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I know the 'nominal' voltage of 110/115/120 is both based on Vrms and that stability @ any of those levels is pretty much a fairy tail.

I also know that most rectifiers will output ~130-150V after re-assembling the waveform & filling a comditioning cap.

What I'm not 100% sure on, and don't really feel like digging out my soldering iron & finding a few diodes to test it today, is the peak-to-peak voltage of the 110VAC 'hot' line.

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    \$\begingroup\$ Place a high resistance (perhaps 1M ohm) from your + to - output. Now there is a closed circuit which includes 2 diodes and a resistor. Where is the source in that loop ? (You didn't say you wanted voltage to GND) \$\endgroup\$
    – Marla
    Feb 6, 2016 at 17:32

3 Answers 3

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Place a high resistance (perhaps 1M ohm) from your + to - output. Now there is a closed circuit which includes 2 diodes and a resistor. Where is the source in that loop ? (You didn't say you wanted voltage to GND)

Then you edited in a capacitor, which is ok.

LT Spice simulator might or might not be trusted.

Along with seeing no source in your loop, LT Spice agrees that voltage is zero.

The voltage at the cathode of D1 (my circuit), with respect to ground is 340 volts peak to peak. ( I used 120 VRMS ).

The OP did state to use 110 VRMS. The voltage at the cathode of D1 (my circuit), with respect to ground is then 310 volts peak to peak.

enter image description here

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  • \$\begingroup\$ The red sine wave goes from +170 to -170 volts. ( I did use 120 VRMS ). 110 VRMS would go from +155 to -155 volts. \$\endgroup\$
    – Marla
    Feb 6, 2016 at 19:01
  • \$\begingroup\$ Which then makes the answer to my intended original question 110VAC = 310Vp-p, correct? \$\endgroup\$ Feb 6, 2016 at 19:04
  • \$\begingroup\$ Yes, the 110 VRMS is 310 volts peak to peak. But not across V+ and V- \$\endgroup\$
    – Marla
    Feb 6, 2016 at 19:05
  • \$\begingroup\$ Thank you. Since the circuit argument really wasn't the intent of the question, I'll mark your answer accepted, since you just answered my original question. With that said, we should probably edit the Q, A, and comments down to just eliminate the controversial circuit altogether...maybe put it in a new question later ;-) \$\endgroup\$ Feb 6, 2016 at 19:08
  • \$\begingroup\$ ...now I still have to scrounge for diodes.....oh well, more to come on that front. \$\endgroup\$ Feb 6, 2016 at 19:14
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I know the 'nominal' voltage of 110/115/120 is both based on Vrms and that stability @ any of those levels is pretty much a fairy tail.

Stability is not based on Vrms. Stability is determined by the ability of generators to maintain voltage despite fluctuating demand. Voltage is measured in Vrms. Utility companies usually comply with their specified voltage with ±10%. It's not a fairy tail. It's a requirement.

I also know that most rectifiers will output ~130-150V after re-assembling the waveform & filling a comditioning cap.

A full-wave bridge rectifier will give out a peak DC voltage of \$ \sqrt 2 \cdot V_{RMS} - 2 ~diode~drops \$ (typically 0.7 V each).

I never heard of a 'conditioning' cap. The smoothing cap will maintain the DC output at peak DC voltage on no load but voltage will droop between peaks depending on load.

What I'm not 100% sure on, and don't really feel like digging out my soldering iron & finding a few diodes to test it today, is the peak-to-peak voltage of just the 'hot' 110VAC line. I.E. Will the circuit below have ~130-150V, or ~260-300V across the + and - output pins?

schematic

simulate this circuit – Schematic created using CircuitLab

Figures 1 and 2 based on original question, "Why is there no voltage reading between Probe+ and Probe-?".

Your circuit will have 0 V between Probe+ and Probe- as there is nothing to induce a potential difference between the probes. Even if you connect NODE2 to mains it just bounces the whole circuit up and down with respect to ground or neutral but still does not induce any potential difference between the two probes.


Edit after OP edit including deletion of the original wonky schematic:

What I'm not 100% sure on, and don't really feel like digging out my soldering iron & finding a few diodes to test it today, is the peak-to-peak voltage of the 110VAC 'hot' line.

\$ V_{P-P} = 2 \sqrt 2 \cdot V_{RMS} = 2 \sqrt 2 \cdot 110 = 311~V\$

Your diodes and capacitor are now irrelevant to this question unless you intended this:

schematic

simulate this circuit

Figure 2. Positive and negative rails generated by half-wave rectifiers.

This circuit will give you the peak to peak voltage of the mains between the two probes.

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  • \$\begingroup\$ I'll allow that my circuit, as laid out might not have worked as intended...but I expect I would not have been the only one surprised. However, that's off-topic for thks question, so I'll do some more research, then start another where that can be the topic. \$\endgroup\$ Feb 6, 2016 at 21:06
  • \$\begingroup\$ @RobhercKV5ROB: So far your commentary has been little more than a dance based on "I'm in charge here" and "Geez, it's not my fault", so why bother adding to the misery? \$\endgroup\$
    – EM Fields
    Feb 6, 2016 at 22:02
  • \$\begingroup\$ @RobhercKV5ROB: OK, but now you've accepted as correct an answer to a question that no longer exists. \$\endgroup\$
    – Transistor
    Feb 6, 2016 at 22:40
  • \$\begingroup\$ @transistor The answer to the question that does still exist, was in a comment that unfortunately no longer exists, but was posted by Marla in comments under the answer I marked. & was the first time that answer was given in this question. In the end, yes your answer was more thorough; just hadn't shown up on my page yet @ that time. \$\endgroup\$ Feb 6, 2016 at 23:34
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You don't have to test anything, all you need to know is that the peak-to-peak value of the nominal (RMS) mains voltage is:

$$ V_{PP} = V_{RMS}\times 2\times \sqrt{2}$$

whether the output is taken across the entire secondary of the distribution transformer or from the center tap (Neutral) to either end, and what's a fairy \$ \style{color:red;font-size:100%}{tale} \$ is that, in the long term, the mains voltage varies more than about +/- 10% of nominal.

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  • \$\begingroup\$ I don't know about any 'universal' effects, but I know I've been working on small-scale generation (5KW-20KW) projects & had ro disconnect from the external grid due to dynamic grid fluctuations of +/- 18% for both voltage potential & frequency. The fluctuations continued for >15minutes while we watched our scopes & equipment monitors all showing agreeing readings; with our generating equipment offline. As for the question, yes Vp-p=Vrms*2 &sqrt;2 is a perfect answer. I had forgotten the formula. \$\endgroup\$ Feb 6, 2016 at 21:04
  • \$\begingroup\$ @RobhercKV5ROB: >15 minutes out of, say, 24 hours is hardly "long term", plus it's hard to believe that there could actually be an 18% variation in nominal frequency coming from the grid. How would you account for the discrepancy? \$\endgroup\$
    – EM Fields
    Feb 6, 2016 at 21:46
  • \$\begingroup\$ It was a frequent phenomenon there. I don't know your exact definition of "long term," but the stations were eventually fully islanded & the power consumed on-site exclusively due to the tremendous grid irregularities in that area of coastal TX. I account for those irregularities as being enescapably tied to the constant (never less than 1/day) brownouts & frequent (usually 2+/week) power outages in the area. I burned up 3 UPS units in the 2 years I lived down there. \$\endgroup\$ Feb 6, 2016 at 21:52

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