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Electrically challenged

825 views 23 replies 2 participants last post by  **DONOTDELETE** 
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#1 ·
Well since it's about 8 degrees outside now and my CJ is stored for a long winter nap I thought I would try to work on the console I've been playing with. I need to figure out some resister formulas and I can't follow the basic electronics book I have. Do any of you know any good electronics hobby bbs's?
Thanks
p.s.- When I'm finished I'll post the schematics.

'83 CJ7 258 i6 31x10.5 3in. lift? /wwwthreads_images/icons/cool.gif
 
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#2 ·
go ahead and post your questions, others probably have the same questions on electrical issues. email me if you like,
big ed

/wwwthreads_images/icons/cool.gif Big Ed
'88 YJ, 4" susp,3" body,35's,283 Chevy V8,TH350,4.11's,D30,D35c
 
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#3 ·
What are you having problems with exactly? If you have any questions, just drop me a line and I can give you a hand /wwwthreads_images/icons/smile.gif I am an Electronics tech for the navy and am currently teaching Systems engineering at the Naval Academy (in other words I access to all of the Prof's and the people in my own community, probally a coupla thousand years of collective experience all together).

I do not know if these will help, I seem to always fall back on them when I forget things. /wwwthreads_images/icons/smile.gif

http://www.tpub.com/neets/
Oh well good luck.

John
Damn the Mud!! Full speed ahead!!!
http://www.Marylandmudbugs.com
 
#4 ·
I am an Electronics tech for the navy and am currently teaching Systems engineering at the Naval Academy

SQUID SQUID SQUID! Heh heh heh, do you know how many times we stole the goat? (Navy's mascot?) Oh what fun the rivalry was, I still have all my old BEAT NAVY paraphernalia. Working on a Navy Base is a little lax up here, no one gets into the spirit, kinda depressing.

USMA '92
Go Army, Beat Navy!

(All in good fun!)

JEEPN
Winter Harbor, Maine
'81 CJ-8 Scrambled!
It's a Jeep, Chevy, IHC kinda thing!
 
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#5 ·
Yeah well you did ya know that "Bill" the goat originally belonged to the ARMY. Anyways we got a winning season this year (lost every game BUT the NAVY-army game)/wwwthreads_images/icons/smile.gif

Boy ya shoulda seen the parties going on then /wwwthreads_images/icons/wink.gif

GO NAVY

John
Damn the Mud!! Full speed ahead!!!
http://www.Marylandmudbugs.com
 
#6 ·
Now that's what I'm talking about! Actually, Navy just assumed we owned the goat, we had it more than they did and it likes us better.

I have to say, that's one problem with Army, they always seem to screw up the A/N game, although I think the record is tied isn't it? Ah, no matter, I don't attend them, too far away and I don't know anyone down there anymore.

Thanks for the ribbing...all in good fun/wwwthreads_images/icons/smile.gif

JEEPN
Winter Harbor, Maine
'81 CJ-8 Scrambled!
It's a Jeep, Chevy, IHC kinda thing!
 
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#7 ·
I should have known you guys could figure out any problem. Here's the nuts and bolts of it. I would also like to learn the formula so I can plug different numbers and get correct info. I want to draw off the CJ's 12 volts and use resisters to bring the voltage down to 3v and 6v (or any other voltage). I have been playing around with a 6v battery and I noticed I can bring it down to 5v with four 1k Ohm resistors. The odd thing is it doesn't seem to make a big difference between the 1k Ohm resistors and the 100 Ohm resistors. Any ideas?
Thanks

'83 CJ7 258 i6 31x10.5 3in. lift? /wwwthreads_images/icons/cool.gif
 
#8 ·
Sorry, I got us off track.

What you're looking for is Ohm's Law (for DC current). The formula is:

Current in Amps = Voltage (Volts)/Resistance (Ohms) = Power (Watts)/Voltage (Volts).

Therefore by manipulating the equation, you can get different scenarios.

Voltage = Current x Resistance
Voltage = Power/Current
Voltage = (Power x resistance)^0.5

Power = Current^2 x Resistance
Power = Voltage x Current
Power = Voltage^2 / Resistance

Resistance = Voltage/Current
Resistance = Power/Current

You also asked about Resistors in series:

Total resistance = Resistance1 + resistance2 + resistance3 + (you get the idea)

Two resistors in parallel:

Total Resistance = (resistance1 x resistance2)/(resistance1 + resistance2)

If you need any other formulas, I have a bunch of Physics links bookmarked, as well as many books of equations and data. Ask and ye shall receive.

