12v to 7.5v

mikeblas said:
Really? What official said that?
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Official? I don't know. EVERYONE has said it for so long that I never even bothered to look to see if it's considered 100% official. If it's not, it's unofficially official since everyone expects it. Well, just a minor example of verification: http://en.wikipedia.org/wiki/Cigarette_lighter EDIT: Oops, this is better: http://en.wikipedia.org/wiki/Car_cigarette_lighter You'll note there are standards for 12V, but, none for 13.6V.
 
Nazo said:
Official? I don't know. EVERYONE has said it for so long that I never even bothered to look to see if it's considered 100% official.
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That's too bad. Maybe you should research it a little more.

Check the charging system section of the shop manual for your car, for example.

You should find that a fully-charged, unloaded car battery shows between 12.6 and 12.7 volts. While charging, it'll be higher; around the 13.6-volt figure that I quoted, which is a value typically used to bench-test automotive electronics. (Those supplies output 13.8 VDC.) In the car, the voltage can be higher; 14.5 VDC, even.

If you made that measurement with the car off and found 12.3 volts, your battery was a little weak. (Maybe just because it was cold outside.) If you made that measurement with your car running, your charging system either wasn't on or is faulty.

The voltage on a car's accessory circuit (which drives the cigarette lighter) isn't regulated, so there's nothing specific can be expected for the voltage found there. (The standards Wikipedia is quoting from the SAE document are about the size of the connector, not the voltage it presents. It if were specifying the voltage, it would include other information, such as a tolerance and ripple limits.)

The volatge at the cigarette lighter is widly variable, is full of noise, and can spike or droop at any time. But, nominally, you will find something higher than twelve volts. (As you yourself did; 12.3 volts is higher than 12.0 volts. How high is too high? How low is too low?)

References:

http://www.skylab.org/~chugga/mpegbox/MPBS1/
http://www.repairfaq.org/ELE/F_Car_Battery.html
http://www.dcbattery.com/optima_self_discharge_curve.html
http://www.escience.ca/circuitT/RENDER/0001/68/3058/11064.html
http://www.termotronic.com/rush/

So a circut designed around an LM317 to deliver 7.5 volts output and driven from a car battery is less likely to have dropout problems than the same circuit driven from a 12.0-volt rail on a computer power supply. The nominal voltage for the power supply is 12.0 volts, while the nominal voltage in a car's electrical system is higher -- not by much, but enough to exceed the regulator's dropout voltage requirement.
 
That or maybe it doesn't really matter. I'm not the one building a circuit, and just tested that much because I needed to know polarity of a plug when I wanted a connector to the 12V of my computer for something that ran off the car power, but, it has it's own voltage handling internally and I didn't have to build any actual circuit so much as just a connector to a molex with the correct polarity.

BTW, in the image, the switch you see at the top is 9V, 7.5V, 6V, 4.5V, 3V, 1.5V, OFF in order from left to right. Just noticed those numbers were cut off of the top there. Also, the diodes say something like IN400/MIC if you should need to know that. The two ceramic capacitors say 104 (just 104, so I suspect that they are a model number, not a rating) and the two ordinary (electrolytic) are 47mF since that's kind of hard to see.
 
ok...back on topic. I am supposed to ignore the bottom part, and the first 22000uF thing. Once those are eliminated, then the only difference between the 2 diagrams is the backflow thing....right? Could I somewhat combine the two diagrams? I want R1 to be 240 and R2 to be 1200, because that should (according to the table) give me 7.5v.
 
fenton06 said:
ok...back on topic. I am supposed to ignore the bottom part, and the first 22000uF thing.
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That "thing" is a capacitor, and "the bottom part" is a step-down transformer and a full-wave bridge recitifer. Yes: if you are sourcing from the 12-volt rail on your power supply, you obviously don't need those parts.

fenton06 said:
Once those are eliminated, then the only difference between the 2 diagrams is the backflow thing....right? Could I somewhat combine the two diagrams? I want R1 to be 240 and R2 to be 1200, because that should (according to the table) give me 7.5v.
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The "backflow thing" is a diode.

Yes, those values should get you a 7.5 volt output -- assuming you've got no dropout voltage problems.

You can double-check your math using the formula given in the data sheet, linked earlier.
 
mikeblas said:
Really? What official said that?
Click to expand...

Alternators SHOULD put out about 13V to V14. If its putting out 12V its dieing or you have to many or to big of a battery in your car and the alternator can't keep up.
 
Nazo said:
The two ceramic capacitors say 104 (just 104, so I suspect that they are a model number, not a rating) and the two ordinary (electrolytic) are 47mF since that's kind of hard to see.
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104 is a rating. It's 10 ^ 4 in picofarads, I think. And it's 47uF (micro Farads).
 
