star882

Would the infamous diode trick to make light bulbs last longer cause the meter to get confused and show no power flow?
For the newbies who never heard of it, the trick is to take an IN5406 diode and connect it in series with the light bulb.
If so, then charge a capacitor instead of lighting a bulb and use the high voltage DC to run a switch mode power supply.

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thec

Hehe, never heard of this trick. How's it working with the lightbulbs? ... I mean, why do they last longer?... I understand that they'll only get the positive part of the DC wave but that really saves lightbulbs? .. Wouldn't you notice them flash a lot more then?

Tell us if you progress any further :-)

//Albert "thec" Sandberg

Phasor

My understanding of light bulbs is that they generally only blow when you switch them on at the peak of the AC wave. I guess the diode cuts out half the peaks, and therefore reduces the chances of blowing a bulb by 50%.

thec

Hmm. thinking about it... that's the way it usually go yes... so mayby there's some truth :-)

Perhaps I should buy some diods and try it out :-)

//Albert "thec" Sandberg

james220557

Hi all,

For what I can expect, a diode is series should apply nothing than a positive 25Hz (or 30Hz) half wave accross light bulb. The positive half wave amplitude should be 320 Vp (or 160 Vp). So, I'm expecting a reduced light with some flickering effects (due to poor retina effects persistence at frequecies less than 50Hz).

About the "Free energy" effects... I don't know. The only reason the meter cannot see energy is when the "cos pi" or phase angle is much less of 1. In fact meter counts just the real power and When cos pi goes to 0, that is a big reactive load exist over unbalance line (for instance, like motors with wrong phasing capacitors), the meter cannot counts a real power flow. This is a condition to strongly avoid and in most of country is illegal.

In this case, however I couldn't understande why the meter got confuse, as diode and light bulb are resistive loads, so (assuming low the line inductance) no phase angle exist. The reducing power is just due to reduced voltage accross the bulb.

To make longers the life of bulbs they are other methods. As Phasor has said the only reason the bulb blows is when switched on the peak of voltage (where the power in max). Electronic switches with zero crossing trigger, generally guarantee a long life to our light bulbs.

james

mechie

Surely the diode will allow only half-waves to the bulb (positive or negative -- its irrelevant) but the frequency will remain at 50 or 60 Hz, only the waveshape will change so RMS = 0.707 * Peak-to-peak will no longer apply.

Assuming we are using an INCANDESCENT :idea: lamp as fuorescent would get really upset with this trick due in part to the inductor recieving harmonics way above design frequency!

The bulb will only benefit from half (approx) its normal power and will therefore be DIMMER, a small decrease in supply voltage will give a disproportionate increase in bulb life (a 10% drop giving some 50% life extension as a guide).

A diode will either pass or block the current flow and so dissipate very little power, allowing it to stay cool - a resistor trying to accomplish the same thing will dissipate the same power as the bulb (30 Watts?) and so be physically big and HOT.

As for your domestic wattmeter being tricked - this is basically the same switching technique that every triac dimmer uses; the meter won't care in the slightest - for such simple electro-mechanical devices these meters are remarkably reliable; I'm sure the local electricity companies would have done something years ago if they thought their meters were that easy to trick.

For free power try (expensive) solar energy or (smelly) biomass.

Phasor

As an employee of Australia's largest electricity distributor, I have to agree with mechie, that this "trick" should not affect the meter one little bit. And I would be willing to attempt the experiment, so I can prove it wrong.