IdahoMan wrote:Just to be clear on the objective: When the PIR triggers, the DVR terminals (A-In and GND) should be momentarily connected. That's it. (Assuming the DVR's circuitry is sensitive enough that it will activate on even a split-second of the terminal's connecting,its programing and OSD can do the rest)
Correct. The transistor effectively becomes a short
between it's collector and emitter, thus grounding A-In to start the recording.
IdahoMan wrote:-The transistor will activate and all the power will then flow from DVR GND to E, through the transistor, and from C to DVR A-In. Consequently, the closing of DVR GND and A-In would deprive power from the PIR and deactivate the transistor. Am I getting this?
Not quite. Recall the underlined above. What would happen if you connected a jumper cable between your cars battery + and - terminals, aka shorted the + supply to car ground ?
Nothing good ! The battery would try to send gobs of current through the low resistance wire (Ohms law
) and both would heat up. The battery might explode (literally) or the wire melt open (like a fuse) or both. Same thing here, the supply to the PIR (not shown but assumed above) would burn out or the transistor might blow open.
Now you might be thinking "Hey I measured 3.3 v at A-In. Why can't that supply power to the PIR, just as shown in my diagram ? No other wire or supply needed." The answer to that lies in (what I believe is) the A-In circuit. It's an input to some logic gate that can't source any current. To put that gate into a known state and prevent false triggers there's also a "pullup" resistor
tied between the A-In input and Vcc (3.3v).
That's why you measured 3.3v. But if you draw current from A-In, it must go through the "pullup" resistor and that means a voltage drop across that resistor (Ohms law again). Therefore the voltage at A-In is;
Va-in = 3.3v - I*Rpu ; I = current draw, Rpu = the pullup resistance, probably 10k to 50 k ohms.
It doesn't take much current draw (30 uA max) by the PIR to drop 3.3v below the min working voltage of 3.0v.
Besides if the transistor did short out the supply to the PIR (and not burn up), where would the voltage come from to drive the base of the transistor ? How could PIR-out go high and stay high ? The DVR stops recording when A-In becomes unshorted, at least that's what I get from the Chinglish.
http://www.foxtechfpv.com/product/minid ... manual.pdf
It'll record when grounding the point(Alarm_in), and stop record when undo.
Is this, if true, a problem ?