"raypsi" wrote in message ...
On Dec 4, 9:26 pm, "NoSPAM" wrote:
"Stev eH"
StevehkhhSDJvhvbjjxbvvbhnvbhzjdnxzvzhzdshbvnjzvnb vnvjbvbcjbvvvvnmxvzjhjzsdgfgsfghgjsghgsljhglhdjfgh fufgfhzysgfhczgugfvzlvufzvllgfzlyfyvlgbylvdfghvbly
wrote in ...
My first amp had 3 PL519s with a lethal power supply using a voltage
tripp{l}er direct off the 240V mains, heaters were direct off mains via a
1n5406 diode. It's still up in the parents loft somewhere, must dig it
out over the holidays and give it a try....
Steve H
Do you realize that the RMS value of half-wave rectified AC is not one half
of the applied RMS voltage but rather 0.707 of the applied RMS voltage. So
your three 40 volt filament PL519s actually saw about 56 volts RMS across
each filament.
Hey OM
I beg to differ. As any casual observer can see that peek is 330V and
the average would be 0.363 of the peak so that's 40 volts in my
calculator all day long, for a 3 tube string.
Lets take a square wave, the average would be 0.5 of peek, far lower
than 0.707 you are used to using. Even if it was full wave 120 cycle
dc, the average would only be 0.636 still lower than 0.707.
I tell you what. Take a diode and rectify the mains with it and
measure the DC voltage with a DVM, VTVM or a VOM and 0.363 of peak is
what you will get for a DC voltage on that meter. Or do the real
integral of the dv/dt and see for yourself.
The TV repair business I was in manufacturers put 50volt peak on a CRT
6.3volt filament, the average voltage was 6.3 volts. Lets see you RMS
that one.
73 OM
n8zu
Ray,
You obviously do not understand the meaning of RMS as you are confusing the average voltage with the RMS voltage. In a resistive circuit, the RMS voltage of any waveform will produce identical _heating_ as a DC voltage of the same value. This is an important point which is often overlooked, and it is why measuring the RMS value of a waveform is difficult. Let's start with some numbers but avoiding the use of integration...
Assume we have an AC voltage described by:
Vac = 339.41 sin[(2*Pi*f)t] where (2*Pi*f) is 376.99 for 60 Hz or 314.159 for 50 Hz power
Vac = 339.41 sin(376.99*t)
Since this is a symmetrical waveform, the _average_ value is 0 while the _peak_ positive voltage is 339.41 volts.
Now let us apply this voltage to a 1000 ohm resistor. Using Ohm's
Law, we can calculate the current, I=V/R.
Iac = 339.41 sin(376.99*t) volts/1000 ohms
Iac = 0.33941 sin(376.99*t) amps
Again this a symmetrical waveform with an average of 0, and a peak positive current of 0.33941 amps.
The instantaneous power dissipated by this resistor can be calculated by multiplying the voltage by the current.
Pac = 339.41 sin(376.99*t) * 0.33941 sin(376.99*t) watts
Pac = 115.20 [sin(376.99*t) * sin(376.99*t)] watts
Using a trigonometric identity (
http://en.wikipedia.org/wiki/List_of...ric_identities) this can be rearranged to:
Pac = 115.20 * [1 - cos(2*376.99*t)]/2 watts
It is important to understand this function. The value of this function goes from zero to a peak of 115.20 watts at twice the frequency of the original waveform. The average value is no longer zero but is now 57.6 watts.
Now let us go back to using RMS values. Since the waveform is sinusoidal, we can use some common formulas.
Vrms = 0.707107 * Vpeak = 0.707107 * 339.41 = 240 volts
Irms = 240/1000 = 0.240 amps
Prms = 240 * 0.24 = 57.6 watts
This is exactly the same as the average of the power calculated above. If we had applied 240 volts DC to the 1000 ohm resistor, the power would again be 57.6 watts.
Now let us half-wave rectify the original AC waveform. The peak voltage will be the same as before, 339.41 volts. But the waveform is no longer symmetrical about zero. The average value is:
Vavg = Vpeak/Pi = 339.41/3.14159 = 108.038 volts
The RMS value is:
Vrms = Vpeak/2 = 169.706 volts
Applying this to the same 1000 ohm resistor gives a current of
Irms = Vrms/1000 = 0.169706 amps
The RMS power dissipated by the resistor is
Prms = Vrms * Irms = 28.8 watts
Note that this exactly _twice_ the power if the waveform had not been rectified.
Now let us do the same calculations for a symmetrical square wave.
Vpeak = 339.41 volts
Vavg = 0 volts
Vrms = 339.41 volts
Ipeak = 0.33941 amps
Iavg = 0 amps
Irms = 0.33941 amps
Prms = 115.2 watts
If the square wave is half wave rectified,
Vpeak = 339.41 volts
Vavg = 169.71 volts
Vrms = 169.71 volts
Ipeak = 0.33941 amps
Iavg = 0.16971 amps
Irms = 0.16971 amps
Prms = 28.80 watts
Note that this is exactly the same as with the sinusoidal waveform. This is exactly the point, half wave rectification produces _one_half_of_the_power_ that would be produced by the original sinusoidal AC waveform. It does not produce half of the original peak voltage. It produces SQRT(2)/2 times the original RMS voltage.
As for your numbers, Ray, you need a refresher course in mathematics. As for DVM, VTVM, or VOM readings, you will have to specify the filtering of the waveform before I can tell you the value. In the case of the CRT filament
(and I really think you might mean the HV rectifier filament) power is supplied by a waveform having a high peak value but an RMS value that is low. Again you would need to specify the waveform exactly.
There is a very simple way to prove that my calculations are right. Get two incandescent light bulbs of the same type and wattage. Wire a diode of adequate voltage and current rating in series with one bulb and connect to the wall outlet. Use a variac to feed the second bulb. With the bulbs side by side, adjust the variac until both bulbs are the same brightness. Measure the variac output voltage. You will find that it is not 50% of the line voltage but instead very close to 70.7% of the line voltage.
In fact, the most accurate RMS voltmeters operate in a similar fashion. They let the unknown voltage heat a resistor. Then they use an adjustable DC voltage to heat an identical resistor. When both resistors are at the same temperature, the DC voltage is the RMS value of the unknown voltage. Consult the Linear Technology LT1088
datasheet for a description of one method of doing this.
http://www.datasheetcatalog.org/data...ogy/lt1088.pdf
The same company also offers several other RMS-to-DC converters that operate on a delta-sigma computation technique.
http://www.linear.com/pc/downloadDoc...54,C1086,D4486
A general discussion of RMS measurements may be found at:
http://www.edn.com/contents/images/46857.pdf.
This last reference shows typical errors for different types of RMS measuring meters. I strongly suggest to anyone still confused to read it. And I really hope Phil tries the experiment I suggested.
73, Barry L. Ornitz, PhD WA4VZQ