Question on Double-Balanced Mixer

[Skywave]

A Q. on a bit of basic theory.

If I have a double-balanced mixer with two RF inputs at F1 and F2, and assuming that this mixer is perfect - and ignoring any filtering on its output (for the time being) - what will be - theoretically - the frequency components of the output signal ?

Al / Skywave.

[electroanorak]

Hi,

This guy has some great tutorials..

http://my.integritynet.com.au/purdic/dbl_bal_mix.htm

Daniel.

[Skywave]

Hi Daniel.

Thanks for the reply.

Yes - I found that Site - & read all that prior to my Post here

He clearly states that the O/P freqs. will be F1+F2 and F1-F2.

However, I have a notion that there's a little bit more to it than that - hence, my Post. To be brief, in a DBM, I am under the impression that the O/P spectra will contain the following components: mF1+nF2 & mF1-nF2, where m and n are only odd integers - unlike a non-balanced mixer, where m and n will be any integer.

I was / am hoping that someone here can confirm / contradict this notion.

Al / Skywave

[GrimJosef]

I believe you're right Al. There's a run-through of single-diode mixers, balanced mixers and double-balanced mixers here http://cp.literature.agilent.com/lit.../5954-2073.pdf although they don't get onto the balanced ones until the next-to-last page.

Many of the mixers in practical use today are made by Macom (now part of Tyco). Their technical glossary page here http://www.macom.com/TechApps/glossary/D.asp defines a Double Balanced Mixer as:

"A frequency translation circuit, which consists of four components with nonlinear impedance, typically Schottky diodes or FET's, connected in a ring quad configuration, with balun transformers at the RF and LO inputs to connect unbalanced transmission lines to the balanced quad. The IF output is via an unbalanced transmission line. The term balanced mixer is used to imply that neither of the input terms will appear at the mixer output. In practice, suppression of these input components is never perfect in an analog mixer circuit. Both types of mixer produce signals at odd harmonics of the carrier frequency, particularly the diode ring mixer. In most cases, these can be easily filtered out."

The point that frequency components which are not [I]supposed[I] to be present in fact sometimes are is well made !

Cheers,

GJ

[ppppenguin]

A theoretically perfect double balanced mixer will just give the sum and difference frequencies. In practice, as others have stated, there will be some leakage of the inputs due to lack of balance and harmonics due to non-linearity.

An alternative form of double balanced mixer for lower frequencies is the analogue multiplier. These are usually based on the 4 transistor Gilbert Cell. The 1496 is not quite a fully linear part but is a very useful modulator. The 1495 has linearising circuits and there are many higher perfirmance devices such as the AD834.

[kalee20]

I think it means here, what you mean by a 'theoretically perfect' DB mixer.

I'd certainly agree with ppppenguin, a perfect analogue multiplier will give just the sum and difference frequencies at its output, and nothing else. Sop, it is truely 'double balanced' But, give either of the input signals a bit of DC bias, and the output will consist of the input frequencies as well! This can be derived from trigonometric identities.

A so-called DB mixer using a ring of 'theoretically perfect' diodes, on the other hand, with just pure sinewave inputs, will contain the sum and difference outputs, plus lots of others as well, in fact exactly as skywave supposes, components at mF1 + nF2, where m and n are any ODD integers (positive or negative). That's because the 'perfect' diodes are either on or off, so the output waveform changes at discrete points throughout its cycle, and so the Fourier series goes on for ever. The important thion is, that it justifies the name 'double balanced' in that neither of the input frequencies appear - whether you want all the other frequencies, is another question entirely!

[Skywave]

Thanks for those above replies. Things are a lot clearer now.

A bit of background; in summary. . .

I have a low-priority design requirement involving a sequence of frequency translations. There is some latitude in the choice of the initial F1 and F2 frequencies. Obviously, the use of filters to reduce / eliminate unwanted products will feature - as will the choice and physical construction of the DBM - which may be MIMC or discrete, (the latter possibly using toroidal transformers & a ring of hot-carrier diodes). I realise that the physical manifestation of a given design will deviate from the 'perfect DBM' - due to a whole raft of factors. But before I get to that stage, I just wanted to get the fundamental (pun not intended) concept of DBM output spectra quite clear in my head. Hence, my initial stated restriction of a 'perfect DBM'.

However, there is still one remaining Q., the answer to which I'm not sure of . . .

On the basis that the O/P spectra = mF1 + / - nF2, (m and n = odd integers only), will there be any output at 2F1, 3F1, 4F1 . . . 2F2, 3F2, 4F2 . . . etc.? Or are all these harmonics balanced out, (as are F1 and F2 as discrete freqs.)? Or are only the even-numbered harmonics of F1 and F2 balanced out ? (I suspect the last - but, again, I'm not sure).

Aside: I have used 4 - quadrant analogue multipliers in the past (e.g. the AD532), but in this case, F1 and F2 are of the range (3 to 10 MHz) where such multipliers are prohibitively expensive for this requirement. As is made clear from the above Posts, an analogue multiplier is not quite the same thing as a DBM.

Thanks guys.

Al / Skywave.

[kalee20]

Hi Al,

Yes, parasitic capacitances and inductances will all conspire to spoil the balance. But, assuming that things are perfect, yes you will get outputs at 2F1, 2F2, 4F1, 4F2 etc but NOT at 3F1. It's the odd harmonics that are balanced out.

At a very simplistic level, think of a 4-diode bridge rectifier, fed by pure AC mains. At the output, you get a waveform that repeats identically 100 times a second (even though alternate peaks are derived from opposite half cycles in the input waveform). So, there is no 50Hz component, but there is 100Hz, and 200Hz, 300Hz etc. in fact all the harmonics of 100Hz (which is the EVEN harmonics of the input signal).

As to whether a 4-quadrant multiplier is a true Double-Balanced Mixer, yes it is, but not the other way round. It's like saying all 6V6's are valves, but not all valves are 6V6's.

[Skywave]

Originally Posted by kalee20

Hi Al,
. . . you will get outputs at 2F1, 2F2, 4F1, 4F2 etc but NOT at 3F1. It's the odd harmonics that are balanced out.

Err . . with respect . . . that doesn't sound right!

Surely you mean the even harmonics will be balanced / phased out, but the odd harmonics will be passed ?

My thinking (which is often wrong ) goes like this:

Start with mF1 + / - nF2; m and n integers.
For m = 0 & n = 0, (regarding 0 as an even number), F1 and F2 will not appear at the O/P: classic DBM theory.
For m = 1 and n = 1 (both odd), we have (again) the classic DBM O/P of F1 + / - F2. Et cetera . . .

Aside: This line of reasoning seems to be supported by the entry in the "Radio Communication Handbook", 6th, Ed. (RSGB) page 6.33.

Al / Skywave.

[Skywave]

Originally Posted by kalee20

. . . think of a 4-diode bridge rectifier, fed by pure AC mains. At the output, you get a waveform that repeats identically 100 times a second (even though alternate peaks are derived from opposite half cycles in the input waveform). So, there is no 50Hz component, but there is 100Hz, 200Hz, 300Hz etc. In fact all the harmonics of 100Hz (which are the EVEN harmonics of the input signal).

Err . . . again, doesn't sound right to me (sorry!)

If the "input signal" is 100 Hz, then 100 Hz and 300 Hz are odd harmonics of 100 Hz.

Aside: Let's not drift into PSU / filtering & harmonics though, as it is OT in this Post: perhaps a new Thread would be appropriate for this Topic.

Al / Skywave.

[kalee20]

Hi Al,

I think you've identified a big flaw in my statements! Reply will follow later, unfortunately it's time for a 'work break' right now!