I = e/r or e =I/R

[DAVID MILTON OLENICK]

I want to drop my 12V auto output to 4.5 volts to operate a small 4.5 V TV.
Can you tell me what size of resistor I would need.The load would be about
300 ma.
I want to wired a resistor in series to drop the voltage.Can you also give
me the appropiate formula for calculating this and an example?
Thanks
David Olenick

[Charles Perry]

I don't think I would use a wired resistor in series if I were you. As the
load drops the voltage across the resistor will drop and the voltage on your
TV will increase.

Charles Perry P.E.

DAVID MILTON OLENICK wrote in message
<01becf5d$8a05d9e0$071647cf@48tango>...

[jak]

DAVID MILTON OLENICK wrote in message
<01becf5d$8a05d9e0$071647cf@48tango>...
>I want to drop my 12V auto output to 4.5 volts to operate a small 4.5 V TV.
>Can you tell me what size of resistor I would need.The load would be about
>300 ma.
> I want to wired a resistor in series to drop the voltage.Can you also give
>me the appropiate formula for calculating this and an example?
>Thanks
>David Olenick
>

You would need a voltage divider. Two resistors in series the some of which
is equal to 4.5/.3=15 ohm.
If the resistor closer to source is R1, and TV is across R2, then choose
resistors such that R2/(R1+R2)=4.5/12=.375. Since you already know the
total R1+R2, you can solve for R2 and then for R1. However, the values you
get may not be standard resistors sizes. I would suggest using one resistor
and a pot in the appropriate range and current rating.

[Greg Fretwell]

Since the load will vary depending on several factors a voltage regulator
is a lot better solution. You can cobble one up with a LM317 and a pass
transistor quite easily and cheaply. These are Radio Shack parts.
Greg

[Don]

Did you know you can buy cigarette lighter adapters (Radio Shack and lots of
truck stops) that will output 3, 4.5, 6 and 9 volts for around $10.

Anyway, the best way to solve this problem (and do-it-yourself) is to build a DC
voltage regulator circuit.

On 16 Jul 1999 10:38:46 GMT, "DAVID MILTON OLENICK" <DAVID_...@JUNO.COM>
wrote:

[Jerry G.]

I would not recomend using resistor networks to drop voltage for
utility use even at low current. It will not be regulated, and the
voltage stability will be fully dependent on load characteristics.

I would recomend a 4.5 volt regulator circuit. There are some IC's
in the LM series, but some knowledge of design and assembly would be
required.

You can find these types of utility cigarette lighter adabtors at
Radio Shack, and some truck stops for under $20.00. They are cheaper
to buy, and would be much more reliable than something home made
unless you came up with a good foolproof design.

--

Jerry Greenberg

======================================
jerr...@hotmail.com
http://www.zoom-one.com
http://www.zoom-one.com/electron.htm
http://www.total.net/~jerryg

=========== Message Separator ===========

DAVID MILTON OLENICK <DAVID_...@JUNO.COM> wrote in message
news:01becf5d$8a05d9e0$071647cf@48tango...

[Ed Schick]

The LM 317 is likely the best approach. And it will
handle 1.5 amps, so you don't need a pass transistor
in this application. A resistor sets the voltage level.
Still, being a belt and suspenders type, I'd use a
2N3055 or equivalent pass transistor too. I'd also
add a couple of caps for spikes. Ed

[Dale Chayes]

Ed Schick wrote:

> The LM 317 is likely the best approach. And it will
> handle 1.5 amps, so you don't need a pass transistor

> in this application. .....

You might want to consider adding heat sink.

-Dale

[Davco]

I= E/R therefore E= IR

Dave

DAVID MILTON OLENICK wrote in message
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