question from reader

[kc7jty]

A reader asks:
Will a 12 volt automotive incandescent dash light bulb give off as much light if it's powered with 12 v ac (60 hz) as opposed to 12 v dc?
...and if there's a difference, how much?

[kc7jty]

hint:
​rms

[W5RB]

It'll burn with the same brightness. The difference is that since the electrons in AC don't actually go anywhere , and only move back and forth , and thus are not used up , AC is 100% efficient .

[kc7jty]

Q2)
Would the same bulb produce equal light on the 12 v ac voltage as opposed to rectified 12 v?

[kc7jty]

It'll burn with the same brightness. The difference is that since the electrons in AC don't actually go anywhere , and only move back and forth , and thus are not used up , AC is 100% efficient .
this seems to conflict with rms or is that just in a resistor producing heat?
shouldn't the dc produce more heat/light?

[w2amr]

I use DC to power the filaments in several radios, I don't think the tubes even noticed.

[kc7jty]

Nice avatar George.
The neocons seem to be getting genuinely agitated with this demonstration stuff. Let's hope it spreads.
This is how real change will be affected. Not that voting shell game crap.

[w2amr]

Nice avatar George.
The neocons seem to be getting genuinely agitated with this demonstration stuff. Let's hope it spreads.
This is how real change will be affected. Not that voting shell game crap.They may be our last hope.

[K7SGJ]

No. While staying at a Howard Johnsons, I saw a documentary where a princess said Obi Wan Kenobi was the last hope.

[ab1ga]

A reader asks:
Will a 12 volt automotive incandescent dash light bulb give off as much light if it's powered with 12 v ac (60 hz) as opposed to 12 v dc?
...and if there's a difference, how much?

The bulb will dissipate the same amount of power in both cases, so if light output were solely dependent on that, the brightness would be the same.

However, the brightness, i.e. amount of visible light emitted is a highly nonlinear function of filament temperature, which in turn depends on filament current in a nonlinear fashion.

It's not inconceivable that the increased light output at ac peaks could offset the reduced light output at zero crossings. In any case, I'd expect the difference to be small.

[N7YA]

No. While staying at a Howard Johnsons, I saw a documentary where a princess said Obi Wan Kenobi was the last hope.


But instead, a short Italian in a red jumpsuit came and saved here.

[kc7jty]

The bulb will dissipate the same amount of power in both cases, so if light output were solely dependent on that, the brightness would be the same.

However, the brightness, i.e. amount of visible light emitted is a highly nonlinear function of filament temperature, which in turn depends on filament current in a nonlinear fashion.

It's not inconceivable that the increased light output at ac peaks could offset the reduced light output at zero crossings. In any case, I'd expect the difference to be small.

I thought the dc would produce more light although probably not noticeable by the human eye. Ac voltage rises and falls, how can a constant, full, non falling voltage not produce more heat in the filament?

[NQ6U]

I thought the dc would produce more light although probably not noticeable by the human eye. Ac voltage rises and falls, how can a constant, full, non falling voltage not produce more heat in the filament?

I think that the hysteresis of the hot filament would result in very little change in brightness at 60Hz.

[kc7jty]

I think that the hysteresis of the hot filament would result in very little change in brightness at 60Hz.
yes, but that's not the question.

[KC2UGV]

But instead, a short Italian in a red jumpsuit came and saved here.

While Lando borrowed Han's clothes...

http://www.jedi-business.com/images/actionFigures/e5/e5_LandoSmuggler_Big_6.jpg

http://www.explore-science-fiction-movies.com/images/han-solo-anti-heroes.jpg

[W3WN]

No. While staying at a Howard Johnsons, I saw a documentary where a princess said Obi Wan Kenobi was the last hope.Oh? I thought the glowing dead guy told the small green muppet that the impulsive kid flying off was the last hope... and the muppet muttered something about there being another.

Although, on second thought, maybe these weren't the "last hopes" you were looking for.

[N7YA]

While Lando borrowed Han's clothes...

http://www.jedi-business.com/images/actionFigures/e5/e5_LandoSmuggler_Big_6.jpg

http://www.explore-science-fiction-movies.com/images/han-solo-anti-heroes.jpg

I would never have noticed that unless i saw that family guy episode...weird!

[ab1ga]

I thought the dc would produce more light although probably not noticeable by the human eye. Ac voltage rises and falls, how can a constant, full, non falling voltage not produce more heat in the filament?

