Calculating VSWR using dual Linear-in-dB RMS Detector

Q uestion : If I’m using a device such as AD8364 or HMC1030 to measure VSWR, how do I actually get from output voltage (or ADC code) to VSWR? Answer : The short answer is that the return loss in dB is proportional to the differential voltage that appears between OUTP and OUTN (i.e. OUTP-OUTN). It’s important to treat this voltage as a differential voltage. So you need to sample it with a ADC that has differential inputs that can handle the common mode voltage ( 2.5 V). The detector must be calibrated. Let’s say that Channel A is going to measure forward power and Channel B is going to measure Reverse power. So you would start by setting the input power (∆ P IN 1) to both channels to be equal. So ∆ P IN 1=0 dB This will give a differential output voltage (OUTP-OUTN) which will be close to 0 V with a common mode level of 2.5V. So you measure this voltage or ADC output code. Let’s call it V1. Next you reduce the power level on Channel B by, say 20 dB. So now your are emulating a return loss of 20 dB. So a ∆ P IN 2 of 20 dB will give you another differential voltage of V2. So the slope is going to be given by the equation: Slope = (V1-V2)/(∆ P IN 1- ∆ P IN 2) The unit for this will be Volts (or codes) per dB. This slope value should be stored for use when the system is operational. When the system is operational, the return loss can be calculated using the equation Return Loss = (OUTP-OUTN)/SLOPE Since the reflected power will always be less than or equal to the incident power on the antenna, (OUTP-OUTN) will always be positive. VSWR is calculated using the equation VSWR = (10 (RL/20) +1)/ (10 (RL/20) -1) Let’s consider an example. During calibration, we first apply two equal input powers. This yields a differential output voltage of 0 V. Next we reduce the input power to channel B by 20 dB. This yields and OUTP voltage of 3.5V and an OUTN voltage of 1.5 V. So the differential voltage is 2V. So the slope comes out to be Slope = (0-2)/(0-20) = 0.1 V/dB. This is the calibration coefficient which must be stored. Now with the system in operation, we measure a voltage on (OUTP-OUTN) of 1V. This converts to a return loss of RL = (1V)/(0.1V/dB) = 10 dB And the VSWR is given by VSWR = (10 (10/20) +1)/ (10 (10/20) -1) = 1.9

Frequency Dependent Power Measurement Error

Question We are using the AD8364 in the band 2.110 to 2.155GHz. We are getting a relatively large frequency dependent power measurement error. For example, if the device is calibrated (intercept and slope) at 2132.5MHz andwe inject a signal at 2112.5MHz and 2152.5MHz the indicated power level variesby +0.38dB/-0.37dB at 2112.5 and 2152MHz from the nominal value at 2132.5MHz. Feedback: It looks like the AD8364 does indeed have significant output voltage variation vs frequency. The attached plot shows how the output would vary vsfrequency if each device was calibrated at a single frequency. For AD8364 (light blue), you can see that it falls off sharply above 2 GHz. HMC1030(another dual rms responding detector), has a much flatter response in this region.You balun seems to have an insertion loss that is increasing slightly vs frequency. This is making things slightly worse. if you could find a balunwhich has insertion loss that was decreasing in this freq range, that might improve things a little. @EOF& Answer

AD8364 Operation at 2.7 GHz

Question: AD8364 is specified to operate up to 2.7 GHz but the datasheet only shows data at 2.5 GHz. Do you have any data showing performance up to 2.7 GHz. Answer: The attached plots show sweeps of Channel A and Channel B at 2.7 GHz. In the first two plots, Channel A and Channel B are swept respectively with the output voltage being measured at OUTA or OUTB. In the third plot, Channel B is held at -25 dBm. Channel A is swept and the voltage difference between OUTP and OUTN is measured. This data was taken on the newer evaluation board (AD8364-EVALZ) which uses a broadband Anaren balun.