Understanding electricity in the TPU.

[gyulasun]

@MrMag

you wrote:

"RMS voltage is .707 of the peak to peak voltage."

Sorry, this is not so.  The correct answer is RMS voltage is .707 of the peak voltage (not peak to peak, only peak). I agree with the rest you wrote.

The same mistake is from GK: he also wrote one 'peak' more, so his answer correctly is:
"A good rule of thumb is r.m.s. of 63% of the AC peak to get the rippled DC."

I can fully accept what teslaalset wrote. 120V AC mains voltage is meant its RMS value and this gives an equivalent 120V DC voltage.
The 91.2V DC voltage measured at the output should give less brightness from a bulb when compared to the 120V AC brightness,  83.4% if you take the 120V AC brightness to be 100%.

rgds,  Gyula

PS: here is a link on comparing RMS, peak, peak to peak and average values, go down nearly to bottom:
http://www.allaboutcircuits.com/vol_2/chpt_1/3.html

[e2matrix]

For what it's worth I used to have some small washer like stick on diodes made for putting on light bulbs (the base of the bulb so it would fit between the center contact on the bulb and the one in the socket).  They ran the regular tungsten bulbs at a light level that was probably 60-80% of the brightness they had on AC.  These were single diodes. 
  wattsup,  it sounds like you are having some interesting results.  I look forward to hearing more!

[gyulasun]

....
I know that when you have a DC voltage output, that the AC equivalent is always lower but I don't know what the formula is.

You see, when SM plugged his bulbs into the wall socket to show them lighting off the house mains, he was getting 110-120 volts AC on those bulbs and the bulbs are supposed to light up to their maximum ability based on that mains power input. He then plugged them onto the OTPU that supplies 91.2 volts DC. So here are some of my questions

1) What does 91.2 volts DC equal in AC terms.

2) If the answer to #1 is less AC then what is available from the mains supply, should the bulb light up less or more then when it was plugged into the mains supply versus when it was plugged onto the OTPU.

3) When the second  bulb was connected to the OTPU the voltage reading of 91.2 vdc did not even budge by a 0.01 vdc. As if the OTPU was impervious (or invisible) to the applied load. Is this normal?

What I would like to do is plug a 60 watts bulb into the mains supply and take a picture. Then set-up a circuit that will provide either 91.2 vdc or the AC equivalent then plug a bulb and take another picture. Then I will compare these two pictures with image grabs I have of the OTPU bulbs that I have included below.

So my next question is what is the best way to set this up. I have AC transformers, bridge rectifiers and the like, but would need a circuit diagram from anyone in the know to show if I need any caps or resistors, etc.
...

Hi wattsup,

The answer for your puzzle in bold above is the peak value of the AC voltage, if you rectify 120V AC with a diode bridge and use a puffer capacitor at the bridge output, then the unloaded DC voltage output is 120V*1.41=169.2V so if you would use only 85.1V AC input to the diode bridge (either from a transformer or from a Variac) then you would get 120V DC voltage across the puffer capacitor in the unloaded case.  (I did not consider the bridge diodes forward voltage drops (minimum 2*.7V=1.4V in case of Si diodes).  And when you introduce any load across the puffer cap, the losses appear right on, AC voltage drop due to the transformer or Variac copper resistances and some forward DC voltage increase across the diodes, the heavier the load, the bigger these losses become of course, so that you have to turn up the Vriac with a few volts more above 85V to get an exact 120V DC output across the load (like a bulb).

I found a schematic on this at random on the web so that no need for drawing: http://www.arabteam2000-forum.com/uploads/monthly_08_2009/post-132119-1250699337.gif

The AC generator symbol can represent either a Variac or a 1:1 mains transformer, both would be the best. The Variac could connect directly to the 120V mains voltage of the wall, the Variac output could be connected to the primary coil of a 1:1 mains transformer and the diode bridge could be connected to the secondary output coil of the transformer.
For the puffer capacitor, you could use as much cap values as you may have at hand, the working voltage rating of the cap should be a minimum of 200V to handle the peak 161V or so and try to use at least a 330 or 470uF electrolytic or even more if you happen to have more to connect them in parallel. This way the ripple voltage can be kept at a minimum and your DC voltage will be 'cleaner' when the 60W bulb shines from the 120V DC output.

A notice on your question 3:  indeed a bit strange the output voltage did not drop down to say 91.1 from the 91.2V: it all depends on the THICKNESS of the output coil wire SM used, if it was a high enough gauge so that the copper resistance was in the milliOhm range, then the voltage drop could have only be measured by 4 or 5 digit DMM resolution, so it could be normal what SM showed with the 3 digit DMM.

rgds,  Gyula

[giantkiller]

'Pressing one wrong button can precipitate our end'. LOL
Thanks and sorry. The fingers fly faster than the mind sometimes.

@MrMag

you wrote:

"RMS voltage is .707 of the peak to peak voltage."

Sorry, this is not so.  The correct answer is RMS voltage is .707 of the peak voltage (not peak to peak, only peak). I agree with the rest you wrote.

The same mistake is from GK: he also wrote one 'peak' more, so his answer correctly is:
"A good rule of thumb is r.m.s. of 63% of the AC peak to get the rippled DC."

I can fully accept what teslaalset wrote. 120V AC mains voltage is meant its RMS value and this gives an equivalent 120V DC voltage.
The 91.2V DC voltage measured at the output should give less brightness from a bulb when compared to the 120V AC brightness,  83.4% if you take the 120V AC brightness to be 100%.

rgds,  Gyula

PS: here is a link on comparing RMS, peak, peak to peak and average values, go down nearly to bottom:
http://www.allaboutcircuits.com/vol_2/chpt_1/3.html

[gyulasun]

Hi Loner,

Well, the regulated output may make sense and can explain why the output did not drop for an additional load, besides my heavy wire hint above. Elsewhere in the forum member 'bolt' mentioned magnetic amplifier in connection with tpu if I am not mistaken and such circuit can easily include feedback/control means IF this is the case indeed.

The filament in a bulb can indeed be considered as a coil and although I have never measured any bulb's self inductance yet, I estimate it has quite a low value, from some hundred nanoHenry to a few microHenry maximum, depending on the make and the power rate of the bulb.  So at 50-60Hz it is negligible and with increasing frequency it surely increases its inductive reactance, though at 5kHz it may still be relatively low: if it has ,say, 3uH filament inductance together with the inner wire connections, the inductive reactance 2Pi*5000*.000003= .094 Ohm and would have 1.8 Ohm at 100kHz, this latter could already modify results.

So these are valid details and it is good we speak and consider them, to ease confusion;  (maybe you could tune the inductive reactance out with a similar reactance value capacitor if your frequency is steadily as high as 100kHz).

rgds, Gyula