Talk:Balanced circuit/Archive 1

Archive 1

There is a really different kind of balanced circuit

There is a really different kind of balanced circuit that is in wide use in electronics, and which has been in use for many decades - going all the way back to the years of the vacuum tube. Balanced electronic circuits are widely used in power amplifiers, modulators, demodulators, mixers. For the basics of a balanced circuit, a good place to start is with learning how a differential amplifier works. It is a rather-simple balanced circuit, but still hard to understand unless you know transistors we;;.98.67.97.108 (talk) 22:42, 12 August 2010 (UTC)

Long-tailed pairs and the like are no different in principle to the meaning of balance described here. It is merely a different circuit implementation. I could also point out that op-amps have just this kind of circuitry internally, making the op-amp example already in the article an example of your claimed "really different" circuit. SpinningSpark 07:14, 13 August 2010 (UTC)

The driving circuit in Fig 6. is not balanced.

The non-inverted output has gain of 2 while the inverted output has a gain of -1 and are driven by the same input.

I might suggest changing the R₁ in the feedback of the inverting op-amp circuit to 2R₁. Spinning, since this is your graphic, can you do that? 70.109.188.83 (talk) 21:59, 16 August 2010 (UTC)

I could easily do it but I am not going to because you are wrong. The gains are +1 and -1, how could they be different, the two circuits are identical. The only difference is that the input of one is inverted. In any case you are confusing balance with symmetry. Even if you were right and the gains were different that would not stop the circuit being balanced, but it would make the output assymetrical. Making the change you suggest would definitely cause an impedance imbalance. SpinningSpark 23:17, 16 August 2010 (UTC)

Spin, the gain of a non-inverting op-amp circuit is 1+R_F/R₁ and the gain of an inverting op-amp circuit is −R_F/R₁, where R_F is the resistor in feedback. Do you not know that?

Now the, what do you mean by an "unbalanced signal" in Fig. 7? Why bother to have both an inverting and non-inverting amp in Fig. 6 if you didn't intend for the output to be "symmetrical" in order to be "balanced"? 70.109.188.83 (talk) 05:18, 17 August 2010 (UTC)

1. Both op-amps are inverting. 2. RFA. SpinningSpark 17:02, 17 August 2010 (UTC)

Spinning, you cannot be serious. The top op-amp circuit is clearly in a non-inverting configuration. How else can you claim it has a gain of +1. The signal goes into the + terminal. Check Operational_amplifier_applications#Non-inverting_amplifier. How can you make such a silly claim out of any non-ignorance? The inverting amplifier in Fig 6 has a gain of -1. And the non-inverting amplifier has a gain of +2. And if you cannot see that, then you need to go to school and learn some electrical engineering. That circuit is not balanced. 70.109.191.39 (talk) 06:32, 19 August 2010 (UTC)

I owe you an apology, you are right in that the two legs have different gains. That had previously escaped me. However, you are still missing the essential point that balanced circuit in the context of driving a line means that the impedances are balanced. That is, the impedance to ground (or anywhere else) from either leg are identical. The sense of balance meaning equal and opposite signal voltages on each leg of the line is signal balance and is better called symmetry. A key sentence in this article is "the line and ciruit remain balanced...whether or not the applied signal is itself balanced (symmetrical)." This circuit succeeds in demonstrating that, albeit accidetally. SpinningSpark 18:58, 19 August 2010 (UTC)

While I still maintain that the diagram is in essence correct, I have now amended it to make the signal symmetrical to avoid future confusion about asymmetry. This was not the intended purpose of the diagram, it disagreed with the text description, and figure 7 adequately makes the point about asymmetry instead. SpinningSpark 14:08, 21 August 2010 (UTC)

Well, since it is there for illustration, I guess it's correct enough. The previous circuit (if that one R₁ was changed to 2R₁) was better for use; you didn't have an extra (and unnecessary) generation of noise in the bottom (then inverted, now it's non-inverted) signal. It also looks nice to have the non-inverted signal on the top rather than the bottom. But, it's your drawing. I won't be fussy. Another way you could have changed the previous circuit to make it symmetrical, is, for the op-amp on the top (which was the non-inverting op-amp), change it to a simple voltage follower; no R₁ to ground and the feedback resistor changed to a wire. Then the gains are +1 and -1, instead of +2 and -2, which was my previous suggestion. Anyway, do it as you wish, Spinning. 72.95.93.187 (talk) 14:23, 21 August 2010 (UTC)

Z_in in section 3.3 not clear.

"Z_in is the input impedance of the line."

Does this mean that Z_in is the:

• Characteristic impedance of a transmission line
• Input impedance of a transmission line
• Input impedance of the receiving amplifier
• The resistance of the line (actually twice the resistance of the line if the equations are correct)

?

