Convert half-wave rectified to AC?

Fri Jul 29, 2011 5:57 pm   

Gentlemen:

I'm an ME, not a EE so be gentle with me.

I have a half-wave rectified signal (half of 120 Vac passed through a
diode), and I would like to convert it to a ~60 Vrms wave that is
clean enough to run a small AC motor.

Is this even possible?

Don

Fri Jul 29, 2011 5:57 pm   

In article <18bb3070-9af7-43ed-b262-
9e34190ba8f5_at_e7g2000vbw.googlegroups.com>,
eromlignod_at_aol.com says...

An Uninterruptible Power Supply from your
local computer store might work.

Not sure how they'd react to pre-rectified
power, but you could always experiment...

Fri Jul 29, 2011 5:57 pm   

On Fri, 29 Jul 2011 08:57:45 -0700, eromlignod wrote:


It'd be much easier if you could reach across that diode and just apply
your 120VAC to a 2:1 step-down transformer.

How small a motor? How much current is available on the half-wave
rectified DC?

I'm not sure that there's an off the shelf device that could do this
(unless the suggestion of a UPS pans out). Filtering the DC and then
converting it to AC may work. How you'd do that without building most of
a UPS, or hacking it, I don't know.

Filter the DC, find a DC-DC converter that can go from that raw DC to
12V, plug the output of that converter into the battery port of a UPS,
plug your motor into that. It's way overkill, but it may work.

Use the DC to actuate a relay, that switches a 2:1 stepdown transformer,
plug the thing into the wall...

If it's really half an AC wave and it stays that way no matter how you
load it, then putting in a honking big blocking cap and applying it
directly to the motor may work. It may not, and if it did it may be more
expensive than custom electronics in quantities more than 10 or so, but
it may work.

--
www.wescottdesign.com

Fri Jul 29, 2011 7:02 pm   

eromlignod wrote:

For a small enough motor, I think it is.
(This would not perhaps be practical for
production.) :)

Push the half wave into a parallel 'tuned tank'
composed of a power transformer primary in parallel
with a lot of 'motor run' capacitors. You have a resonant
transformer that synthesizes the negative half
cycles to create a full A.C. sine wave output
on the secondary.

http://en.wikipedia.org/wiki/LC_circuit
http://www.play-hookey.com/ac_theory/ac_transformers.html

Plug 7.036201 microfarads and 1 Henry
into this calculator to see how close you can
get to 60 Hz:
http://www.calculatoredge.com/new/resonance.htm

--Winston

Fri Jul 29, 2011 8:52 pm   

Winston wrote:

http://www.surpluscenter.com/sort.asp?catname=electric&keyword=EMRC

Fri Jul 29, 2011 10:49 pm   

"eromlignod"


** No

Rectified AC is DC, the ripple frequency is 30Hz.

Run your motor via a suitable cap from the 120V.


..... Phil

Sat Jul 30, 2011 1:02 am   

On Sat, 30 Jul 2011 08:49:30 +1000, "Phil Allison" <phil_a_at_tpg.com.au>
wrote:


Uh, would you please tell us how a diode divides the frequency by two?

John

Sat Jul 30, 2011 1:06 am   

eromlignod expressed precisely :

Wots your REAL Problem?
This sounds like a silly thing to want to do so you will only get silly
answers.
If your starting SIGNAL is only a SIGNAL i.e. no real power, then you
are not going to get anywhere without some form of amplification to get
the power and in that you could generate whatever type of output you
need.

Hypothetical questions are very difficult to answer.
Please tell us your REAL problem. :-?

--
John G.

Sat Jul 30, 2011 1:06 am   

eromlignod wrote:

of the 120v to the motor lead while the other lead is connected to the
common. Mean while as this is taking place, it is also
charging a large cap.. WHen the incoming sine wave drops near 0 volts
you then would have a triggered circuit to switch that bridge so that
now the charged cap is in a reverse manner and thus generating a (-)
signal. The start of the sine wave regenerating from the incoming
will simply shut this switch off and go back to repeat the cycle.

Actually, as I think about it, you wouldn't even need a full bridge, a
half bridge would do it, you only need to short the (+) of the cap to
the common lead of the motor and open the common of the cap from the
common of the system coming in. You can do this because the DIODE you
are using will block the (-) coming in and there for, shorting this
will not cause any unwanted blue smoke!

If you look at charge pump voltage generating circuits, you can
see where opposite polarities of power can be obtained. Many times

Why not just use a step down transformer?

