Balancing an unbalanced circuit

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ruffrecords

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Thanks Ian for thinking about my original question.
I have to admit that I'm lost with all this extremely technical turn this thread has taken. Not complaining or anything. I know how fora work. I just can't follow the thread any more - I mean IC technology is beyond me. Maybe I will one day.
Could you please clarify with plain words what you mean by "simple impedance balancing". In my case (and I guess many other future and current readers) simpler terms help way more than precise technical terminology. For example instr. signal impedence (hi z), or line signal impedence etc.
As others have pointed out in subsequent posts, impedance balancing may not always work. It depends on the original circuit. If the output device is an op amp then it should work. All you need to do is connect the original output to the hot pin of the balanced connector. Then connect a resistor from the cold pin to analogue 0V. The value of this resistor should be equal to the output impedance of the original output. The hard part is working out the value of that output impedance and this is what depends on the particular circuit. Can you give us an example of a circuit you want to do this to?

Cheers

Ian
 

k brown

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Refering to the attachment, for a device with an opamp output, R2 would be the resistor already installed between the amp and the out jack (which would be replaced with an XLRM, or TRS jack). One simply adds R3 (same value as R2) from ground to the other pin of the XLR or TRS. R4 and C1 are just icing on the cake (read text). From Rod Elliott's site.

(One can do this on the output of a BJT follower, as in the Okatve MK012, but the resistor to ground would be larger than the one on the transistor, by the value of the Q's output impedance [as seen in the '012 schem]).
 

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abbey road d enfer

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Pedals include all sorts of simple-to-complex devices nowadays - more needs to be known about the specific pedal in question - not all have pots on their outputs.
Agreed. In the absence of specific data and considering the subject was open to various pedals, I think it is wise to assume the worst.
I've run up to 100' of unbalanced cable - under all sorts of comditions in many different locations - utterly silent;
In the case of inserting pedals between a guitar and an amp, probably. However, using pedals in conjunction with studio equipment is bound to create ground loops. The best way to minimize the negative effects of such ground loops is to run connections that don't use the shields to carry signal.
 

ruffrecords

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Refering to the attachment, for a device with an opamp output, R2 would be the resistor already installed between the amp and the out jack (which would be replaced with an XLRM, or TRS jack). One simply adds R3 (same value as R2) from ground to the other pin of the XLR or TRS. R4 and C1 are just icing on the cake (read text). From Rod Elliott's site.

(One can do this on the output of a BJT follower, as in the Okatve MK012, but the resistor to ground would be larger than the one on the transistor, by the value of the Q's output impedance [as seen in the '012 schem]).
Rod's right about the basic modul operandi but wrong about the 6dB loss in gain. There is no loss in gain but there is a loss of headroom compared with a differential output.

Cheers

Ian
 

57sputnik

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Gary Hebert, THAT CTO, provided these sims to me years ago.

THAT1646 noise vs. Rsource:

1646_Noise_vs_Source.jpg


THAT1646 CMR vs. Rsource:

1646_Output_CMR_vs_Rs_18K.jpg


The above show why all of these devices (1646, 134, 2142) should be driven from a low impedance source. If AC coupling is used at the 1646 input its reactance should be included.

There is another difference between the THAT1646 and the DRV134 that hasn't been mentioned. The DRV134/SSM2142 are designed for +6 dB differential gain into a 600Ω load. The THAT1646 is designed for +6dB differential gain into 18KΩ.

Another thing to remember about all of them (1646, 134, 2142) is that a single-ended output taken from them should always have the opposing output leg grounded or they will be noisy. The devices have a large amount of common mode gain: Upsetting the internal resistive bridge balance, which a floating output does, increases noise.

Some other 1646 observations:

The input bias current is about 1 uA. No bias resistor is required: The internal network supplies Ib if the input floats.
Avoid any stray inductance on the ground pin. The ground trace should be short and never telescoped to a distant location. It doesn't take much trace inductance on pin 3 to make a 1646 oscillate around 15-20 MHz.
Always have local bypass at the device power pins.
 
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ccaudle

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Rod's right about the basic modul operandi but wrong about the 6dB loss in gain.

