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I'm experimenting with a SE612 mixer to learn the basis of RF. I read that the mixer can either be single-ended or balanced, and a double-balanced mixer has the best performance in terms of LO-RF isolation, as both LO and RF are suppressed at the output.

However, my circuit suffers from strong LO leakage despite using two transformers at my input and output for a balanced, untuned input. To investigate the problem, I reduced the circuit to bare minimum and performs three experiments.

Setup

Input is a signal generator, sine wave, 15 mV (peak-to-peak), 10 MHz signal, 50R output impedance.

Output is a coax connected directly to a 100 MHz oscilloscope, high-impedance input.

Local Oscillator frequency is 24 MHz, using the built-in SA612 LO driver.

No attempt was made to match or terminate the impedance, I don't think matching is important for this experiment (note: the actual circuit I was building, was a 3-30 MHz shortwave upconverter for RTL-SDR receiver, this application calls for a wideband, untuned input, which is difficult if not impossible to match anyway...).

Experiment 1: Single-Ended Mixer

Schematic

A bare-minimum circuit. The mixer drives a 24MHz crystal oscillator Y1 and a loading capacitor C5, enough for the crystal to start oscillating.

Input is single-endend, with two DC-blocking capacitors. Output is single-endend, with a DC-blocking capacitors. Since its single-endend, only IN_A is used, IN_B is AC-coupled to ground, similarly, only one output is used, the unused output is not connected.

Power is supplied by a 9V battery.

Single-Ended Mixer Schematic

Photo

The circuit is built on a piece of copper board, using dead-bug construction for a solid ground plane.

Single-Ended Mixer Photo

Measurements

When input is not connected, a strong 24 MHz LO signal 30 dB greater than the noise floor of the oscilloscope FFT is visible at the output. Higher-order harmonics at 48 MHz, 72 MHz, etc, are also visible.

Output FFT with no input

Feeding a 15 mV (peak-to-peak), 10 MHz signal to the input, now 14 MHz and 34 MHz are generated by the mixer, 10 dB stronger LO.

Somehow, the LO magnitude decreased by 10 dB. But LO and its harmonics are still clearly visible.

Output FFT with 10 MHz input

Experiment 2: Double-Balanced Mixer with 1:4 Transformer

Schematic

DC-blocking capacitors at the input and output are removed, and two 1:4 (4T:8T) transformers are added to convert the signal between single-ended and balanced. Both input/output pins are connected directly to the transformer.

The transformers are winded on a toroid core, the high-impedance side is 8 turns of magnetic wire evenly winded across the core, the secondary side is 4 turns of wire winded on one side. No center tap.

The toroid core is a general-purpose Ni-Zn core for radio electronics in China, its initial permeability is 1000.

Double-Balanced Mixer with 1:4 Transformer Schematic

Photo

Double-Balanced Mixer with 1:4 Transformer Photo

Measurements

When input is not connected, a strong 24 MHz LO signal 30 dB greater than the noise floor of the oscilloscope FFT is visible at the output. Higher-order harmonics at 48 MHz, 72 MHz, etc, are also visible.

Output FFT with no input

Feeding a 15 mV (peak-to-peak), 10 MHz signal to the input, now 14 MHz and 34 MHz are generated by the mixer, 10 dB stronger LO. The magnitude of the LO remains the same, still 30 dB stronger than the noise floor.

All mixer products are 10 dB stronger than the last experiment, due to less mismatch loss.

Output FFT with 10 MHz input

Experiment 3: Double-Balanced Mixer with 1:1 Transformer

Schematic

Similar to Experiment 2, the only difference is that two 1:1 transformers are used.

The transformers are winded on a toroid core, 3 turns on one side, 3 turns on the other side. No center tap.

The toroid core is FT-37-43, a common American core, its initial permeability is 850.

Double-Balanced Mixer with 1:1 Transformer

Photo

Double-Balanced Mixer with 1:1 Transformer Photo

Measurements

When input is not connected, a strong 24 MHz LO signal 15 dB greater than the noise floor of the oscilloscope FFT is visible at the output. Higher-order harmonics are below the noise floor and invisible.

Output FFT with no input

Feeding a 15 mV (peak-to-peak), 10 MHz signal to the input, now 14 MHz and 34 MHz are generated by the mixer, 10 dB stronger LO. The magnitude of the LO remains the same, still 15 dB stronger than the noise floor.

All mixer products are 15 dB weaker than the last experiment, due to increased mismatch loss at input and output.

Output FFT with 10 MHz input

Question

In the experiments, the relative strength between LO and mixer products remains the same, no reduction of LO power level is observed at all, despite a double-balanced topology is used.

What is the problem?

I know a center-tapped balun should have better performance, but even without a center-tap, the signal should be more balanced than an unbalanced output, and I expect a reduction of LO power level at the output, but it is not the case. Is winding a balun considered rocket science?

Or the problem lies somewhere else? Or I have some foundational misunderstandings about a mixer?

比尔盖子
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4 Answers4

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Is winding a balun considered rocket science? Or the problem lies somewhere else? Or I have some foundational misunderstandings about a mixer?