JEEPN
Winter Harbor, Maine
'81 CJ-8 Scrambled!
It's a Jeep, Chevy, IHC kinda thing!
 
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#9 ·
What are you trying to power with that 6v and 3v?
Have you thought about using a transformer with a tapoffs.
One of the big problems I have with using resistors to cause a voltage drop is Heat. Depending an the amount of current you put though them,you could possiblly burn the resistors up(or in Technical terms "let the smoke out")/wwwthreads_images/icons/wink.gif and anything else in the area.
post your diagram and well try to give ya hand.

Whadda ya mean "lick test" the Maggie/wwwthreads_images/icons/smile.gif

John
Damn the Mud!! Full speed ahead!!!
http://www.Marylandmudbugs.com
 
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#10 ·
O.k. I know the theory but I'm not sure how I can put it into the real world. I made this spreadsheet to hold some of the formulas I've been playing with. One is Ohm's Law. Let's say I have a 6v battery and I want to bring it down with resistors to use a 3v light what type of resistors do I need? I attached a simple diagram, please feel free to draw on it and send it back.
Thanks for all the help.

'83 CJ7 258 i6 31x10.5 3in. lift? /wwwthreads_images/icons/cool.gif
 
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#12 ·
Here's a half done diagram of what I want to do. More importantly I want to understand the theories. I want to be able to understand the formula to lower the voltage for any project I'm working on. Not just this one. Thanks for the help.
Hey, what's a tapoff? And what does it do?

'83 CJ7 258 i6 31x10.5 3in. lift? /wwwthreads_images/icons/cool.gif
 

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#13 ·
The eqation you need is called the voltage divider law and is as fallows v = V* r/(R+r). Where V is the input voltage, v is the voltage you want, r is the resistence of the part in question, and R is the resistence you needed to put in serice with the part.

In your case V = 12, and v =3, r can be found by turning the part on and using your ohm meter across the power terminals. solving the voltage divider law for R we get R = 3*r. so you will need a resister of 3*r placed in line with the part.

12 v R 3 v
----------------/\/\/\-----------!
!
!
!
ground --------------/\/\/\------!
r

I hope this helps. I have a picture but can't find how to attach it.

bandhmo

 
#14 ·
Holy shinola jeep head, is that a bomb! /wwwthreads_images/icons/smile.gif
Looks like a detailed project.
Save Jeepn's formulas, they will come in handy. Also, band's voltage divider is correct. IMPORTANT, whenever you put a resistor in series with a device (in your example the one that is requiring 3V) the same amout of current that the will flow through the device will flow through your resistor. If the resistor is not rated at the correct power lever (watts) the resistor will get hot and blow. Voltage dividers are very inefficient btw! They are quick and dirty. If I need to step something down I used a linear regulator with a variable output, a whole 'nother discussion.

Example,
say you have a light bulb that has R = 5 ohms but for some reason it can only operate at 5 volts, if you hooked a 5V battery to it you would have 1A flowing, I=V/R=5/5=1. You have a 10 volt battery, and want to step it down to 5V so you put a r = 5 ohms resistor in series with the light , you do indeed have 5V at the light now [using the voltage divider eqn, v = V*r(r+R)=10*5/(5+5)=5V, i.e. anytime the resistances are of the same value you cut the voltage in half]. Now lets see how much current is running in the system. I = V/R = 10 / (5+5) = 1A The current is the same but the voltage source has doubled! Where is the extra power being absorbed, the external resistor r! So the r has to be rated at greater than P=IV = 1*10 = 10W.
anyhow, hope I was somewhat clear, mostly just babbling though!!
big ed

/wwwthreads_images/icons/cool.gif Big Ed
'88 YJ, 4" susp,3" body,35's,283 Chevy V8,TH350,4.11's,D30,D35c
 
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#15 ·
Wow, you guys never cease to amaze me with the stuff you know. Bandhmo, that was exactly the formula I was looking for. I really appreciate all the answers I got.
V8JEEP, you are hilarious, a bomb, I laughed out loud. I couldn't make a bomb, I would have the wrong resistors and the switch would burn out. So, in your scenario where the resistor needs to be rated for 10watts, if I used 2 resistors does the watt rating get cut in half?

