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.


  • Not new but still learner *
  • \$\begingroup\$ You could use an NPN but you'd be better off using a P channel MOSFET as a high side switch ( +12V side) . How are you intending to operate this switch? \$\endgroup\$ May 23 '16 at 9:25
  • \$\begingroup\$ @JImDearden: if he used a P channel MOSFET as a high side switch he's have to connect the MOSFET's source to +12. Then, since his PIR outputs only either 3.3V or zero volts he'd never be able to turn the MOSFET OFF. \$\endgroup\$
    – EM Fields
    May 23 '16 at 9:42
  • \$\begingroup\$ @EMFields My mistake - I mis-read the question \$\endgroup\$ May 23 '16 at 9:45
  • \$\begingroup\$ Probably the best solution would be an N-channel logic level MOSFET wired as a low side switch, but what does the circuit you want to switch look like? \$\endgroup\$
    – EM Fields
    May 23 '16 at 9:51

enter image description here

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;

enter image description here

You could use a low side N channel switch or a high side PNP switch.

  • \$\begingroup\$ Hi Jim. Thanks. Your circuit looks promising. Can I know what MOSFET to use? \$\endgroup\$
    – Amit Kumar
    May 23 '16 at 11:46
  • \$\begingroup\$ @AmitKumar What your looking for on the p mosfet is something with an Ids (current drain-source) of 2A or more - there are lots of them. For the N mosfet you must choose a logic level device (switched by a low voltage, 3v3, on the gate), again anything that can switch 2A or more is good.. The 2A minimum will give you a good safety margin for a 1A load. (if you go the transistor route I'd be looking for a 3A min. rating for the collector current. Its a good exercise to do the search for yourself (you'll learn how to select an appropriate device by looking at datasheets) \$\endgroup\$ May 23 '16 at 12:45
  • \$\begingroup\$ Thanks a lot. I tried to find the MOSFET at jaycar.com.au/Active-Components/c/… but couldn't found higher Amps. My findings are 2N7000 and VN10KM. \$\endgroup\$
    – Amit Kumar
    May 25 '16 at 0:36
  • \$\begingroup\$ N7000 too small, not enough current capacity, VN10KM not a logic level device. You need to search for a 'logic level N channel mosfet' Here's an example of the sort of datasheet you should expect. fairchildsemi.com/datasheets/RF/RFP12N10L.pdf. The device is rated at 12A , 100V, which exceeds the minimum requirements (good). The important thing is the statement " facilitating true on-off power control directly from logic circuit supply voltages." Most mosfets need higher inputs (about 10v) so its really important to select the correct type. \$\endgroup\$ May 25 '16 at 14:50
  • \$\begingroup\$ Hi Jim. In the datasheet for VN10KM (jaycar.com.au/Active-Components/Discrete/FETs/…) the rds(on) is VGS=5V, ID=0.2A so I thought it should work. I don't think my local store (JayCar) has many MOSFETs for my need. I may have to look in Ebay or somewhere else. And yes, I have learned something new. Thanks to you. \$\endgroup\$
    – Amit Kumar
    May 26 '16 at 11:27

You could use two transistor, one npn and one pnp. A circuit like this should work (from the venerable "Horowitz").

enter image description here

Don't worry about voltages, work also with your 12V and 3.3.

I used \$Q_2=\$2SC4604 and \$Q_3\$=2SA1761.


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