Fundamental to the Gilbert Mixer is that you must have a low impedance band stop and high impedance bandpass filter on the output to achieve the SNR expected with the mixer gain and to prevent saturation effects (Vce<2V)as well as keep collector current within a 10:1 range to optimize on distortion performance. (opinion)

The datasheet gives you 4 options. Use the last one with a tuned Cap to the Balun transformer for better harmonic results. enter image description here Also keep in mind the LO level and rejection ratio in the output need to meet your RF mixer SNR requirements.

enter image description here

Tony Stewart EE75
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  • I have no personal experience with this IC per se, but I do have experience with transistor mixers. (+1) Good work with your test methods. – Tony Stewart EE75 Nov 17 '19 at 01:16
  • In other words, for a Gilbert mixer, using a double-balanced topology by itself does not necessarily guarantee better harmonic results compared to a single-ended topology, and either a tuned output or a filtered output must be used? – 比尔盖子 Nov 17 '19 at 06:03
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    Correct , it must be a tuned IF filter output to function at all. – Tony Stewart EE75 Nov 17 '19 at 06:06
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My understanding is that baluns such as those you wound are used on the RF and IR ports of a double-balanced passive mixer (i.e., diode ring), and sometimes also on the LO port for a double-double balanced diode ring, but not on gain-type mixers such as the SA602/612 devices.

If your purpose at this point is to learn about RF and mixers, I wonder if you might want to experiment first with diode-ring mixers and their associated baluns before tackling the less-straightforward application of Gilbert cells. You'd have to provide the LO signal currently supplied by the 612's on-chip oscillator, but that's not a big deal and it has other benefits as well (frequency agility being just one).

  • You seems to be correct. The SA612 datasheet and its application notes say nothing about carrier suppression and port isolation (and the SA605 design guide only says a balanced topology reduces second-order products), as if it's not even designed to be good for this aspect (on the other hand, a Linear Technology 2 GHz Gilbert mixer guarantees 40-50 dB isolation if it's balanced and matched). It seems the SA612 is only designed for the 455 kHz IF stage, not a good choice for broadband applications to begin with. I'll try diode ring mixers. – 比尔盖子 Nov 23 '19 at 04:21
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Have here a crude SPICE model of that mixer. With inputs single-ended (as in your first circuit), and looking at single-ended pin 4 as output, unloaded. 24 MHz oscillator is not modelled properly, it is a simple single-ended square wave large enough to switch the upper quad transistors fully on-off.
The model uses 2N3904 transistors everywhere, with output @ pin 4 and pin 5 fed from +8V through 1500 ohm resistors. It is not surprising that higher frequencies are somewhat attenuated (the 34 MHz component should be the same amplitude as the 14 MHz component: single-ended SA612 output pin 4
You can see that the 24 MHz carrier dominates. 10 MHz is prominent too.

Looking at the same setup differentially (between pin 4 and pin 5 as output) does satisfactorily reduce carrier amplitude, making the mixing outputs at 14 and 34 MHz dominant: differential SA612 output pin 4 to pin 5
This demo suggests that your output transformer could be improved. Since the output stage is around 1500 ohms from pin 4 to Vcc, and 1500 ohms from pin 5 to Vcc, it will be somewhat difficult to wind a wide-band transformer for such high Z. Much easier when impedances are lower, like 50 ohms.
For FT-37-43 cores, you should get \$ 420 nH/t^2\$
I'd suggest winding at least 7 trifilar turns on your core. Twist three wires together and wind as if it is one wire. That gives you a transmission-line type transformer with 3 windings. You combine two of those windings to give a centre-tap (which goes to Vcc). Be careful to tie the far end of one winding to the near end of the other.

The unconnected third winding is output, which should be terminated in a resistor load. Output stage looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

glen_geek
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  • So winding a balun *is* rocket science?! Anyway, I've already winded tested 10+ transformers for a related project, and I almost broke my fingers doing this, so I won't wind any new transformer with a high turn ratio anymore. I'll purchase a balun with a center-tap from Mini-Circuit and report back within 15 business days of shipping ;-(. Another question: What does LO rejection have anything to do with impedance matching? If I keep the impedance unmatched, but achieve perfectly balance, I would be able to see improvements, wouldn't I? – 比尔盖子 Nov 17 '19 at 18:51
  • Tried again with sine instead of square 24MHz. Similar results. Then dropped *both frequencies* by ten. That helped reduce carrier feed-through at pin4 output (single-ended). Your oscillator looks somewhat strange (missing capacitor from pin 7 to ground), You could be over-driving those upper Gilbert Cell switches. No more than 300mV p-p on pin 6. Try also with reduced amplitude going in. – glen_geek Nov 17 '19 at 21:36
  • Thanks. Overdriven is possible, I'll check tomorrow and report back (5 AM now in CST). I just put a random capacitor and the crystal started oscillating, and stopped thinking about it. (My actual circuit used an injected LO and I carefully tweaked it to 200-300 mV p-p, which I didn't bother to replicate for this demo). I'll try following the capacitor values from the reference design tomorrow. – 比尔盖子 Nov 17 '19 at 22:15
  • Your breadboard looks OK. You *should* be able to get carrier suppression with single-ended output, but not when inputs are overdriven, or when the oscillator amplitude is too large. A lower oscillator frequency may help with balance: don't expect any more than about 30dB. – glen_geek Nov 18 '19 at 03:15
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    Confirmed. LO was probably overdriven. I replaced the LO with an external signal generator and solved the problem, the LO suppression is exactly 30 dB down. I'll update the question with more test results soon. – 比尔盖子 Nov 24 '19 at 09:13
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Try reducing the supply to just under 8 volts, say 7.7V and adding a 500k trim potentiometer across input pins 1 and 2 with the wiper to ground. You should be able to get quite a nice balance to brighten your day.

toolic
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