'83 CJ7 258 i6 31x10.5 3in. lift? /wwwthreads_images/icons/cool.gif
 
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#16 ·
Sir,
I'm 1/c at the Academy and I saw your post. Haven't seen a post from anyone at the academy before. I was just wondering who specifically you were, and where your office was. Also, how often and where you go wheeling at. Thanks for your time,
Very Respectfully
MIDN 1/c Jarrod Groves

 
#18 ·
Yep, if two 5watt resistors are twisted in parallel and then put in series to the line, they could handle 10watts total. Remember that if the voltage divider eqn called for a 200 ohm R capable of 10W, you'd have to twist up two 400ohm of 5W rating.
I have an unfair advantage at this stuff since I do it for a living! I'm an elect engineer for a semiconductor company so it stays fresh in my head. I still think your schematic is for a bomb though!
big ed

/wwwthreads_images/icons/cool.gif Big Ed
'88 YJ, 4" susp,3" body,35's,283 Chevy V8,TH350,4.11's,D30,D35c
 
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#19 ·
No bomb, the only thing that goes boom is my Jeep. Someone suggested a "tapoff". What's that and how does it work? Someone else was telling me to use a regulator. How do these work.
Again, thanks for all the help.

'83 CJ7 258 i6 31x10.5 3in. lift? /wwwthreads_images/icons/cool.gif
 
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#20 ·
O.k. I wiped this formula up according to the info you gave me and I'm confused like crazy. I attached my MS Excel spreadsheet. Can you guys look at this and tell me if I'm doing anything wrong? According to the formula as r (resistance of the part in question) goes up so does R (Resistance you needed to put in service with the part). This makes no sense to me. I would think as one resistance goes up the other would go down?
p.s.- To add an attachment to a post in this BBS. After you write the message you wish to post, check off "I want to preview my post and/or attach a file". Then click on the Continue button.
Thanks

'83 CJ7 258 i6 31x10.5 3in. lift? /wwwthreads_images/icons/cool.gif
 

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#21 ·
Sorry about the pore drawing last time here is a better one.

As some have said the voltage divider is wasteful. But at low currents (high resistence) the waste is minimal. LED draw very little current and in this case the voltage divider is ideal.

I can't make this thing do subscripts so I will write V sub L for V subscript L. Look on the pict for where these values are found.

You are correct in that as one R subscript L goes up R sub. 1 goes up (see attached). This is due to ohm's law V=IR. The voltage drop is in the voltage divider is dependent on the ratio of the R sub. L and R sub 1. To conferm that this is true let me do an example.

lets let R sub L be 10 ohms. Then by the formula (attached and in my last post (R sub 1 equals R sub L times three)) We will get a R sub 1 of 30 ohms.

Now using ohms law lets find the current the circuit we will add the values of the two resisters (resisters in series add). We will now have a total resistence of 30 + 10 =40 ohms. Now appling ohms law (V=IR) and solving for current I we get (I=V/R). putting in numbers we get I=12/40 =.3 amps. Now to find the voltage in the LED we will again use ohm's law V sub L equals I times R sub L. Putting in numbers we get V sub L = .3*10=3 volts. just what we wanted. tri this from the begining for different values of R sub L and it will always give you 3 volts.

I hope this helps I will send a better picture when I can get to a scaner.

bandhmo.

 

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#23 ·
I was suggesting using a transformer. The "tapoffs" I mentioned are just the points on the secondary coil where you "tap off" the voltage you are trying to get.

http://www.tpub.com/neets/book2/5h.htm

here is a link to the exact page with the formulas for figuring oout how many windings you need for the secondary and primary /wwwthreads_images/icons/wink.gif that is if you Ever feel like making your own transformers. Personnally the only time I have ever had to do that though is in the middle of the south atlantic (no radio shack nearby).

Midn 1/C Groves
I am in Rickover 54
ext 2920
if you would like to conversate on jeeping. /wwwthreads_images/icons/smile.gif

good luck and have a FINE NAVY DAY

John
Damn the Mud!! Full speed ahead!!!
http://www.Marylandmudbugs.com
 
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#24 ·
tranformers are a more efficient way to lower voltage. However they can only be used with alternating current, or with an intermitent current as in the ingition system. to us a transformer the current would need to pulsed or inverted while there are ways to do this it is not IMHO nessitary to run some LED's. There current draw is so small that the voltage divider will no waste enough energy to mater.

bandhmo

 
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