Blinking led with PIC16F84A multisim

pic16f84a led

I test MULTISIM circuit make blinking led with PIC 16f84.
Assembler code is ok and working good but if i attach one LED with resitor over port RB0 pic pic16f84 this led dont blink becouse the level in OUT over port go to 0. Is same led and resitor make short circuit over port. If i attach other display device or logic probe all is OK, blinking, and i see level V move to 0 at 5 V.
Same circuit make in PROTEUS working good also with LED and resistor and blinking OK without problem.
I think MULTISIM when make simulation PIC dont make possible drive led device over PORTS but only possible use display with external power source.
I want know if this is bug of MULTISIM ???? or maybe need setup some option for make possible PIC working same real device ???? and driving also LED with resistor.
Many thanks.
Edgar

pic16f84 led

Just to give an answer to this question for someone who reads this question:

Inversing the logic can be a solution, if the OUT port do not give enough current (mA) to drive the LED.

So, you can plug a resistor (maybe 500 ohms) in the +5V => then your LED => and you plug it into the OUT port. In your ASM program, instead of lighting it with a 1, light it with a 0 and turn it of with a 1. If you don't know, "0", in logic circuit, acts like a ground and 1 as a +5V opposition to another +5V if you plug it in the way I describe, which is common anode (meaning all plugged in the same +) instead of common cathode (all plugged in the same - ). To understand these two methods is very important when you're a beginner and want to design (ie: PIC16) circuits in Multisim and/or on a breadboard. Also, sometime this is easier and maybe funnier to design the real PIC16 circuit on a breadboard, since you take the time to make the good connections and an acurate ASM code.

If you still want to use common cathode method, Out-> Resistor -> Led -> Gnd, then you can choose in the led properties in Multisim how many current your LED need to light, but this is less real. And (i think) you're better not to take current in your circuit, anyway, it takes its current in the same source.