Wiring Unbalanced Inserts to a Patchbay?

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JamieOxford

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May 7, 2017
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I'm currently wiring a TT patchbay and need to connect unbalanced inserts from a mixer to it.

I've been using the Rane site for info on wiring other cables but I couldn't find a solution on there for this.

The patchbay will be half-normalled to my audio interface which has balanced I/O.

I found the below diagram on another thread. Is this the correct/best way to wire it to the patchbay? ie, TT Send wired with the ring tied to ground, and TT Return ungrounded?

I want to avoid full balancing or anything too costly.

Any help is appreciated!
 

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It's ok if you don run very long cables from console to patch and audio interface.

You'll probably loose 6dB compared to a balanced signal
 
I know this was a few months ago but I have some info that could help you out.

From the TRS Jack on the console, Send (usually Tip) goes to Tip on Top (output) row of patchbay.  Return (usually Ring) Also goes to Tip on Bottom Row of Patchbay. Both Top and Bottom row "rings" need to be tied to shield(Ground) in order to not lose 6dB of signal.  The way to do this is use two unballanced (TS) wires both tied to Tip and Sleeve on Top and Bottom Rows of patchbay.  When you solder the TRS Jack that plugs into your insert jack, solder the bottom row wire to the return, and top row to the send. combine and solder both shields from the two unballanced wires to the sleeve of the TRS jack.  Then run a jumper from Ring to Shield on both top and bottom rows.  Again check your mixers manual to see which is send and return.. most consoles/mixers are Tip - send, Ring - return, but there are a few consoles that are the opposite :  Ring - send, Tip - return.  Hope this helps.

-Will
 
[email protected] said:
I know this was a few months ago but I have some info that could help you out.

From the TRS Jack on the console, Send (usually Tip) goes to Tip on Top (output) row of patchbay.  Return (usually Ring) Also goes to Tip on Bottom Row of Patchbay. Both Top and Bottom row "rings" need to be tied to shield(Ground) in order to not lose 6dB of signal.  The way to do this is use two unballanced (TS) wires both tied to Tip and Sleeve on Top and Bottom Rows of patchbay.  When you solder the TRS Jack that plugs into your insert jack, solder the bottom row wire to the return, and top row to the send. combine and solder both shields from the two unballanced wires to the sleeve of the TRS jack.  Then run a jumper from Ring to Shield on both top and bottom rows.  Again check your mixers manual to see which is send and return.. most consoles/mixers are Tip - send, Ring - return, but there are a few consoles that are the opposite :  Ring - send, Tip - return.  Hope this helps.

-Will
Click to expand...
That is not the correct way to do it. One should use balanced (2cond. + shiled) cable everywhere. So the patchbay jacks should be wired balanced. Unbalancing should be done at the mixer's insert jack. This allows benefitting interference rejection when connected to a balanced outboard, in particular hum loops.
 
abbey road d enfer said:
That is not the correct way to do it. One should use balanced (2cond. + shiled) cable everywhere. So the patchbay jacks should be wired balanced. Unbalancing should be done at the mixer's insert jack. This allows benefitting interference rejection when connected to a balanced outboard, in particular hum loops.
Click to expand...
I was going to say the same thing. Not just for getting some rejection but also so that if you want to rewire a connector you can do it on the end. And you can more easily try an unbalanced connector with the shield connected to "ground" or left unconnected.

However, there is one wart. If the insert is tip-send and ring-return, then you'll have two separate wires running into one jack which will be somewhat odd. But if you use proper snake cable with multiple bundles of balanced cables in one jacket then with a little shrink tube it's less strange.

The ideal scenario would be to have balanced inserts. Unbalanced may not be good for long runs which is somewhat implied by running it through the bay.
 

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