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I'm estimating the Vrms of millivolt level (less than a diode drop) sinusoidal signals in the 3 to 30 MHz frequency range, by eye-balling the peak-to-peak height of the sine wave on my 100 MHz oscilloscope and multiplying by sqrt(2)/2 . The voltage is measured across a 1% 50 ohm resistor using a 10X scope probe. But that method might be only maybe accurate to ±10%.

Is there a more accurate way (better than ±10%) to measure millivolt Vrms, maybe using a simple circuit plus generic bench instruments (DVM, scope, etc.)?

hotpaw2
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  • I assume you mean you divide the peak-to-peak by 2*sqrt(2) (ie 2.828) not 1.414. If it is a noise like signal then dividing by 6 will get the RMS (assumes the display persistence gives about 2 x standard deviation on the display). – Kevin White Jun 05 '21 at 17:39
  • @Hotpaw2: Are you using an analog scope? – JRE Jun 05 '21 at 17:49
  • @JRE : Using a Hantek DSO5102P scope – hotpaw2 Jun 05 '21 at 17:58
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    If you need extremely high accuracy and know what the sinusoides are (e.g. you are generating them in the first place), lock in detection can accurately measure nanovolt class signals. – user1850479 Jun 05 '21 at 18:03
  • I had a look at the DSO5102P manual. I was surprised that a DSO doesn't have a way to measure $V_{RMS}$. – JRE Jun 05 '21 at 18:08
  • @JRE : The scope does have both Vpp and Vrms, but those measurements seem to jump around by +- 10%, even when the waveform seems stable. I'm looking for maybe closer to +- 2%. – hotpaw2 Jun 05 '21 at 18:11
  • Get hold of a webcam and download this software. Follow the instructions in the wiki for measuring RMS voltage. – JRE Jun 05 '21 at 18:12
  • Possibly answering my own question: It looks like there's open source software in Python that will allow me to download samples from the Hantek scope. So I could dump samples and do a regression curve fit to remove sampling noise. – hotpaw2 Jun 05 '21 at 18:21
  • @hotpaw2: Regardless of how you do it, you need to get at least 8 cycles to calculate RMS accurately. – JRE Jun 05 '21 at 18:25
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    I recommend not to trust the ground for such tiny voltages. Use two probes or a differential probe. – Janka Jun 05 '21 at 18:42

2 Answers2

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Yes use AC coupled to coax to a BNC T connector on scope if it does not have 50 Ohm internal option. Then insert a leaded 50 ohm 10% or 49.9 ohm resistor and measure on X1 scale.

But measure Vpp in a 20 MHz filter on, which should not make much difference and don’t scale incorrectly as you did in the question, just use Vpp.

That’s the proper way to measure DC supply ripple. If you don’t have test point that can be easily soldered with coax, make some with 2 turrets. Keep both leads very short ~ 1cm. That should be as accurate as your scope after calibration with the test loop.

Tony Stewart EE75
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....by eye-balling the peak-to-peak height of the sine wave on my 100 MHz oscilloscope and multiplying by 2/sqrt(2)

That doesn't yield RMS. You need to divide by 2 to get peak then divide again by \$\sqrt2\$ to get RMS.

Is there a more accurate way (better than +- 10%) to measure milliVolt Vrms, maybe using a simple circuit plus generic bench instruments (DVM, scope, etc.).

You could make a peak detector circuit using a fast op-amp wired as an ideal rectifier but make sure you use a really fast op-amp (circa 500 MHz GBP) and a really fast reverse recovery diode like the BAS16 or 1N4148. The DC output will be the peak value and then you can use a voltage divider to produce the RMS. That then makes it a peak detector that is useful.

Begin with a precision rectifier (circuit from from wiki): -

enter image description here

I've built one that was good for 40 MHz using a fast op-amp and fast diode. You'll need to add a small capacitor on the output to 0 volts (circa 100 pF) to "catch" the peak and, maybe a parallel resistor with the capacitor to slowly discharge the peak after the event has gone.

If I remember correctly I actually used a fast comparator (MAX999) instead of an op-amp but, bear in mind that the MAX999 has input hysteresis of a few milli-volts that will create a tiny error however, it does slightly simplify the precision rectifier to this: -

enter image description here

If you make \$R_L\$ around 10 kΩ and use a parallel 100 pF capacitor, you get a time constant of 1 μs which is good enough as a peak detector for 30 MHz but may be a bit sloppy for 3 MHz so, make the output capacitor more like 1 nF.

Andy aka
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  • Sorry, typo, multiplying should have been dividing. But multiplying by the reciprocal is better. – hotpaw2 Jun 05 '21 at 18:04
  • What's a circuit or schematic for a peak detector (for those levels and frequencies) look like? Resources (parts, blogs)? – hotpaw2 Jun 05 '21 at 18:06