e.g. 1% or better). Finally, choose a gain that provides +6dB or whatever is needed to properly driver your amplifier, as before. Now you can use the balanced inputs on your amplifier and get the benefits of hum and noise cancellation.
This seems like a great way to re-use an existing circuit, adapted for a balanced line, and with the ability to tailor the gain to suit my needs. I'd be interested to hear thoughts about this approach, especially if there are any downsides to balanced circuits in which only one conductor is actively driven, for instance from a DC offset voltage, etc.
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It turns out that it's even SIMPLER than what is explained in the post above - you don't need ANY additional components to use the outputs of my gain stage circuit to drive balanced inputs! Here's why that is:
It turns out that I (serendipitously) chose to use a topology put forth by Douglas Self in his book "The Design of Active Crossovers" that he calls a "Zero Impedance Output" (see page 509, Figure 17.2a). This topology results in an output impedance of only 0.24 Ohm at 1kHz. With the hot (+) line having such a low output impedance you can simply connect the cold (-) line to the ground of the gain stage circuit to get a very well balanced line pair. What could be simpler? I had forgotten about this little benefit, but I am glad that I re-discovered it now.
So if you are using my gain stage circuit, as-is it is also a great balanced line driver and gain stage.
It turns out that I (serendipitously) chose to use a topology put forth by Douglas Self in his book "The Design of Active Crossovers" that he calls a "Zero Impedance Output" (see page 509, Figure 17.2a). This topology results in an output impedance of only 0.24 Ohm at 1kHz. With the hot (+) line having such a low output impedance you can simply connect the cold (-) line to the ground of the gain stage circuit to get a very well balanced line pair. What could be simpler? I had forgotten about this little benefit, but I am glad that I re-discovered it now.
So if you are using my gain stage circuit, as-is it is also a great balanced line driver and gain stage.
That Jensen appnote which AndrewT references is an important one. Balanced signal interfaces have nothing inherently to do with differential signaling. Being balanced has only to do with interface impedance term matching so that common-mode noise transfer is nulled across the interface. An balanced interface essentially operates as a Wheatstone bridge which nulls common-mode noise at it's cross nodes.
Some audio industry professionals still incorrectly conflate an balanced interface with differential (anti-phase) signaling. I'm surprised that so few (if any) commercial components appear to utilize the balanced interface shown in fig. 2.4 that Jensen appnote, especially tube based components.
Some audio industry professionals still incorrectly conflate an balanced interface with differential (anti-phase) signaling. I'm surprised that so few (if any) commercial components appear to utilize the balanced interface shown in fig. 2.4 that Jensen appnote, especially tube based components.
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Since my circuit has essentially a zero Ohm output for the driven line (the hot line) simply grounding the cold line at the source provides a balanced pair. Closely matched impedance for each signal line is the defining characteristic of a balanced connection.
I am not aware of any reason to insert an expensive line level transformer that in a home audio setting would result in any better performance that what my circuit (a la Self) can achieve. Although a transformer can give TRUE galvanic isolation, I think that is outweighed by the additional cost, LF distortion, and HF resonance characteristics of a line transformer, and I do not see a need for galvanic isolation in a home setting if a balanced line is being used already. If you know of additional benefits of a transformer coupled connection I would like to know about them.
The non-transformer solution in section 2.4 of AN003 from Jensen is fine but you need to match the two 470R resistors quite closely, e.g. 0.1% or better to keep up CMR. This is more or less what I had originally proposed in post 1. But since I do not need to add these components to my circuit to achieve balanced lines, there is no motivation to do so (as far as I know). Again, if you can point out a flaw in this line of thought I would appreciate it.
I am not aware of any reason to insert an expensive line level transformer that in a home audio setting would result in any better performance that what my circuit (a la Self) can achieve. Although a transformer can give TRUE galvanic isolation, I think that is outweighed by the additional cost, LF distortion, and HF resonance characteristics of a line transformer, and I do not see a need for galvanic isolation in a home setting if a balanced line is being used already. If you know of additional benefits of a transformer coupled connection I would like to know about them.
