Resistor ........?

Question

if i want to lower down my voltage for example ac 240volt to dc 12 volt and i just want to use the diode & resistor. how do i Calculate the value for the resistor and for the diode which value is the best ( 1n4001 or 1n4007 )

Answer 1

Transformer is a must. if you want exact 12Vdc, use a 7812 Regulator also.

Below is the umm.., "block diagram" of the required ckt.

240Vac ~ 0-12VTransformer ~ BridgeRectifier ~ 25VCapacitor ~ 7812Regulator ~ yourCircuit.

1N4001 has reverse breakdown voltage of 50V. 1N4007 has reverse breakdown voltage of 1000V. since both cost same, use the latter.

if you're using only diodes, you'll need h**l lots of them. each Si diode will introduce only 0.7V drop. so, for 240Vac to 12Vac, you'll need 240 / (2*0.7) = 172 (2 diodes are in series at any time in Bridge Rectifier) in each arm of the Bridge!! Also, 240Vac comes usually along with 2Amps. Diodes will fry right in front of you the moment you switch on.

Answer 2

It would help to know what you are trying to power, as one other answerer said, an amp of current will be impractical, but a milliamp of two might not be too much of a problem, but one  _must_ know the current in order to calculate the resistance (also the current must be constant if you are just using a resistor, though you could add a zener regulator).

You would not use a 1N4001 because the reverse voltage rating is only 50V, the 1V4007 (1000V) is overkill, but they are so cheap you might as well use it, though a 4004 should work, but the 4007 (or a 4006) will give you a better safety factor.

Capacitors are also sometimes used to drop the voltage in some line powered applications where tiny amounts of current are needed.

Generally a transformer would be a better option, "wall warts" (plug in supplies) are readily available and fairly inexpensive, well, 240V input ones would not be so common in the US, but I imagine they should be in some other countries.

Answer 3

It would be best to use a transformer to step down the voltage from 240 to 12 V. Using resistors becomes tricky because you need to take into account the total load and current requirements of the circuit. And high power resistors are very expensive and get very hot. Either diode type will work but you'll need at least 4 to construct a full wave bridge rectifier. You also need a conditioning capacitor in parallel with the load to minimize the ripple voltage.

If you really want to do it with a resistor, heres how:

Lets say the power supply your constructing need to be able to deliver 1 Amp to the load. If each diode has a voltage drop of 1 V and you use the bridge configuration you'll need two resistors that produce a voltage drop of 240V - 14V= 226V

At 1 amp the resistors will dissipate 226 watts so at the very minimum they need to be rated at 500 W. They will probably cost several hundred dollars a pop and run hot enough to cook off of.