Well, I meant micro. I didn't want to put the u since the micro symbol isn't easily entered, but, I suppose it's not accurate to put m for micro since in SI it means milli I suppose.

Anyway, if anyone takes an interest in using those images, now you know for certain what the capacitors are. Might be kind of handy to compare the diagram to a real circuit. BTW, you'll note that there is still a fuse in there, only it's on the input inside the car adaptor plug, so that's why you don't see it in the images.
 
Fenris_Ulf said:
104 is a rating. It's 10 ^ 4 in picofarads, I think. And it's 47uF (micro Farads).
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You're almost right. The numbers work the exact same way as the color bands on a resistor--except the units are picofarads instead of Ohms. So you get a digit 1, a digit 0, (put them together to get 10), and the 4 is the power of 10. So you get 10 x 10^4 = 100,000pF or .1uF.

 
The LM317 or similar linear regulator is your best bet...but if your device really does just draw 250 mA, you can use a 20 ohm resistor in a pinch. You can't use a standard quarter-watt resistor; you'll need a 2 watt or larger power resistor.
 
OK, I am going to build this for a school project now because:

a. my mom will pay for it
b. I will get credit

OK, so if I make this in a circuit design program, is there anywhere I can send the design so they can make the circuit board? I don't really want to etch my own board, but if I need to I suppose I could.
 
If your ok with running 9 volts, then use a LM7809, can run off 12volts and drive up to 1Amp

http://www.eidusa.com/Electronics_Voltage_Regulator.htm

Edit, if you can find a LM7808 that would be even better, these also support short circuit protection, thermal overload protection.

I can't find it through my local dealer right now, but will look at a few others dealers that i know of tomarrow.
 
masher said:
The LM317 or similar linear regulator is your best bet...but if your device really does just draw 250 mA, you can use a 20 ohm resistor in a pinch. You can't use a standard quarter-watt resistor; you'll need a 2 watt or larger power resistor.
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You could, but I would recommend against it. The power draw will not always be constant, so if you use a resistor, the supply voltage will vary. Which may not be dangerous per se, but has a very good chance of causing undesired operation. Stick with a regulator--it's not that complex, it can handle higher loads, and it'll give you a better supply.

 
does anyone have a suggestion for the layout and where to send it? Or do I need to research etching my own circuit board.
 
fenton06 said:
does anyone have a suggestion for the layout and where to send it? Or do I need to research etching my own circuit board.
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I wouldn't even etch a board for something this trivial. Just put it on a square of perfboard, and solder wire interconnects on the backside.
 
thank you all very much... :D I relly appreciate it, this will make one hell of a project :D I will try and post pictures of the progress, since everyone on the [H] loves pics. :D
 
fenton06 said:
I will try and post pictures of the progress, since everyone on the [H] loves pics. :D
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Especially the ones of the topless model holding your newly-etched circuit board...
 
lol....yeah...I'll work on that one. I have run into a kind of snag though, 2 of the electronics schematics programs I have tried do not have the parts in wither of those schematics! I don't know enough about electronics to just substitute stuff, any input is appreciated.
 
I can't believe they don't have parts that common in their database. But from a pcb perspective, all you need to match on a device substitution is the form factor. If the device size and pin locations match, you're good...no matter what the electrical characteristics are.

The last page of the item's datasheet should have this info. Just print it out, and compare to a test printout of your PCB before submitting it.
 
aherm...ok...now I need some help with the components. Here are the questions:

a) what kind of capacitors should I use...electrolytic, ceramic, tantalum...or does it matter

b) what kind of resistors do I use? 1/2 watt, 1/4 watt, or 1/8 watt?

I have the diode and the regulators, those were easy, since they had a part number.

Masher - thanks for the design program help, I was confoozled why they didn't have them also, but since it doesn't matter, that helps a lot :D

 
Which circuit is it that you decided to build?

If it's this one, then:

a) It hardly matters. C1 can be ceramic or tantalum. C2 can be electrolytic or tantalum. Tantalum is more expensive. Make sure both capacitors are rated for at least twice the voltage they'll see. C1 should have a rating of 24 volts or more, C2 should have a rating of 15 volts or more.

b) For the resistors, you can use 1/4 watt or 1/8 watt resistors, whichever is cheaper. The figures in that listing are a little silly, since many of the values are standard resistor values. In particuar, since these resistances are going to set the output value of the circuit, you might want to use metal film resistors with a 1% tolerance. You can find a list of standard values on the web, or browse your suppliers catalog. You mgiht want to go back and do the math to figure out how to get 7.5 volts out of standard vaues for 1% resistors, or stick with the values given-- which seem to be OK for you for 5% resistors.