I think what he was referring to was that outputs which have symmetric signals usually have the original signal level on the hot pin, and then an inverted copy of that signal on the cold pin, which implicitly gives +6dB compared to the original signal on the hot pin, and just an impedance balanced connection to reference on the cold pin.
 

ccaudle

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I think balanced has a better signal to noise ratio (at least for studio use)

"Signal to noise ratio" is a wide ranging term. Using properly connected balanced interfaces should give lower power line related noise (60Hz and 120Hz hum, also higher frequency harmonics if implemented with high quality), but does nothing about thermal noise (hiss), or noise generated or picked up in the signal processing circuitry. I point this out only to make sure you have realistic expectations about what you can achieve by adding balanced interfaces to existing circuitry.
 

57sputnik

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I found a sim of the THAT1646 output common mode impedance versus frequency.

There are two plots one with Csense at 10µF (red) and a second with Csense shorted. (Green.)

THAT1646_Output_Zcm_vs_Frequency.JPG


Another 1646 tip to offer is that if an op amp inverter is used to buffer the 1646 input one can correct the polarity inversion by simply redefining the 1646 outputs.

The 1646 "+" output can be redefined to be "-" to drive pin 3 of the XLR and the 1646 "-" output redefined to drive XLR pin 2.
 
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ruffrecords

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I think what he was referring to was that outputs which have symmetric signals usually have the original signal level on the hot pin, and then an inverted copy of that signal on the cold pin, which implicitly gives +6dB compared to the original signal on the hot pin, and just an impedance balanced connection to reference on the cold pin.
I agree that was his thinking but it ignores the fact the required signal level between hot and cold remains the same so for this to be so you need to reduce gain somewhere else in the chain. But he ignores the fact that the maximum output level from a symmetric output can be 6dB higher so the headroom has increased by 6dB which is a real benefit.

Cheers

Ian
 

tedsorvino

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As others have pointed out in subsequent posts, impedance balancing may not always work. It depends on the original circuit. If the output device is an op amp then it should work. All you need to do is connect the original output to the hot pin of the balanced connector. Then connect a resistor from the cold pin to analogue 0V. The value of this resistor should be equal to the output impedance of the original output. The hard part is working out the value of that output impedance and this is what depends on the particular circuit. Can you give us an example of a circuit you want to do this to?

Cheers

Ian


Thanks Ian. Here's a project I'm thinking of transforming from unbalanced to balanced.

PS. IMO out of (a 30 year studio and live) experience, I tend to use a really good noise suppressors (like ISP decimators) with most of my pedals, but I never need such things (that any way are not ideal for "real" tones) with balanced devices.
I never use any kind of noise suppressor with fuzzboxes or heavy drive whatever the amount of noise is- If i choose a noise DIRT device for a job, the noise is part of the beauty of it. I mean I love my "real" and "lo fi" sounds (never a fan of "clinical" stuff anyways). But real is real and annoying is annoying. Of course it's absolutelly subjective.

Anyways here is an example of a circuit (designed by bow_and_error and taken from the great freestompboxes.org forum)
 

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ruffrecords

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This one is fairly straightforward because the output is a TL072 op amp which has a very low output impedance. For some reason this particular design does not include a build out resistor in series with the op amp output. These are often included with op amp outputs because cable capacitance can cause op amp instability without it. So first I would add a 47 ohm build out resistor in series with the existing output. This becomes the hot output. Then connect the cold output via another 47 ohm resistor in series with a 10uF capacitor to GND. To be super accurate? you could also connect a 1Meg ohm resistor across the 10uF. Job done.

Cheers

Ian
 

Newmarket

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This one is fairly straightforward because the output is a TL072 op amp which has a very low output impedance. For some reason this particular design does not include a build out resistor in series with the op amp output. These are often included with op amp outputs because cable capacitance can cause op amp instability without it. So first I would add a 47 ohm build out resistor in series with the existing output. This becomes the hot output. Then connect the cold output via another 47 ohm resistor in series with a 10uF capacitor to GND. To be super accurate? you could also connect a 1Meg ohm resistor across the 10uF. Job done.

Cheers

Ian

Interesting that the AC coupling capacitor is often omitted in commercial designs. Typical Electrolytic of +/-10% if not 20% may make this a moot point. Too late here to run the numbers tbh. But I suspect actual values usually a lot tighter.
Alternatively go for it big time and use a tight tolerance film cap if you have the space.
Or go further and use the Signal Ground as an input to an opamp so replicating the impedance variations as frequency increases. Together with an exact duplication of the output components.
In reality though simple resistive balancing works well.
 

tedsorvino

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Very clear suggestions Ian about the output.
But what about the input stage? Is it ok to power it with just simple single rail 9vDC?
Newmarket could you please be more specific about the placement of the op amp (TL 072 or something other???) would go and what would it replace in this circuit.
 

