Power transistor switch and opto-coupler isolation

A power transistor is used as a switch by driving it into saturation (ON) or cut-off (OFF) with its base current, so it behaves like a closed or open switch with very small power loss.

Power transistor as a 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 make sure the transistor saturates, 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 the storage time).

• The active region is avoided because both $V_{CE}$ and $I_C$ are large there, giving a high power loss $V_{CE} I_C$. In 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

Isolation of the drive circuit using an opto-coupler

Opto-coupler isolation: the gate (base) signal generator (logic level, grounded to the control circuit) and the power circuit (hundreds of volts, emitter floating) must be electrically separated. An opto-coupler does this using light.

   control side     |      power side
                    |          + Vaux
    R1              |          |
 o--/\/\--+         |          R2
 pulse    |         |          |
         LED ~~~~>  |   photo-transistor
          |  light  |   (C top, E below)
 o--------+         |          |
 GND                |          +------> B (base of power transistor)
                    |          |
                    |          R3
                    |          |
                    |  Vaux(-) +------> E (emitter of power transistor)

1. The control pulse drives a current through the LED via resistor $R_1$.
2. The LED emits infrared light, which falls on the phototransistor (or photo-darlington) inside the same package.
3. The phototransistor turns ON and connects the auxiliary supply $V_{aux}$ (referenced to the emitter of the power transistor) through $R_2$ to its base, supplying base current. $R_3$ prevents false turn-on by noise.
4. When the pulse ends, the LED goes dark, the phototransistor turns OFF and the base current stops.

There is no electrical path between the two sides, so the isolation can withstand a few kV, and noise from the power circuit does not reach the logic circuit.