Test: Transistor Switching Times - Electrical Engineering (EE) MCQ

When a pulse is given, the collector current will not reach the steady state value instantaneously because of stray capacitances. The charging and discharging of capacitance makes the current to reach a steady state value after a given time constant.

 At t < 0, the BJT is OFF in cut off region. I_B=0 as β=∞, so I_C=I_E. When t > 0, switch opens and BJT is ON. The voltage across capacitor increases. From the input loop, -5-V_BE-I(4.3K)+10=0 and gives I=1mA. I_C1=1-0.5=0.5mA. V_C1=0.7+4.3+10=-5V. I_C1=C₁dV_C1/dt. From this equation, we get t=50ms.

The technique used to quickly switch off a transistor is by reverse biasing its base to collector junction. It is demonstrated in a high voltage switching circuit. The advantage of this circuit is that it is not necessary to have high voltage control signal.

This method is used to quickly switch off a transistor is by reverse biasing its base to collector junction. It is demonstrated in a high voltage switching circuit. The disadvantage of using the method of reverse biasing base emitter junction is that the output does not switch completely to GND due to forward voltage drop of the diode.

This is an inverter, in which the transistor in the circuit is switched between cut off and saturation. The load, for example, can be a motor or a light emitting diode or any other electrical device.

Connecting a resistor connected from base of a transistor to ground/negative voltage helps in reducing the switching the switching time of the transistor. When transistor saturate, there is stored charge in the base that must be removed before it turns off.

When sufficient charge carriers exist, the transistor goes into saturation. When the switch is turned off, in order to go into cut off, the charge carriers in the base region need to leave. The longer it takes to leave, the longer it takes for a transistor to turn from saturation to cut off.

Sometimes DC current gain of a bipolar transistor is too low to directly switch the load current or voltage, so multiple switching transistors is used. The load is connected to ground and the transistor switches the power to it.

From the cut off characteristics, the base emitter voltage (V_BE) in a cut off region is less than 0.7V. The cut off region can be considered as ‘off mode’. Here, V_BE < 0.7 and I_C=0. For a PNP transistor, the emitter potential must be negative with respect to the base.

Switching leads to move holes in P region to N region as minority carriers. Removal of this accumulation determines switching speed. P+ regards to a diode in which the p type is doped excessively.