Double balanced mixer question - Page 1
Forum section: electronics / rf-microwave

author: 5065AGuru
author_rank: Frequent Contributor
timestamp: July 23, 2020, 05:59:27 pm
content: Hi,I'm feeding 5MHZ and 5.000001MHz into an HP10514A mixer.A buffer and a 12dB attenuator feed each input and a 50 Ohm buffer amp (10Mhz) is on the output.I get a nice sine output but get the 1Hz as amplitude variations.Playing with input levels I can minimize the variations but the best I can get is a 3.2 V P-P with a .4 V P-P amplitude modulation.Are there mixer schemes I can use that will eliminate the amplitude variations?Cheers,Corby

author: edpalmer42
author_rank: Super Contributor
timestamp: July 23, 2020, 07:12:58 pm
content: I replicated your test as follows:10514A mixer5 MHz and 5.000 001 MHz @ +7 dBm no buffers or filters - source was 5065A (of course!) then into two PTS X10 synthesizersIF output buffered by an EG&G Parc 113 Preamp ( 100 Mohm input impedance, DC - 300 KHz response, X10 gain)My output is as shown - no trace of any AM.Random thoughts:- What did you use as an output filter?- Why a 10 MHz output buffer when the signal is only 1 Hz?- You mentioned 12 dB attenuators. What were the actual levels into the mixer? Remember that the 10514A is a +7 dBm mixer.Ed

author: fourfathom
author_rank: Super Contributor
timestamp: July 23, 2020, 07:57:50 pm
content: Quote from: edpalmer42 on July 23, 2020, 07:12:58 pmI think the OP was trying to get the sum frequency (10.000 001 MHz) and not the 1 Hz difference.The output of an ideal double-balanced mixer will include the sum and the difference frequencies (the two input frequencies will be suppressed). Is your buffer clipping on the 1 Hz component of the output? To eliminate the 1 Hz difference you will want a high-pass filter on the output of the mixer. There should be no amplitude modulation unless there is clipping or some other nonlinear element after the mixer.

author: TimFox
author_rank: Super Contributor
timestamp: July 23, 2020, 08:00:51 pm
content: A "+7 dBm" mixer normally requires that power level at one port so that the diodes switch fully. The lower level applied to the other input port will then give a proportional response at the mixing products from the output. The output level when driven by this LO will be relatively insensitive to the actual LO level.7 dBm = 5 mW into 50 ohms is 0.5 V rms, or 1.4 V pk-pk, which should switch two series diodes (of the four Schottkys in the bridge) fully on when driven from a 50 ohm source.Higher power mixers use more diodes and series resistors to handle the higher LO voltage.One of the three ports in a normal double-balanced mixer (labeled "IF") is DC-coupled, and the other two are transformer primaries. Therefore, if you want the 1 Hz difference, it must come from the IF port.

author: tkamiya
author_rank: Super Contributor
timestamp: July 23, 2020, 08:55:13 pm
content: Thanks for this discussion. I'm trying this same thing in very near future.I'm curious also on filter detail.

author: 5065AGuru
author_rank: Frequent Contributor
timestamp: July 24, 2020, 04:01:21 am
content: Yes it is the 10Mhz I want, not the 1Hz.My mixer feeds a 10Mhz tuned buffer amplifier.The output of this amplifier is the one that has the 1HZ AM on it.I varied the two input levels to the mixer for the lowest AM level.Just wondering if a different mixer architecture might help.I did also think about trying an LTC6957 per this app note:https://www.analog.com/en/technical-articles/halve-the-jitter-with-the-ltc6957-filters-while-buffering-10mhz.htmland then feeding the CMOS output into my tuned amp to get back to a sine wave!?!?Cheers,Corby

