Resisitor to 1/2 6.5 volts dc ?

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

1. o{O}oGuest

What resistor would I use to halve 6.5 volts dc ?

2. MCGuest

Eh ?
Q1 : What do you want to use the 3.25 Volts for ?
Q2 : What is supplying the 6.5 V ?

3. Alan RutlidgeGuest

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

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

Cheers,
Alan

4. o{O}oGuest

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.

Thankyou

5. o{O}oGuest

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

Alan Rutlidge .iinet.net.au> wrote in message

6. MCGuest

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)
Hokay..
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}oGuest

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

MC wrote in message ...