Power transistor V-I and static switch

A power transistor (BJT) is a three-layer, two-junction device (NPN or PNP) with a vertical structure suited to high voltage and current. Its V-I (output) characteristic is the graph of collector current $i_C$ against collector-emitter voltage $v_{CE}$ for different base currents $I_B$.

V-I characteristics

   iC
    ^ saturation
    | |     IB4
    | |----------------------
    | |     IB3
    | |----------------------  active
    | |     IB2               region
    | |----------------------
    | |     IB1
    | |----------------------
    |/_______________________  IB = 0
    +----------------------------> vCE
            cut-off region

• Cut-off region: $I_B = 0$; both junctions reverse biased; only a small leakage current $I_{CEO}$ flows. The transistor is OFF.
• Active region: base-emitter junction forward biased, collector-base reverse biased; $i_C = \beta I_B$. The device acts as an amplifier, but the power loss $v_{CE} i_C$ is large, so power circuits avoid this region.
• Saturation region: both junctions forward biased. $i_C$ is set by the external load and $v_{CE(sat)}$ is very small (0.2 to 1 V). The transistor is fully ON.
• Quasi-saturation: in power BJTs, the lightly doped collector drift region gives a region between active and hard saturation; operating here keeps storage time small.
• A primary breakdown ($BV_{CEO}$) and a second breakdown limit the safe operating area.

Use as a static switch

                 +Vcc
                  |
                 RL (load)
                  |
          Rb     C|  iC
   vB o--/\/\/---B|/
                  |\  NPN
                 E|
                  |
                 GND

• OFF state (cut-off): when $v_B \le 0$, $I_B = 0$, both junctions are reverse biased and $I_C \approx 0$ (only leakage). The transistor acts as an open switch and the full supply voltage $V_{CC}$ appears across C-E.
• ON state (saturation): when enough base current is supplied, both junctions are forward biased. $V_{CE(sat)}$ is only about 0.2 to 1 V, so the transistor acts as a closed switch and $I_C = \frac{V_{CC} - V_{CE(sat)}}{R_L}$.
• To be sure of saturation, the base current is made larger than the minimum value:

$$I_B > I_{BS} = \frac{I_{CS}}{\beta_{min}}, \qquad \text{ODF} = \frac{I_B}{I_{BS}}$$

where ODF (overdrive factor) is usually 1.5 to 3 (too much overdrive increases storage time).

• The active region is avoided because both $V_{CE}$ and $I_C$ are large there, giving high power loss $V_{CE} I_C$. In the switching mode the loss is small: in cut-off the current is nearly zero, and in saturation the voltage is nearly zero.