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'[EE] Re: : Modified sine timing?'
2005\10\02@124413 by

On 10/2/05, Philip Pemberton <philpemdsl.pipex.com> wrote:
> Hi,
>   Has anyone got any info on the timing specs for modified sine waves? I'm
> using a 6-step (Z/1/Z/Z/0/Z) modified sine to drive my homebrew transformer
> and it doesn't seem to be running very well. Each step is 3.333uS, BTW
> (6*3.333 = 20mS = 50Hz).
>   Am I doing something wrong here?

Hi Philip

Your pseudo sinusoidal signal it's a variable amplitude rectangular
one (multiple rectangular steps with different amplitude and constant
step size ?) or other sort of signal with constant amplitude like PWM
?

cheers,
Vasile

>
> Thanks.
> --
> Phil.                              | Acorn RiscPC600 SA220 64MB+6GB
> 100baseT
> philpemphilpem.me.uk              | Athlon64 3200+ A8VDeluxe R2
> 512MB+100GB
> http://www.philpem.me.uk/          | Panasonic CF-25 Mk.2 Toughbook
> Wanted: Heisenberg. Reason uncertain.
> -

In message <5eeda4c20510020944u53a561dcgd8da86737fe2772email.gmail.com>>          Vasile Surducan <piclist9gmail.com> wrote:

> Your pseudo sinusoidal signal it's a variable amplitude rectangular
> one (multiple rectangular steps with different amplitude and constant
> step size ?) or other sort of signal with constant amplitude like PWM
> ?

It looks like this on a scope:
+    ___
|   |
0 __|   |__     __
|   |
-          |___|

I think I've worked out the timing - the 0V segments last for t/6, the + and
- pulses last for t/3. "t" taken to mean "time for one cycle"

What I'm trying to work out is a way to find the resonant point of the
transformer. The usual method is to feed a sine/square wave to the primary
and watch the secondary on an oscilloscope. The problem being I have a scope
but no signal generator. I was thinking something along the lines of a PIC
with internal ADC, a diode/capacitor filter, then some code to sweep the
entire frequency range and store the timing figures that produce the highest
voltage. Which means I also need to find some resistors to build a voltage
divider for the A/D input, and maybe a voltage reference chip. This is going
to take some planning.

I can get 7Vish out of the transformer with normal square wave driving, but
the transformer is as noisy as hell when I do that. Using a modified square
wave gets the noise down, but the voltage goes through the floor.

The transformer primary is 270 turns of 28SWG Enamelled Copper wire on an RM
core, secondary is 150 turns on the same RM core. Design frequency was 50Hz,
though that produced a rather lame 2V output. It should - in theory - work up
to 500mA. Of course, theory almost never matches real life.

What I'm trying to do is convert 12V DC into 6.3V RMS AC or 6.3V DC, with a
10% tolerance (about 5.6 to 7V) for a CRT heater. One end of the heater is
connected to the cathode, which floats at around -600V (IIRC). Much fun.

Later.
--
Phil.                              | Acorn RiscPC600 SA220 64MB+6GB 100baseT
philpemphilpem.me.uk              | Athlon64 3200+ A8VDeluxe R2 512MB+100GB
http://www.philpem.me.uk/          | Panasonic CF-25 Mk.2 Toughbook
... 24 hours in a day, 24 beers in a case, Hmmm.....

part 1 1058 bytes content-type:text/plain;Not directly answering your question but here is an 12V to 5V
DC-DC converter. You can change the post regulator to get 6V.
The transformer is a simple ring transformer with ratio of
76:76:50 (can be adjusted). This is one of the cheapest
isolated DC-DC converter. The self-oscillating circuit can
be replaced by a CD4047 based circuit as well.

I've done some design with similar circuits. The CD4047 based
design is very low power. I have done one design with the
whole isolated DC-DC converter and a 16LF872 based level
sensor with less than 1mA power consumption when inactive.
It was a tough design though.