author: fourfathom
author_rank: Super Contributor
timestamp: July 24, 2020, 03:21:47 pm
content: Quote from: 5065AGuru on July 24, 2020, 04:01:21 amThere are many mixer topologies, and they will all give you sum and difference frequencies at the output. The DBM (double-balanced mixer) does a good job of suppressing the two inputs, which in your case is a good thing. With many other mixer designs your two 5 MHz inputs will also show up at the output (they also will show up with the DBM, but at a greatly reduced level). You want to reduce the difference frequency amplitude (1 Hz), so you need a filter at the mixer output. I suggest a 10 MHz bandpass filter between the mixer output and the amplifier input.That app note you linked shows how a low-noise buffer can introduce less clock jitter, but this is not your problem here.So exactly why are you creating the 10.000 001 MHz signal in this way? Have you considered using a frequency doubler and feeding it with a 5.000 000 5 MHz OCXO? (I assume your OCXO has this amount of tuning range.) You will still want a 10 MHz filter before the amplifier, since the doubler will have some 5MHz leaking into the output. I'm pretty sure you can use the same DBM as a doubler by feeding the OCXO signal to both inputs.

author: 5065AGuru
author_rank: Frequent Contributor
timestamp: July 24, 2020, 04:20:35 pm
content: OK,A bit more info, and thanks so far for the ideas, The 2 5Mhz signals will actually be within parts in 10 to the -9th of each other. You still get the amplitude variation but at a super low rate.I'm just using the 1Hz offset as it's faster to see while trying to figure out how to reduce it.I do feed the mixer output into a 10Mhz tuned amp. This amps output has the variation on it.Since the LTC6957 uses limiting I thought that might work?Cheers,Corby

author: fourfathom
author_rank: Super Contributor
timestamp: July 24, 2020, 05:39:32 pm
content: Quote from: 5065AGuru on July 24, 2020, 04:20:35 pmWhere is the filtering in your tuned amp? It needs to be before the amplification element, otherwise the difference frequency component will effectively phase-shift and duty-cycle modulate the sum frequency by shifting the zero-crossing point. This may be what you are seeing, and will see if you send the sum+diff into a buffer without pre-filtering. I wouldn't try to fix this by reducing the mixer input levels since this might cause other problems (as long as the mixer inputs aren't so large as to cause distortion.) If you have a decent filter at the output of the mixer and good input levels then there should be virtually no difference frequency component going into your amplifier.

author: radiolistener
author_rank: Super Contributor
timestamp: July 26, 2020, 08:09:59 am
content: Quote from: 5065AGuru on July 24, 2020, 04:01:21 amThey all give you 1 Hz component on the output if you put RF and LO with 1 Hz difference. This is the essence of the mixer, this is how it works.Mixer is analog multiplier of two waveforms. For example, if you have two sine with frequency A and B, it will multiply two sine:sin(x) * sin(y) = (cos(x - y) - cos(x + y)) / 2It means that on the output of mixer you will get two new components: A-B and A+B.In your case:A = 5'000'000 HzB = 5'000'001 HzIt means that you will get two new components on the output of mixer:A-B = -1 HzA+B = 10'000'001 HzSince this is not quadrature, we cannot distinguish direction of vector rotation, so negative frequencies will be equals to positive.As result you will have two new components on the mixer output: 1 Hz and 10'000'001 Hz.This is expected behavior of mixer.Quote from: 5065AGuru on July 23, 2020, 05:59:27 pmjust add high pass filter on the output to cut-off all frequencies below 7-9 MHz. That's it.

author: TimFox
author_rank: Super Contributor
timestamp: July 26, 2020, 12:58:47 pm
content: Two double-balanced mixers operated in phase quadrature can be used as an "imageless mixer", to select only one sideband. Seehttps://www.rfcafe.com/References/articles/wj-tech-notes/ImageRej_n_SSB_mixers.pdf

author: fourfathom
author_rank: Super Contributor
timestamp: July 26, 2020, 02:31:35 pm
content: Yes, with a quadrature mixer design you can isolate the sum and difference components. In the OP's case, given the relatively fixed-frequency of the two inputs they can create the phase-shifted inputs using simple L/C circuits. But since there is such a large difference between the sum and difference frequencies I think that the simple 10 MHz filter at the mixer output will be the way to go. I would use a bandpass rather than a highpass to clean up any harmonics, although these shouldn't affect the resulting clock signal.