I am trying to create an electrical diagram of what I know, so bear with me...
I will edit this post to include the image later okay.
So I created like two schematics thus far.
The problem is that I want to use a capacitor to delay the activation of the circuit, but because of the high resistances involved there may not be enough current to trigger the gates.
That is a recurring problem in my schematic :p.
The numbers aren't there yet but this is a 12V
starts flowing as soon as its base is connected (closed).
doesn't flow then P2
is open because it offloads some current through its base through R3 and into ground.
So what is it? It is a circuit that is activated when a certain voltage is not there, in other words when the main load of the system (not shown) draws very little energy. Load Q is a transmitter that will shut off a relay that powers the whole thing.
As soon as P2[b] is open [b]C1
will start accumulating charge (voltage) until it reaches a certain amount equal to Z1[b]. At that point a little bit of voltage (and current) will get into [b]N2
, probably enough to trigger it because N2
should block if it can't open its gate.
(Of course in electronics that is called closing a circuit).
As N2 opens (closes) P3
will start to allow current (at 12V) into Q that I hope won't block. Through N2, I mean.
As current flows through P3
into its base, it will also flow into P4
. P4 then opens (closes) because it can offload about 10µA through R4 and R5, and more through R4 and into C2. (I assume the voltage drop would be 12V for R4 and so would take at least 52 µA at the designated ohmage). The gate has an ampliciation factor of 190 at 53µA and would yield some 10mA through the gate).
"opens" 12V now arrives at the base of 'gate' N2
. This will throw it wide open (if it wasn't already) and all parts of that picture are now saturated with 12V until C2 also becomes saturated with some 11V.
Because R5 > R4 (at least 4 x greater) current accumulates in C2 and doesn't flow out R5 so easily.
As C2 fills up the base of P4
saturates with 12V and closes it. (Actually just 11V but that is enough).
By closing P4 the feedback loop to N2
stops. Unfortunately C1 is now also saturated with 12V and will take so long to drain
, that C2 will have drained before it through R5 :p. Which would activate the cycle again and it would never stop :p.
All it needs though is a forward diode at Z1
to stop the current from flowing there. The voltage at the base of N2 should then quickly dissipate to Q. However all of this won't work probably :p.
What I really think will happen though is that P3 and N2 will keep each other in an equilibrium where the actual voltage is 11.4V after P3, at the base of N2, and at Q. How well it will conduct I do not know :p.
So, not as certain as I thought I would be :p.