[simpits-tech] electrical ??
Chris Woodul
simpits-tech@simpits.org
Mon, 16 Sep 2002 19:09:22 -0500
Here is my solution for all electrical problems:
Call-Email-Invite Matt Wietlespach down to Texas!
I Put the F-111 on a custom base stand and had to remove all the control
boxes that power and drive everything( 115 400 hz, 28 volt, and 5 Volt MEGA
AMPS) and all the special boxes that Matt built.
The F-111 Crewstation has nine million edge-lit panels and internally
lighted instruments. It takes a bunch of 5volt power supplies to drive them
and cooling fans to keep them and all the power supplies in happy working
order.
I need Matts help to reconnect everything and not fry myself or any of the
goodies.
I did enjoy , howevever, reading all the electrical advice in this thread.
Its nice to have a source to get imediate help from when ever a question
arises.
Thanks,
Chris W
----- Original Message -----
From: "Heidner, Troy W" <theidner@ku.edu>
To: <simpits-tech@simpits.org>
Sent: Monday, September 16, 2002 12:00 PM
Subject: RE: [simpits-tech] electrical ??
> Dave,
>
> I just finished reading all the posts in this thread. I decided that I
have
> a couple of comments that might be helpful and haven't been specifically
> addressed yet. I'm an electrical engineer. I thought I might offer a
> couple of insights.
>
> The current rating of the individual bulb gives you most of the
information
> that you need, but you also need to know if the lighting circuit that the
> bulbs are used in is wired in series, parallel or some combination
thereof.
> Most likely it is wired entirely in parallel. You can pretty much
determine
> that from the facts. You said the supply voltage was 5V and the bulbs are
> rated for 5V. The only way to get 5V to each bulb with a 5V supply is to
> wire them in parallel. That being the case, then your calculation (0.115A
x
> 10 bulbs = 1.15A) is correct.
>
> As a side note, in case you're interested or encounter different wiring,
> here are two other possibilities. If they were wired in series then the
> total current draw would be equal to the current draw of a single bulb
> (0.115A). Sometimes lighting circuits are wired in a combination of
series
> and parallel. It's done sometimes to simplify wiring and sometimes to
meet
> certain voltage requirements. (I used to be an avionics tech in the Air
> Force and I remember some panel lighting like this. Mostly in older
> aircraft.) You can tell if there is any series wired lighting by removing
> single bulbs. If any other bulbs go out when you pull any single bulb,
all
> of those bulbs are on the same series circuit. To find the total current
> draw in a situation like this you need to find the total number of series
> bulb groups and then multiply that number by the current draw of a single
> bulb. Each series group draws the same current as a single bulb. If that
> doesn't make sense I wouldn't worry about it, it sounds like you are
dealing
> with a purely parallel circuit anyway.
>
> I have two other comments that might be helpful. One has to do with
making
> current measurements with a multimeter. I taught electronics for about 5
> years at a community college. I saw more multimeters with blown fuses
from
> incorrectly made current measurements that you would believe. The good
news
> is that all multimeters have fuses to protect them from too much current
in
> a current measurement so you shouldn't be able to hurt it even if you do
it
> wrong (but if after making a measurement you find that you're reading 0
Amps
> all the time you might check the fuse). The current measurement in a
> multimeter is actually a separate circuit than the one for voltage and
> resistance measurements. You can blow the current measurement fuse and
> still make voltage and resistance measurements with your meter. Anyway,
the
> natural tendency when using a meter is to first select the meter function
> and scale that you want and then just put your test probes ACROSS the
> terminals or device that you want to measure the current through. That is
> ***NOT*** the correct way to do it. That IS the correct way to blow a
fuse
> in your meter though, if such is your objective. :-) As Roy stated in a
> reply, you must make current measurements in series with the object you
want
> to measure the current flow in. That means that you actually have break
the
> circuit somewhere and insert the meter IN LINE with the device to be
> measured. Not ACROSS. Usually the easiest way to do that would be at the
> power supply. Just hook one side (it doesn't matter which, let's make it
> the negative side for reference purposes) of the power source up to your
> light circuit as normal. Then hook the red lead of your meter to the
> positive side of the power source and hook your black test lead to the
other
> contact point of your lighting circuit. That should measure the total
> current draw. One last thing to be aware of with current measurements.
> Some meters have very small current measurement capabilities. 200mA is
> common. Your circuit probably draws in the neighborhood of 1.15A or
1150mA.
> Obviously that won't work. Also, most (but not all) multimeters with
> current measurement capabilities over 200mA have a separate test lead
> connection point for higher current measurements. Just make sure you have
> your test lead plugged into the right hole!
>
> Okay last point. (Sorry if this is dragging on.) This has to do with
> resistance measurements and the application of Ohm's Law. First of all,
> bravo to you, and all the other message repliers, everyone has made
correct
> theoretical applications of Ohm's law in all it's forms! But... There is
a
> But... There is an application here that is still incorrect in practice.
