Electronics nerds?

[Disco Dan]

I have a big pile of resistors, now I can identify each one (big range of values) but which one would I use to change voltage from 10v (sometimes 11v) down to 5v ?

Any help would be much appreciated!!

What I have got going on here is making a charger for my satnav - I have found a cell phone car charger and cut the end off. The two wires are putting out 10v and i need to put a mini USB plug on the end which should have only 5v going into the device. (it is already stepping the voltage down from 12v to 10v inside the charger unit)

[Disco Dan]

I have some bread-board lying around if a simple circuit is required rather than just an inline resister.

[The Pastor]

voltage = resistance x current.

What you want is a voltage divider

http://hyperphysics.phy-astr.gsu.edu...ic/voldiv.html

[avgas]

ok without the super detail, do you need a specified current?
I have attached a basic drawings.
To half the voltage you need 2 of the same resistor (R) and a cap (220uF?) and a 5V (or 5.1 i think it is) Zenner diode.
The rail along the bottom is ground, top is 10v.
Clear as mud?

[Disco Dan]

voltage = resistance x current.

What you want is a voltage divider

http://hyperphysics.phy-astr.gsu.edu...ic/voldiv.html

um ok

so i have two wires - red and black. They read 10v on multimeter.

So do I need to measure the resistance using these two wires? Because all I get on the screen is "1" regardless of what ohm setting i click it too?

[Disco Dan]

ok without the super detail, do you need a specified current?
I have attached a basic drawings.
To half the voltage you need 2 of the same resistor (R) and a cap (220uF?) and a 5V (or 5.1 i think it is) Zenner diode.
The rail along the bottom is ground, top is 10v.
Clear as mud?

I only have resistors and a few assorted salvaged capacitors.

Any way to do it without a diode?

[Disco Dan]

Oh and I found a few tiny transistors too if that helps?

Tried putting a few resistors inline but it is not changing the voltage

I obviously did not pay enough attention in electronics at high school!

[Disco Dan]

Yikes...

Practical Voltage Division
Don't fall prey to the temptation to use a voltage divider instead of a voltage regulator. Voltage dividers are totally dependent on their supply voltage, so don't try to "regulate" 12V to 5 with a divider. However, there are cases where you might want to use a divider. For example, perhaps you have a 5V A/D converter, but you need to measure 0-10V with it. Of course, you will lose precision, but maybe that's OK.
Converting 10V to 5V is easy, right? Two 10K resistors and you have a 50% voltage divider. Except it isn't that simple. If you build a voltage divider in the lab, you'll look at the output with a voltmeter or a scope. All modern DVMs have input impedances of 10M or more, so for practical purposes it isn't there as far as the circuit is concerned. But when you draw real power from the center node, you are effectively putting another resistor in parallel with R2. This changes the circuit and the output voltage.
Suppose your A/D convert has an input resistance of 25K. Now you don't have two 10K resistors. You have R1=10K and R2=10K and 25K in parallel. Resistors in parallel add their admittance, so:
Rtotal = 1/(1/R1+1/R2+.....1/Rn)
If you only have two resistors, this can be simplified to: Rtotal = R1*R2/(R1+R2). So in the above example R2 is effectively 7143 ohms. The divider's ratio is then about 42% (7143/17143). Not the 50% you were looking for. This even happens with a meter, but the effect is so small you don't care. Consider if R2 was equal to 10K and 10M in parallel. That's 9990 ohms and the ratio is 49.97% -- hardly any difference at all, and probably less error than the 5% or 10% resistors you are using.
So when calculating your A/D converter input divider, you want to make R2's total value 10K. If the input resistance is 25K, you have to solve for the parallel resistor with 25K that gives you 10K. So:
10K = 25K*Rx/(25K+Rx)
A little algebra results in:
250K/15K = Rx or Rx=16.667K (you can verify that by computing 16.667K in parallel with 25K).
So the actual voltage divider would be R1=10K, R2=16.7K, but you can't really find that value more than likely. Substitute a 15K resistor and you'll find the true division ratio is just over 48% (can you get the same result?). However, 16K is a standard value and provides a ratio of 49.4% which is pretty close.

