Drop Voltage using Resistor: is it practical?

Question

I should probably say that this a basic question, I don't know much about this stuff, please don't be too hard on me.

Ok so I kind of thought it would be fun if I could light up 12V halogen lamps using 220V power supply, so I did a few calculations: The bulb is rated at 60W, so according to P=VI, it should draw 5A current. Now since the input voltage is 220V and I need 12V, the voltage drop across the resistor will have to be 220-12=208V. The bulb draws 5A current so according to R=V/I, so I would need a 41.6 Ohm resistor, say approx. 40 Ohm.

I wanted to ask if my calculations are correct and if it is indeed safe and practical to drop voltage using resistors. I would also like some details on what kind of resistor I should use. Thanks.

Answer 1

Your calculation is correct, however you also have to consider the power rating of the resistor.

Power is \$I^2 \cdot R\$ which is 1040W. That's a physically huge and expensive resistor and your circuit would be wasting 95% of the energy that you put in before it even gets to the bulb.

Here's a typical resistor style capable of that level of power dissipation:

It's 300mm long, 60mm in diameter and costs more than $40 US.

Would it ever make sense? Possibly- if you needed a 1kW heater for some reason as well as the lamp then it could conceivably.

Another consideration is that the socket of the halogen lamp may not be designed to keep fingers away from the supply. No big deal with 12V- it's pretty hard to electrocute a human with 12V, but 220V mains could be lethal. Most approaches to supplying low voltage lamps provide galvanic isolation which protects the user.

A better choice would be a 220:12 transformer or an 'electronic transformer' which uses switching power supply techniques to reduce the voltage.

Electronic transformer:

For a hobbyist, an old PC power supply could provide regulated 12V at 5A without breaking a sweat and it should be completely safe. See articles on the internet on how to get it to turn on (you may have add a dummy load). It provides DC rather than AC, however halogen bulbs of that voltage and power don't care much- the life may be slightly reduced on DC.

Answer 2

The calculations are right mathematically but just see this:

The voltage drop across the resistor is 208 V. So power dissipation is

P=V²/R

P= 1.04 kW !!!!!!!!

Now that is some power you are wasting, 17.33 times what you actually need. So this is not a practically feasible method. Instead you can go for voltage regulators and transformers which are much more feasible and efficient .

Answer 3

For the lamp current:

$$I = \frac{P}{E} = \frac{60W}{12V} = 5 \text{ amperes} $$

For the ballast resistor value:

$$ R =\frac {Vt-Vl}{I} =\frac {220V - 12V}{5A} \approx 42 \text{ ohms} $$

For the ballast resistor dissipation:

$$ P = (Vt-Vl) \times I = \style{color:red} {1040\text{ watts}} $$

Answer 4

Another difficulty with using a resistor is that light bulbs have a resistance which increases with temperature. If one is driving a 12V light bulb with a 12V source, the lower resistance will cause the light bulb to draw more current and thus consume more power until it gets hot. This behavior causes light bulbs to turn on quickly--in some cases more quickly than would be optimal for the lifetime of the filament.

Adding the monster-sized resistor in series with the light bulb will mean that the bulb will effectively be fed with a five-amp constant current source. When driving a cold bulb from a 5A current source, the low resistance of the bulb will cause it to drop far less than 12V and thus consume a lot less than 60 watts. If it consumes enough power and generates enough heat to warm the bulb to the point where its resistance increases significantly, the bulb might be able to warm up to its normal operating voltage. On the other hand, the hotter the bulb gets the more heat it will produce. The rate at which power consumption increases will probably be slower than the rate at which radiative dissipation increases, thus preventing thermal runaway, but behavior will likely be far less stable than when driving a bulb with a stable voltage source.

Answer 5

Essentially all you calculated for was the required resistance on a 220V source to dissipate 60W.

I would try to avoid using 220V to light up 12V halogen bulbs without some form of an adapter. The other thing to consider, is the 220V AC or DC? I am assuming the 12V required by the bulb is DC, so if 220V is supplied in AC, you would need to do an AC to DC conversion.

This would simplify a lot if you used an existed adapter and just plugged it in. Just make sure that the current supplied isn't over it's current spec.

Hope that helps. Also, could you post the model # or data sheet of your bulb? And can you clarify as to whether the 220V is AC or DC?