resistors
I have a power supply of 5 volts @ 1A. I want to reduce the current from 1A to 250mA. Which resistor should be use?
I tried this equation(Ohm's Law):
R = V / I
= 5/(0.75)
= 6.66 Ohms.
I tried the 6.66 Ohm resistor but it only reduce few mili Amps.
There are a few ways to get 5V from a 12V supply. Each has its advantages and disadvantages, so I've drawn up 5 basic circuits to show their pros and cons.
It works, BUT it only works at one value of load current and it wastes most of the power supplied. If the load value changes, the voltage will change, since there is no regulation. However, it will survive a short circuit at the output and protect the 12V source from shorting out.
It works, BUT most of the power is dissipated by the Zener diode. Not very efficient! On the other hand it does give a degree of regulation if the load changes. However, if you short circuit the output, the magic blue smoke will break free from the Zener... Such a short circuit may also damage the 12V source once the Zener is destroyed.
It works, BUT most of the power has to be dissipated by the transistor and it isn't short circuit proof. Like circuit 2, you could end up damaging the 12V source. On the other hand, regulation will be improved (due to the current amplifying effect of the transistor). The Zener diode no longer has to take the full load current, so a much cheaper/smaller/lower power Zener or other voltage reference device can be used. This circuit is actually less efficient than circuits 1 and 2, because extra current is needed for the Zener and its associated resistor.
It works, BUT the device (or circuit) has to dissipate more power than is supplied to the load. It is even more inefficient than circuits 1 and 2, because the extra electronics take additional current. On the other hand, it would survive a short circuit and so is an improvement on circuits 2 and 3. It also limits the maximum current that would be taken under short circuit conditions, protecting the 12v source.
It works, BUT the output can be a bit spikey due to the high frequency switching nature of the device. However, it's very efficient because it uses stored energy (in an inductor and a capacitor) to convert the voltage. It has reasonable voltage regulation and output current limiting. It will survive a short circuit and protect the battery.
These 5 circuits all work (i.e. they all produce 5V across a load) and they all have their pros and cons. Some work better than others in terms of protection, regulation and efficiency. Like most engineering problems, it's a trade off between simplicity, cost, efficiency, reliability etc.
Regarding 'constant current' - you cannot have a fixed (constant) voltage and a constant current with a variable load. You have to choose - constant voltage OR constant current. If you choose constant voltage, you can add some form of circuit to limit the maximum current to a safe maximum value - such as in circuits 4 and 5.
I'm going to take a WAG and assume this LED you have is designed for automotive use on a "12VDC" source (typically more like 14V), so it already has a resistor in there. Be sure to verify this before hooking it up to a supply directly or you could damage the LED!
In that case, you will only need to add to the resistor(s) that are in there.
You can attach it to a power supply that outputs something like 14V (preferably a bench supply) and measure the current Im. Take that current Im, let's say it's 100mA, and calculate the value of the extra resistor:
Rs = \$\frac{22V-14V}{Im}\$ so for our example, it would be 80 ohms.
Pick a standard E24 value that's the same or a bit higher (say 82 Ohms in this case).
Power will be \$\frac{(22V-14V)^2}{R}\$, so for our example 0.8W. Pick one that's a bit higher, say 1W or 2W.
Go to your favorite distributor and find a resistor with the required specs.
Best Answer
The 1A rating on the power supply is simply it's maximum, not the actual amount of current drawn by whatever you hook up to it. That is the amount you must ensure the load stays below.
If the "load" is a short circuit, then your calculation is nearly right. $$ R = {{V}\over{I}} = {{5V}\over{250mA}} = 20\Omega$$ If the load is something else, a resistor of lesser resistance is required.