Resistor on XLR?

Just built a ribbon mic and I was wondering if adding a resistor to bridge pin 1 and 3 would make the transformer (work harder).
Any thoughts?

Pin 1? No. You mean between pins 2 and 3? That would load the transformer. Post a schematic.

You are correct. I misspoke on the pins.

Is a basic passive ribbon signal. Motor>transformer>xlr

Impedance at the xlr Jack is 125 ohms.

But why would you want to loose signal?

Adding a resistor to the output of a transformer does not make it work harder. The only thing that makes it work harder is a bigger signal voltage.

Cheers

Ian

ruffrecords said:

Adding a resistor to the output of a transformer does not make it work harder. The only thing that makes it work harder is a bigger signal voltage.

Click to expand...

Eeehh, if by "work" we mean power then increasing the load might increase the current and therefore power. Presumably the intent is to push the transformer to emit higher distortion. Although I suppose the current will actually depend on the source impedance of the supply which is presumably phantom which is not particularly low and therefore the voltage could simply drop and therefore it won't really increase power much or at least not nearly enough to cause the transformer to emit higher harmonics and this is a run-on sentence. Of course this is all wild speculation without a schematic of further explaination.

squarewave said:

Eeehh, if by "work" we mean power then increasing the load might increase the current and therefore power. Presumably the intent is to push the transformer to emit higher distortion. Although I suppose the current will actually depend on the source impedance of the supply which is presumably phantom which is not particularly low and therefore the voltage could simply drop and therefore it won't really increase power much or at least not nearly enough to cause the transformer to emit higher harmonics and this is a run-on sentence. Of course this is all wild speculation without a schematic of further explanation.

Click to expand...

Transformer distortion depends primarily on the voltage applied to the windings. Increasing the load will simply increase the losses in the transformer.

Cheers

Ian

ruffrecords said:

Transformer distortion depends primarily on the voltage applied to the windings. Increasing the load will simply increase the losses in the transformer.

Click to expand...

Interesting. I didn't know that. I thought distortion was a function of how much of a magnetic field has been built up and more specifically how close that magnetic field is to saturating the core. I also thought that magnetic fields are a function of current and not voltage.

So if drive 600:600 transformer with a very low impedance such that the voltage across the primary is constant regardless of load and then a) load the secondary with 600 ohms and measure the distortion and then b) load it with 150 ohms and measure the distortion again, there will be no increase in distortion between cases a and b even though there will be 4x as much current going through it in case b?

squarewave said:

Interesting. I didn't know that. I thought distortion was a function of how much of a magnetic field has been built up and more specifically how close that magnetic field is to saturating the core. I also thought that magnetic fields are a function of current and not voltage.

So if drive 600:600 transformer with a very low impedance such that the voltage across the primary is constant regardless of load and then a) load the secondary with 600 ohms and measure the distortion and then b) load it with 150 ohms and measure the distortion again, there will be no increase in distortion between cases a and b even though there will be 4x as much current going through it in case b?

Click to expand...

Basically yes. I found this strange at first but it is true. Remember the secondary current creates a back emf which creates a flux that opposes the flux in the primary. The result is any increase in primary flux is opposed by an equal and opposite flux in the secondary so the total flux remains nearly constant. Because there is more current the I2R and possibly the iron losses do increase and the output voltage will drop. The main things that determine a transformer size are the voltage it operates at, the core cross section and the heat generated by losses. If you had zero ohm wire and a lossless core you could make a 1KVA transformer an inch cube in size.

I will try and find the book where I read this.

Cheers

Ian

Edit: I found it. It was in Bill Whitlock's seminal paper on audio transformers:

https://www.jensen-transformers.com/wp-content/uploads/2014/08/Audio-Transformers-Chapter.pdf

Jump to paragraph 1.1.3 where Bill says:  "Because secondary current flow is in the opposite direction, it creates magnetic flux which opposes the excitation flux. This causes the impedance of the primary winding to drop, resulting in additional current being drawn from the driving source, which creates additional flux just sufficient to completely cancel that created by the secondary. The result,which may surprise some, is that flux density in a transformer is not increased by load current. This also illustrates how load current on the secondary is reflected to the primary"

And it is flux density that determines distortion (see section 1.3.1)

ian

ruffrecords said:

Jump to paragraph 1.1.3 where Bill says:  "Because secondary current flow is in the opposite direction, it creates magnetic flux which opposes the excitation flux. This causes the impedance of the primary winding to drop, resulting in additional current being drawn from the driving source, which creates additional flux just sufficient to completely cancel that created by the secondary. The result,which may surprise some, is that flux density in a transformer is not increased by load current. This also illustrates how load current on the secondary is reflected to the primary"

And it is flux density that determines distortion (see section 1.3.1)

Click to expand...

Ahh, that is interesting. And it kinda makes sense because the increased load current is going into the load and so it cannot contribute to increasing flux density.

That link looks like a good read. Thanks.

So, just to tie this into the post, if one wanted to make a transformer "work harder" for the purpose of deliberately increasing distortion, just strapping a resistor across the secondary will not work. However, if the transformer had a split secondary, then you could parallel them, load with Zout / 4 and then put 4x more signal into the primary. But for a ribbon mic circuit, presumably you cannot increase the gain.