npnpirswitches
I am trying to switch on a circuit using PIR motion detection.
I have a PIR module which outputs 3.3v 1A DC when it detects a motion. I want to switch on a circuit rated at 12v 1A DC. I read some articles that we can use NPN transistor to switch on the load.
But I am not able to figure out the value of transistor to use and resistors if required. Can someone please help.
A complete circuit diagram will be really helpful.
Regards
Add a fly-back diode.
When switching the current off, a coil will reduce the voltage until something starts to conduct so it can continue to let the current flow, as long as it has magnetic energy inside. Sadly, the conductive path is provided by your transistor -- after it was fried.
That negative pulse can reach thousands of volts. Seriously!
The TIP120 is a Darlington transistor pair, which consists of two transistors with the Emitter of the first one connected to the Base of the second one. This provides very high current gain, but doubles the required Base bias voltage and increases the Collector saturation voltage.
10mA Base current is plenty enough to 'fully switch' the TIP120, but it has a minimum Collector-Emitter saturation voltage of about 0.75V (rising to 1V at 2A). You will never get the full voltage out of your battery using this transistor.
If you are switching the positive battery lead with an NPN transistor then it has to be wired in Emitter Follower configuration, and the output voltage will be less than the control voltage (3.3V in your case). To get full output voltage you must use a PNP transistor or a P-channel MOSFET, with level shifting to ensure that it is correctly biased. A MOSFET has the advantages of negligible bias current and lower saturation voltage than a typical bipolar transistor.
The level shifting circuit below uses a general purpose NPN transistor to drive either a PNP power transistor or a P-Channel MOSFET. With R1 at 150Ω it delivers ~20mA bias current to the PNP transistor. The MOSFET requires no bias current, but its Gate threshold voltage must be 1V or less to fully turn on with 3.7V.
simulate this circuit – Schematic created using CircuitLab
Best Answer
The circuits show two different ways of switching the load.
The first uses NPN transistors connected as a Darlington pair as a low side switch. This simple circuit has the advantage of high current gain and high output current.
When the output from the PIR exceeds 1.2V (2 x Vbe drops) the transistors turn ON and current will flow through the load. If the load is inductive (relay coil, motor) then add a diode across the load to prevent damage from back emf (negative voltage spike when the load current is turned off).
The second circuit shows a high side p channel mosfet switch. In this case when the output from the PIR goes high it turns on T1 which pulls the gate of Q2 low, turning the mosfet on. This circuit has the advantage that one side of the load is connected to ground. Again, if the load is inductive, add a diode across it.
There are lots of suitably rated transistors and mosfets that can be substituted for the ones shown. You can make up your own Darlington pair using separate transistors (e.g. 2N3904 + 2N4922 (medium power))
Just for completeness;
You could use a low side N channel switch or a high side PNP switch.