LM13700 voltage controlled resistor using linearizing diodes

[Beo]

The Ross Phaser uses the LM13700 configured as a VCR for phase stages. I was looking a an unverified schematic of the Moog Phaser, and it also seems to be doing this, but using a different input method. The Ross leaves pins 2 and 15 unconnected and mates the positive and negative inputs with a resistor. The Moog trace has a fixed voltage into the Diode Bias (pins 2 and 15) and uses resistors to ground to balance the +/- inputs. You can find the Ross Schematic on Tonepad, and if interested, the unverified Moog trace is on FSB.

Here's the applicable application diagrams from the datasheet.


My question is, what is the advantage of using the linearizing diode configuration, particularly for a Phaser Stage application.
Thanks,
Travis

[R.G.]

OTAs in the 3080 tradition can only deal with signals in the 10-25mV range without distorting. The linearizing diodes pre-distort the incoming signal in the reverse direction that the inputs will distort them, so with the combination, much larger signals can be handled at lower distortion.

[Beo]

OTAs in the 3080 tradition can only deal with signals in the 10-25mV range without distorting. The linearizing diodes pre-distort the incoming signal in the reverse direction that the inputs will distort them, so with the combination, much larger signals can be handled at lower distortion.

Curious. Guitar signals exceed that voltage range easily. The CA3080 doesn't have linearizing diode capability, but is used for the Ross/Dynacomp compressors. And the Ross Phaser leaves the diode bias unused. Perhaps the amount of distortion we're talking about is pretty insignificant, particularly for guitar applications. I guess I could breadboard a basic gain circuit with the 13700 and see if I can hear the difference between the two input methods.

[bool]

The ability to handle larger input signal will also project to a (ever so slightly) better s/n ratio.

But hey, we want to make some noise, do we?

[R.G.]

Curious. Guitar signals exceed that voltage range easily. The CA3080 doesn't have linearizing diode capability, but is used for the Ross/Dynacomp compressors. And the Ross Phaser leaves the diode bias unused. Perhaps the amount of distortion we're talking about is pretty insignificant, particularly for guitar applications. I guess I could breadboard a basic gain circuit with the 13700 and see if I can hear the difference between the two input methods.

The National Semiconductor datasheet outlines this in its applications section. It has some graphs of distortion versus input level with and without linearizing diodes. Now that TI has eaten National Semiconductor, the datasheet may or may not have that info. I haven't looked for a while.

I suspect that the 2K trimpot - which can almost always be replaced by a pair of 1K resistors - has something to do with it. The incoming signal drags both + and - inputs around together, separated by the 2K. The input "load", a pair of caps and a 150K resistor, forms the other side of a very odd, inverted voltage divider, I think. So maybe the actual difference between + and - inputs is much smaller than the input signal.

Most 13700 and other OTA applications that don't use feedback to make the input differential very small tend to have heavy voltage dividers on the incoming signal to get the total level down. That's one reason OTAs get a rep for being noisy - the input signal is attenuated down so much, then must be amplified back up. As bool says, you get better S/N if you can run in a hotter signal to almost everything. Padding signals down is nearly always making S/N worse.

[merlinb]

Curious. Guitar signals exceed that voltage range easily. The CA3080 doesn't have linearizing diode capability, but is used for the Ross/Dynacomp compressors.

Only a small portion of the signal actually appears between the two inputs in the Dynacomp. In practice an OTA will take about 100mVpp before serious distortion, and the linearizing diode are, on the whole, not worth the trouble.