Balanced out

[Diynot]

So is implementing a balanced out (XLR/TRS) as easy as simply splitting the mono Tip signal to the Tip and Ring of a TRS jack (then using a TRS->XLR adapter cable)?

 

[Bricksnbeatles]

No. A balanced signal is phase inverted. It works on the principle that they’ll receive the same interference on both sides of the signal, so when it reaches its end destination, one of the signals is inverted again, maintaining the original signal intact while fully canceling out any interferences introduced along the cable run.

 

[Diynot]

No. A balanced signal is phase inverted. It works on the principle that they’ll receive the same interference on both sides of the signal, so when it reaches its end destination, one of the signals is inverted again, maintaining the original signal intact while fully canceling out any interferences introduced along the cable run.
View attachment 51132

Thanks. Suspected it wasn’t that simple, but had hoped.

 

[JTEX]

If you don't mind their price, there are some balanced line driver ICs that will do a great job with virtually no external components (just power decoupling caps):

THAT1606/1646
DRV134/135

 

[Diynot]

If you don't mind their price, there are some balanced line driver ICs that will do a great job with virtually no external components (just power decoupling caps):

THAT1606/1646
DRV134/135

Thanks for the heads up. I could just buy a DI box, but where’s the fun in that.

 

[JTEX]

Thanks for the heads up. I could just buy a DI box, but where’s the fun in that.

... or you could use an impedance-balanced output, which is the cheapest and easiest of them all to implement. But, you have to know the output impedance of your existing circuit. It has to be fixed, low, and not frequency-dependant (i.e. coming straight from an opamp buffer, not from the wiper of a volume pot). Then you just add a resistor from the "cold" pin of the balanced out connector to ground, with a value matching the impedance from the "hot" (driven) pin to ground. Lots of manufacturers do precisely this, often in lower-priced "balanced out" audio interfaces.

 

[Diynot]

... or you could use an impedance-balanced output, which is the cheapest and easiest of them all to implement. But, you have to know the output impedance of your existing circuit. It has to be fixed, low, and not frequency-dependant (i.e. coming straight from an opamp buffer, not from the wiper of a volume pot). Then you just add a resistor from the "cold" pin of the balanced out connector to ground, with a value matching the impedance from the "hot" (driven) pin to ground. Lots of manufacturers do precisely this, often in lower-priced "balanced out" audio interfaces.

Looks like Mouser has the THAT1646 (wouldn’t think I would need anything more for building a little DI box) for $6. I’ve paid way more for BBDs. So looking at the THAT16X6 data sheet it looks like it requires a VCC/VEE supply voltage, how do I accomplish that? I know I’ve seen it in circuits that I have built, I just can’t remember which ones or what IC/components were used?

 

[benny_profane]

Looks like Mouser has the THAT1646 (wouldn’t think I would need anything more for building a little DI box) for $6. I’ve paid way more for BBDs. So looking at the THAT16X6 data sheet it looks like it requires a VCC/VEE supply voltage, how do I accomplish that? I know I’ve seen it in circuits that I have built, I just can’t remember which ones or what IC/components were used?

Vcc and Vee are just the positive and negative supply voltages. For a split rail design, that could be +9 and -9V, referenced to GND. If you’re running a single supply, it’d be +9 and 0V, but you’d also have to provide a (1/2)Vcc reference voltage.

 

[Diynot]

Vcc and Vee are just the positive and negative supply voltages. For a split rail design, that could be +9 and -9V, referenced to GND. If you’re running a single supply, it’d be +9 and 0V, but you’d also have to provide a (1/2)Vcc reference voltage.

So this is way out of my comfort zone, not necessarily in respect to the terminology but in application. Given this schematic from the data sheet

(I chose the offset reduction one simply based on the idea that we typically work to eliminate any DC at the output. Correct me if that is not applicable in this context), using a single 9v supply as is customary for pedal usage, where would Vref (achieved through a voltage divider I presume) and Vee fit?

 

[JTEX]

So this is way out of my comfort zone, not necessarily in respect to the terminology but in application. Given this schematic from the data sheet
View attachment 51153
(I chose the offset reduction one simply based on the idea that we typically work to eliminate any DC at the output. Correct me if that is not applicable in this context), using a single 9v supply as is customary for pedal usage, where would Vref (achieved through a voltage divider I presume) and Vee fit?

Offset reduction is not usually necessary. You're going into a balanced input on a mixer or audio interface, which is likely to be AC-coupled anyway, so a few mV of DC offset does not matter.

In any case, for single supply, you'll need DC-blocking caps on input and outputs, which would take care of the offset. Your Vref (mid-supply) would go into the ground pin (3), and it needs to be low impedance. A resistor divider will not do - you need to use an op amp as a rail splitter, or if you prefer, a dedicated rail splitter chip, such as TLE2426. You'd probably end up with something like this:

THAT1646 on a single supply?

I have + 24V and i need to power a THAT1646 ....what would be the best way?? to get the 1646 to work? At some point i have +18V and tought about using this on a single supply... Jorge Aristondo.

groupdiy.com

It's probably less work to just generate the missing negative supply with a charge pump.

 

[Diynot]

It's probably less work to just generate the missing negative supply with a charge pump.

Thanks! Beginning to think that I should just buy a DI box……..

Edit: OR, I use the MBP road rage to supply both 9v Vcc and Vee….