3
\$\begingroup\$

I've tried making a ring modulator like the one in following picture but without using transformers (as the right ones are quite a bit harder to get and more expensive compared to op amps).

I assume that both signals (Ux=V1 and Uy=V2) are measured with respect to ground. (I hope that this is a good assumption.)

Then "translating" the left transformer is rather easy, basically we can assume that the center tap must be our V2, and the signal in going to the left (lower left) node of the diode bridge must be proportional to V2-V1, and the one to the right (upper right) node must be proportional to V2+V1.

But now I have a hard time implementing the transformer on the right side in the picture. My first approach was just using a "floating" resistor (R15), but that doesn't seem to work very well (see brown curve).

Is there a way to improve this output signal (within the given constraints of not using transformers, just standard op-amps)?

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
3
  • \$\begingroup\$ A ring modulator is a crude analog multiplier. Making one out of op amps won't be easy. \$\endgroup\$ – Hearth Jun 14 '20 at 21:56
  • \$\begingroup\$ @Hearth But that would be my goal:) So far it does seem to work at least for the sign, but the amplitudes still need a little bit of work. \$\endgroup\$ – flawr Jun 14 '20 at 22:09
  • \$\begingroup\$ Yes, it’s feasible. \$\endgroup\$ – Andy aka Jun 14 '20 at 22:16
1
\$\begingroup\$

The diode modulator works well when one signal source (likely 20 kHz) saturates the diodes, so that they act as switches...this signal source would be called the "local oscillator" in R.F. parlance. The other signal source in the modulator application (likely 1 kHz) remains linear, smaller than the forward voltage of the diodes.

The alternative uses an actual analog switch. There are lots of JFET-based, or MOSfet-based analog switches that make excellent modulators that work equivalently to the 4-diode modulator. No transformers are needed. A single-pole double-throw switch is shown in the conceptual circuit below. SPST switches plus an inverter can substitute.

schematic

simulate this circuit – Schematic created using CircuitLab
This modulator can deliver larger output voltage than the diode version.

\$\endgroup\$
1
\$\begingroup\$

I found following an alternative circuit by Sebastian Azevendo on the EDN-Network website. A video demonstration can be found here. Apparently it does need well matched resistors, and well matched diodes.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
0
\$\begingroup\$

There is a nice simpler example from the analog devices wiki, here W1 is the difference between the two signals, and W2 the sum (or vice versa), and the output is going to be the difference between +1 and -1. It does have a little bit of through-zero distortion if one of the inputs is close to zero. So it seems from the original attempt only a center tap of R15 to ground is needed.

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.