possible 1N007 fault

[MTJB]

How would 0 to V+ be the same as V- to 0 to V+?, obviously it's only half. And assuming 240V mains, the output of a half wave rectifier is 120V RMS.

And it's not changing the mains (unless you have a high current draw on it), it's simply changing what appears on the output of the diode.

Historically TV's used half wave rectifiers, but after a while the electricity companies insisted that they used full wave ones - back in the old two pin plug days it didn't matter, it was random choice if a TV used the positive half cycle, or the negative one, so it balanced out. But once 13A 3 pin plugs were in universal use the VAST majority of TV's were running off the positive half cycles, leaving the negative half cycles unused - and this created a negative bias on the electrical system which the electrical companies didn't like

Nigel, so to be clear? Your Euro 240v 3 pin plug home electrical wall socket service is actually 2 hots and 1 neutral, and NOT 1 hot, 1 neutral, and 1 separate ground, as it is here in the States? We call your type of service here as, 220v single phase power.

 

[Nigel Goodwin]

It still applies. For the raw AC, W for one half cycle is exactly the same as W for the next half cycle, so if you remove alternate half cycles (which is what the half-wave rectification does) you end up with W/2 average.

Except you're mixing up volts and watts - adding a half wave rectifier reduces the RMS VOLTAGE by half, not the wattage. Simple calculations shows that this means the wattage is only a quarter.

 

[Nigel Goodwin]

Nigel, so to be clear? Your Euro 240v 3 pin plug home electrical wall socket service is actually 2 hots and 1 neutral, and NOT 1 hot, 1 neutral, and 1 separate ground, as it is here in the States? We call your type of service here as, 220v single phase power.

You're wrong in every respect, the UK doesn't use the inferior Euro plugs, and our mains is 240V single phase to domestic premises (split off from 3 phase, with alternate houses on alternate phases). This balancing allows the use of a thinner neutral wire in the distribution system, as it reduces neutral current greatly (to zero if perfectly balanced).

The UK 13A sockets used are live, neutral and ground.

 

[Pommie]

As this totally confuses me I set it up in LTspace using 240V AC (looks like LTspice uses peak voltage) and a 240R resistor. As expected the current through the resistor is a sign wave with a peak of 1A. Adding a diode just takes away half the current and leaves the peak unchanged. So, it's got to be half the power.

Mike.

 

[alec_t]

adding a half wave rectifier reduces the RMS VOLTAGE by half

No, it doesn't. The result is 0.707 times the original.

 

[Pommie]

No, it doesn't. The result is 0.707 times the original.

OK, confused again. Why isn't it simply 50%? How does the on cycle vary because there is an off cycle?

Mike.

 

[alec_t]

OK, confused again.

Don't be. The on half-cycle is unaffected. The off half-cycle reduces the long-term RMS voltage and power. The 0.707 figure (half the square root of 2) and the posted sim relate to long-term RMS voltage, not power. Long-term average power gets halved.

 

[Pommie]

Don't be. The on half-cycle is unaffected. The off half-cycle reduces the long-term RMS voltage and power. The 0.707 figure (half the square root of 2) and the posted sim relate to long-term RMS voltage, not power. Long-term average power gets halved.

Yay, happy again. And maybe had a little too much wine. Time for bed.

Mike.

 

[MTJB]

It still applies. For the raw AC, W for one half cycle is exactly the same as W for the next half cycle, so if you remove alternate half cycles (which is what the half-wave rectification does) you end up with W/2 average.

Considering a resistor as the load here, wattage is measured as the amount of heat, (work and/or horse power), produced or emitted from the load. With the half wave rectifier inline there is NO heating going on during the periods of no pulses; so there IS 1/2 less wattage than with either the full wave rectified signal that has the same amount of pulses as the full AC sine wave. The full AC sine wave, also has double the voltage in the circuit, because the voltage is measured top to bottom of any waveform, that is applied across the load. In this case (for the full 240v AC sine wave), 120v for the top + swinging side of the waveform, and 120v for the - swinging side of the waveform. (+)120v + (-)120v= 240v total. Where as, in the half wave or full wave rectified waveforms only swing + 120v from the "0" reference line. (Disregarding whether we are measuring the voltage at the peak, average, or RMS voltage level.)
So, for the half wave rectified signal or waveform. There is half the heat or wattage produced at the load, (the foot heating pads in this discussion), because there is half the voltage (120v) of the 240v input, and there is also only half of that lower 120v voltage heating going on at the load because, there are 50% less, or half of the "on" time periods or pulses to produce that heat. So, 1/2 of 1/2, equals 1/4 of the heat or wattage produced at the load.
As per ohm's law. P= (E x E) ÷ R, and/or P = (V x V ) ÷ R, where as "P" equals power in watts.
P= watts
E or V= voltage
R= load resistance

Also, these pictorial waveform drawings can portray either, the voltage or current swings in the circuit, because you need both voltage and current in a circuit to produce power. P= E x I.
ie. P (watts)= E (volts) x I (current/amps)

 

[MTJB]

You're wrong in every respect, the UK doesn't use the inferior Euro plugs, and our mains is 240V single phase to domestic premises (split off from 3 phase, with alternate houses on alternate phases). This balancing allows the use of a thinner neutral wire in the distribution system, as it reduces neutral current greatly (to zero if perfectly balanced).

The UK 13A sockets used are live, neutral and ground.

So that means 480v approx. or slightly less (416v) across two different phase hot or "live" line wires?

 

[Nigel Goodwin]

So that means 480v approx. or slightly less (416v) across two different phase hot or "live" line wires?