Here's how my reasoning goes so far:

After applying a constant voltage, the current through the filament, and therefore the power dissipated in the filament, will reach a steady-state value once filament temperature stabilizes. This temperature will be such that the heat lost through radiation, conduction, and convection will just equal the amount of energy dissipated in the filament. Assuming the filament is a blackbody radiator, the emitted spectrum of light will depend only on that temperature, and the radiant flux will depend on the fourth power of the absolute temperature of the filament. The question is just how much of the emitted light is visible: light bulb filaments can reach temperatures of 3000K. At that temperature, the peak radiance is at a wavelength of one micrometer, so the vast bulk of the light radiated is not visible to the human eye.

If we apply a sine wave with an rms value equal to the dc voltage in the previous case, then the thermal environment, because both conduction and convection require time to remove heat away from the filament, but heating is faster, since the energy is deposited within the filament. So for a short time interval, at the peak of the waveform, the applied voltage is about 1.4 times the DC case, doubling the amount of energy deposited within a length of filament. The filament heats up faster than convection can remove the excess heat, so the peak temperature reached by the filament would be higher for a short time. During that time interval, the amount of radiant flux increases by the fourth power of the ratio of temperatures, and more of the light emitted falls in the visible spectrum, which would be perceived as an increase in brightness.

At the lower values of the AC waveform, the question is just how much does the filament cool off during "low power" intervals. Since convection and conduction out of the filament could be relatively slow, the temperature might not drop so much. Even if the temperature did vary uniformly around a central value, the nonlinearity of the blackbody emission means that you gain more brightness on the up side than you lose on the down side.

At least, that's how I've thought it through so far. If my basement weren't piled with junk, I'd try and see if I could actually measure the effect, but the best I can do is add it to the Bucket List. I could, of course, be completely wrong, which would be oddly reassuring because it would be so familiar.

73,

[kc7jty]

If we apply a sine wave with an rms value equal to the dc voltage in the previous case
No! 12 v p to p sinewave.

[ab1ga]

No! 12 v p to p sinewave.

Whoa! DC voltage definitely the winner there!

[WØTKX]

Are we talking 12vac RMS? :chin: If this is correct, the 12vac would be brighter. That would be umm, like maybe 17 volts p-p, right? That's higher than 12 volts, and even though it's just a peak... and won't be there long.

There is a kind of hysterisis in the heating and glowing of the filament. It's not just an electronic problem, because you are dealing with the physical properties of the filament... which heats up and glows, eh?

Now, a "coily" filament might have more impedance to the AC vs a straighter one. It might stay warmer and glow longer between peaks as well. Depending on the design of the filament, it might change the brightness with the AC as well. Guessing it would be a small difference tho.

It's fun to get confuzzed, and try to think outside the "box of the electrical issues"? I'm prolly wrong after all this. :dunno: :spin:

Let's do an experiment Timmy! Gee Mr Wizard, can we!

http://4.bp.blogspot.com/_IoU3bEFUwWc/SjtsUecBDWI/AAAAAAAAFtY/sdJmtNGP6xs/s400/Mr.+Wizard+NBC.jpg

[WØTKX]

Oh, OK. 12v p-p is like maybe 8.5 volts. Well DC then.

So much for my silly tangential rant. :rofl:

http://www.clinicalgaitanalysis.com/art/silywalk.gif

[W5RB]

12v p-p equals about 4.25 v AC RMS . You guys trust your memories too much .

[kc7jty]

Whoa! DC voltage definitely the winner there!
​BINGO!

[WØTKX]

Umm, yea. How the hell did I double it?

[W5RB]

Umm, yea. How the hell did I double it?

Your calculation was valid for peak voltage , not peak-to-peak . For those of us out of practice on such things , its an easy error . I might've done the same thing , but I've had to refresh on that stuff since starting to teach an electronics / test prep course here .

[KC2UGV]

12v p-p equals about 4.25 v AC RMS . You guys trust your memories too much .

RMS is root mean square ( y=x sin (2 pi f) ), right?

[ab1ga]

In general, to calculate the rms value of a function (voltage, current, etc.) you perform the equivalent of the following steps:

1. Square the function, with the interesting consequence that it is only the magnitude of the value which is relevant, not the sign.

2. Calculate the average value of the squared function over the interval of interest, from t to t + dt. For time-limited signals, the interval is the duration of the signal, for periodic signals, it's the duration of a single period. Generally, this is done by integrating the squared function from t to t+dt, and dividing that value by dt.

