Electronic – Usage of a transistor configured as diode


I faced lots of circuits that has transistors connected as diodes (gate connected to the drain). I know some of these circuits employ such transistor for safety reason but I couldn't figure out the reasons of others.

My question: Is there any reason behind connecting the transistors as diodes other than the one I mentioned? Some of my colleagues suggest that they may be used to realize high resistance but I think connecting the transistor in such configuration (\$V_{g} = V_{d}\$) will force the transistor to work in saturation region, not in linear region. Am I right?

Best Answer

NMOS connected in diode configuration:


simulate this circuit – Schematic created using CircuitLab

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


This means that once \$V_{DS}>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.