The usual figure is given as 415V, and is what is supplied to industrial users. Where I used to work we had a small goods hoist/lift and that rather bizarrely used three phase, even though it wasn't a very large motor. Interesting though because different parts of the building used different phases, and if you lost power in part of the building you could tell if was an external or internal problem by trying the lift - if one phase was lost it just buzzed, so you knew the problem was external. The service department where I used to work had three phase coming in, but only a single phase was used - presumably they had to dig outside to run the cable, so it made sense to run a 3 phase one 'just in case'.

 

[Nigel Goodwin]

No, it doesn't. The result is 0.707 times the original.

0.707 is simply the multiplication factor for peak to RMS (divide by 1,414), where 0.3535 is the figure for peak-to-peak to RMS (divide by 2.828).

 

[alec_t]

0.707 is simply the multiplication factor for peak to RMS

Indeed it is.

 

[Visitor]

So Nigel maintains that operating a light bulb from 220V AC RMS would be twice as bright as operating the same bulb from full-wave rectified DC with the same RMS voltage because the peak-peak voltage is twice as large as the 0-peak voltage? Owww. That makes my head hurt.

If you see the ridiculous of that statement (RMS level <> RMS level), then it must follow that you'll get half the energy for a half-wave rectified voltage, because a half-wave rectified voltage has half the RMS level of a full wave rectified voltage.

 

[MTJB]

The usual figure is given as 415V, and is what is supplied to industrial users. Where I used to work we had a small goods hoist/lift and that rather bizarrely used three phase, even though it wasn't a very large motor. Interesting though because different parts of the building used different phases, and if you lost power in part of the building you could tell if was an external or internal problem by trying the lift - if one phase was lost it just buzzed, so you knew the problem was external. The service department where I used to work had three phase coming in, but only a single phase was used - presumably they had to dig outside to run the cable, so it made sense to run a 3 phase one 'just in case'.

Roger that! I use to work for my dad in his machine shop business as the shop forman, before I went into electronics, and the building had both single phase and 3 phase 220v service. When we added 3 single phase air conditioning units to the building a few years later, it was more convenient to wire them into the 3 phase copper buss bars we had installed in the building to power all of the heavier machinery; instead of running individual single phase lines. We had to wire them in a fashion so as not to upset the balance of the 3 phase, or some times certain motors would not start, or blow a fuse or breaker. So I am very familiar with the BUZZ! There were also other single phase machinery wired into the buss bars too.

 

[MTJB]

So Nigel maintains that operating a light bulb from 220V AC RMS would be twice as bright as operating the same bulb from full-wave rectified DC with the same RMS voltage because the peak-peak voltage is twice as large as the 0-peak voltage? Owww. That makes my head hurt.

If you see the ridiculous of that statement (RMS level <> RMS level), then it must follow that you'll get half the energy for a half-wave rectified voltage, because a half-wave rectified voltage has half the RMS level of a full wave rectified voltage.

That depends upon whether you are talking about having either a full wave (2 diode) rectifier circuit that is connected to a center tapped transformer input supply, or using a full wave (4 diode) bridge rectifier circuit with the same AC voltage potential at it's input. The full wave bridge rectifier circuit would have twice the DC voltage on it's output compared to the center tapped full wave rectifier circuit version.

 

[Nigel Goodwin]

So Nigel maintains that operating a light bulb from 220V AC RMS would be twice as bright as operating the same bulb from full-wave rectified DC with the same RMS voltage because the peak-peak voltage is twice as large as the 0-peak voltage? Owww. That makes my head hurt.

If you see the ridiculous of that statement (RMS level <> RMS level), then it must follow that you'll get half the energy for a half-wave rectified voltage, because a half-wave rectified voltage has half the RMS level of a full wave rectified voltage.

Perhaps you ought to try reading the posts?, which all refer to a half wave rectifier (single diode), which is this thread is used for alow power setting on small heating elements.

Obviously for a full wave rectifier the RMS voltage is the same as no rectifier (less the insignificant diode voltage drop of course).

And your 'conclusion' is similarly confused, as you're mixing VOLTAGE and POWER, which seems to be the common error through out this thread. Half the voltage means quarter of the power, it's hardly difficult - apply ohms law!.

I must admit I'm absolutely amazed that there's any confusion over simple basic electrics (not even electronics).

 

[rjenkinsgb]

And your 'conclusion' is similarly confused, as you're mixing VOLTAGE and POWER, which seems to be the common error through out this thread. Half the voltage means quarter of the power, it's hardly difficult - apply ohms law!.

I must admit I'm absolutely amazed that there's any confusion over simple basic electrics (not even electronics).

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.

240V is 240V RMS - adding the diode makes it 120V RMS (only using alternate half cycles), and halving the voltage quarters the power.

You're confusing volts and watts.

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 temperatures].

 

[Nigel Goodwin]

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?

Again, you've leapt to POWER instead of VOLTAGE - adding the diode halves the voltage. Stick a rectifier in circuit and measure the RMS voltage out, preferably with a load - 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.

The long term average over many cycles (or indeed one) is then 50% of full VOLTAGE - can anyone argue that?.

 

[rjenkinsgb]

Again, you've leapt to POWER instead of VOLTAGE

No.

Just think of the waveform - it alternates each half cycle with the same (full) voltage or no voltage.
When it's the same (full) voltage, the current (and therefore power) for each half cycle is unchanged, whether it's one individual half cycle or many.

The power transfer per half cycle on a fixed load is a constant, in other words.

eg. With a 240V AC 50Hz supply [so 100 half cycles per second] and a 100W nominal power resistive load (non inductive and zero temperature coefficient), the power (energy) transfer will be one joule per half cycle.

That does not change just because some other half cycle periods are at zero volts - but it would have to be a different for your argument to be true!

The duty cycle is changed, not the instantaneous power.

(The long term average voltage will read as 50% but power is instantaneous square law, with a PWM type on-off voltage, so different from the average voltage power equivalent).