How do I measure the VSWR of this video detector? - Page 1
Forum section: electronics / rf-microwave

author: Michael Craft
author_rank: Contributor
timestamp: October 25, 2022, 04:48:04 pm
content: We received a video detector from someone, and she asked us if we knew how to measure the VSWR of it.It's a little thing, with an SMA connector on each end.I am not an RF engineer, but I do have alittlebit of experience in the area. (I did microwave engineering as a co-op student many centuries ago.) So I asked a few RF engineers how to do it, and, well... they didn't seem to know, either.Unfortunately I don't think it's as simple as connecting it to a VNA and measuring the VSWR. The spec says there must be a 300 Ω "video load" connected to the output, and the video detector must be powered using a constant current source. Which Ithinkmeans I need to measure the VSWR using a modulated signal.The best way to describe the problem is with some images. I put together some PP slides. Attached is a pdf of the slides. I have also linked to png images of the slides.Any help/guidance/assistance you can provide would bemuchappreciated.Thank you

author: tggzzz
author_rank: Super Contributor
timestamp: October 25, 2022, 05:34:06 pm
content: Quote from: Michael Craft on October 25, 2022, 04:48:04 pmFirstlywhydo you need tomeasurethe VSWR, and to what accuracy? Isn't knowing it is <4:1 sufficient? If 4:1 isn't good enough, then put a 3dB attenuator (with better than 4:1 VSWR!) in directly in front of it.Whattypeof detector is it? If a diode, the the current is probably used to bias it into the correct part of its V-I curve. If the output is a DC voltage related to the signal amplitude, then you don't want a blocking capacitor. The data sheet and/or application notes should explain this; I'm not going to double guess the data sheet!I don't see why signal modulation would affect the VSWR measurement, unless it is so wideband that the VSWR varies significantly across the bandwidth.I would want to test whether the bias current caused a DC voltage visible at the detector input. It might be that such a voltage would damage a VNA.If the data sheet says a 300ohm load is required, then a 300ohm load is required whatever signal source is used.

author: Michael Craft
author_rank: Contributor
timestamp: October 25, 2022, 11:53:52 pm
content: Thankstggzzz.QuoteI didn't ask, but am assuming it's a requirement of some sort.QuoteYes, it is a Schottky Diodes, and the 270 μA is used to bias the diode. The problem withnotusing a blocking capacitor (as I see it) is that the current will be split: some of the current will be going into the detector's output terminal, and some of the current will be going in to the 300 Ω resistor. In other words, I would be sourcing current into a diode and resistor that are connected in parallel. I'm not sure if this is an issue or not. It also means I would need to use an adjustable current source: I guess I would have to monitor the the voltage across the 300 Ω resistor to determine the current through the 300 Ω resistor, and then subtract this value from whatever the current source is producing, with and keep adjusting it with the hope that the difference will become 270 μA. It just seems messy. And the datasheet gives no guidance on it.

author: tggzzz
author_rank: Super Contributor
timestamp: October 26, 2022, 07:50:23 am
content: Quote from: Michael Craft on October 25, 2022, 11:53:52 pmAs I put it here:https://entertaininghacks.wordpress.com/library-2/good-questions-pique-our-interest-and-dont-waste-our-time-2/QuoteThat's just one example of the "XY problem".QuoteWhoever will use the detector will need to use it correctly as specified - and you will test that configuration. Hence ask them; if they don't know then the measurement is moot!

author: virtualparticles
author_rank: Regular Contributor
timestamp: October 26, 2022, 01:19:50 pm
content: You don't need to modulate the RF. Simply apply the bias shown in your schematic and set the VNA output level to the operating input level of the diode and measure the VSWR. The input impedance is probably not terribly linear with input power so you might set the VNA to do a power sweep at a few frequencies and see what the VSWR does.A diode like this is only useful for stripping off the modulation envelope of course so you can see pulse modulation or the AM component of the signal.Cheers,Brian

author: radiolistener
author_rank: Super Contributor
timestamp: October 26, 2022, 01:41:05 pm
content: you can measure it's impedance with VNA (vector network analyzer).Since your DUT needs bias tee, you can calibrate VNA through bias tee. I didn't tried it, but I think it should work.VNA can show you complex impedance of your DUT and VSWR.Don't forgot to put 300 Ω terminator on the second DUT port during measurement.And be careful, check that DC offset is not going to VNA, otherwise you can burn it's port.

