Measuring output power with oscilloscope & attenuator - Page 1
Forum section: electronics / rf-microwave

author: isso
author_rank: Contributor
timestamp: October 24, 2021, 04:58:06 am
content: Hi, I'm trying to measure the power output of an RF power amplifier by connecting a 20 dB attenuator to the amplifier output and then connecting the output of the attenuator straight to the oscilloscope input via a BNC cable.My question: should the oscilloscope input be set to 50 Ohm mode or to high impedance for correct measurements? Those options give pretty different results. Thanks!

author: bob91343
author_rank: Super Contributor
timestamp: October 24, 2021, 05:00:28 am
content: 50 Ohms.

author: isso
author_rank: Contributor
timestamp: October 24, 2021, 05:06:41 am
content: Thanks for the fast answer! Do you mind to explain why 50 Ohm should be chosen? I mean the attenuator already acts like 50 Ohm load for the PA, why do we need the oscilliscope input to be 50 Ohm too? Thanks

author: radiolistener
author_rank: Super Contributor
timestamp: October 24, 2021, 05:19:17 am
content: The cable should be terminated with 50 Ω pass-through dummy load at oscilloscope connector. If your oscilloscope has integrated pass-through 50 Ω dummy load, you can use it. Otherwise you will needs to use external pass-through dummy load on oscilloscope connector. But integrated pass-through dummy load has pretty weak limit.It is limited with about 5 Vrms (0.5 W or 26.9 dBm). So, it is more safe to disable internal pass-through dummy load and use external one on oscilloscope connector. In such way you can avoid burning out of of your oscilloscope input.If your oscilloscope has 5 Vrms for internal pass-through dummy load, you can still burn it if you put 50 W through 20 dB attenuator.

author: isso
author_rank: Contributor
timestamp: October 24, 2021, 05:27:40 am
content: Thanks for the explanation! One more question - what if I use the standard oscilloscope probe (as opposed to BNC cable) that gets connected at the attenuator output? Should the oscilloscope input still be set at 50 Ohm? Thanks

author: radiolistener
author_rank: Super Contributor
timestamp: October 24, 2021, 05:44:39 am
content: Quote from: isso on October 24, 2021, 05:27:40 amNo, if you use oscilloscope probe it has 1 MΩ output.But note that probe max voltage rating for RF signal is much less than specified for DC!Usual 1:10 probe has absolute maximum rating about 50 V at 1 MHz and 25 V at 100 MHz. If you exceed that value, your probe can be burned out. This is often mistake, when someone trying to measure RF power amplifier output with oscilloscope probe. Because 100 W amplifier has about 100 V amplitude on the output, which exceed absolute maximum rating of the probe at frequency above 500 kHz.Dummy load is required if you connect oscilloscope with coax cable. But note, dummy load should be placed as close as possible to oscilloscope connector, this is important and this is why it needs to use pass-through dummy load placed on the oscilloscope connector.Internal dummy load that you can enable from oscilloscope menu allows to see signal with less reflections, but it is very weak, so don't apply powerful signal when internal dummy load is enabled. Usually absolute max rating of internal dummy load is about 0.5W = 26.9 dBm = 5 Vrms = 7 Vpk. Don't exceed max input rating!

author: isso
author_rank: Contributor
timestamp: October 24, 2021, 06:13:52 am
content: Thanks a lot, appreciate your detailed answer!

author: ahbushnell
author_rank: Frequent Contributor
timestamp: October 24, 2021, 10:33:18 am
content: i assume your attenuator is rated for the RF power? How much power do you expect?

author: bob91343
author_rank: Super Contributor
timestamp: October 24, 2021, 11:32:28 pm
content: I will try to clear the air. The proper measurement technique requires the transmitter be loaded with 50 Ohms. The attenuator does that, although it also needs to be loaded with 50 Ohms for it to realize its marked attenuation. Of course, the attenuator needs to be rated for the power you intend to put into it.Coming out of the attenuator, into the 50 Ohm load (either the scope's built in one or a pass through external one) will be 1% of the power going into the attenuator. So if the transmitter is putting out 50 Watts, the load on the attenuator will dissipate 0.5 Watts. If you use an external-at-the-scope load, it will dissipate 0.5 Watts. You could probe that external load with a common scope probe and the scope can handle it because it's only seeing one-tenth of the transmitter load voltage, or 5 Volts. Or you can use the scope's internal 50 Ohms and skip the probe.If the transmitter puts out more than 50 Watts you will be overstressing one or more components. This statement ignores the attenuator's power rating, which should be respected.

