Help: how to convert the power gain to the voltage gain?
The power gain of a mixer is measured from an apparatus with a 50 ohms input impedance while the mixer doesn't matched to 50 ohms. Now, I want to get the voltage gain. So, anyone knows how to convert the power gain to the voltage gain? Supposing the impedance and reactance of the mixer are R and X respectively. (The input frequency of the mixer is above gigahertz while the converted output frequency is 5 MHz.)
I'm waiting for your reply.
sandy
Hello,
Conversion power gain is probably specified based on the maximum available power from the RF source and a specified IF (real) load impedance (maybe 50 Ohms too).
The maximum available power is when the 50 Ohms source is terminated with 50 Ohms. The EMF of the source is twice as high. So a source with 10 dBm (1Vp) output has a EMF = 2Vp (1.414Vrms).
The actual input voltage you can derive from a voltage divider
Vactual/V50 = 2*Zmix /(Zmix + 50).
Vactual = actual voltage at input of mixer
V50 = voltage when source is terminated with 50 Ohms.
Zmix = input impedance of mixer.
You may use a spice simulator (or other product that you know).
The voltage gain of the mixer will be:
Voltage gain = (V50/Vactual) * sqrt( (powergain)*Rif/50 )
Power gain as a number, not in dB's. Rif = IF load impedance (real). for a passive mixer, power gain will be betwen 0 and 1.
Thanks for WimRFP's reply.
While I still have something in doubt: should the reflection of the signal be considered? For the singal is above gigahertz.
Hello,
If you can live with defining the voltage gain based on the maximum available generator power, you can ignore input reflection from the mixer.
Maximum available power conditions means that the RF source is terminated with its own impedance to get maximum power transfer. In that case the actual voltage at the source's terminals is half the EMF.
In the RF world, voltages are mostly taken with respect to maximum available power condition. That voltage is mostly called forward voltage or incident voltage in case of S-parameters.
Using the voltage under maxim available power condition will be:
Voltage gain = sqrt({power gain}*Rif/50).
Power gain NOT in dB, Rif in Ohms.
Assuming 50 Ohms RF source impedance.
In this case you cannot measure the voltage gain by probing the actual input and output voltage. Because of unknown input reflection coefficient, the actual input voltage may not be equal to the voltage under max available generator power.
Got it. Thanks.