You may consider using a diode on the 1st output pin. This should keep the signal from routing from the 2nd output pin into the 1st.
But all depending on the reason you would need to do this (Which doesnt seem to make sense to me yet)
Why would you want to connect 2 output pins to 1 input pin on the same IC?
Is this a microcontroller?
If so, you dont want to do this. The input pins can typically sink 25ma and each output can source 25ma. So you will be trying to sink 50ma into one input. Not a good idea.
I'll explain why I want to connect two output pins to an input pin.
The input pin is the RSTn pin of an 8051 Uc.
Output pin #1 is a pull-up switch (when the switch is held down, it grounds the RSTn pin of the Uc).
Output pin #2 is a RSTn signal of an evaluation board, which I use in order to burn firmware into the Uc and to debug the Uc.
The EV board has 5 signals going to the Uc - RSTn, DD (Debug Data), DC, GND, VDD.
I may not be the best person for this thread, as I have 0 experience with the 8051.
But, It the #1 pin is a pull-up, and not a pull-down, It will not pull the rst pin to ground. Now if it is both, cool, ive never seen that.
If you need RST held low, you should just tie it to ground with a pull-down resistor. If you need variability, I would connect it to ground through a transistor. The transistor can be switched by the pull up switch. When it is active, pin 2 will be pulled to ground.
There may be an internal need for the 8051 to have things this way, but it seems unlikely.
Are you tying VDD to the reset pin and ground?
I think you may be confusing which pins are set as inputs and outputs as a default before anything is programmed.
Well, the button is normally open, so normally, no current flows through the pull-up resistor.
Meaning, currently, when no one presses the button, the voltage of the Uc's RSTn pin is VDD.
However, when the button is pressed, current is flowing through the resistor (I = VDD / 43KΩ) and therefore the voltage of the Uc's RSTn pin is 0V (GND).
I don't see how the RSTn pin of the Uc is floating, as its either connected to VDD (throught a pull-up resistor) or connected to ground when the switch is pressed.
I already tested the functionality of this circuit and it works well.
The problem is that ORCAD issues an error as 2 output pins are connected to each other in the schematics.
Well, I think that we agreed that adding a resistor won't prevent the two output pins from being connected to each other.
Moreover, the RSTn pin isn't floating so I dont see why a pull-down resistor is needed.
Also, I'm intending to send the ORCAD schematics to a PCB designer and I want the schematics to be with no errors (and everything that is placed there will appear on the PCB eventually).