Transistor switch with opto-coupler base drive

A transistor acts as a static switch (no moving parts) when it is operated only in cut-off (OFF) and saturation (ON), switched 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.

Base current signal generating circuit using an opto-coupler

 control side      |          power side
 +--------+        |         +--------------+
 | pulse  |  LED   | photo-  | totem pole   |-Rb-> B
 | (PWM)  |--|>~~~~~~> tran- | Q1 to +Vaux  |   (power
 +--------+        |  sistor | Q2 to -Vaux  |    BJT)
                   |         +--------------+---> E
     only light crosses the isolation barrier

1. The control pulse (PWM from a 555 timer or microcontroller) lights the LED of the opto-coupler through a current-limiting resistor $R_1$.
2. The phototransistor turns ON and drives the totem-pole stage $Q_1$-$Q_2$, which is powered by an isolated auxiliary supply referenced to the emitter of the power BJT.
3. When the pulse is high, $Q_1$ conducts and supplies positive base current $I_{B1}$ through $R_b$; the power transistor saturates.
4. When the pulse is low, $Q_2$ conducts and connects the base to $-V_{aux}$, giving a negative base current that removes stored charge and turns the power BJT OFF quickly.
5. Since the only link between the two sides is light, the logic circuit is fully isolated from the high-voltage power circuit.