possible 1N007 fault

Pommie · Dec 8, 2019

Visitor said:
It appears that some members here need to refresh their knowledge of the meaning of RMS levels. The Wikipedia article may be helpful.

If they've not got it after 5 pages of posts, I'm not sure a wiki article will help.

Mike.

MTJB · Dec 9, 2019

Visitor said:
It appears that some members here need to refresh their knowledge of the meaning of RMS levels. The Wikipedia article may be helpful.

But here is my point, (and also most likely Nigel's, tho I can not actually speak for him). The 240v "value" (although actually should be 230v) he originally had been discussing, is already the RMS "value" of voltage. Please see the below text from the WIKI article you referred.

Because of their usefulness in carrying out power calculations, listed voltages for power outlets (for example, 120 V in the USA, or 230 V in Europe) are almost always quoted in RMS values, and not peak values. Peak values can be calculated from RMS values from the above formula, which implies VP = VRMS × √2, assuming the source is a pure sine wave. Thus the peak value of the mains voltage in the USA is about 120 × √2, or about 170 volts. The peak-to-peak voltage, being double this, is about 340 volts. A similar calculation indicates that the peak mains voltage in Europe is about 325 volts, and the peak-to-peak mains voltage, about 650 volts.

Although I do believe the stated 120v here in the states, is of the "split" phase value, of the 240v single phase service we have available on our residential mains here in the states.

Pommie · Dec 9, 2019

MTJB said:
assuming the source is a pure sine wave.

This is the pertinent part of your quote. Half wave rectified AC is not a sine wave.

Mike.
Edit, you do realize that you're the only one still arguing this point. Everyone else realized their error.

Visitor · Dec 9, 2019

Pommie said:
Edit, you do realize that you're the only one still arguing this point. Everyone else realized their error.

....or have simply dropped out of the conversation....

tvtech · Dec 9, 2019

Where is Ratchit when you need him lol. He would of loved being involved in this thread. Miss you Ratch you clever man

MTJB · Dec 9, 2019

alec_t said:
Post deleted.

Question. How does one delete a post of there's here in this forum if mis-spoken, other then editing it for correction???

alec_t · Dec 9, 2019

MTJB said:
How does one delete a post

You can ask the moderators to delete it, but they may decline to do so.

unclejed613 · Dec 9, 2019

this might be a bit of help... http://www.nessengr.com/technical-data/waveform-rms-and-average-values/

MTJB · Dec 9, 2019

unclejed613 said:
this might be a bit of help... http://www.nessengr.com/technical-data/waveform-rms-and-average-values/

So, since we were actually discussing POWER in watts of the circuit, what exact parameters would need to be added to the POWER equation, P= (E x E) ÷ R, that would represent the insertion of a single half wave rectifier diode? Explaining the 50% reduction in the duty cycle, the voltage drop across the diode, and then solving for the results. Where:

Input voltage E= 240v RMS @ 50hz
Load R= 240 ohms

Pommie · Dec 9, 2019

You don't need to do any math to prove 50%. Consider the instantaneous voltage and current with AC - the power is the same at any particular moment. You then take away half the cycles. As always, assuming a perfect diode.

If you really want the maths with the diode drop, P= 0.5*(V-0.7)^2/R

Mike.

possible 1N007 fault

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