Step 3: Approach 1: Digitized Sensor Output - Simplest way
1. Using a #62 drill bit drill a hole through the circuit board from the copper side at
the locations marked Sd and Jmp1 in the photo. The locations are indicated
by a yellow dot. Both locations have a convenient little circular pad that
we can use. If you can not find a #62 drill bit, see if your hardware store
carries a #60, this will work. A drill bit especially made for PCB drilling
works the best.
2. Cut the trace that is next to Jmp 1. You might have to click the "i" at the corner
of the photo to see the location clearly.
3. Place a wire jumper on the non-copper side of the PCB between Sd and
Jmp1 and solder.
4. Remove the switch from the black female connector's wires. The black
male connector now has the conditioned digital sensor signal and ground.
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I believe that your question has 4 parts:
O1: Can the Vcc for the module be 5V?
A1: Yes, I have used the module in circuits from 3V to 6V. I do not know
the BVCDO of the transistors or the BV rating of the capacitors
in the module, but they are greater than 6V.
Q2: Can I drive a 5V reed relay directly from the Sd node on the
module?
A2: I am going to make 2 assumptions here;
1. You are using a reed relay similar to Radio Shack part
275-232 (i.e., 5V, 250 ohm coil). So you are asking
can the Sd node source 20 mA.
2. Since the Sd node goes low when the PIR is upset,
the coil would be energized most of the time.
The Sd node is the collector of transistor Q4, which is connected
to Vcc through resistor R14. R14 is ~100 ohms. At 20 mA, the
voltage at Sd would be 3V (i.e., Vsd = 5V - (100 ohms * 20 mA)).
I do not know if this is enough to energize the relay's coil. I would
have to try it to find out. If you use the diode protection I mention
below, I do not think that you will damage the module.
Q3: Would I need to do anything else to the circuit?
A3: You need to put a reverse diode across the relay coil to
handle the inductive kick when the relay switches.
Q4: What if that doesn't work?
A4; I would go through the Instructible a little farther and make the
power transistor and its pre-amp accessible (node R). It was
designed for this type of purpose and has the necessary diode
in place. It would also have the additional advantage of inverting
the signal so that the relay coil would not be energized except
when the PIR sensor was upset.
.
I just rethought my answer a bit. If you...
1. Connect one end of the relay coil to Vcc and the other end to the SD
node.
2. Put in the reverse biased diode to handle the inductive kick across
the coil.
3. Disconnect or remove R14.
Q4 should be able to handle a small reed relay directly. It would also
only be activated when a disturbance occurred.
Hope this helps.
The PIR will give multiple trips as someone moves or walks across its path. It will detect movement or sudden changes in IR radiation (i.e., an object in shade suddenly becomes an object in direct sun light or a warm object, say leaves on a tree in sunlight, start to wave with the wind).
The Sd node doesn't latch or lock itself out for a delay. It wil report the continous status of the PIR sensor. When the sensor is unbalanced, the node will go from Vcc to ~gnd. See the last photo
on step 3.
You might be interested in http://www.instructables.com/id/Cheap-Motion-Detection-Wildlife-Camera/ for some ideas how to handle these issues.