$V_{+}$ and $V_{-}$ are voltage dividing equations. I don't think anybody would actually calculate them or use them as such. It would seem these equations are here to make a point about the voltages in the circuit. When the statement is made that "These are clearly not symmetrical since V- is much smaller than V+." way too many assumptions are made about the values in the equations and absolutely no explanation is given about them (both the values and the assumptions).

Zin in these equations is trying to put together several quite different views and complex interpretations of the circuit into one small variable, with any and all interpretations/explanations of Zin being swept under the rug and dismissed nonchalantly.

The input impedance to differential instrumentation amplifiers (and plain op-amps) nowadays is extremely large, almost an open. If we see this circuit in DC then we can forget about Zin and just make it Rin. Rin being the input impedance of the amplifier. From an AC point of view the input impedance is capacitive at lower frequencies maybe having inductive components at higher frequencies. In AC I don't think anybody would write these equations as voltage dividers.

@Spinningspark: Jordsan (talk) 18:38, 5 October 2015 (UTC)

Where does it say that the amplifier is an op amp? It does not matter how complex the impedance function at the line input is. The expressions will remain correct. SpinningSpark 19:41, 5 October 2015 (UTC)

How can the claim be made that "These are clearly not symmetrical since V- is much smaller than V+."? In general terms if an impedance is complex there is no way of telling that V- is clearly much smaller than V+ with respect to Vin. V- could be larger than V+, or they could be of similar magnitudes. The point could be made for telephony audio or other very low frequencies, but a general interpretation of the circuit requires a transmission line for the balanced line, and a complex impedance between V+ and V-. There is no way to generalize the statement and these circuits are used in high frequency applications (MHz and even GHz).

The equations are being used to prove that the line signal is not balanced. Maybe the voltage dividers are not the best approach.

@Spinningspark: Jordsan (talk) 14:02, 6 October 2015 (UTC)

Jordsan, please. Think about what you are claiming. Can you suggest a value of Zin that will result in V- being greater than V+ ? The ratio of V+ to V- is 1+Zin/R1. For any passive structure of Zin, no matter how complex, the magnitude of this expression is always greater than one. SpinningSpark 14:28, 6 October 2015 (UTC)

A simple passive two element low-pass LC filter has a complex impedance of $1/(1-LC\omega ^{2})$. The magnitude of its output can be larger, smaller, or the same as the input. Plugging that Z into 1+Z/R and plotting the magnitude every value between 0 and $\infty$ is obtained.

The V+ and V- equations are incorrect. The correct ratio of V+ to V- is $1+Z_{ad}/(Z_{in}+R_{1})$, where Zad is the impedance from V+ to V- looking into the receiving amplifier. The ratio is guaranteed to be anything. The application of a voltage divider in this situation is at least questionable, therefore, at the very least, the equations should be written correctly.

@Spinningspark: Jordsan (talk) 21:34, 6 October 2015 (UTC)

Your expression for the impedance of an LC circuit is incorrect. It is not even in units of impedance and an LC circuit can never have an impedance that is a real number, let alone a negative one. Your definition of Zad is the same as Zin. SpinningSpark 00:20, 7 October 2015 (UTC)

Difficult wording and missing definitions

I would like to suggest that several terms which appear in the first paragraph should provide a link to a definition as they are highly specific: impedance (and "matched impedance"), interference, and circuit blocks

I've also spent a few hours making the 2nd section easier to understand. Just one example is as follows:

Begin the section by stating the basic difference between balance and symmetry: balance refers to a comparison between the components of 2 different conductors on either side of their connection point (not AROUND their connection point - this is a binary statement), while symmetry is the comparison between components of a single conductor on either side of a specific point.

I feel that making this clear at the start of this section is critical. Immediately after adding this change it was removed. Perhaps and third party could make a decision about this???

The author seems to be assuming that the reader will be well versed in the schematics of electrical circuits, thus referring to points of comparison as "vertical" and "horizontal". I absolutely think the schematics should be included, but they must separately explained (or at least provide a link to a clear explanation), and only then compare and contrast schematic, physical and functional features. That a difference exists between these 3 elements of a circuit is only alluded to one at the end of this section, when stating that in fact a schematically unsymmetrical circuit may nevertheless be symmetrical. I think this is highly confusing to anyone not well versed in electronics. PeacefulPlanet3 (talk) 21:21, 20 May 2018 (UTC) PeacefulPlanet3 (talk) 21:20, 20 May 2018 (UTC) PeacefulPlanet3 (talk) 21:19, 20 May 2018 (UTC)

 — Preceding unsigned comment added by PeacefulPlanet3 (talk • contribs) 21:07, 20 May 2018 (UTC)