Newmarket

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Very clear suggestions Ian about the output.
But what about the input stage? Is it ok to power it with just simple single rail 9vDC?
Newmarket could you please be more specific about the placement of the op amp (TL 072 or something other???) would go and what would it replace in this circuit.

I'm not sure what I can offer beyond what Ian has said.
For the simplest solution simply add a resistor (suggest 47R, 68R, 100R or anything in that range) immediately before the 10uF output capacitor.
That is your 'T' signal of a TRS output jack socket as currently.
Then connect an identical same value resistor between 0V on the pcb and the 'R' connection of the output socket.
That is the simplest solution and similar is commonly implemented on commercial kit.
Taking that a stage further as per Ian's suggestion you can also fit a 10uF capacitor to better replicate the source impedance of the 'T' signal and optionally the 1M0 resistor across that.
Since you're 'DIY' I suggest it's a nice idea to use precision resistors for better matching and temperature coefficient.
eg 0.1% tolerance 20ppm/degC compared to a 'standard' 1% tolerance 200ppm part. Whether you'll experience any tangible benefit is debatable but you might get a warm fuzzy feeling of knowing you did it :)

Not wanting to muddy the waters but depending on what is connected on input and output you might still get a 'ground loop' problem from a "Pin 1" problem. So allow for a ground loop defeat solution. 100R in parallel with 100n in the Screen connection usually works well (often used eg Bo Hansen DI).
 

tedsorvino

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I'm not sure what I can offer beyond what Ian has said.
For the simplest solution simply add a resistor (suggest 47R, 68R, 100R or anything in that range) immediately before the 10uF output capacitor.
That is your 'T' signal of a TRS output jack socket as currently.
Then connect an identical same value resistor between 0V on the pcb and the 'R' connection of the output socket.
That is the simplest solution and similar is commonly implemented on commercial kit.
Taking that a stage further as per Ian's suggestion you can also fit a 10uF capacitor to better replicate the source impedance of the 'T' signal and optionally the 1M0 resistor across that.
Since you're 'DIY' I suggest it's a nice idea to use precision resistors for better matching and temperature coefficient.
eg 0.1% tolerance 20ppm/degC compared to a 'standard' 1% tolerance 200ppm part. Whether you'll experience any tangible benefit is debatable but you might get a warm fuzzy feeling of knowing you did it :)

Not wanting to muddy the waters but depending on what is connected on input and output you might still get a 'ground loop' problem from a "Pin 1" problem. So allow for a ground loop defeat solution. 100R in parallel with 100n in the Screen connection usually works well (often used eg Bo Hansen DI).
Actually never really use 1% resistors in pedal builds. Expensive for no actual sonic difference. Since I've never experience any real difference (and as a lo fi fan and after a lot of pedal builds) I tend to go for whatever is available close to me- at the end of the day I've seen several "boutique" pedal on the inside and sometimes what I'm using is some kind of luxury.
 

ccaudle

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Taking that a stage further as per Ian's suggestion you can also fit a 10uF capacitor to better replicate the source impedance of the 'T' signal

Not really optional. The reactance of a 10uF capacitor at 60Hz is 265 Ohms, so is by far the dominant impedance at power line frequencies.
That also makes capacitor mismatch the dominant source of impedance variation at power line frequencies.
 

Newmarket

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Not really optional. The reactance of a 10uF capacitor at 60Hz is 265 Ohms, so is by far the dominant impedance at power line frequencies.
That also makes capacitor mismatch the dominant source of impedance variation at power line frequencies.

Fair enough. I didn't pay much attention to the detail. 10uF is low value for output cap'. I'm more use to values at least a decade higher. Realistic solution may be either duplicate the capacitor. Or increase the existing capacitor value to where it becomes insignificant.
 

Newmarket

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Actually never really use 1% resistors in pedal builds. Expensive for no actual sonic difference. Since I've never experience any real difference (and as a lo fi fan and after a lot of pedal builds) I tend to go for whatever is available close to me- at the end of the day I've seen several "boutique" pedal on the inside and sometimes what I'm using is some kind of luxury.

If not 1% then what - 5% ? 1% tolerance Metal Film is inexpensive now and basically the standard thing for non-precision commercial application.
 
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