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possible 1N007 fault

MTJB

Member
In any half cycle the voltage doesn't change therefore the current doesn't change. If only half the cycles are present then half the energy is used.
Which part of this statement do you disagree with?

Mike.
Edit, I need a beer.
Yes, the voltage does change. So does the current at the same rate. There is a gradual sloped/sloping increase to the voltage, and then a gradual drop off of the voltage in a sine wave. We do not have an instant on/off as in a digital square wave. That's why Nigel's explanation of printing out the sine wave on grid paper, and then counting the squares, or square area, within the half cycles, above and below the center "0" reference line works to explain the power in the circuit. Does your simulation schematic drawing, not show both R1 (w/diode) & R2 circuits wired parallel across the input voltage source??? I am confused by that. Do Nigel's simple bench test to see what you get.
 

alec_t

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Visitor

Active Member
Meaning of the term RMS from Wikipedia :

SmartSelect_20191207-041617_Firefox.jpg
Thus, 240VAC RMS provides the same amount of energy as a full-wave rectified DC voltage. The "flipping over" of half the peaks doesn't cut the energy in half because the peak-peak voltage has changed.

Going from full-wave rectification to half-wave rectification eliminates half the RMS entry in a resistive circuit. A resistive load will receive half the energy.
 

Pommie

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Yes, the voltage does change. So does the current at the same rate. There is a gradual sloped/sloping increase to the voltage, and then a gradual drop off of the voltage in a sine wave. We do not have an instant on/off as in a digital square wave. That's why Nigel's explanation of printing out the sine wave on grid paper, and then counting the squares, or square area, within the half cycles, above and below the center "0" reference line works to explain the power in the circuit. Does your simulation schematic drawing, not show both R1 (w/diode) & R2 circuits wired parallel across the input voltage source??? I am confused by that. Do Nigel's simple bench test to see what you get.
With your logic, a fully rectified AC current/Voltage is also 1/2 the peak to peak voltage. So is a full wave rectified voltage also 1/4 of the power?

Mike.
 

MTJB

Member
The p-p voltage of the full sine is double that of the half wave rectified sine. The RMS voltage is thus halved, and the RMS current quartered.
Here is where I think the problem, or confusion lies, that they can not understand in the point we are trying to make. Nigel is comparing the input 240v "peak to peak" "value" of the AC FULL sine wave input voltage, as referenced to the neutral wire, (which the input voltage "value" IS 2x compared to the output), to the pulsating half wave rectified DC output. Which no longer actually can be referred to as a "peak to peak" value; only as either a peak and/or RMS voltage value, because the resulting sine wave still has an AC component, and is not pure DC.
 

MTJB

Member
With your logic, a fully rectified AC current/Voltage is also 1/2 the peak to peak voltage. So is a full wave rectified voltage also 1/4 of the power?

Mike.
No. The full AC sine wave, and the full wave bridge rectified pulsating DC output are the same. But please note, that with the bridge rectified pulsating DC wave, we are no longer referencing the output voltage to the neutral wire; because of the extra diodes. Only the single diode, half wave rectified, pulsating DC wave, (for which the output voltage is referenced to the neutral wire), is different, because the peak output voltage is 1/2 the "peak to peak" input voltage. Minus the obvious voltage drop in the diode. I also would like to point out or mention, that the difference that Nigel may have experienced with his simple "bench test", (not getting less than 1/2 the voltage), may be due to the fact that most VOMs I do believe, display "RMS" voltage and not peak or P-P; and he did not make it clear whether he was reading AC or DC voltage on the output of the circuit. Or whether he was using a VOM or scope? Some VOMs & DMMs may not give an accurate AC voltage reading, if it is not a pure sine wave. But I'm sure he knows that!;)
 
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rjenkinsgb

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Does your simulation schematic drawing, not show both R1 (w/diode) & R2 circuits wired parallel across the input voltage source??? I am confused by that.
Totally irrelevant.
If you do not agree with that, duplicate the sim file and delete one or the other load from each copy.

The results will be identical to having both loads in the same simulation.
 

MTJB

Member
Don't you believe the post #45 sim Nigel?
That has a 240V RMS input and the half-wave rectified output is ~170V RMS, not 120V RMS as you think.
You said "I did it yesterday without a load (couldn't find anything suitable). RMS voltage with the diode shorted out was 25V, with the diode in-circuit it was 14V." Doesn't that rather disprove your theory? Per your theory the rectified output would be <12.5V (allowing for the diode drop).
Your 240v value here is stated wrong! 240v "peak" times(x) 0.707 equals(=) 170v RMS. 240v RMS half wave rectified, would be 240v RMS? NO? "Peak to Peak " in this case now would be 480v. Hi, Hi. :nailbiting: Now I am getting a headache and need a FEW beers!
 
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alec_t

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240v RMS half wave rectified, would be 240v RMS? NO?
Definitely NO. 240v RMS full-wave rectified would be 240v RMS.
 

MTJB

Member
Totally irrelevant.
If you do not agree with that, duplicate the sim file and delete one or the other load from each copy.

The results will be identical to having both loads in the same simulation.
Would not R 1 be loading the circuit 100% of the time (100% duty cycle), and R2 (with the diode in series), only be loading the circuit 50% of the time (50% duty cycle) if they are in parallel with the AC voltage source???
 
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Pommie

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Would not R2 be loading the circuit 100% of the time (100% duty cycle), and R1 (with the diode in series), only be loading the circuit 50% of the time (50% duty cycle) if there are in parallel with the AC voltage source???
YES, maybe you've finally got it. It will load the circuit 50% of the time and consume 50% of the energy.

Mike.
I know, but couldn't stop myself.
 

MTJB

Member
YES, maybe you've finally got it. It will load the circuit 50% of the time and consume 50% of the energy.

Mike.
I know, but couldn't stop myself.
But would they not ONLY be in parallel, 50% of the time, when the diode is forward biased; as the circuit is drawn? The circuit we have been discussing all along, or started out with; only involves 1 diode which is added to the circuit by a switch, and 1 resistive heating element!
 
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alec_t

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Post deleted.
 
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Pommie

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But would they not ONLY be in parallel, 50% of the time, when the diode is forward biased; as the circuit is drawn? The circuit we have been discussing all along, or started out with; only involves 1 diode which is added to the circuit by a switch, and 1 resistive heating element!
Think about the voltage of the half cycle, it's no different to the voltage when both cycles are present. Therefore the current during that half cycle will be the same. The only thing that has changed is it's only on for 50% of the time. Another way to think about it is, HOW does the on cycle know to reduce power because the other half cycle is off?

Mike.
 

rjenkinsgb

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Most Helpful Member
But would they not ONLY be in parallel, 50% of the time, when the diode is forward biased; as the circuit is drawn? The circuit we have been discussing all along, or started out with; only involves 1 diode which is added to the circuit by a switch, and 1 resistive heating element!
It's simply showing both states, the power with the switch open (diode in circuit) and switch closed (direct connection), side by side in the same simulation rather than two different ones.

As I said, you can easily split it to two if you wish - the results are the same either way.
 

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