Consider this double balanced mixer:

enter image description here

Its output contains only the frequencies generated by the product of those of LO and RF signals, where LO is supposed to be a large signal (which switches ON and OFF the diodes) and RF a small signal.

I read that the fact that it is double balanced is due to its symmetry. Precisely, LO frequency is not present at the output (IF) terminals since the the corresponding BALUN splits its voltage on a central node put to GND and so points A and B are virtually at ground at the LO frequency (since half of VLO drops on a diode and the other half on the other).

But I do not understand why at the output there is not the frequency of RF (I read it is also due to symmetry but it is not clear to me.

Reference: http://mwl.diet.uniroma1.it/people/pisa/RFELSYS/L07_MixerModDemod.pdf, page 6.

  • \$\begingroup\$ apparently that transformer is wound, to be balanced with respect to ground, so RF input energy is cancelling. \$\endgroup\$ Commented Jun 25, 2019 at 17:46

2 Answers 2


RF port and LO port are inter-changeable. If you imagine that LO-to-IF is balanced, then RF-to-IF is balanced as well.
But any output at the IF port does cause unbalance. In normal mixer use, both RF signal level and IF signal levels are small...much smaller than LO signal level.
Any signal developed at the IF port causes unbalance that will transfer a fraction of that large LO power to the RF port. That back-flowing LO power can do no good in the common mixer scenario.
The unbalance goes both ways: some RF port signal is transferred to LO port, but that scenario isn't usually a worry (because RF port power is tiny compared to LO port power).

This DBM circuit is sometimes used as a variable attenuator. In this case, the unbalance is put to good use: a DC current applied to the IF port controls signal transfer between RF port and LO port.
The other application is a modulator, where the baseband modulation signal is applied to the IF port. The RF port becomes the modulator output. Were it not for the unbalance @ IF port, this wouldn't be possible.

My use of the term "unbalance" above is the process by which an RF port signal provides desired IF port output. A mixer-maker defines unbalance differently: it is what destroys symmetry. It is caused by diode mis-match, transformer irregularity, stray capacitance and inductance.
Since the LO port has the largest signal by far, a mixer-maker may concentrate on measuring LO port -to- RF port isolation, and LO port -to- IF port isolation. The LO port is driven with a large signal, at the proper impedance (usually 50 ohms). The RF port is terminated with 50 ohms, and IF port with 50 ohms passively. Perfect symmetry would transfer no power from LO to RF, and would transfer no power from LO to IF.
But be aware that this test setup meant to measure port-to-port isolation isn't doing any useful mixing.

  • \$\begingroup\$ Is it correct saying that "the points between D2-D3 and D1-D4 (let's call them C and D) are at 0 voltage with respect to the IF pin (because of symmetry) and so, since C and D are connected to GND through the LO transformer, from an RF point of view, we do not have signals between IF pin and GND"? \$\endgroup\$
    – Kinka-Byo
    Commented Jun 26, 2019 at 0:41
  • \$\begingroup\$ What you say sounds OK: symmetry works to isolate all three ports. Symmetry still applies when the LO port is driven hard, so long as the RF port isn't driven with a signal at the same time. Perhaps my use of the term "unbalance" causes confusion. I am editing my answer, defining how a port-to-port isolation measurement is done. \$\endgroup\$
    – glen_geek
    Commented Jun 26, 2019 at 1:21

A & B are common to both sides which use the balanced neutral for isolation and are commutated by LO so a +/-difference frequency and some IMD harmonics result in the output. Any mismatch in balance is seen in the output which includes ferrite distribution tolerances, leakage, stray capacitance, geometric tolerances and some port impedance mismatch.

The metrics commonly used are port isolation and output power level vs. frequency. Costs increase with some k* log frequency, power level [dBm] and isolation in dB and sometimes 3OI distortion power level.

So a cheap 1GHz TV Splitter for $10 might be $1000 for a good 10GHz mixer in roughly the same size.

They also make triple balanced mixers.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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