I am trying to trigger a 12 V relay with an arduino input. All I am doing in the Arduino sketch is to make pin 13 go high and low every 1 sec (1000 ms). The relay refuses to get triggered. Can someone please tell me what am I doing wrong? Struggling with this for way too long.
Adding to what other people said in their answers. You must ensure that the current needed to power the coil of your relay doesn't exceed BC548's maximum current rating (with a suitable safety margin).
The maximum collector current for a BC548 is 100mA, so (to be really on the safe side, even when its temperature rises), I wouldn't want to have it switch a current above, say, 50mA. If your relay needs more than 50mA of current, you should choose a BJT with a greater maximum collector current rating.
To check your relay current absorption connect the relay directly across a power supply (12V) and measure the current it draws from it.
0.03A/150=0.0002A=0.2mA. Besides that small computation error, that is the base current that drives the BJT at the edge of the linear region (assuming an hFE=150, but you should design for the worst case, i.e. hFE=110, according to the datasheet). What you need is to switch the BJT completely ON, i.e. to drive it in full saturation independently of any other condition. This means you have to choose a base current which is much greater than that value you have computed. Say ten times bigger: Ib=2mA. This leads you to Rb=(5V-0.7V)/2mA=2.15kOhm (a standard 2.2kOhm will do).
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Commented
Dec 30, 2014 at 12:35
I assume you're actually trying to drive it with an Arduino output- an input won't drive anything.
A lot of power 12V relays need around 100mA (or more) coil current. That means your base current should not be much less than about 5mA to guarantee the transistor is well saturated. If you have a 5V Arduino it might drive to 4.5V at higher current, the transistor base needs 0.7V so that's about 270uA (a lot worse if the micro is 3.3V), so it will only drive about a 5mA relay safely. From the transistor datasheet:
Reduce the 14K resistor to something more like (4.5V - 0.7V)/0.005 = 750\$\Omega\$ and you'll be able to drive a 100mA relay safely (assuming a 5V micro).
Most likely the transistor can't sink enough current to turn on the relay. Look up how much current the relay needs with 12 V applied. Divide that by the minimum guaranteed gain of the transistor. That tells you how much base current you need. If that is more than your digital output can source, then you need more amplification in there somewhere. If the digital output can source the minimum required base current, then size the base resistor accordingly.
Figure 700 mV B-E drop in the transistor. If the digital output drives to 5 V when high, then that leaves 4.3 V across the base resistor. (4.3 V)/(14 kΩ) = 300 µA. If the transistor gain is 50, for example, then that can only support 15 mA of relay current.
Most likely your digital output can source at least 5-10 mA. A few mA base current should be possible, which should be able to support the coil current of a modest relay.
