Implement a simple circuit using logic doors

Thread Starter

LordVader12

Joined Nov 10, 2019
2
I need to implement a circuit but Im having some trouble since Im novice on electronics.
I have 4 inputs. A being ON/OFF switch. B is an Inibitor. C is an Emergency and D is not an adequate ocasion.
In order to turn off the lamp I have to meet 2 conditions.
  • Condition 1: Switch is ON (A), Inibitor is OFF (B) and the ocasion must be adequate (NOT D).
  • Condition 2: Emergency is ON (C). This one is independent and will always turn off the light as long as it is activated.
I came up with this:


I did this schematic. NOR only outputs 1 when both are 0, which means INIBITOR off and D is adequate. Then join with On from Switch with AND and light is ON.

This is represented by A(NOT)B(NOT)D. Now I only need to add the C variable which is independent.

The final equation is A(NOT)B(NOT)D + C. Am I thinking right? Sorry if this sounds stupid but Im still learning and a bit confused!

Thanks in advance.
 

Thread Starter

LordVader12

Joined Nov 10, 2019
2
TYPO: Condition 2: Emergency is ON (C). This one is independent and will always turn ON the light as long as it is activated.
 

ci139

Joined Jul 11, 2016
791
  • Condition 1: Switch is ON (A), Inhibitor is OFF (B) and the occasion must be adequate (NOT D).
  • Condition 2: Emergency is ON (C). This one is independent and will always turn off the light as long as it is activated.
    ((((is this a text game or an Advanced English Technical Poetry))))
  • ... so you need to preffer the Condition 2 over Condition 1 ← there are several ways to achieve/implement such ...
otherwise Light.state.OFF = ( (A=1) , (B=0) , (D=0) , (C=1) ) or (C=1) -- if i'm not stupid the control is done by C=1 only ...
... why → (the simplified condition) Y = EC + C = EC + C(E + E̅) = EC + EC + E̅C = EC + E̅C = (E + E̅)C = C // about : Boolean identities -- you need to be logged out or open it in defferent browser to read
rules change "independent and will always turn ON the light as long as it is activated. "
changes the Light.state.OFF = ( (A=1) , (B=0) , (D=0) , (C=0) ) and not (C=1) ← the easiest way to implement it is likely
- - - - - -- - Light.state.OFF = ( ( not(A)=0 ) , (B=0) , (D=0) , (C=0) ) // 0=False , 1=True , (expression of comparison) has the range of [0,1]∈Z


however , if it's "one of the following conditions has to be valid"

then Light.state.OFF = ( (A=1) , (B=0) , (D=0) , (C=C) ) or (C=1)
 
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