Message-ID: <39794957.6C7ED709@regandesigns.com>
Date: Sat, 22 Jul 2000 00:12:23 -0700
From: "Hamid A. Wasti" <hamid@regandesigns.com>
To: lancair.list@olsusa.com
Subject: Re: Diode Voltage Drop
References: <FDD045B6B83FD311A9880090273053E2CF0B17@CDIM-PTS-MAIL>
Reply-To: lancair.list@olsusa.com
Mime-Version: 1.0

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"Rumburg, William" wrote:

>         Wow, am I getting a lot of flak regarding my "Keep It Simple" design
> for electrical backup power!

Actually a lot of people are pointing out that your system is so simple that it
will not work as you are expecting it to.

> My VMS 1000 indicates a steady-state charging
> system voltage of 14.0 V DC.

What is the accuracy specification on the VMS?  When it says "14.0V" the voltage
can be somewhat lower or somewhat higher.  By how much?  Could it be that the
14.0V that the VMS is reporting is actually 13.8V? 13.5V? or even 13.0V?  In
normal operation when the charging voltage can be anywhere within a 0.5V range,
a 0.2V uncertainty in the charging voltage is not an issue.  But when you start
cutting it close, all of this little stuff starts becoming relevant.

Next, where is the VMS measuring the 14.0V?  Is it at the output of the
alternator?  At the main battery?  At the main bus?  With the magnitude of
currents being shuttled around, I would not be surprised to see 0.5V or more
voltage difference between these various points.  The location of the ground
lead for the measurement is also important.  If you measure the voltage between
different "ground" locations in your airplane, you may find several 100mV of
voltage difference.

> I didn't bother to look up a curve for the
> voltage drop across a silicon diode vs. current, just assumed a worse case
> of 0.6 V DC.

Lets rephrase the above for the real world:  "... just assumed a best case of
0.6VDC"  If you are using a silicon diode, 0.6V is almost your the best case.
In real life, you may be looking at something close to or over 1V for silicon
diodes.  It depends on the particular diode you are using and anything that I or
anyone else has to say is just a guess.  Look at the specs.

As Bob Jude pointed out, at lower currents a diode will drop less than 0.6V.
The I/V (voltage vs. current) curve has a knee with a very sharp slope to the
left and an almost flat region to the right.  To the left of the knee, the diode
behaves almost like a large resistor while to the right, it looks like a
constant voltage with a small series resistor.  It is my opinion (and I disagree
with Bob on his opinion) that in order to charge a battery, the current required
would be high enough that you will be well past the knee, in the constant
voltage region.  Once again, neither of our opinions matter -- look at the specs
for the battery and the diode actually being used to get the current and voltage
numbers.

> So, at worst, the charging voltage [14.0 - 0.6] = 13.4 V DC
> (minimum) to the backup battery is borderline for maintaning full charge.

That will be true: IF you are actually getting at least 14.0V to the diode; IF
the diode is dropping only 0.6V; IF there is no appreciable voltage drop in the
cable between the diode and the battery; IF there is no appreciable voltage drop
in the ground path between the battery and the alternator.  That is a lot of big
IFs that need to be looked into.  But assuming all your numbers are correct and
the battery is actually seeing 13.5V, how much charge will it accumulate at the
lower voltage?  Is it 90%? 75%? or could it be 50%?  Is the manufacturer able to
provide those numbers?

>         It wouldn't be difficult to periodically check the backup battery's
> charge state.

And how do you propose to do that?  Switching over to the backup system for a
few minutes only tells you that the battery is healthy enough to operate the
system for a couple of minutes.  It does not tell you anything about how long
the battery will run the system.

The only way I can think of to truly test the system is to start operating on
the backup battery and then let it run down under normal load and see how long
that takes.  Ofcourse, then you are leaving yourself without a backup for as
long as it takes your charging system to charge the battery to whatever state it
manages to charge the battery to.

Then there is the little issue that Jim Frantz pointed out.  Diodes tend to fail
shorted rather than open.  If your diode fails, there goes your backup too.  If
you need your backup system, it is because your primary system died.  Why did it
die?  If it was due to a catastrophic event, it is quite possible that the same
event would also take out your diode.

The bottom line, Bill, is that many people are trying to point out that in
trying to keep it simple, you may have ignored a few issues that can not be
ignored or overlooked.

Hamid

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