Ring Mod question

[Dylan]

Hi and Howdy Doody.

I found the schematic for a ring modulator made from an LM13700 IC and I have a couple questions.

First off (see schematic attached) it has two inputs, a signal input and a modulation in put; I'm pretty sure the signal input would just be  your guitar/synth/whatever, but I was wondering what the modulation input is and what it hooks up to.

Second question, from the OP Amp symbol there are those two intertwined circles, what does that mean?

thanks in advance.

[Circuitbenders]

Second question, from the OP Amp symbol there are those two intertwined circles, what does that mean?

The triangle, diodes and two circles are the symbol for an 'operationsal transconductance amplifier' or OTA.

it has two inputs, a signal input and a modulation in put; I'm pretty sure the signal input would just be  your guitar/synth/whatever, but I was wondering what the modulation input is and what it hooks up to.

You need two inputs to a Ring Modulator. According to wikipedia 'Ring modulators frequency mix or heterodyne two waveforms, and output the sum and difference of the frequencies present in each waveform'

http://en.wikipedia.org/wiki/Ring_modulator

[Bogus Noise]

Traditionally the modulator input will be an oscillator. You can plug any audio signal in there but a simple waveform like a sine will get you better results. More complex modulators give you more frequencies at the output, which can start to get a bit noisy!

[Dylan]

Thanks fellas. Makes a helluva lot more sense now.

[Gordonjcp]

A high-precision "ring modulator" is called a "four-quadrant multiplier", and thereby lies a clue to its operation!

Effectively what you're doing is multiplying the modulator with the carrier, so when the carrier is at +1 the output is passed straight through, when it's at -1 the output is inverted, and when it's 0 there's no output at all.

See this -> http://www.gjcp.net/~gordonjcp/ringmod.jpg

The red high-frequency sine is the carrier and the green low-frequency sine is the modulator.  Notice how as the modulator rises the blue output trace gets larger and starts to fall once the modulator passes the peak, then as it crosses the zero the output follows the carrier again - but inverted?

The "four-quadrant" part comes from the way that it will apply modulation around the positive and negative halves of a cycle.  A two-quadrant multiplier would give you AM instead of ring modulation, as you can see here -> http://www.gjcp.net/~gordonjcp/am.jpg