possible 1N007 fault

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