author: TimFox
author_rank: Super Contributor
timestamp: October 25, 2021, 03:36:40 am
content: A matched 50 ohm attenuator will show 50 ohms at its input when the output is terminated with 50 ohms.If the attenuator has a very high attenuation, the impedance into the input is almost independent of the load impedance.However, to achieve the advertised attenuation, and give the correct output, the output must be terminated with 50 ohms.If terminated in a high impedance, >> 50 ohms, the output will be 6 dB higher (twice the voltage).A high-power attenuator is a very good idea for measuring amplifier output power, since the low-power output can be measured easily with a 50 ohm meter or oscilloscope; it's far easier than using a high-power dummy load (resistor) and somehow measuring the voltage at the input.

author: isso
author_rank: Contributor
timestamp: October 25, 2021, 12:28:19 pm
content: Thanks again to everyone for the detailed information. I proceeded to do some tests and I'm getting some strange results that I home someone can explain.So I have a 20 dB attenuator connected to the power amplifier output. The PA is supposed to produce around 8W of power at 433 MHz which I'm trying to measure. Here's the results:1. Attenuator connected to the scope via coax (scope input at 1 M): 6V peak to peak2. Attenuator connected to the scope via coax (scope input at 50 Ohm): 4.8V peak to peak3. Attenuator connected to the scope probe via an adapter (scope input at 1 M): 6.1V peak to peakAccording to the responses that I received above I should have received similar results with options 2 and 3, however that's not the case. Can someone explain which peak to peak value is correct and why?Thanks

author: dl6lr
author_rank: Frequent Contributor
timestamp: October 25, 2021, 02:21:12 pm
content: Quote from: isso on October 25, 2021, 12:28:19 pmThe second one is correct. From 0.08W (8W through 20dB attenuation) one would expect 2Vrms. 4.8Vpp is 1.7Vrms, so your transmitter emits 6.8W.

author: isso
author_rank: Contributor
timestamp: October 25, 2021, 02:39:09 pm
content: Thanks, but why is option 3 showing 6.1V then?

author: Hamelec
author_rank: Regular Contributor
timestamp: October 25, 2021, 03:20:33 pm
content: what a scope you are using measuring @ 433 MHz?with a 50 Ohm Input or 1MOhm with e.g. 15pF?

author: TimFox
author_rank: Super Contributor
timestamp: October 25, 2021, 03:39:01 pm
content: Note that 15 pF (a typical oscilloscope input capacitance) is -j(25) at 433 MHz.

author: isso
author_rank: Contributor
timestamp: October 25, 2021, 03:44:18 pm
content: I'm using Rigol DS4054. As for the inputs - I tried both 50 Ohm and 1 M - please see my post above.For capacitance - can you please elaborate as what does that mean?

author: Marsupilami
author_rank: Frequent Contributor
timestamp: October 25, 2021, 04:03:12 pm
content: Quote from: isso on October 25, 2021, 02:39:09 pmhttps://rahsoft.com/2021/04/11/maximum-power-transfer-in-rf-circuits/It's derived here how the power delivered to the load depends on the load and you get maximum power when the load is equal to the source impedance.In your case the amplifier is terminated well enough, but the same concept applies between the attenuator and the scope. Think of the attenuator output as it's equivalent ideal voltage generator with a source impedance of 50Ohms. You have to load this part of the circuit with 50Ohms too (at the scope) to get the maximum power. You could math your way out of this but cabling makes it even more complicated so it's a whole lot simpler just to terminate properly with 50Ohm at the scope end.

author: isso
author_rank: Contributor
timestamp: October 25, 2021, 04:36:39 pm
content: Thanks, but I'm actually getting the opposite results - the minimum power when terminating it with 50 Ohms.

