Why does oscilloscope dispaly 200mV Vpp?

tony_lth said:

oscilloscope to measure a -16dBm/700KHz sine wave signal

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-16dBm into what load impedance?
If you don't have the proper load impedance connected to output of a signal generator than the output will be twice what you calculated.

Hi,

Crutschow is correct.
I'm doing some math with 50 Ohms (just guessing)

-16dBm means -16dB × 1mW = 0.025mW
P = V × V / R --> V = sqrt(P × R) = sqrt (0.025mW × 50 Ohms) = 35.4 mV RMS.
Vpp = 2 × sqrt(2) × V_RMS = 100 mV

So your math seems to be correct.(for 50 Ohms)

*******

So what's happening?
The generator has a source impedance of 50 Ohms (to match the characteristic cable impedance and to match the termination resistance = load tesistance)
And it expects to drive a 50 Ohms load, thus it expects a voltage divider with 50 Ohms and 50 Ohms.
(Mind: the characteristic cable impedance does not need to be used, since it does not cause a voltage drop, just a delay in time)
The output voltage of this voltage divider is half of the input voltage
Or in other words the input voltage has to be twice of the (expected) output voltage.

So, by expecting a load impedance of 50 Ohms the generator drives 200mVpp into it's internal 50 Ohms source impedance.

But you did not connect (most probably) a 50 Ohms load, thus there is negligible current and thus negligible voltage drop across the source impedance.
--> You see the full generator voltage (200mVpp) at the output.

Klaus

Switch the oscilloscope input termination to 50 ohms to get expected 100 mVpp.

The moral is simple.

In common application, dB is useful to measure power levels. But it is a logarithmic measure.

You can take log of a number only; hence we use dB as a ratio with respect to the reference unit.

In this case, you call for -16 dBm; that m after dB refers to 1 mW of power used as a reference.

Hence -16 dBm = 10^(-1.6) mW ---> 0.0251 mW; what does it mean?

But you are actually measuring a voltage with a voltmeter (right?)

At a constant load, the power and voltages are NOT linear; they are quadratic relation. In log scale, that means a factor of 2 comes up!

Of course you can use dBV or dBmV but they are less common.

Unless you specify the waveform, the peak volt cannot be correlated with the rms (which is related to power) volt.

c_mitra said:

Unless you specify the waveform, the peak volt cannot be correlated with the rms

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The waveform is shown in the scope picture.

Klaus

KlausST said:

The waveform is shown in the scope picture.

Klaus

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Right. He also mentioned in the original post that it is a sine wave. I was talking in general.

I met a strange result with oscilloscope to measure a -16dBm/700KHz sine wave signal.

Click to expand...

I was talking about the general case. It can be messy to calculate the power if the load and the waveform are not specified. For dissipated power the load can be considered purely resistive but there can be radiated power that is also to be considered.