Hi I am wondering for what the relation is of the power decrease if you cap a sinus through a transformer.

a transformer can only transform energy if the field is moving.

so if you cap the sinus for example with one third from the peak than the output peak will also be 1/3 smaller but the time that the field is moving halved so the energy transformation is than also halfed??

if i have a transformer that is 220V/36V 250VA and i cap the sinus with roughly 1/3 to 150V.

the output voltage will than be about 24 a 25 V but is the power rating than 166.6 VA, or is the power rating 125VA because the time that the field is changing is only half than original?

second question is how do i achief this caping of the sinus with the netpower

what i did draw, I know that that is probably not the way I should do it.


what i have in mind is that if i have 1 type of transformer
for 2 situations.

First situation is having 3 lamps of 12V in serie and with reducing the voltage i would have a dimm option as over the 3 lights only 24V fals.

second situation is if i want to conect only 2 lamps of 12V in series, 2/3 of the 36 volt will make 24 volt

first i thought to solve this with a dimmer but that will result in the second situation that the peak will always be 36V and it would blow the lights

can sombody give me some insight on this?

Robert-Jan

 

I forgot to post the drawing sorry

Robert-Jan

Attached Images

TMP15.JPG (148.8 KB, 22 views)

 

Ok the 2 situations discribed in my earlyer post are solved

a can use a dimmer that works with cuting the phase of the sinus the only thing that i have to do is rectify the power on the secondery side of the transformer and flaten the ripple with capacitors, that will give me a stable and smooth (smooth enough for a lamp) voltage

still i want to know what the relation is of caping the sinus on the power output you cut 1/3 of the voltage but it takes 1/2 of the time for the full sinus

Robert-jan

 

Hi RJ.
I do not fully understand you question.

In the earlier drawing with antiparallel diodes and series resistor, any clamping of the waveform would depend upon the source impedance of the driving voltage.
As the mains impedance is very low, the sine wave would not be clamped.

EDIT:
I am refering to the connections to the transformer on the rightside of the drawing.
The voltage across the resistor would be clipped.

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/

 

What is "caping"?

Each zener diode has a forward biased diode across it so its clamping voltage is only 0.7V, not 75V.
Then the resistor in series with the diodes "shorts" the transformer and wastes a lot of power.

__________________
Uncle $crooge

 

as i said in my post the way i draw it is probably not the way to do it

at this time it's from lesser importance how to achieve this form

as i thougt it could be tha solution on the 2 situations but that one is tackeled

caping (I would not know another term for it )

is cap (put a limit on the max voltage) the top of the sinus at a sertain level

the drawing should give you the idea what i mean

still the theoretical question is what impact does it have if you have a wave form like this?


Robert-Jan

 

hi RJ,
I would using 'clipping' or 'clamping'.

The antiparallel zeners are going to have about 0.7V drop across them.
Even though a zener has a 'zener voltage' when a positive voltage is applied to the cathode, when a postive voltage is applied to the anode it behaves as a normal diode.

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/