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Resisitor to 1/2 6.5 volts dc ?

Discussion in 'Hobby Electronics' started by o{O}o, Sep 27, 2004.

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  1. o{O}o

    o{O}o Guest

    What resistor would I use to halve 6.5 volts dc ?
  2. MC

    MC Guest

    Eh ?
    Q1 : What do you want to use the 3.25 Volts for ?
    Q2 : What is supplying the 6.5 V ?
  3. There is a lot more to it than simply wanting to know the supply voltage.

    Let's for simplicity assume the source is 6.5 volts and you want to reduce
    this to 3.25 volts. To use a simple resistor to drop the voltage you need
    to assume the current flow is going to be constant. If that was the case
    and you knew the amount of current the circuit drew you could easily work
    out the ohmic value of the resistor by applying Ohm's Law. R = V / I
    where R is in ohms, V is the voltage that would be dropped across the
    resistor in volts and, I is the current drawn by the circuit in Amperes.

    So if the circuit drew 120mA (0.12A) of current, the ohmic value of the
    resistor would be -

    R = V / I
    = 3.25 / 0.12
    = 27.08 ohms
    A 27 ohm resistor would be a close enough value.

    Another consideration would be to determine how much power the resistor
    would be dissipating so that a resistor with a sufficiently high enough
    power rating could be selected.

    A simple calculation of power in Watts based on the voltage dropped across
    the resistor and the current flowing through it could be determined from the
    following formula -

    P = V x I where P is the power in Watts, V is the voltage dropped across
    the resistor, and I is the current flowing through it in Amperes. So based
    on our previous example -

    P = V x I
    = 3.25 x 0.12
    = 0.39 Watts
    A resistor of 0.6W power rating would suffice.

    All of the above assumes the power supply source is capable of providing a
    relatively stable terminal voltage when the required current is drawn from
    it by the load.

    If the current drawn by the load varies then a simple series dropping
    resistor as described above would not be a suitable arrangement. In this
    case your application may be better served by using some form of series
    electronic regulation (simple 3 terminal regulator IC), or alternatively a
    series resistor and a shunt voltage regulation device such as a zener diode
    (which is old hat technology by today's standards).

    Hope this simple explanation is of some assistance. I suggest you get a
    book on basic electrical concepts from you local library and do some further

  4. o{O}o

    o{O}o Guest

    The transformer is DC 6.5V 150ma.
    It is powering an indoor sensor night light.
    I want to use a 5 or 10mm high brightness 9000 mcd LED
    as a spot light instead of the diffused krypton globe.
    I could simply add 2 LEDs but wanted to try 1.
    I have already got one of these units operating with 2 LEDs.

    Also i made a mistake saying 6.5 v.
    I need to halve 4.5v,
    This is the power closest to the voltage sent to the krypton globe.

  5. o{O}o

    o{O}o Guest

    Thanks Alan for your answer.

    Giving $change is the only math this brain wants to know.
    Otherwise I use the force ......... of just tinkering.

    Alan Rutlidge> wrote in message
  6. MC

    MC Guest

    If you had two white LEDs operating with that transformer then
    it is producing more than 7.3Volts DC.
    (If you had no resistor in series with the LEDs then you might
    have damaged the LEDs with excessive current)
    Assuming that the transformer is producing about 7 Volts DC with
    no load, and that the white LED is intended to be operated at 20mA
    at about 3.6V, then you need a ( 7 - 3.6 ) / 20 kohm series resistor.
    Buy a 180 ohm, 0.25 Watt resistor and put it in series with the
    white LED.
  7. o{O}o

    o{O}o Guest

    OK thanks.
    I will go to Dick Smith and get resistor/s.

    MC wrote in message ...
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