> You stated that you measured the resistance of the circuit and it came out
> to be around .5 Ohms. That was probably a correct measurement (or close
> anyway, many multimeters aren't tremendously accurate at resistance ranges
> that small). With that correct measurement you also made a correct
> calculation of 10 amps of current draw. You said that sounded high. Well
> it should sound high. We have already established via specifications and
> circuit deductions that the current draw is more like 1.15 amps. Where is
> the problem here? I'll tell you. The filament resistance of an
> incandescent light bulb goes way up when the light bulb is on because the
> heat of the filament is so much greater. The resistance of a typical low
> voltage incandescent light bulb is around 50 ohms when it's lit. In fact
> from the specifications of this bulb, you can calculate the approximate
> resistance that they should have when lit. According to Ohm's law.
> Resistance = Voltage / Current. From the specs. this works out to be
> 5V/0.115A = 43.5 Ohms. Then when you put ten 43.5 Ohm bulbs in parallel
the
> total circuit resistance becomes 4.35 Ohms. Then using Ohm's law again
> (just to validate) Current = Voltage / Resistance 5V / 4.35 Ohms = 1.15
> Amps. Total current flow. So, that's where the trouble is. Your
> resistance measurement is made with the bulbs off and the resistance is
very
> different with the bulbs on. But don't try to measure the resistance with
> your meter with power applied to the circuit! Multimeters are not
designed
> to be able to make resistance measurements that way.
>
> Well, that's all I have to add. I hope it was helpful. I also hope it
> wasn't too long-winded. I've been accused of that before! ;-) My wife
> says it's the teacher in me. ;-) If you have any other questions that I
> could help with, I'd be happy to answer.
>
> Troy
>
> > -----Original Message-----
> > From: Dave Hensley [mailto:dphens@nc.rr.com]
> > Sent: Saturday, September 14, 2002 1:57 PM
> > To: simpits-tech@simpits.org
> > Subject: Re: [simpits-tech] electrical ??
> >
> >
> > Thanks! I did find this, which may help. On the back of the
> > panel it has
> > "10#7152 LAMPS". I'm assuming this means there are 10 lamps
> > in the panel,
> > part number 7152. I searched on this and found a PDF doc with
> > lamp part
> > numbers.
> >
> > Line Part
> > Filament
> > Life
> >
> > No. No. Volts Amps M.S.C.P. Type
> > Hours
> >
> > 7 7152 5.0 0.115 0.15 C-2R 40,000
> >
> >
> > So, IF that is the lamp being used then could I figure the amps by
> > 0.115*10(number of lamps)=1.15. I'm starting to feel in over my head a
> > little, so I may do some studying and come back to this later. I still
> > appreciate any help you guys can give.
> >
> > Dave
> >
> > ----- Original Message -----
> > From: "Andreas Fransson" <andreas.fransson@post.utfors.se>
> > To: <simpits-tech@simpits.org>
> > Sent: Saturday, September 14, 2002 2:05 PM
> > Subject: Re: [simpits-tech] electrical ??
> >
> >
> > > (disclamer: my high school electrical skills may be a bit rusty...)
> > >
> > > IIRC...
> > >
> > > P=UI (watts = volts x amps)
> > > I=U/R (amps = volts / ohms)
> > > =>
> > > P=U^2/R (watts = volts(square) / ohms)
> > >
> > > You have already concluded that the panel has a 0.5 ohm
> > resistance, and is
> > > rated at 5 volts. This means that it draws 5^2/0.5=50 watts at that
> > voltage.
> > > That seems to me to be a whole lot...
> > >
> > > At 5 volts, the current should be 5/0.5=10 amps...
> > >
> > > Someone please correct me if I have gotten all of this backwards. ;)
> > >
> > > Andreas
> > > http://valhallainc.d2g.com
> > >
> > >
> > > From: "Dave Hensley" <dphens@nc.rr.com>
> > > > I need a little assistance with a power problem. I am trying to
> > determine
> > > > the amount of amps I need going through the panels I have
> > to light them
> > > up.
> > > > The panels are rated for 5vdc. I plugged in a 6vdc
> > adapter which is
> > > putting
> > > > out 7.70v before anything is connected, measured with a
> > mutlimeter. Then
> > I
> > > > connected leads to the adapter and the panel, while
> > leaving the leads
> > for
> > > > the multimeter connected also. The volts in the
> > multimeter dropped to
> > > 4.41v.
> > > > Next I connected only the meter to the panel set to
> > measure resistance
> > and
> > > > had a reading that fluctuated between .4 to .6, so I split the
> > difference
> > > at
> > > > .5. So, using V=I*R formula, 7.70/.5=3.85. So, the lights
> > are pulling
> > > almost
> > > > 4 amps? Am I even close to having that right?
> > > >
> > > > Obviously you guys are seeing that I know very little
> > about this and am
> > in
> > > > the process of learning. I am playing with the panel in
> > an attempt to
> > > > understand the relationship of volts, current and
> > resistance. If am I
> > way
> > > > off at this point, I'll stop and do some more reading. I
> > am just hoping
> > I
> > > am
> > > > on the right track. :-)
> > > >
> > > > Thanks!
> > > >
> > > > Dave
> > >
> > >
> > >
> > > _______________________________________________
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> > >
> >
> >
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