Edit: Thank you for your input @Kevin White, much better results with 100 ohm injection resistors, a 1K resistor for a DC pathway and a 1K bias resistor from Anode to +5V.

Revised results


I am an RF enthusiast that is building a superhet FM reciever. I have previously made a functioning FM Slope detector radio, but that had poor selectivity. I want to take it to the next level.

I have a working RF amplifier and variable LO. I want to mix these signals and filter off the difference peak. I would like to use an unbalanced single diode mixer. I have attempted this but it seems quite inefficient. My difference peak is tiny given the strength of my LO and RF signals.

How can I optimize my mixer circuit? Should R1 and R2 be larger or smaller than their current value of 1K perhaps? Am I making any blatant mistakes?

This is my plan:

Block diagram

Here is what I have so far:

1.RF Amplifier

RF Amp RF Amp Output RF Amp Schematic

2.Variable Local Oscillator (top) LO Circuit LO Schematic


Mixer Circuit Mixer Schematic

4.Mixer output

Probing the Cathode of the diode

I am putting my probe directly on the cathode of the 1N34A mixer diode. Given the unbalanced nature of the mixer, I expect to see my LO and RF peaks in addition to my sum and difference peaks. Also some miscellaneous peaks. The tall unlabelled peak is a harmonic of my RF. It appears when I use just the amplifier circuit alone.

The peak labeled "122" in red is my LO signal.

The peak labelled 98.6 is my RF signal.

The peak labelled "23.4" is my difference signal.

My sum and difference peaks have almost no amplitude, is that to be expected? Can anything be changed to make them larger amplitude eg.resistor values? I intend to use a crystal ladder to isolate the difference frequency. I do not know how well that will work out if the difference peak is very low amplitude.

Any insight is welcome. Thank you for your help.

  • 1
    \$\begingroup\$ The 1k injection resistors seem high for 100MHz. Is there a DC path to ground for the diode? Have you tried adding some forward DC bias for the diode? \$\endgroup\$ May 14, 2022 at 21:17
  • \$\begingroup\$ Good question with the DC path! I left the cathode floating. A falstad simulation tells me that I should see a big difference with a 1K resistor to ground. Let me try that, and using 100 ohm resistors instead of 1k- I have also not looked into biasing the diode, good idea! \$\endgroup\$ May 14, 2022 at 21:27
  • \$\begingroup\$ @Kevin White, feel free to post your reply as an answer. You have answered my question spot on! \$\endgroup\$ May 14, 2022 at 22:32
  • \$\begingroup\$ Once you get that working ,you can try to get one of these for an upgrade. nxp.com/docs/en/data-sheet/SA639.pdf  92 dB of IF amplifier/limiter power gain \$\endgroup\$ May 15, 2022 at 1:40
  • \$\begingroup\$ Well that is certainly an upgrade! \$\endgroup\$ May 15, 2022 at 2:16

1 Answer 1


In an unbalanced single diode mixer it is important to provide a DC path through the diode to ensure that the bias conditions are correct.

There should be significant conduction occurring during the half-cycle of the LO waveform.

With no DC path the coupling capacitors will charge up with the rectified local oscillator signal until the diode is only conducting a very small amount to make up for diode and leakage. Under these conditions very little mixing will occur.


simulate this circuit – Schematic created using CircuitLab

With a DC path such as in this arrangement below not only can any rectified DC ensure conduction through the diode but a DC bias can be provided as well. Mixing efficiency can be improved by passing a small DC current through the diode.

enter image description here

The Diode Mixer

  • \$\begingroup\$ Interesting points and schematic, I did not think about using DC blocking capacitors to isolate the diode bias. \$\endgroup\$ May 15, 2022 at 1:32

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