[sci.electronics] Running a Mac/SE from batteries and inverter

murray@jumbo.dec.com (Hal Murray) (04/04/89)

I'm thinking about taking my Mac out into the field where there aren't any
power lines. Has anybody run one from batteries?

A friend has scrounged an inverter from an old ambulance. It says "square
wave output". I haven't put it on a scope yet. The Macintosh Family
Hardware Reference says the SE needs 85-135 V (rms), 47-73 Hz. I don't see
anything about the shape of the waveform.

I'd assume that the vanilla line => bridge => big caps power supply front
end would be happy with a squarish wave as long as it didn't brown out.
This type of supply normally only uses a small fraction of each half
cycle near the peak. I can't reverse engineer a power supply and convert
from rms to peak and guess the waveform from the inverter and ... all at
the same time.

Anybody know the answer and/or where to look/ask for more info?. Trial and
error could be expensive.

PS: Anybody know if the fan runs on DC or AC?

Bernard.Aboba@f444.n204.z1.FIDONET.ORG (Bernard Aboba) (04/04/89)

My suspicion is that you will not be able to run a Mac SE for long off of 
your inverter.  The SE will probably consume as much as 1.5-2 amps, 
particularly if it has a hard drive in it.  Depending on your battery 
life, you will probably have to consider a means of turning the SE on and 
off.
 
I'm assuming you're looking to operate the SE remotely, rather than for 
personal use, but there is a device called a Mr. MOX, which will turn on 
a PC in response to a phone call. Then one can use a cellular phone modem 
(again run off the battery) to do this.  It's kind of expensive, but it 
does work fairly well.



--  
INTERNET:  bmug!User.Name@apple.COM UUCP: apple!bmug!User.Name

postmaster@mailcom.FIDONET.ORG (Bernard Aboba) (04/04/89)

My suspicion is that you will not be able to run a Mac SE for long off of 
your inverter.  The SE will probably consume as much as 1.5-2 amps, 
particularly if it has a hard drive in it.  Depending on your battery 
life, you will probably have to consider a means of turning the SE on and 
off.
 
I'm assuming you're looking to operate the SE remotely, rather than for 
personal use, but there is a device called a Mr. MOX, which will turn on 
a PC in response to a phone call. Then one can use a cellular phone modem 
(again run off the battery) to do this.  It's kind of expensive, but it 
does work fairly well.



--  
Sometimes when I've got the blues,
And woo the shape I'm in,
At least I'm not
A tubby crooner with a voice of tin.
== From "Elvis didn't die(t) in vain"
Via  apple!mailcom, Jailhouse Rock BBS, Fido 1:204/444

jimc@iscuva.ISCS.COM (Jim Cathey) (04/06/89)

In article <13671@jumbo.dec.com> murray@jumbo.dec.com (Hal Murray) writes:
>I'd assume that the vanilla line => bridge => big caps power supply front
>end would be happy with a squarish wave as long as it didn't brown out.

Switching power supplies might very well object to square(ish) waves as input.
The problem is inrush current through the diodes and into the capacitors.
This can cause premature failure on both components.  If you do need to feed
the supply this nasty current, you could probably put some sort of series 
resistor in the power feed to tame the waveshape a little.  Unfortunately
this beastie will waste some power.

If you're into hacking, what the Mac supply really wants is 180 VDC (a guess, 
easy enough to measure it) at its internal power supply point.  The switcher
runs off of this.  In fact, you might be able to feed the Mac DC directly.
I can't remember all the details, although I did trace out the supply once.

If this is so, all you need to do is think up a rectifier scheme that _will_
survive the square waves, and save your poor Mac the trouble!  Peripherals
are another question.

+----------------+
! II      CCCCCC !  Jim Cathey
! II  SSSSCC     !  ISC Systems Corp.
! II      CC     !  TAF-C8;  Spokane, WA  99220
! IISSSS  CC     !  UUCP: uunet!iscuva!jimc
! II      CCCCCC !  (509) 927-5757
+----------------+
			"With excitement like this, who is needing enemas?"

jeffw@midas.STS.TEK.COM (Jeff Winslow) (04/08/89)

In article <2446@iscuva.ISCS.COM> jimc@iscuva.ISCS.COM (Jim Cathey) writes:

>Switching power supplies might very well object to square(ish) waves as input.
>The problem is inrush current through the diodes and into the capacitors.
>This can cause premature failure on both components. 

What problem?   :-)

First of all, if you really mean inrush current as in what you get when you
first turn the switch on: Since switching supplies don't try to synchronize
the application of line power to the capacitors with a zero-crossing (at least,
the cheap ones don't), you could just as easily be turning on the switch at
the peak of the sinewave as near the zero - more likely in fact, considering
the shape of the sine curve, especially after it's been chopped off by the
load of all those *other* switching supplies out there. For the same peak
input voltage, which is what switchers care about, the peak inrush current
is the same for sine or square.

Secondly, rms running current will be lower for a square wave, since the
conduction angle is much greater than for a sine wave, even the chopped-off
version. The rectifiers will run cooler, last longer. The capacitors will,
too - lower rms current -> lower core temperature, and the single most telling
variable affecting electrolytic capacitor life (in normal use :-)) is
temperature.

						Jeff Winslow

wordy@cup.portal.com (Steven K Roberts) (04/08/89)

The inverter you want is from Statpower -- 100 W continuous, 200 peak
from any 12V source... very clean.  Works fine with Mac SE, Tek scopes,
etc...

the company is at 7012 Lougheed Hwy, Burnaby, B.C., Canada V5A 1W2
604-420-1585.  Ask for Cynthia.

I'm using one on my bicycle from solar power....

