# BJT triode region

As we know in JFET or MOSFET there is a triode or Linear region, Where for a fixed gate-source voltage, increasing drain source voltage is equivalent to applying drain source voltage across a resistor (approximately) . My question is whether there is a similar region for BJT in which the bjt would work as if its a resistor, though I know only about cut-off, active, and saturation region in BJT.

• No a BJT does not have such a "linear" region. That's why in some circuits where an adjustable resistor component is needed, often a JFET is used. Mar 20, 2021 at 13:17
• @Bimpelrekkie comment -> answer? Mar 20, 2021 at 13:18
• The BJT saturation region has the same role, it's just not all that linear though.
– user16324
Mar 20, 2021 at 13:57

As @Brian says, in the saturation region this is true. The transistor has a dynamic resistance (for very small voltage changes) of approximately Vt/Ib.

So, for example, the collector of an NPN transistor with 50uA of base current will behave approximately like a 500 ohm resistor to ground for very small voltage changes (mV or less) that are AC-coupled. Remember that dynamic resistance is $$\r_d=\Delta V/\Delta I\$$.

simulate this circuit – Schematic created using CircuitLab

For larger voltage changes the nonlinearity will become very obvious.

• So you are assuming this is independent of hFE.? Mar 20, 2021 at 14:33
• @TonyStewartSunnyskyguyEE75 Yes, it is. For hFE >> 1 Mar 20, 2021 at 14:33
• Over what range of Ib? Mar 20, 2021 at 14:34
• Should be pretty broad so long as the transistor is well into saturation. Similar to the dynamic resistance of a diode. Mar 20, 2021 at 14:38
• Yes saturation is the keyword here where hFE drops rapidly and becomes less sensitive using hFE/10 then Rce is the limiting factor as the 500 Ohms drops to Vce(sat) /Ic or 2 Ohms (bulk resistance)for a PN2222A which is a function of <1/Pmax rating and super beta types of the chip. (Stewart’s law) Mar 20, 2021 at 15:14

No - the "quasilinear" region of the BJT cannot be used as a (grounded) resistor (as it is the case for the FET). Two main reasons:

• The Ic=f(Vce) set of curves does NOT cross the origin

• Even in the vicinity of the origin in the 3rd quadrant, these curves are NOT symmetrical to the "normal" first quadrant (for positive values of Ic and Vce).

If the load voltage is fixed and there is an emitter resistor greater than Rbe to linearize the input impedance hFE*(Rbe+Re) then using a fixed input voltage and variable Rb will amplify conduction to appear as Rb/hFE as a common emitter amplifier.

This is one way to make an Active Load . The other way is to use a fixed Rb and vary the Vin to control load current as a fixed resistance Ic ~= Ib * hFE, where Ib = (Vin - 0.7(~) )/Rb. So the load resistance is proportion to Vin or Rin. Current feedback can make this even more linear.