• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

possible 1N007 fault

Pommie

Well-Known Member
Most Helpful Member
The long term average over many cycles (or indeed one) is then 50% of full VOLTAGE - can anyone argue that?.
Yup, It's the FULL voltage for half the time. A totally different thing to half the RMS voltage. I convinced myself of that as explained in post #44. The power is halved (ignoring diode drop).

Mike.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
In fact a better example :
Imagine a PWM with a 10% "on" cycle at 10V and a 1 Ohm load.

The average voltage is 1V, which would give 1W dissipation if a steady voltage.

The on time with 10V means 10 Amps through the one ohm load and 100W power for that 10% time, so 10W average power out.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
Obviously for a full wave rectifier the RMS voltage is the same as no rectifier (less the insignificant diode voltage drop of course).
Leaving out half cycles does not change the voltage of other half cycles.

Again, imagine the AC scope waveform. Omitting some half cycles does not affect the voltage between live and neutral of the half cycles that are not omitted, so does not change the current during those half cycles, or the power transfer for each.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Leaving out half cycles does not change the voltage of other half cycles.

Again, imagine the AC scope waveform. Omitting some half cycles does not affect the voltage between live and neutral of the half cycles that are not omitted, so does not change the current during those half cycles, or the power transfer for each.
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.
 

Pommie

Well-Known Member
Most Helpful Member
Some here think your way, some think my way - my crude no-load test yesterday certainly supported my way (where's a nice load when you need one?).
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.
 

Nigel Goodwin

Super Moderator
Most Helpful 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?
That you're completely ignoring RMS - you're suggesting the power halves, I'm suggesting the voltage does.
 

Pommie

Well-Known Member
Most Helpful Member
You keep stating that the RMS voltage halves. It doesn't. Which part of my previous statement are you saying is wrong?

Mike.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
That you're completely ignoring RMS - you're suggesting the power halves, I'm suggesting the voltage does.
The BOTH halve, on average - but it's simply a situation where the two averages are not related by Ohm's Law; see my 10% PWM example a few posts back for another try at an explanation.
 

alec_t

Well-Known Member
Most Helpful Member
It's never the full voltage, only half of it, either p-p or RMS.
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).
 
Last edited:

MTJB

Member
Are you maintaining that the RMS value of a full bridge rectified waveform is different than the RMS value of a sine wave (minus the small loss in the diodes)?
This pictorial is for a full wave rectification of an AC sine wave. We are discussing half wave rectification here, (using a single diode in the circuit), so you would need to remove every other upward pulse in the lower wave drawing.
 

Visitor

Member
This pictorial is for a full wave rectification of an AC sine wave. We are discussing half wave rectification here, (using a single diode in the circuit), so you would need to remove every other upward pulse in the lower wave drawing.
Yes, the point is the RMS voltage of the AC signal and full-wave rectified signal is the same, and therefore the power used by a resistive load is the same. This key point of understanding is absent from this discussion.

The next comparison, which I thought would be trivial to explain and should be understood without explanation was that a half-wave rectified voltage would eliminate half of the sine wave, resulting in half the energy being delivered over time. I should have explained the obvious.
 

MTJB

Member
I have to agree with that last line, but I'm afraid it's you who is confused, in this specific instance; you just seem to be looking too deep in to the technicalities rather than basic principles.



Forget the time scale, just stick to 50% duty cycle.

Imagine it's on for eg. ten seconds, then off for ten seconds, repeating.
The long term average over many cycles is then 50% of full power - can anyone argue that?

As long as the load is purely resistive, each "on" cycle (or half cycle) is at full voltage and the current is proportional to that voltage & so full power; or no power for off cycles or half cycles - never more or less, whether half cycles or half minutes timescale.
There is no "half voltage", only only half duty cycle.


[The 40W <> 60W output proportion rather than 2:1 ratio in the original post is presumably due to the positive temperature coefficient of the heating element, it's not a "perfect" resistor, the resistance is lower at lower temperature.
 
Last edited:

MTJB

Member
Yes, the point is the RMS voltage of the AC signal and full-wave rectified signal is the same, and therefore the power used by a resistive load is the same. This key point of understanding is absent from this discussion.

The next comparison, which I thought would be trivial to explain and should be understood without explanation was that a half-wave rectified voltage would eliminate half of the sine wave, resulting in half the energy being delivered over time. I should have explained the obvious.
By inserting the series single diode into the circuit, artificially moves the "0" reference point (nuetral wire) for measuring the voltage, (or current for that matter), in the waveforms, by eliminating either the positive, or negative swings of the waveform; depending upon the polarity of the diode. There by becoming 1/2 of the 240 P-P AC sine wave input voltage, and 1/2 of the current to create heat (light, horsepower, and/or 'work') in the load. It also reduces the duty cycle (my reference to frequency of the pulses) by 1/2 (50%). Correct? So since POWER(P) is the product(x) of VOLTAGE(E) and CURRENT(I), E x I =P; hence, 1/2(E) x 1/2(I) = 1/4(P). As per my earlier post. It doesn't really matter what the actually values really are. Which I think is Nigel's point. Current(I) is the other factor left out of the equation, or discussion, but is created by the resistive load. As per the equation: P= (E x E) ÷ R, and/or P= (I x I) x R. But if you do just reduce the voltage by 50%, which will also reduce the current by 50%, you will end up with 1/4 the power. As per Nigel's test circuit, the VOM becomes the inherent hidden, current producing, load resistance!
 
Last edited:

Pommie

Well-Known Member
Most Helpful Member
By inserting the series single diode into the circuit, artificially moves the "0" reference point (nuetral wire) for measuring the voltage, (or current for that matter), in the waveforms, by eliminating either the positive, or negative swings of the waveform; depending upon the polarity of the diode. There by becoming 1/2 of the 240 P-P AC sine wave input voltage, and 1/2 of the current to create heat (light, horsepower, and/or 'work') in the load. It also reduces the duty cycle (my reference to frequency of the pulses) by 1/2 (50%). Correct? So since POWER(P) is the product(x) of VOLTAGE(E) and CURRENT(I), E x I =P; hence, 1/2(E) x 1/2(I) = 1/4(P). As per my earlier post. It doesn't really matter what the actually values really are. Which I think is Nigel's point. Current(I) is the other factor left out of the equation, or discussion, but is created by the resistive load. As per the equation: P= (E x E) ÷ R, and/or P= (I x I) x R. But if you do just reduce the voltage by 50%, which will also reduce the current by 50%, you will end up with 1/4 the power. As per Nigel's test circuit, the VOM becomes the inherent hidden, current producing, load resistance!
This is complete nonesense. The peak voltage of the on cycle is the same and therefore the current is the same. The duty cycle is the only thing changed.

Mike.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
OK, I had to find out if I was cracking up or not....

LTSpice sim attached.
100V RMS 50Hz sine and 100 Ohm load resistors. Showing six half cycles as a convenient plot size.

On full wave 100W is dissipated in the load resistor and one joule is transferred each half cycle. (OK, 99.99something watts).
The load resistor with the diode in series dissipates 50W and transfers one joule per full cycle in round figures, just a fraction under due to the diode voltage drop.

The load fed through the rec definitely dissipates half the full cycle power. Check it for yourself.

(To get the power display popup in the simulator, control-click the label of a wattage trace).


half_cycle_P_R2.jpghalf_cycle_P_R1.jpg

[Edit - six half cycles not full cycles.]
 

Attachments

Last edited:

Latest threads

EE World Online Articles

Loading

 
Top