How does noise Vrms and affect measurable peak-to-peak voltages of sine waves

[ddr_controller]

I am having trouble understanding how noise affects the signals you can measure.

For example, imagine the real signal I want to measure of a sinewave with a peak-to-peak amplitude of 5uV.

This signal goes through a resistor before being amplified by an ideal amplifier. I don't understand the meaning or how to use the $V_{rms}$ that has $V/\sqrt(Hz)$, how do I calculate the bandwidth? What does V_rms have to do with the 5uV peak-to-peak of the sine wave?

My question is whether due to resistor noise, I will be able to measure the signal.

[BillyO]

We don't have enough information to provide a meaningful answer about bandwidth.

1) The way the circuit is drawn, it looks like it will ideally have an infinite bandwidth.

2) The noise in the resistor would be reflected in the current through it.  If the source impedance was very low or 0, the noise through the resistor would not be a factor.

3) We know nothing of the resistor.  What's it made of and how is it made.  What are it's noise characteristics?

4) If the circuit is not ideal, what are the bandwidth limiting aspects?  Stray capacitance?  Inductance? The op-amp?

Now, as to you other question V_rms of a pure sine wave is V_p-p / 2^1/2:

V_rms = V_p-p/2^1/2   or   V_rms = V_p-p/1.4142   or   V_rms = 0.7071 * V_p-p

This relationship is only true of a ideal pure sine wave though.

[David Hess]

The conversion between Vrms and V/Sqrt(Hz) is:

Vrms = Sqrt(bandwidth Hz) * V/Sqrt(Hz)

And the peak-to-peak value is roughly 5 or 6 times the RMS value.

So working backwards, 5uVpp is about 1uVrms, and then to find the bandwidth, if the input referred noise is 5nV/Sqrt(Hz), which is typical for a low noise operational amplifier, then the bandwidth where noise is equal to the signal is 40 kHz.  If the noise was higher, then the bandwidth would have to be reduced, and of course in practice the noise must be lower than the signal.

That input referred noise could be from any source, and does not account for excess noise like flicker noise.  As shown, it is not possible to determine the noise contribution from the resistor because the source impedance is not known.

[BillyO]

The conversion between Vrms and V/Sqrt(Hz) is:

Vrms = Sqrt(bandwidth Hz) * V/Sqrt(Hz)

And the peak-to-peak value is roughly 5 or 6 times the RMS value.

So working backwards, 5uVpp is about 1uVrms, and then to find the bandwidth,

?

We know neither the bandwidth nor the frequency.  Or are you just giving a random example?

[TimFox]

In general, when you add two signals that are uncorrelated, such as your sine wave and the random noise voltage, the rms voltages add "in quadrature".
That is, V_{tot rms}² = V_{1 rms}² + V_{2 rms}².
As discussed above, you need to specify the bandwidth to get a finite answer for the rms noise voltage:  this is covered in textbooks that discuss noise.

[David Hess]

The conversion between Vrms and V/Sqrt(Hz) is:

Vrms = Sqrt(bandwidth Hz) * V/Sqrt(Hz)

And the peak-to-peak value is roughly 5 or 6 times the RMS value.

So working backwards, 5uVpp is about 1uVrms, and then to find the bandwidth,

?

We know neither the bandwidth nor the frequency.  Or are you just giving a random example?

I gave an example with values that would be typical for a similar application.

With realistic parts, dealing with a 5 microvolt peak-to-peak signal will limit bandwidth to 10s of kilohertz at most.  DC accuracy would limit it even further.