[Mikkel]

Just grab a green one... Green is good! Green will do the trick

[Hitcher]

Go to Duck Smuth and buy one of their $20 cigarette lighter-USB adaptors. Then buy an extension cord with a USB plug one end and a USB-B five-pin plug on the other. Connect it all up, and Bob's your father.

I have done this, and it works.

[Ixion]

'Taint simple. Mainly cos your car charger is outputting DC. If it was AC it would be much easier you could just take the 0 to 5v + a bit part of the AC curve and regulate and smooth it.

I looked at EXACTLY the same issue for my Navman. WTF Navman don't provide a hard wire cable GOK. You can buy them in the UK but they are a horrendous price.

So, what I did, I popped into Surplustronics at the top of Queer St, and bought a little unit that does exactly that - converts 12v to 5v . Hard wire it in, sod the silly cigarette lighter thingie. About $30 ish. About the size of a match box. Explain what you need to the dude, they are on the rigjht hand wall as you enter the shop, sort of behind the counter.

I'll look and see if mine has a name on it.

[avgas]

Ok will clear this all up after work, a few misconceptions here - but am flat stick right now.
yes you can do this setup without the diode, and no AC does not affect the design i did before. Yes it is better to use a regulator (LM, 5V range - LM3501? Sorry its been a while)

[CookMySock]

I have a big pile of resistors, now I can identify each one (big range of values) but which one would I use to change voltage from 10v (sometimes 11v) down to 5v ?

Thats not a smart way to do that.

go here http://search.dse.co.nz/ and search for "7805"

The LM7805 is a 5V regulator. It costs $1.50

Scroll down abit theres a link called "A Typical 78xx Power Supply Circuit". You only need the two capacitors to make it work.

Trim all the wires really short and solder it together xmas-tree sylez and pot it in some epoxy inside a matchbox or smaller.

Easy. Fast. Reliable. Cheap.

DB

[Disco Dan]

The OEM Navman charger is available at DSE for $49.

The idea is to do this as cheaply and simply as possible.

Go to Duck Smuth and buy one of their $20 cigarette lighter-USB adaptors. Then buy an extension cord with a USB plug one end and a USB-B five-pin plug on the other. Connect it all up, and Bob's your father.

I have done this, and it works.

Will go and check it out - I have already wired up a cig lighter socket under the seat (so i can also charge my phone).

'Taint simple. Mainly cos your car charger is outputting DC. If it was AC it would be much easier you could just take the 0 to 5v + a bit part of the AC curve and regulate and smooth it.

I looked at EXACTLY the same issue for my Navman. WTF Navman don't provide a hard wire cable GOK. You can buy them in the UK but they are a horrendous price.

So, what I did, I popped into Surplustronics at the top of Queer St, and bought a little unit that does exactly that - converts 12v to 5v . Hard wire it in, sod the silly cigarette lighter thingie. About $30 ish. About the size of a match box. Explain what you need to the dude, they are on the rigjht hand wall as you enter the shop, sort of behind the counter.

I'll look and see if mine has a name on it.

Will have to remember that! Cheers

[dhunt]

I have a big pile of resistors, now I can identify each one (big range of values) but which one would I use to change voltage from 10v (sometimes 11v) down to 5v ?

Any help would be much appreciated!!

What I have got going on here is making a charger for my satnav - I have found a cell phone car charger and cut the end off. The two wires are putting out 10v and i need to put a mini USB plug on the end which should have only 5v going into the device. (it is already stepping the voltage down from 12v to 10v inside the charger unit)

For what you want it for I would say a resistor divider is not suitable at all for what you want.
What you want to do is got to DSE and buy a LM7805 (5V voltage regular) and a couple 100n ceramic capacitors and make a properly voltage regulator. Depending on current consumption and voltage input you may need to use a heat sink (Piece of Metal). If you need more details let me know.