The non-transformer solution in section 2.4 of AN003 from Jensen is fine but you need to match the two 470R resistors quite closely, e.g. 0.1% or better to keep up CMR. This is more or less what I had originally proposed in post 1. But since I do not need to add these components to my circuit to achieve balanced lines, there is no motivation to do so (as far as I know). Again, if you can point out a flaw in this line of thought I would appreciate it.
You are right - it increases with frequency. From 0.24 Ohm at 1kHz to 2.4 Ohms at 10k and 4.8 Ohms at 20kHz using Self's circuit but this is because (according to him) the open loop gain of the 5532 he uses is falling off. That is still quite low and should not degrade the CMRR very much.
The op-amp has a resistance (e.g. 68R) in series with the output. Self has two feeback paths. One branches before the series output resistance and has a small cap. This shorts at HF. The other is taken AFTER the series output resistor and has the typical larger (e.g. 10k) feedback resistor. Because the feedback for audio frequencies is taken after the series output resistor, its voltage drop is factored out and you get essentially a zero output impedance even though the amplifier output sees 68R in series with the load, e.g. a capacitive one.
It's similar to this circuit I found via Google, but Self adds another resistor from the inverting input to ground to create the gain stage:
An externally hosted image should be here but it was not working when we last tested it.
Not exactly. It's matching the RATIO of the impedances for each path that is the defining characteristic, not the impedances themselves. A balanced signal interface essentially forms a Wheatstone bridge. So, for, example, one path could have a 10 ohm output impedance and a 10k input impedance, while the other path could validly have a 100 ohm output impedance and an 100k input impedance. Both paths have an impedance ratio of 1:10, and so will null the propagation of common mode noise across the interface even though none of the four impedance terms is anywhere close to being the same.
Of course, the compatibility of randomly interconnected commercial equipment dictates that the output impedance terms be equal, and the input impedance terms be equal for common-mode noise nulling to occur across any random interconnection of equipment featuring balanced interfaces.
That depends on the amount of CM rejection one desires. CMRRs in excess of 100dB are not a trivial matter to obtain, especially over the entire audio band. Audiophiles regularly expect SNRs specs. in excess of 100dB for their active gain devices, so an over 100dB CMRR spec. doesn't seem that crazy an expectation.
A well designed signal transformer, while imperfect, still is superior to any active input solution I'm aware of, for delivering the highest, wideband CMRR. Transformers provide such high CMRRs because they present a very high common-mode input impedence. Generally speaking, the higher the common-mode input impedance the higher the balanced interface's CMRR.
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Thanks for your reply, but I am not familiar with your example equating the balanced 'output > line > input' circuit to a Wheatstone bridge. In my mind I visualize the input stage of the downstream equipment as a differential amplifier with the output impedance of the source equipment that it is connected to in series with the input impedance of the balanced input. If that is not correct, then please tell me how or (better) point me to a reference that explains that in detail.
I assumed a differential amplifier as input, with 0 Ohms in series with the cold line and 4.8 Ohms in series with the hot line. I assumed perfectly equal resistors in the input diff amp. Then I calculated the differential and common mode gain, and then the CMRR. The result is a worst-case CMRR of over 85dB at 20kHz, increasing to 92dB at 10kHz and getting better as frequecy decreases. The limiting factor will be how well matched the input resistances are in the diff input amp. As you mentioned some people will want to see 100dB or more across the audio band, but you need to work hard to get sufficient impedance matching to reach that level of CMRR. I can get at least 85dB for free, more given that I use an op-amp with a higher open loop gain than the 5532.
For my calculations I used the equations presented in this presentation.
