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I am interested in using the following mixer by mini-circuits: RMS25-MH+. In the datasheet it is specified that the required LO drive power is 13dBm.

As i understand it double balanced mixers use diodes and so are non-linear devices. I only have available a signal generator that I can set the peak voltage to. My question is how can I determine the required LO input voltage. Do I assume that the input impedance looking into the LO port is 50 ohms and calculate from there?

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  • \$\begingroup\$ You're quite right that those diodes present a non-linear load to the LO driver... anything BUT 50 ohms. You substitute a 50 ohm load to the driver instead of the mixer, and deliver 20mW to that 50-ohm load. When you then substitute the mixer for that 50-ohm resistor, no doubt the peak voltage will be different (and most likely non-sinusoidal too). \$\endgroup\$
    – glen_geek
    Commented Aug 1, 2022 at 18:08

2 Answers 2

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My question is how can I determine the required LO input voltage. Do I assume that the input impedance looking into the LO port is 50 ohms and calculate from there?

Yes you do. The clue is the VSWR specification versus frequency; this will be for a 50 Ω input/output. 13 dBm is 20 mW and, relative to 50 Ω this is a voltage of 1 volt RMS. Don't forget to use a series resistor of 50 Ω to drive the signal to the device.

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If the output impedance of your generator is also 50 Ohm (like the input impedance of your mixer) and if your transmission line on your PCB is also matched to 50 Ohm you can simply calculate the required peak voltage:

\$ P = \frac{V_{\mathrm{rms}}^2}{Z_0} \Leftrightarrow V_{\mathrm{peak}} = \sqrt{2} \sqrt{P \cdot Z_0} = \sqrt{2 \cdot 20~\mathrm{mW} \cdot 50~\Omega} = 1.41~\mathrm{V} \$

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