BJT as a switch in power circuit

A BJT is used as a switch in power circuits by driving it either fully OFF (cut-off) or fully ON (saturation) with its base current, never in the active region. In this way it behaves like an ideal ON/OFF switch with very small power loss.

Circuit and operation

                 +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.

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

Switching waveforms and losses

 vB   ^ ____          ____
      ||    |        |    |
      |+    +--------+    +----
 vCE  ^     ______        ______
      |    |      |      |
 Vcc  |----'      |------'
 ~0.2 |           '__ON__
 iC   ^ ___        ___
      ||   |      |   |
      |'   '------'   '-----> t

• Conduction loss $= V_{CE(sat)} I_C \times$ duty ratio (small because $V_{CE(sat)}$ is small).
• Switching loss occurs during the short turn-on and turn-off intervals; it rises with frequency.

Example

For $V_{CC} = 200$ V, $R_L = 10\ \Omega$, $V_{CE(sat)} = 1$ V and $\beta_{min} = 20$: $I_{CS} = (200 - 1)/10 = 19.9$ A, $I_{BS} = 19.9/20 = 0.995$ A. With ODF = 2, the base drive is designed for about 2 A.

Points to note

• A freewheeling diode is connected across inductive loads to protect the BJT from voltage spikes at turn-off.
• A negative base current at turn-off reduces storage time.
• The BJT is a current-controlled device, so its driver must supply a continuous base current; for higher frequency a MOSFET or IGBT is often preferred.