3. Take the square root of the integrated value, and you're done.

When you think about (1) above, you realize that a sine wave and a pulsating dc waveform made up of a series of positive half-sine waves have the same rms value.
Also, you realize that the rms value of a waveform is at most equal to the peak value of the waveform, and this only happens when the waveform has a constant value, i.e. dc. Since 12V pk-pk is 6V peak, you know the rms value has to be less than 6V, which is a handy sanity check.

The original post was also a bit confusing, since when AC voltages are stated, the convention is to assume rms value unless otherwise specified. Power line voltage is called "120V" AC, but the peak value is about 170V.

[K7SGJ]

Are we talking 12vac RMS? :chin: If this is correct, the 12vac would be brighter. That would be umm, like maybe 17 volts p-p, right? That's higher than 12 volts, and even though it's just a peak... and won't be there long.

There is a kind of hysterisis in the heating and glowing of the filament. It's not just an electronic problem, because you are dealing with the physical properties of the filament... which heats up and glows, eh?

Now, a "coily" filament might have more impedance to the AC vs a straighter one. It might stay warmer and glow longer between peaks as well. Depending on the design of the filament, it might change the brightness with the AC as well. Guessing it would be a small difference tho.

It's fun to get confuzzed, and try to think outside the "box of the electrical issues"? I'm prolly wrong after all this. :dunno: :spin:

Let's do an experiment Timmy! Gee Mr Wizard, can we!

http://4.bp.blogspot.com/_IoU3bEFUwWc/SjtsUecBDWI/AAAAAAAAFtY/sdJmtNGP6xs/s400/Mr.+Wizard+NBC.jpg

Don Herbert ruled on Sat mornings.

[n2ize]

In general, to calculate the rms value of a function (voltage, current, etc.) you perform the equivalent of the following steps:

1. Square the function, with the interesting consequence that it is only the magnitude of the value which is relevant, not the sign.

2. Calculate the average value of the squared function over the interval of interest, from t to t + dt. For time-limited signals, the interval is the duration of the signal, for periodic signals, it's the duration of a single period. Generally, this is done by integrating the squared function from t to t+dt, and dividing that value by dt.

3. Take the square root of the integrated value, and you're done.

When you think about (1) above, you realize that a sine wave and a pulsating dc waveform made up of a series of positive half-sine waves have the same rms value.
Also, you realize that the rms value of a waveform is at most equal to the peak value of the waveform, and this only happens when the waveform has a constant value, i.e. dc. Since 12V pk-pk is 6V peak, you know the rms value has to be less than 6V, which is a handy sanity check.

The original post was also a bit confusing, since when AC voltages are stated, the convention is to assume rms value unless otherwise specified. Power line voltage is called "120V" AC, but the peak value is about 170V.

That's right !!! :)

[n2ize]

RMS is root mean square ( y=x sin (2 pi f) ), right?

RMS is Richard Stallman !!;)

[kc7jty]

4748

remenber this one?

[n2ize]

RMS is root mean square ( y=x sin (2 pi f) ), right?

You forgot the time variable...

y = xsin(2(pi)ft)

RMS is a mathematical average in which we take the square root of the average of a set of values which have been squared. It can be computed over any set of discrete set of values or, over any continuum, such as a sine wave. In the case of a continuum, such as a sine waveform we use a technique called integration to determine the average (RMS) over the continuum (waveform). The formula above is the voltage at a given time for a sine wave of constant amplitude. If we square it and integrate it with respect to time (t) over a cycle i.e, from 0 to pi (as the wave goes from 0 volts to peak and back to zero) and then take the square root we get the formula for RMS in terms of ampllitude which simplifies to

RMS = X/sqrt(2) or approx RMS = 0.7071(X) where X is the amplitude of the sine wave form.

So for a sine wave of 170 V peak amplitude we have (0.7071)(170 V) = 120V(rms). For a 12 V peak we get (0.7071)(12) = 8.48 V. If we are talking about peak-peak voltage just divide by 2 so 12v P2P = 8.48/2 = 4.24 V

Bear in mind this ONLY applies to sine wave forms. For square, sawtooth or other complex wave forms this does not hold and we have to go back to the original process of averaging and integrating the values over the waveform to get a formula or a suitable approximation. Or use the formulas already derived for at least some of those wave forms.