Regards,
Xiaofan

-----Original Message-----
From: Philip Pemberton [philpemdsl.pipex.com]
Sent: Monday, October 03, 2005 3:40 AM
To: piclistmit.edu
Subject: Re: [EE] Re: : Modified sine timing?
...
What I'm trying to do is convert 12V DC into 6.3V RMS AC or 6.3V DC, with a
10% tolerance (about 5.6 to 7V) for a CRT heater. One end of the heater is
connected to the cathode, which floats at around -600V (IIRC). Much fun.

Later.

part 2 8722 bytes content-type:image/gif; (decode)

part 3 35 bytes content-type:text/plain; charset="us-ascii"
(decoded 7bit)

On 10/2/05, Philip Pemberton <philpemdsl.pipex.com> wrote:
> In message <5eeda4c20510020944u53a561dcgd8da86737fe2772email.gmail.com>> >           Vasile Surducan <piclist9gmail.com> wrote:
>
> > Your pseudo sinusoidal signal it's a variable amplitude rectangular
> > one (multiple rectangular steps with different amplitude and constant
> > step size ?) or other sort of signal with constant amplitude like PWM
> > ?
>
> It looks like this on a scope:
> +    ___
>     |   |
> 0 __|   |__     __
>            |   |
> -          |___|

This ASCII looks like hell, but I think I've understood, you have
three different voltage levels counting your 0 level too.

> I think I've worked out the timing - the 0V segments last for t/6, the +
> and
> - pulses last for t/3. "t" taken to mean "time for one cycle"
>
> What I'm trying to work out is a way to find the resonant point of the
> transformer.

You are on the right direction.

The usual method is to feed a sine/square wave to the primary

Unfortunately a square signal on input will generate a lot of noise on
the output. The noise is proportional with the signal speed variation
(the time of the rising and falling edges) Sometime (usually) there
are also multi-resonant spikes on both rising and falling edges of the
input square signal.

{Quote hidden}

Phil, the simplest way for getting RMS is to use a simple square
driving signal and to found the secondary transformer resonance.
You'll get a perfect sinusoidal signal ( would be harmonics also but
much smaller ).

or 6.3V DC

OF COURSE, this is much simple: Square input, square output, fast
rectifier (two diodes) a capacitive filter and an inductive dual
filter (C+differential inductance+C)
This will kill most of the switching spikes to a low level ripple bellow 10 mV.

, with a
> 10% tolerance (about 5.6 to 7V) for a CRT heater. One end of the heater is
> connected to the cathode, which floats at around -600V (IIRC). Much fun.

Absolutely no problem, use a rectangular switching transformer on ferite core..
Do you remember the Faraday law isn't it ? You don't need more.

If you need smalles noise (I have doubts, usually 100mV ripple on
heater is excellent) then you need a voltage/current slew rate
controlled driver. Take a look for example at LT1533/LT1534 family.

cheers,
Vasile

2005\10\03@133105 by
Philip,

An RM core isn't designed for operation at 50Hz.  You probably can't get
enough turns on the core to make it work there.  Try 50kHz, you'll have
better luck there (and you'll need less turns).

Dave

Philip Pemberton wrote:

{Quote hidden}

Yes the circuit I attached in the previous post works
at about 50KHz. The transformer is using a small ring
core and use solid isolation (>=1mm plastic between
the primary and secondary) to achieve reinforce (double)
insulation. Therefore the efficiency can not be very
high (at about 70%) but it is good for normal low power
use.

Regards,
Xiaofan

-----Original Message-----
From: David Minkler [minkluxtron.com]
Sent: Tuesday, October 04, 2005 1:32 AM
To: Microcontroller discussion list - Public.
Subject: Re: [EE] Re: : Modified sine timing?

Philip,

An RM core isn't designed for operation at 50Hz.  You probably can't get
enough turns on the core to make it work there.  Try 50kHz, you'll have
better luck there (and you'll need less turns).

Dave

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