I have been looking around for an easy way to convert 12V to 5V. I have seen some people saying that a simple resistor is all that is needed.
$$ Volts = Ohms \cdot Amps $$
$$ Amps = \frac{Volts}{Ohms} $$
$$Ohms = \frac{Volts}{Amps}$$
So applying a resistor will diminish the voltage of the circuit. That should mean that an appropriately sized resistor could simply be placed in the path of a 12V circuit, converting it to 5v.
- If this is the case how would one reduce amps?
- Would series vs parallel make a difference in this area?
I have seen designs that include a regulator IC and some capacitors, but if a simple resistor/fuse/diode setup will do the trick, I would really prefer that.
Best Answer
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.