I learned that a balanced interface is in essence a Wheatstone bridge (With respect to common-mode noise) from a Bill Whitlock presention. Whitlock is head of Jensen Transformers and a true expert on the subject. Slide #90 begins the Wheatstone bridge explanation, but the whole presentation is excellent and worth reviewing. Below, is a link:
http://www.bennettprescott.com/downloads/grounding_tutorial.pdf
http://www.bennettprescott.com/downloads/grounding_tutorial.pdf
Thanks for the links guys. I will study those in more detail, especially the "Wheatstone bridge" analogy. But after taking a look at Rod Elliot's page I still don't think that it changes the results of my calculations. I know that the equipment I will be interfacing to has an input impedance of 50k Ohms and that is the value I used to determine the CMRR. In the Wheatstone bridge analogy, the bridge is only slightly unbalanced by the drifting output impedance (WRT frequency) of my output circuit. It's unbalanced by no more than 4.8R versus 50k, 1 part per 10,000 or 100ppm and that is an upper bound. At 1k the output impedance mismatch would result in a CMRR of 120dB. The balanced input is likely becoming the limiting factor to the overall CMRR at that point, and for lower frequencies.
I think I am coming from the point of view that what I have is simple and sufficient, and not from the point of it being a new or superior approach.
I think I am coming from the point of view that what I have is simple and sufficient, and not from the point of it being a new or superior approach.
Just did a full read through of the Whitlock presentation. Wow that is packed with good info, from a source I trust.
I noticed on page 132 he describes a "Universal" output that can drive unbalanced or balanced lines and do each well (just not at the same time). This is exactly the situation that I have in my circuit. If you look at his illustration he shows a "Z is approx. zero" pointing to where the feedback loop leaves the amp output. My circuit does not have the other passive components shown after this node, that is no series resistor R1, no DC blocking cap C1, and no drain resistor R2. In my circuit the amp remains stable because the series output resistor is still there, just tucked INSIDE the feedback loop. Without the output network (R1, C1, R2) shown in Whitlock's ciruit there is no need to have an identical (matched) output network connected to the grounded (cold) signal line to achieve balanced impedances. That's why my circuit is just a version of Whitlock's "Universal" output.
I noticed on page 132 he describes a "Universal" output that can drive unbalanced or balanced lines and do each well (just not at the same time). This is exactly the situation that I have in my circuit. If you look at his illustration he shows a "Z is approx. zero" pointing to where the feedback loop leaves the amp output. My circuit does not have the other passive components shown after this node, that is no series resistor R1, no DC blocking cap C1, and no drain resistor R2. In my circuit the amp remains stable because the series output resistor is still there, just tucked INSIDE the feedback loop. Without the output network (R1, C1, R2) shown in Whitlock's ciruit there is no need to have an identical (matched) output network connected to the grounded (cold) signal line to achieve balanced impedances. That's why my circuit is just a version of Whitlock's "Universal" output.
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the "Wheatstone Bridge" analogy is in the preceding Jensen app note AN002
This is discussed in a paper by W.Jung and by a few others.
Walt's discussion suggested that the receive end could with advantage be pushed beyond 1M input impedance.
It's here that transformers operate, they achieve 10M to 15M of common mode input impedance and why they have superior CMRR to all the electronic versions..
This is discussed in a paper by W.Jung and by a few others.
Walt's discussion suggested that the receive end could with advantage be pushed beyond 1M input impedance.
It's here that transformers operate, they achieve 10M to 15M of common mode input impedance and why they have superior CMRR to all the electronic versions..
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I'm wondering if the increasing output impedance on the hot line can be 'balanced' by an inductor on the cold line. In Self's example, he states 0.24 Ohm at 1k Hz and 2.4 Ohm at 10k Hz. An equivalent impedance would result if the cold line had a series inductance of 0.038 milli Henries, assuming negligible DC resistance. I'm not sure what kind of low inductance, low DC resistance coils are available but for example, this 0.05mH crossover inductor has a DC impedance of 0.08 Ohms:
https://www.parts-express.com/dayton-audio-005mh-16-awg-copper-foil-inductor-crossover-coil--257-600
I imagine that an inductor for line-level applications would have much lower DC resistance...
Perhaps it's as easy as winding some wire on a former to make an air core inductor with low DCR. That could provide very good impedance matching to better than 0.1 Ohms.
https://www.parts-express.com/dayton-audio-005mh-16-awg-copper-foil-inductor-crossover-coil--257-600
I imagine that an inductor for line-level applications would have much lower DC resistance...
Perhaps it's as easy as winding some wire on a former to make an air core inductor with low DCR. That could provide very good impedance matching to better than 0.1 Ohms.
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