I am connecting a mixer to a microwave source and need to be careful of how much power I am giving to the LO and RF of the mixer.

If I just stay in the unit of dBm, it would be ok but I became curious as to how to convert the source output in Volts to dBm.

If the signal source outputs 10mV, is there a way I can convert that to dBm, so that I know how much power the source is outputting?

One way I thought of is the diagram below. The MW source is basically outputting some voltage value and the power transmitted from the MW source is going to depend on the load impedance. (most measuring devices are 50 Ohm impedance so I guess if I set the MW source output to 10mV, given the load impedance of 50 Ohm, it's power output is 500mW?

I am not sure if the resistor value I need to put in is the 50 ohm resistor in the measuring device (oscilloscope) or the 50 ohm resistor in the MW source.

This is the mixer I am using. The datasheet says RF power should not exceed 50mW. I

enter image description here


I guess if I set the MW source output to 10mV, given the load impedance of 50 Ohm, it's power output is 500mW?

Power is \$V^2/R\$ so 10 mV into 50 ohms is a power of 2 micro watt or 0.002 mW or -26.99 dBm. The final conversion takes the log\$_{10}\$ of 0.002 then multiplies by 10 to produce dBm.

  • \$\begingroup\$ With the caveat that the above equations work for voltage in RMS -- if you're looking at peak-peak values, or the peak of a sine wave, or whatever, you need to convert. \$\endgroup\$ – TimWescott Jan 9 at 19:15
  • \$\begingroup\$ @TimWescott the rule is that if there is nothing to clarify whether a voltage is peak, RMS or p-p, the RMS must be assumed. This is standard in engineering. \$\endgroup\$ – Andy aka Jan 9 at 19:21
  • \$\begingroup\$ @Andyaka thank you for the answer. Regarding \$R\$ in \$\frac{V^2}{R}\$ though, is it the resistor in the MW source or in the oscilloscope? \$\endgroup\$ – Blackwidow Jan 9 at 19:40
  • \$\begingroup\$ @Andyaka Well, yes. But they teach people everything they need to go from dBm to voltage in engineering, too -- if the OP is missing the one, you can't assume they're not missing the other. \$\endgroup\$ – TimWescott Jan 9 at 20:48
  • \$\begingroup\$ In a 50 Ohm system, 0dBm is 0.223 volts RMS, or 0.632 volts PP. Thus the thermal noise floor for 250,000Hz bandwidth is -174dBm/Hz + 10*log10(250,000) = -174 + 10*5.4 = -174 + 54 = -120dBm. Which in volts RMS is 0.223/1,000,000 = 0.223 microVolts RMS. \$\endgroup\$ – analogsystemsrf Jan 10 at 4:16

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