While studying for his U.S. General class license, my son wondered about the "balanced modulator" referred to by a few questions in the FCC pool.

Quoting KB6NU's General Class study guide to give a sense of the two questions:

Filters are also used in amateur radio transmitters. A filter is used to process signals from the balanced modulator and send them to the mixer in a single-sideband phone transmitter. (G7C01) A balanced modulator is the circuit used to combine signals from the carrier oscillator and speech amplifier and send the result to the filter in a typical single-sideband phone transmitter. (G7C02)

Coming from the SDR world, I'm much more familiar with "mixers" being used for frequency conversion, and I tend to think of SSB from that perspective (i.e. raw baseband that's simply been upconverted). Other test pool questions want me to conceive of SSB as generated by an AM signal that's somehow had its carrier [sharply!?] filtered out afterward. Neither of those sounds like this.

In short, the term "balanced modulator" is unknown to me (despite working through the same exam pool a few years ago ;-] ) and my initial web searches on a small screen didn't turn up very much information either. Among Google results on a bigger screen today, the Armstrong phase modulator does come up but that seems more for FM than SSB, and, as far as I can tell, only uses a balanced modulator rather than explaining what one is….

What are the core principle(s) of a balanced modulator? What is the significance of them being "balanced"? In broad strokes how would one design/build one in practice? Are they still used in commercial SSB transceivers and/or modern kit radio circuitry?

  • $\begingroup$ Noting another reference to "balanced modulator" that I found while cleaning up browser tabs: hamwhisperer.com/2011/01/… — this page seems put the "filtered AM" concept to the modulator component but I'm not sure if that's just for teaching/memorization aid vs. a fully accurate technical explanation. $\endgroup$ Commented Sep 6, 2018 at 23:36

3 Answers 3


To put it simply, a balanced modulator is a mixer which has two inputs and two outputs.

The outputs are (input1 + input2) and (input1 - input2). So, if the inputs are 100kHz and 1kHz, the outputs are 101kHz and 99kHz.

For transmitting: If the inputs are 1MHz and (voice), the outputs are the lower sideband of the voice and the upper sideband of the voice, centered on 1MHz. Effectively a "double sideband, suppressed carrier" signal of voice modulated on a 1MHz (suppressed) carrier. One of the sidebands can be filtered out (if you want to transmit USB or LSB) when it's transmitted.

For receiving: Working the other way, you mix the received single sideband with a 1MHz signal, to generate enough of the original voice to be intelligible (since the lower and upper sidebands are mirrors of each other, you only need one to rebuild the original signal, it's just slightly trickier to tune by ear).


Partial answer, because I don't know the actual electronics theory, but I hope which will help make progress towards a complete answer: A balanced modulator is a mixer with a particular feature.

Basic analog mixer designs tend to include the carrier in the output, whereas a balanced modulator is one which is designed to "suppress" the carrier.

Hence, if you feed an audio signal and a RF carrier into a mixer that is not "balanced", you will get AM output (possibly with unstable carrier power level?), whereas a balanced modulator will give you DSB-SC.

If you're generating SSB by filtering the output of a mixer fed by the audio signal, this means that you only need to filter out the opposite sideband rather than also the carrier — twice the frequency difference and hence less demanding on the filter.

  • $\begingroup$ Ah, yes! While studying for the GROL late at night I came across a similar mention in Gordon West WB6NOA's book. Appreciate the memory jog AND the explanation/affirmation that a balanced modulator and balanced mixer are indeed the same. $\endgroup$ Commented Sep 6, 2018 at 23:32
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    $\begingroup$ Specifically, per this, a balanced modulator feeds two AM modulators with opposite phases of the AF signal (and the same LO) and takes their difference. The carrier outputs of the two modulators cancel, but the sidebands reinforce. $\endgroup$ Commented Sep 7, 2018 at 0:51
  • $\begingroup$ Ah! @hobbs-N2EON that page has a great explanation and the "balanced" name now makes complete sense. Great simple concept too… a summary of that page would make a great answer if anyone wants to take a swing at it. $\endgroup$ Commented Sep 7, 2018 at 4:37
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    $\begingroup$ Yes! Any old diode has a non-linear response and will work as a mixer. But a ring of four diodes, balanced just right, can do a four quadrant multiplication of the two input signals, which suppresses the carrier and leaves only the sidebands. $\endgroup$
    – tomnexus
    Commented Sep 8, 2018 at 4:40
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    $\begingroup$ Folks, it looks like you're putting together a better answer in these comments. Consider writing that answer instead. $\endgroup$
    – Kevin Reid AG6YO
    Commented Sep 8, 2018 at 4:41

A balanced modulator is a kind of mixer. Specifically, it's one that works in all four "quadrants", that is combinations of each of the two input voltages being positive or negative. Along with that come certain other expectations about the device.

We could draw a graph of the two input voltages $v_1$ and $v_2$, and it would look like this:

enter image description here

Any possible input to the mixer at an instant is a point somewhere on this graph. If both voltages are zero, that's the point in the middle. As $v_1$ increases, we go up. As $v_2$ increases, we go right.

Let's consider simple amplitude modulation where $v_1$ is the baseband signal, and $v_2$ is the carrier, at 50%, 100%, and 150% modulation. On the left for each case is the ideal mixer output, and on the right a shaded region showing the region in which the inputs will fall on the graph under these conditions.

enter image description here

At 50% modulation, the baseband signal is always significantly more than zero, so only the upper parts of the first and second quadrant are encountered.

At 100% modulation, the baseband signal touches zero but is never negative, so the first and second quadrant are used all the way down to the origin.

At 150% modulation, the baseband signal goes negative, which means an inversion of the output as operation enters the 3rd and 4th quadrants.

A mixer which can operate in all four quadrants, producing the inverted signal as would be expected of an "ideal" mixer where $v_\text{out} = v_1 \times v_2$, is a balanced modulator.

Some mixers can't produce $v_\text{out} = v_1 \times v_2$ in all four quadrants. They may clip the output rather than inverting it when crossing into the 3rd and 4th quadrants:

enter image description here

Such mixers are not balanced modulators.

Why does this matter?

Consider: the difference between 50% and 100% modulation in AM is that the latter has a decreased carrier power in relation to the sidebands. As the modulation increases beyond 100%, the sideband power continues to increase and the carrier power decreases. When the modulating signal is "balanced", meaning it spends as much time being positive as it does negative, the carrier power reaches zero and you're left with just the sidebands.

For generating AM this is no good, but for SSB it's great since there's no need to somehow remove the carrier. Only the unwanted sideband needs to be removed, and that can be done with a sharp filter, or by combining with a second modulator but at a 90 degree phase shift.


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