Transistor as static switch and base drive

A static switch is a switch with no moving parts. A power transistor works as a static switch by operating only in two states: cut-off (OFF, open switch) and saturation (ON, closed switch), and it is moved between them by its base current.

Transistor 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.

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

The switch has no arcing, no wear, and can be operated at tens of kHz, which is why transistors are used in choppers, inverters and SMPS.

Generation of the base (gate) signal

The base signal is a train of pulses whose width and frequency are set by the control circuit (for example a PWM signal from a 555 timer, comparator or microcontroller). It is amplified and isolated before it reaches the base.

 +------------+   +-----------+   +--------+   +-------+
 | PWM / pulse|-->| isolation |-->| driver |-->| base  |
 | generator  |   | (opto or  |   | amp    |   | of    |
 | (555, uC)  |   | pulse tr.)|   | +/-V   |   | BJT   |
 +------------+   +-----------+   +--------+   +-------+

Base drive (gate signal) circuit: the base drive must (i) give a high base current quickly at turn-on, (ii) keep just enough base current to hold saturation during conduction, and (iii) pull out the stored charge with a negative base current at turn-off.

               C1
          +----||----+
   vin    |          |    R2
   o------+--/\/\/---+--/\/\/--> B
             R1                 (power
                                 BJT)
   GND -------------------------> E

1. Turn-on control: at the rising edge of $v_{in}$ (high level $V_1$), capacitor $C_1$ acts as a short across $R_1$, so a large initial base current $I_{B1} \approx \frac{V_1 - V_{BE}}{R_2}$ flows and the BJT turns ON fast (short $t_d$ and $t_r$).
2. Conduction: $C_1$ charges and the base current settles to $I_B = \frac{V_1 - V_{BE}}{R_1 + R_2}$, which is just enough to keep the transistor in (near) saturation, so the storage time stays small.
3. Turn-off control: when $v_{in}$ goes to zero or negative, the charged $C_1$ forces a reverse base current $-I_{B2}$, which sweeps out the stored base charge quickly and reduces $t_s$ and $t_f$.

Other refinements: a Baker (anti-saturation) clamp diode from base to collector keeps the BJT just out of hard saturation; proportional base drive makes $I_B$ proportional to $I_C$; and an opto-coupler or pulse transformer isolates the logic circuit from the power circuit.