Convert half-wave rectified to AC?

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...

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

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.

Plug 7.036201 microfarads and 1 Henry into this calculator to see how close you can get to 60 Hz:

--Winston

"eromlignod"

** No

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

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

.... Phil

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

John

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.

I suppose you could construct a bridge that would pass the ON time 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

"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

"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

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

--
?? 100% natural

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

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

Rather:

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

One transformer, one capacitor.

That's pretty simple. :)

--Winston

(...)

Efficiency. Add a tuning cap to the primary of that transformer and input power falls by about 50%.

--Winston

I can think of a magic black box that would work. The front-end would have just the right size fuze across the input, and the first thing it would do is blow out his half-wave rectifier, shorted.

John

The second rule is that they post the question and never return, leaving the regulars to debate the issues for some weeks.

John

Actually, you can use a diode to generate subharmonics:

ftp://66.117.156.8/SubHarmonic.gif

John