Of course it makes a difference. When you learn about digital logic then you will see that when the inputs states of a logic device are high or is low then the output state is determined.When unused CMOS input pins are tied to +VCC or ground does it make a difference if the unused pins are tied to +VCC or ground?
When you learn about Cmos logic and read a Cmos datasheet then you will see that it makes no difference because the supply current of an unloaded Cmos logic device is almost nothing when its inputs are high or low.Does tied unused pins to +VCC (or ground) draw more current from the power supply?
Don't you mean to say " the logic input" of that particular pin? You do not want to short an active output.its depends about the logic output of that particular pin. If the pin is active low need to always pulled-up and if it is active high then need to pulled down
That will not work if you have an AND or NAND gate because if any one (or more) input is low then the output locks up.For the 4000 Series of CMos IC's, It is usually Recommended to Ground, UNUSED INPUT PINS.
That will not work if you have an AND or NAND gate because if any one (or more) input is low then the output locks up.
An example is when a CD4011 two inputs NAND gate is used as an inverter. The unused input must be high because if it is low then the output will always be high.
An unused input on an AND or NAND gate must be high (not grounded) or the gate or resulting inverter (NOT gate?) will not work.
You are wrong.There are NO unused input pins, on ANY used gate.
Code Verilog - [expand] 1 2 3 4 5 6 7 8 9 10 11 module inv ( output o, input i ); nand3 nand3_inst ( .O (o) .I0 (i), .I1 (i), .I2 (i) ); endmodule
Don't you mean to say " the logic input" of that particular pin? You do not want to short an active output.
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