hi Eric sorry to but in another persons post again but i am trying to improve my circuit reading skills(not doing very well though),i wanted to ask what are the diodes in your diagram for?i mean it will send a high when the other side is grounded?rite
And why the RC circuit on the manual switch side i mean the signal wont go anyways if the switch is not pressed?
hi,
The two diodes and the resistor up to +V form an AND gate, both inputs
[cathodes have to be high if the junction of the two anodes is able to go high].
The transistor input [base] has to be low so that AND gate can be enabled.
So, when 'c & g' are high and 'a' is low, pins 1 and 2 of the 4011 will be high.
Pin 3 will go low and the Cap discharged, the RESET pulse is via pins8/9 thru a differentiator C/R to the 40110 RESET pin.
This RESET will clear the inputs to pin 1/2 of the 4011 and the RESET pulse will disappear.[the diode AND gate will open]
As pin3 goes low, pins 5/6 also go low, so pin4 goes high and the clock pulse goes low on pin 11.
As the R/C on pins 5/6 charge towards +V, eventually pin4 will go low and pin 11 go high,, its this high edge which clocks the 40110.
The R/C delay isnt on the manual switch side.?
Its down 'stream' of the R/C and RESET.
Do you follow, if not please ask.
EDIT:
Added image of 2inp AND, using diodes and a resistor.
Both A and B have to be high in order for the Output to go high