Mixer (product detector) question.

[firewalker]

Can someone explain how the following mixer works? I know the mathematic aspect of the subject.

A really simple circuit to produce the difference of the two frequencies. 10 MHz at the one input 10.001 MHz, 1 kHz after the Low Pass Filter.





As I can understand it; the signal F2 is responsible for the biasing of the diodes. The two signal beat each other. Due to the frequency difference the dc offset of the produced signal (the signal at point A) moves up and down following this difference. The low pass filter rejects everything else except this dc bias movement.

Am I correct?

Alexander.

[fcb]

It's just a unbalanced mixer (look up double balanced mixer for theory).

[valentinc]

        The mixer relies on the nonlinear I-V characteristic of the diodes to produce other frequencies (f1 + f2 and f1 -f2)... Since it's an unbalanced mixer, a part of f1 and f2 are also present at the output...
The low-pass filter removes f1, f2 and f1+f2 from the output and only passes f1-f2 (which is 1Khz in this case)

[jimmc]

The diodes are used as switches controlled by the local oscillator signal applied to the F1 input.
When the F1 input is sufficiently +ve wrt Gnd both diodes are on and the F2 input is connected to 'A',
when the F1 input is -ve the diodes are off and there is no connection between F2 and 'A'.

The mixer is singly balanced i.e. if the diodes and transformer secondaries are perfectly matched then there is no F1 signal fed to 'A'.

Try running your simulation with V1 = say 1v pk and a 50ohm resistor in series to limit the diode current. Use a square wave to make things clearer.
V2 should be much lower than V1 (say100mV pk) so that it does not affect the diode currents much.

Jim

[firewalker]

Thanks you guys.

Alexander.