Usage of a transistor configured as diode
NMOS connected in diode configuration:







simulate this circuit &ndash; Schematic created using CircuitLab

Since Gate and Drain are shorted, the following saturation condition always holds:

$$V_{DS}&gt;V_{GS}-V_T$$

This means that once \$V_{DS}&gt;V_T\$ the transistor both begins to conduct and enters saturation.

In saturation (after substitution \$V_{GS}=V_{DS}\$ for diode mode):

$$I_{DS}=\mu C_{ox} \frac{W}{2L} (V_{DS}-V_T)^2$$

The equivalent resistance of this device is:

$$R=\frac {V_{DS}}{I_{DS}}=\frac{2L}{W} \frac{1}{\mu C_{ox}} \frac{V_{DS}}{(V_{DS}-V_T)^2}$$

Now you can see that the equivalent resistance can be controlled by changing the dimensions of the transistor (\$W\$, \$L\$). 

However, this resistance is not constant - it depends on the applied bias. This is bad, but it is not that you have too many alternatives in integrated circuits (you can implement precision resisters by various techniques, but they are usually costly). 

On the positive side - there are many application which do not require precision in resistances.

Can you implement a big resistor with diode connected transistor? Yes. There are two approaches:


Long and narrow transistor
Ensure that \$V_{DS}\$ does not rise much over \$V_T\$


However, "big" resistor in integrated circuit is not the same as big resistor as discrete component - in integrated circuit all resistances are relatively low.