Or just use a fixed value for R1 (say, 270 ohms) and then use a 5 kilo-ohm, ten-turn potentiometer for R2.
 
yes, I chose to use that one adding the the back flow diode. If that doesn't work out, then I'll simply take it out. Looks like I will be using tantalum for the capacitors. :D

I do have a question about the potentiometer though. I would be able to fine tune how much voltage is output right? - or - I could just use a 240 ohm and a 1200 ohm resistor and get 7.5 volts like it says in the chart, right?

 
My bad.. i read the problem wrong..

you cna use diodes in series as someone else stated

you cna also get a variable voltage reguler.. take the input of the reguler (12V).. and then use a Potnetioameter to adjust it so that the output is 7.5V or whatever it is you need..

so to do this..

you would need.. A voltage regulator, two capacitors, a potentiometer, and i guess you can go with some shitty proto type board from radio shack.. or you can design a PCB and send it out :D It wouldn't take more than 2 inchs by 1 inch of space to do..

i didn't read the whole thread.. but if no one gave you this answer by now... damn... i'm ashamed of you guys lol..

the only problem with that circuit is.. that i've been told that it's not good to chain regulators (your PSU is a regulate supply.. so you're gonna chain it iwth the PSU).... i don't know why it's considered not a good thing..

there are other methods i cna't think of any right now..
 
I'd forget the PCB, the regulator and everything else, and just use a zener diode.

Microsemi 1N5336B. Reverse drop of 4.3 volts (gives 7.7 volts) and it can dissipate 5 watts. Just install it directly in the power cable.
 
gee said:
I'd forget the PCB, the regulator and everything else, and just use a zener diode.
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That takes all the fun out of it, and then I wouldn't have a project anymore now would I. :D
 
gee said:
I'd forget the PCB, the regulator and everything else, and just use a zener diode.

Microsemi 1N5336B. Reverse drop of 4.3 volts (gives 7.7 volts) and it can dissipate 5 watts. Just install it directly in the power cable.
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LOL! let the cat have his fun..
 
fenton06 said:
That takes all the fun out of it, and then I wouldn't have a project anymore now would I. :D
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ok, forget everything I just said.

And once that works, get your hands on a MAX724 or something and do the same thing over again but switchmode. Fun will lead to frustration, and eventually insanity. The good kind. :D
 
gee said:
Microsemi 1N5336B. Reverse drop of 4.3 volts (gives 7.7 volts) and it can dissipate 5 watts. Just install it directly in the power cable.
Click to expand...
Is that right? Are you making a shunt regulator, or just relying on the voltage drop through the zener?
 
mikeblas said:
Is that right? Are you making a shunt regulator, or just relying on the voltage drop through the zener?
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sounds like he's relying on the voltage drop provided by the zener...

but IIRC.. you're supposed to put a few in series (however many you need to get the voltage you want) and then run you're circuit parrallel to the diodes no?
 
ok...here is the schematic....how does it look, and will this work? Pins 3 and 4 are not connected, even though they look like they are.

VoltageRegulator2.jpg


I'm not really sure how I want to connect the power to the LCD though...any ideas? What I really need is a pinout of where the power and the RCA go, because then I could incorporate this onto the board. I would have a molex connector and RCA connector on the board I am building, and then I could just unplug the RCA and power, and take the LCD out of necessary. I'll try and post a picture of what I'm talking about soon.
 
mikeblas said:
Is that right? Are you making a shunt regulator, or just relying on the voltage drop through the zener?
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Using it as a series voltage drop.
 
gee said:
Using it as a series voltage drop.
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Then you're achieving no regulation, which seems pretty bad. The Zv will take 4.3 volts off the input, and the input isn't regulated -- it'll vary between 0 and 14.8 volts or so.
 
ok...I found out that the LCD draws 6.8 watts according to Sony, so that should help some of you electrical people out :D Oh, will the schematic work?

 
fenton06 said:
ok...I found out that the LCD draws 6.8 watts according to Sony, so that should help some of you electrical people out :D Oh, will the schematic work?
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6.8 watts at 7.5 volts is 0.91 amps. Make sure you get the correct LM317 part, and the right heat-sink for it.

You've got the backflow diode's polarity backward.

I'm not sure why you've chosen a four-pin plug for the power input. I'd use a 3-pin regulator, and assign the pins so that you're guaranteed to not short or damage anything if something is plugged-in backwards.

fenton06 said:
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What's all that crap, by the way? I see it in many people's posts, and it's annoying to have to delete it every time I respond by quoting.
 
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