Polarity Inversion circuit

[ashok]

I have been using the polarity inversion circuit for more than 10 years ! Surely this is not a proprietary circuit.
I saw it in some (old) electronic design magazine - could have been Wireless World. I find that a very large proportion of the circuits we use today and some which we think we have invented have actually been around for umpteen years and no one has taken notice of them. The basic groundwork has already been laid out for us by some really smart people - long ago !

To the moderators:
So is it wrong to put up a new schematic and circuit description?
It's a very useful circuit.
Cheers

[Tube_Dude]

I find that a very large proportion of the circuits we use today and some which we think we have invented have actually been around for umpteen years and no one has taken notice of them

Nothing new under the sun...

jorge

[Steve Eddy]

Originally posted by ashok
So is it wrong to put up a new schematic and circuit description?

I'm not a moderator, but as long as the drawing isn't copyrighted, there should be no problem. If someone would care to send me the original I'd be happy to draw it up.

In the meantime, for those who aren't allergic to high permeability magnetic materials, here's another approach:

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<img src="http://www.q-audio.com/images/doitall.jpg">
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Certainly not as cheap as an opamp, but in addition to switching polarity without requiring additional active circuitry, you can mix and match balanced and unbalanced inputs and outputs.

Only two real caveats: DC output offset from source components and the coupling capacitance of any source components which are capacitively coupled.

To avoid any low frequency resonance due to the relatively low primary inductance of an output transformer, you need several hundred uF of coupling capacitance.

Of course you could use an input transformer such as the JT-11P-1 or JT-11P-1HPC, but you'd lose the ability to easily do unbalanced to balanced conversion (the input however will accept either balanced or unbalanced sources).

se

[ashok]

The following circuit is ancient and as useful today as it ever was. With an op amp it is implemented very easily as shown in the diagram below. The resistor values are arbitrary but R1=R2=R3 and R4 must be much higher than R1 , 2 or 3.
The resistor R4 is only used to form a current path for the positive input of the opamp. Reducing its value will decrease the output voltage in non inverting mode from the ideal of 1 to less than that because it acts as a divider with R3.

Assumptions – the opamp has infinite gain , zero output impedance , no input current as the input impedance is infinite. It follows that the differential voltage between + and – inputs is zero. Also assume that R4 is infinite.

An open switch condition:
The input voltage (say V) appears at the + input terminal because there is no voltage drop in R3 as there is no input current . Voltage between the + and – input terminals is zero because of infinite voltage gain. So the – input is also at V volts. Since there is no input current ( in R1 and R3) , there will be no current in R2 and no voltage drop across it. Hence V will also appear at the output. Hence there is no phase inversion. Voltage gain is also unity.

Closed switch condition:
The + input is now at ground potential or zero volts. There is V volts across R3 and the current flows to ground. The – input is also at zero volts ( infinite gain and so Vin+ is = Vin- = 0V). Hence there is a current flow in R1 due to V volts across it. This does not enter the opamp ( zero input current) and hence flows out through R2. Note the current direction. It enters at the – input and flows towards the opamp’s output. Hence the opamps output is more negative than the – input. The current in R1 is V/R1 = the current in R2 . Voltage across R2 is R2x V/R1 = V. Since one end is at zero volts ( - input of the opamp) the output end is at –V as explained above. So we get an inverted unity gain output.

Real world opamps have high input impedance and gain and low Zout as compared to the resistors we are using , so the assumptions we made will hold easily.
Cheers.

[planet10]

Originally posted by ashok
To the moderators:
So is it wrong to put up a new schematic and circuit description?
It's a very useful circuit.

The problem was not one of the circuit being posted, but the posting of someone else's picture expressly against the imbedded copyright notice. It was then compounded by the removal of the copyright notice.

dave

[Steve Eddy]

Originally posted by planet10
The problem was not one of the circuit being posted...

Or more to the point, it wasn't the circuit itself but the particular illustration of it. The circuit itself is not patented (and copyright doesn't cover electronic circuits) and is in the public domain.

se

[mrfeedback]

Ashok,
thankyou for writing such a circuit description.

I originally posted this circuit in order that it be of benefit to DIYers who may not have seen it previously.

This selectable polarity line level circuit is very useful for switching audio in-room Absoloute Polarity, and in my experience is required equipment given that the recorded polarity of music cd's is quite random, especially compilation cd's, and also albums where the 'hit' track is often in opposite polarity to the remainder of tracks.

By inverting both the signal into an amplifier and inverting the speaker connections, an amplifier can be compared to the non-inverted throughput condition.

I have found many amplifiers to differ in sonics according to the throughput signal polarity, especially when driven into temporary overload.

Regards, Eric.
I expect that this sonic difference may be due in part to different peak overload behaviours, and speaker/cable back EMF being handled subtly differently.

This circuit would be quite useful for SET amplifiers given their well known +ve half wave/-ve half wave non-linearity.

[mrfeedback]

Steve,
watchout, are you sure that the graphic that you posted from a Jensen app note is not copyrighted ?.

I have mentioned the transformer circuit exactly as drawn several times here on the forum, and I agree that it is a fine solution, within the caveats that you advise.

Have you tried this on your systems, and what sonic results have you noted ?.

Regards, Eric.

[fdegrove]

Hi,

especially compilation cd's, and also albums where the 'hit' track is often in opposite polarity to the remainder of tracks.

Ah, you too.And you know whys these s*ckers do that don't you?

By inverting both the signal into an amplifier and inverting the speaker connections, an amplifier can be compared to the non-inverted throughput condition.

Not sure I understand you here....

Cheers,

[mrfeedback]

Hi Frank,

In my uderstanding the 'hit track' inversion wrt the rest of the album is likely caused because it has been mixed down in a different studio, or that that it has been 'finalised' in a different facility and hence the opportunity for a polarity inversion.

I have thought about whether this inversion is deliberate or not, and in discussions with other listeners it has been noted that according to (random) overall system polarity, the hit track may sound 'wrong', and the remaining tracks sound correct.

With this knowledge in mind, I hesitate to think that this is deliberate - moreso I think the final mastering guys are not hearing AP.

"By inverting both the signal into an amplifier and inverting the speaker connections, an amplifier can be compared to the non-inverted throughput condition."

By this I mean for the same given in room AP for a particular track, by means of flipping the amplifier input polarity and speaker connection polarity together, the amplifier signal throughput polarity behaviour can be compared.

I find that this can make a sometimes substantial differece to the in-room sonics, despite the fact that the in room AP is correct in both cases.

In my view, any polarity inversion control should be at the very souce end (in the cdp), and the speaker connection polarity should be optimised according to the behaviour of the amplifier.

Eric.