author: Bud
author_rank: Super Contributor
timestamp: October 26, 2022, 02:05:20 pm
content: If the output impefance of the detector is specified as 300 Ohm, do not use a blocking capacitor, connect a 300 Ohm load resistor to the output and double the current source bias current. The current will split equally between the device and the load.VNAs operate in continuous waves, they have no modulation capabilities as far as i know. SWR and s-parameters are measured at a single frequency at a time.What the input impedance of the detector is spec-d to? If it is also 300 Ohm then using a 50 Ohm VNA will produce incorrect result. The VNA will need to be reconfigured to 300 Ohm reference or manual translation will be needed.

author: radiolistener
author_rank: Super Contributor
timestamp: October 26, 2022, 02:40:45 pm
content: Quote from: Bud on October 26, 2022, 02:05:20 pmyes, VNA shows VSWR for z0=50 Ω by default, but you can get measured complex impedance and calculate VSWR for 300 Ω manually:gamma = (z - z0) / (z + z0);VSWR = (1 + abs(gamma)) / (1 - abs(gamma));wheregamma - reflection coefficientz0 - base impedance, in this case z0 = 300z - measured impedance

author: Joel_Dunsmore
author_rank: Regular Contributor
timestamp: October 27, 2022, 02:30:43 am
content: Quote from: Bud on October 26, 2022, 02:05:20 pmThey do now: S93072BBut to the question: the picture shows an SMA connector so let's presume the reference Z is 50 ohms. VSWR is a format on most VNAs, but the underlying measurement is S11. You can do this by injecting the current as you show and loading with a 300 ohm resistor, blocked as you show, making a 1 port measurement. Or you can make a 2-port measurement (after calibrating in 50 ohms) and use port impedance transformation (on modern VNAs) to show the S11 in the condition that port 2 would have been terminated in 50 ohms. And bias through the test ports (which are DC blocked to the RF bits). you might need to pull the current out the port 1 input if the design does not have a shunt resistance element inside (VNA's are almost always DC blocked so can absorb DC except through the bias tees inside).As to needing modulation, typically VSWR is a small signal and so no modulation is needed or wanted, but if the modulation were to cause the impedance to change, due to detector diode turn-on biasing, then you should do as mentioned and measure S11 with varying power levels and perhaps with modulation. You can create square-wave modulation by using pulsed RF, or you can combine 2 sources in a 4 port VNA to create a pulse on top of an average RF signal.But, most likely, the detector is intended to be used with a large bypass capacitor to hold the Detected voltage (might even be integrated in) so it can be sensed on the top of the 300 ohm resistor; this bypass will look like a short circuit at 13 GHz, and the termination is mainly to be a reference Z for the detected voltage spec. In such a case I think you need to pull the current out through a bias tee, which means a blocking cap before the 300 ohm resistor. in fact, if the detector is internally bypassed by a capacitor at the 300 ohm port, then you probably don't need the 300 ohm at all and can just pull 270uA. Best if you had the spec sheet that shows the proper use configuration, as already mentioned.

author: virtualparticles
author_rank: Regular Contributor
timestamp: October 27, 2022, 04:17:37 pm
content: Quote from: Bud on October 26, 2022, 02:05:20 pmRespectfully, I would leave the blocking cap in place. The RF side of the diode is 50 ohms, so a 50 ohm VNA will give a good VSWR measurement. The video output side of the diode expects the 300 ohm load. Without a blocking cap, you can't know how the current will split, it must be left in.