Why measuring Voltage and Current becomes difficult at Microwave frequencies? Why measure Power?

Question

I have been reading some Classic books viz. Techniques of Microwave Measurements - Carol Montgomery, Foundations of Microwave Engineering - Collins, and some other google searches and they all claim following:

the fundamental quantities, current and voltage, cease to be significant in the microwave region, and the power and the phase of the waves become important.

and

At microwave frequencies instruments for the direct absolute measurement of voltage or current are difficult to construct and use.

But they don't go in lengths to explain "Why".

Any insights on why is measuring Voltage/Current difficult at Microwave frequencies and Power is preferred instead? Isn't Power just proportional to Vrms which is instead 0.707xVp-p?

I also read somewhere due to distributed L-C elements in a transmission line which have a constant characteristic impedance of 50 ohms, if the they are probed to perform measurement the lead effects on the impedance may make the measurements less accurate. But what if we make use of a matched 50 ohm impedance lead instead?

Answer 1

In RF context, voltage is not a size that exists. It simply does not appear in any place that describes a wave, directly.

What is "voltage," for an electromagnetic wave, eecs, before we can talk about how to measure it? You need to first convert your wave into something you can observe directly, which involves termination with a specific impedance, which makes the measurement frequency- and waveguide-specific.

Also, as Kevin said, proper observation will change the impedance of your system, so that a good measurement cannot be done in-system l, because it affects the system so significantly. (But this also applies to power measurements).

Power on the other hand is always easily definable: how much hotter does this 1g of material get if it absorbs all the power for 1s? Can directly derive the power sunk from that.

Even when we are ok with having to unplug the system, and know the characteristic impedance of the port we attach to well enough (we usually can't know) so that we can match our measurement equipment to it, there are still technological challenges that make observing power easier than observing voltage or current:

1. Power is always positive. So if I have an analog circuit that converts power to a current (and quite possible also incorporates correction factors for different frequencies, because the real world is never an all-pass), then I can just integrate that power with a capacitor. Read the voltage over the capacitor every so and so many seconds and to get an estimate is how much energy passed by in that time. Energy per time is power.
   if you were to just build a circuit that takes the integral (or average) of current or voltage as found through converting some wave to that using a specific impedance feed, that integral would mostly be 0, because we already know the signal is zero-mean, no matter how high the amplitude are.
2. A pure electromagnetic wave without modulation has constant power over time. The current or voltage you could measure changes as often as its frequency dictates.
   For modulated carriers, the power changes at the speed at which the modulating signal changes power, not at the frequency of the carrier wave.

Answer 2

Isn't Power just proportional to Vrms …

Yes.

… which is instead 0.707xVp-p?

No!

For one thing, you've mixed up peak voltage and peak-to-peak voltage — if you use peak-to-peak then you need to divide that coefficient by two. But assuming you meant peak voltage, that formula is still only true if the incoming signal is a single sine wave. As soon as the signal contains components of multiple frequencies, you cannot use it.

For example, imagine a square wave. Its instantaneous value is always equal to its peak value (of one sign or the other). So it has more power for the same peak voltage:

$$V_{rms} = V_{p}$$

If you used the sine wave formula to convert power to voltage you'd be under-measuring the voltage.

(This is an actual problem if you're buying a multimeter, let alone RF test equipment — low-end models have AC modes that measure peak voltage and divide it by $\sqrt{2}$ instead of actually measuring the AC waveform, and so they will give wrong answers if the waveform is not a sine.)

I also read somewhere due to distributed L-C elements in a transmission line which have a constant characteristic impedance of 50 ohms, if the they are probed to perform measurement the lead effects on the impedance may make the measurements less accurate. But what if we make use of a matched 50 ohm impedance lead instead?

1. You cannot just “tap” a transmission line, like you can a low-frequency circuit, off to the measurement instrument. If you add a second 50 Ω line in parallel with the existing line that's part of the device you're testing, you've made an impedance of 25 Ω — just like if you paralleled two 50 Ω resistors. That will have all those bad effects you're trying to avoid.

2. You need to consider transmission line effects in the measurement instrument — in whatever wiring and circuits the signal passes through on its way to the detector that finally turns the RF into some kind of DC or low-frequency signal. This is possible, yes, but it is not simple — and it it was especially not simple in the days when measurements were typically made with analog meters.