biasing npn transistor without a resistor

Im designing a board with a pic microcontroller. In the past i have used darlington arrays, which can be driven directly from the microcontroller. I now need to use a single npn transistor as a switch. In most applications that i have seen there is a biasing resistor being used.

I am wondering if i can bias the transistor without using a resistor to drive it directly into saturation. With a maximum output of 20ma from the ucontroller this seems resonable, but i am not sure.

If this does not pose a problem to the transistor or microcontroller can i use this practice in general, even when there is a non regulated amount of current that can flow into the base, where i dont care what the collector current will be and only care that the transistor is in saturation?

In general i would like to know what my limitations are and when i may end up damaging the transistor??

Thanks for your input.

If you connect a transistor directly to a pin, the current will be limited by the internal pullup resistor ( or P-channel transistor ) in the output stage of your uC.
This way your uC pin output voltage will never exceed the Vbc of the transistor, which is c.a. 0.7 V. The voltage drop thru the internal uC resistance will be VCC -Vbe. the power dissipation for this pin will be U*I. for 20mA and Vcc=5V P=0.086W.
Several pins configured like that and your uC will be fried up very quickly.
Another factor is total current your uC can handle for all its pins. Usually it is not more than 100mA. Read Absolute Maximum Ratings for your uC.
All in all it is much better to use even a small resistor in the transistor base.
Else you can't be sure when your uC will fail.