# Input impedance oscilloscope

I am using Keysight MSOX3102T oscilloscope which has 1 GHz bandwidth. At these frequencies, $1~M\Omega$ input impedance is not suitable due to reflections. Therefore, it has the possibility to select $50 \Omega$ input port. simulate this circuit – Schematic created using CircuitLab

Assuming the configuration shown in the figure in which I measure a RMS voltage of $V_{rms} = 2.2V$ on the oscilloscope, is it correct to compute the power provided by the signal generator as: $$P = \frac{V^2_{rms}}{R}= \frac{2.2^2}{50}= 0.0968 W$$ where $V_{rms}$ is the RMS voltage measured on the oscilloscope and R the input port impedance?

UPDATE: Thank you everyone for the replies. Maybe I have not been clear enough explaining my doubt. It can be summarized as, setting the input impedance at $50 \Omega$, can I take the load impedance as just $50 \Omega$ or does it not work that way in an oscilloscope?

• Usually the AWG has the settings to select the downstream input impedance to compensate the internal 50ohm. E.g. if you set down stream impedance to 50ohm then the displayed voltage on the AWG will be half of v1. – user3528438 Jan 20 '17 at 15:20
• " setting the input impedance at 50Ω, can I take the load impedance as just 50Ω?" yes of course – Sunnyskyguy EE75 Jan 20 '17 at 16:22
• The measurement scenario you've presented delivers maximum available power to the 'scope. 96.8 mW is the most you can deliver to any load impedance. – glen_geek Jan 20 '17 at 16:28

If you measure 2.2 volts RMS on the scope then that equates to a power taken by the scope's 50 ohm internal impedance of 96.8 mW. The power provided by the signal generator shares equally into it's own 50 ohms and the scope's 50 ohms so, V1 (in your pictures) provides a power of 193.6 mW or, put another way, V1 has an RMS output of 4.4 V RMS.

setting the input impedance at 50Ω, can I take the load impedance as just 50Ω or does it not work that way in an oscilloscope?

If the scope didn't make its input impedance 50 ohm then you would get signal reflections and people would be up-in-arms.

Regardless of what the scope's impedance is, it still shows you the voltage at its input. The power the scope is taking from that signal is V2/R. In your case that comes out to 97 mW, as you calculated.

So this is a long way of answering "yes" to your question.

Note that the power going into the scope is not the same as the power put out by the voltage source inside the signal generator (V1 in your diagram). It's actually double the 97 mW since R1 is also dissipating 97 mW. However, that's all stuff that happens inside the black box, and is irrelevant from a outside point of view.

• Just as a complement, think that if the scope is "seeing" 2.2V between the source's and its own 50 Ohm resistors, the whole circuit acts like a resistor divider. This means that the actual voltage source is outputting 4.4V at a 100 Ohm load, which translates to the double power Olin is talking about. – Ronan Paixão Jan 20 '17 at 15:51

The oscilloscope input resistance is matched to the impedance of the signal generator. The internal voltage of the signal generator V1 is 4.4 V and the oscillograph measures 2.2 V. The voltage drop over R1 is 2.2 V too and the power dissipated in R1 and the oscilloscope input resistance is both 0.0968 W. The power generated by the source V1 is 0.1936 V.

There are sine generators for lower frequencies with a 50 Ohm impedance too. You may connect an oscilloscope to it and switch the input impedance to 1 MOhm or 50 Ohm. At frequencies below 1 kHz you won't see any reflections. If you set the generators output voltage to 5 V you will measure 5 V with the oscilloscope if the input is switched to 50 Ohm. But when you use 1 MOhm input, you measure 10 V amplitude. The internal source voltage of the generator is indeed 10 V. But using the 50 Ohm input resistance of the oscilloscope you have build a voltage divider of two 50 Ohm resistors. The voltage is divided by two: 2 = 50/(50 + 50).