author: TimFox
author_rank: Super Contributor
timestamp: October 25, 2021, 04:41:06 pm
content: WRT capacitance:Your oscilloscope has a finite input capacitance at the BNC connector, which should be specified by the manufacturer. It should also be posted on the front panel, so that you know if it is compatible with a given probe. 15 pF is typical, which can be a problem at high frequencies.Attenuator termination: assuming a perfect attenuator (and very good ones are available into the GHz range from the usual vendors), the output impedance from the attenuator itself should be 50 ohms (at the reference plane). If you terminate the attenuator with a perfect 50 ohm resistor (again, good ones are available as coaxial terminations), the impedance driving the oscilloscope input will be 25 ohms (two parallel 50 ohm resistors). The actual output impedance of your amplifier is probably not 50 ohms, but after a 20 dB attenuator the output impedance will be only a weak function of the source impedance.Now, if your oscilloscope had a true 1 megohm input impedance, it would have a negligible impact when driven by 25 ohms. However, 15 pF is -j(25) at 433 MHz, which is very significant. After you find the capacitance value, then do the complex-value arithmetic to see how much attenuation is caused by it.

author: Marsupilami
author_rank: Frequent Contributor
timestamp: October 25, 2021, 05:14:44 pm
content: Quote from: isso on October 25, 2021, 04:36:39 pmNo you don't. You're getting minimum voltage, not minimum power.You get(2.1Vrms)^2/1MOhm= ~4uW for HiZvs(1.7Vrms)^2/50Ohm= ~57mWAssuming no cabling loss and that your 20dB attenuator is accurate (which is 20dB = 100x power ratio) then your amp is outputting 5.7W.

author: Marsupilami
author_rank: Frequent Contributor
timestamp: October 25, 2021, 06:16:24 pm
content: Theoretically you should see 2x the voltage in the hi impedance case, but it's not going to be true because of the cabling. The 50Ohm terminated case should get you a reasonably good idea of what's coming out of the attenuator.

author: isso
author_rank: Contributor
timestamp: October 25, 2021, 07:47:07 pm
content: Thanks a lot, appreciate the time that you took to explain this!

author: TimFox
author_rank: Super Contributor
timestamp: October 25, 2021, 08:16:53 pm
content: Note that in the useful diagram above, the generator has an actual output impedance of 50 ohms.That is not a necessary assumption for this problem--the assumption is that the generator (transmitter output) is designed for optimum use into a 50 ohm load.Laboratory RF generators usually have an actual 50 ohm output impedance (provided by an actual resistor), but power amplifiers can have whatever output impedance the design produces, either higher or lower than the optimal load.(The "maximum signal transfer theorem", with active amplifiers, is relevant to gain, but not necessarily applicable to power output of a practical amplifier.)Again, the attenuator with a good 50 ohm termination presents 50 ohms to the generator, and a large-attenuation attenuator will present close to 50 ohms to the generator, regardless of termination.Therefore, in his diagram, measuring the actual voltage and current at the attenuator input will give a ratio of 50 ohms and a product equal to the power from the generator.One can calculate that from his diagram of the "pi" attenuator, using his resistor values: with a 50-ohm termination, the input impedance is 51.6 ohms; with a short-circuit termination, the input impedance is 50.6 ohms; with an open-circuit termination, the input impedance is 52.6 ohms; or, 51.6 ohms +/- 1 ohm.For a generator optimized for 50 ohms load, one needs to tune or adjust the actual load (antenna or whatever) to get 50 ohms at the cable connector that will connect to the generator output.

author: Marsupilami
author_rank: Frequent Contributor
timestamp: October 25, 2021, 10:06:36 pm
content: TimFox is right. Eventually it all comes down to how accurately do you want to know and how good the pieces are in your signal chain.If you detail your application I'm sure people can give further guidance.

author: David Hess
author_rank: Super Contributor
timestamp: October 26, 2021, 12:46:14 am
content: Quote from: isso on October 24, 2021, 05:06:41 amThe attenuator is designed to operate into a 50 ohm load.Quote from: isso on October 24, 2021, 05:27:40 amThen a feedthrough termination should be attached between the attenuator and oscilloscope probe. Whether the oscilloscope should be set to 50 ohms depends on if the probe is a high impedance probe or a low impedance probe.With all of the above in mind, you can likely get away with not having the attenuator properly terminated if lead lengths are short, however it will no longer attenuate by 20 dB. Whether the attenuator is properly terminated or not, the system should be calibrated against a known level because the oscilloscope's response will likely have considerable error unless it is a sampling oscilloscope.

author: isso
author_rank: Contributor
timestamp: October 26, 2021, 06:05:22 am
content: Thank you very much everyone, now all is clear!