Cheers!
  Steve Roberts
  Computing Across America

jimc@iscuva.ISCS.COM (Jim Cathey) (04/11/89)

In article <4224@midas.STS.TEK.COM> jeffw@midas.STS.TEK.COM (Jeff Winslow) writes:
>First of all, if you really mean inrush current as in what you get when you
>first turn the switch on: Since switching supplies don't try to synchronize

Bad choice of words on my part.  I meant to refer to the peak current through
the diode during each cycle.  This will be worse for the squarish wave because
if the capacitors sag during the dead part of the input cycle the voltage
differential across the diode will be much higher (or at least it'll try to
be).  This can result in hot spots in the diode, leading to premature failure.


>Secondly, rms running current will be lower for a square wave, since the
>conduction angle is much greater than for a sine wave, even the chopped-off
>version. The rectifiers will run cooler, last longer. The capacitors will,
>too - lower rms current -> lower core temperature, and the single most telling
>variable affecting electrolytic capacitor life (in normal use :-)) is
>temperature.

I disagree, the conduction angle may well be larger, but the _peak_ current
will be very high in the early part of the cycle.  Thus higher RMS current,
and the exact opposite results.

Or maybe the math says otherwise.  I'm just quoting what the data sponge in
my head says it saw several years ago.  Original source was a power supply
design article (quoth the sponge).  Any real experts care to comment?

+----------------+
! II      CCCCCC !  Jim Cathey
! II  SSSSCC     !  ISC Systems Corp.
! II      CC     !  TAF-C8;  Spokane, WA  99220
! IISSSS  CC     !  UUCP: uunet!iscuva!jimc
! II      CCCCCC !  (509) 927-5757
+----------------+
			"With excitement like this, who is needing enemas?"

jeffw@midas.STS.TEK.COM (Jeff Winslow) (04/12/89)

Question is, will a typical off-line switcher's input rectifiers and 
capacitors have a reduced life for square-wave input compared to sine input?

In article <2452@iscuva.ISCS.COM> jimc@iscuva.ISCS.COM (Jim Cathey) writes:
>In article <4224@midas.STS.TEK.COM> jeffw@midas.STS.TEK.COM (Jeff Winslow) writes:

>>Secondly, rms running current will be lower for a square wave, since the
>>conduction angle is much greater than for a sine wave, even the chopped-off
>>version. The rectifiers will run cooler, last longer. The capacitors too...

>I disagree, the conduction angle may well be larger, but the _peak_ current
>will be very high in the early part of the cycle.  Thus higher RMS current,
>and the exact opposite results.

I agree that the peak current will be higher, but I don't think it will be
*enough* higher for long enough to counteract the reducing effect of the
wider conduction angle, especially when you consider that the inverter is
going to have some non-zero output impedance. But I honestly couldn't tell
you for sure without doing some measurement, simulation, or equation-bashing.

If the square does turn out to be worse, one could experiment with putting
an inductor bridged with a power resistor between the inverter and the
switcher to knock off that initial peak.

>Or maybe the math says otherwise.  I'm just quoting what the data sponge in
>my head says it saw several years ago.  Original source was a power supply
>design article (quoth the sponge).  Any real experts care to comment?

Well, I designed a few of the buggers. Doesn't that count? :-)
I hope the article wasn't from Power Conversion International or its
precursor, Solid-State Powerconversion. I remember seeing some rather
amusing things in there while I was subscribing to it. Unintentionally
amusing, I mean.

						Jeff Winslow

strong@tc.fluke.COM (Norm Strong) (04/12/89)

In article <2452@iscuva.ISCS.COM> jimc@iscuva.ISCS.COM (Jim Cathey) writes:
}In article <4224@midas.STS.TEK.COM> jeffw@midas.STS.TEK.COM (Jeff Winslow) writes:
}>First of all, if you really mean inrush current as in what you get when you
}>first turn the switch on: Since switching supplies don't try to synchronize
}
}Bad choice of words on my part.  I meant to refer to the peak current through
}the diode during each cycle.  This will be worse for the squarish wave because
}if the capacitors sag during the dead part of the input cycle the voltage
}differential across the diode will be much higher (or at least it'll try to
}be).  This can result in hot spots in the diode, leading to premature failure.
}
}
}>Secondly, rms running current will be lower for a square wave, since the
}>conduction angle is much greater than for a sine wave, even the chopped-off
}>version. The rectifiers will run cooler, last longer. The capacitors will,
}>too - lower rms current -> lower core temperature, and the single most telling
}>variable affecting electrolytic capacitor life (in normal use :-)) is
}>temperature.
}
}I disagree, the conduction angle may well be larger, but the _peak_ current
}will be very high in the early part of the cycle.  Thus higher RMS current,
}and the exact opposite results.
}

I assume this is a full-wave inverter.  If so, the peak voltage is the same as
the averge; the capacitor is quite small; the peak rectifier current is the
same as the average; there is no "sag" in the capacitor voltage.  

If you don't believe this, draw a picture of a full wave rectified square
wave.  
-- 

Norm   (strong@tc.fluke.com)

jeffw@midas.STS.TEK.COM (Jeff Winslow) (04/14/89)

In article <7687@fluke.COM> strong@tc.fluke.COM (Norm Strong) writes:

>I assume this is a full-wave inverter.  If so, the peak voltage is the same as
>the averge; the capacitor is quite small; the peak rectifier current is the
>same as the average; there is no "sag" in the capacitor voltage.  

Duh - me stupid. Of course. Actually, the capacitor is still big, since I
assume the Mac/SE's switcher is designed to run off of sinewaves, but there's
still no sag, and little if any peak on the input current.

						Jeff Winslow