Jamie

Sat Jul 30, 2011 1:09 am   

"John Larkin"
"Phil Allison"


** My bad.

Was thinking about half and full wave rectification.

You cannot connect an AC motor via diode or diode plus cap in series, it
will fail to spin and or burn.



..... Phil

Sat Jul 30, 2011 1:14 am   

"John Git"

** The OP has stated it.




** Read the OP's first line.

He has no idea of the special meaning of the term "signal" in electronics.




** The unwritten rules of usenet require that posters never reveal their
actual purpose.

Cos that would spoil the magic.



..... Phil

Sat Jul 30, 2011 2:17 am   

On 2011-07-29, eromlignod <eromlignod_at_aol.com> wrote:

the functional opposite of a rectifier is an inverter, so yes it's
possible. it's not simple however.





--
⚂⚃ 100% natural

Sat Jul 30, 2011 2:36 am   

On 2011-07-29, eromlignod <eromlignod_at_aol.com> wrote:

Usualy you need an inverter to get AC from DC. someone else suggested
a UPS, Common Uninterruptable power supplies use an inverters to convert DC from
a secondary battery to AC in th event of a power falure.


If your DC supply is also poweing a DC load you may be able to extract
a small amount of AC by puttign a choke in series with the rectifier

||
+---||----------
| ||
| AC load
|
----->|-+--wwwww--+-----
===== |
|
in ===== DC load
|
|
-----------------+-----


The waveform that the AC load sees will have lots of harmonics and
the voltage will vary depending on the size of the DC load and the AC
load

it may be a easier idea change to a motor that can handle DC
(eg BLDC, PM, or universal)

--
⚂⚃ 100% natural

Sat Jul 30, 2011 5:06 am   

Winston wrote:

Notice how the current in the primary becomes
'almost symmetrical' within about half a second
of startup:

Version 4
SHEET 1 880 680
WIRE -64 16 -128 16
WIRE 224 16 0 16
WIRE 448 16 224 16
WIRE -128 80 -128 16
WIRE 656 80 560 80
WIRE 448 96 448 16
WIRE 560 96 560 80
WIRE 224 112 224 16
WIRE 656 112 656 80
WIRE 560 224 560 176
WIRE 656 224 656 192
WIRE 656 224 560 224
WIRE 448 240 448 176
WIRE 560 240 560 224
WIRE -128 256 -128 160
WIRE 224 256 224 176
FLAG 224 256 0
FLAG -128 256 0
FLAG 448 240 0
FLAG 560 240 0
SYMBOL cap 208 112 R0
SYMATTR InstName C2
SYMATTR Value 7.036201µF
SYMBOL ind2 432 80 R0
SYMATTR InstName L1
SYMATTR Value 1
SYMATTR SpiceLine Rser=.01
SYMATTR Type ind

Sat Jul 30, 2011 5:14 am   

Winston wrote:


Version 4
SHEET 1 880 680
WIRE -64 16 -128 16
WIRE 224 16 0 16
WIRE 448 16 224 16
WIRE -128 80 -128 16
WIRE 656 80 560 80
WIRE 448 96 448 16
WIRE 560 96 560 80
WIRE 224 112 224 16
WIRE 656 112 656 80
WIRE 560 224 560 176
WIRE 656 224 656 192
WIRE 656 224 560 224
WIRE 448 240 448 176
WIRE 560 240 560 224
WIRE -128 256 -128 160
WIRE 224 256 224 176
FLAG 224 256 0
FLAG -128 256 0
FLAG 448 240 0
FLAG 560 240 0
SYMBOL cap 208 112 R0
SYMATTR InstName C2
SYMATTR Value 7.036201µF
SYMBOL ind2 432 80 R0
SYMATTR InstName L1
SYMATTR Value 1
SYMATTR SpiceLine Rser=.01
SYMATTR Type ind
SYMBOL voltage -128 64 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value SINE(0 170 60 0 0 0 100000)
SYMBOL diode -64 32 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D1
SYMATTR Value MUR460
SYMBOL ind2 576 192 R180
WINDOW 0 36 80 Left 0
WINDOW 3 36 40 Left 0
SYMATTR InstName L2
SYMATTR Value .2525
SYMATTR SpiceLine Rser=.01
SYMATTR Type ind
SYMBOL res 640 96 R0
SYMATTR InstName R1
SYMATTR Value 600
TEXT -160 472 Left 0 !.tran 0 10 0
TEXT 464 56 Left 0 !K L1 L2 0.99