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Bias Supply and Temperature Compensation for Transistors

Intrinsic Conduction | Extrinsic Conduction | Diffusion Currents in Semiconductors | The Capacitances of a Semiconductor Diode | Semiconductor Diodes as Rectifiers | Structures of Semiconductor Diodes | The Tunnel and Inversed Diodes | Microwave Semiconductor Diodes | Physical Processes in a Transistor | The Basic Circuit Configurations of Transistors |


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It is usual to employ a single source, that of the output circuit, to energize a transistor stage EC. For the transistor to operate normally, it is essential that a d. с voltage of a few tenths of a volt

(UBE = 0.7V), known as the base bias voltage, be maintained between the emitter and base.

Figure 4.16,shows how the bias voltage is derived with the aid of a divider R1R2 in a CE stage. Here, the greater proportion of EC is dropped across R1, and the smaller part, serving as the bias voltage UBE is dropped across R2 which is placed in shunt with the transistor input.

.

The d. с. blocking capacitor Cb serves to pass the a. c. voltage to be amplified to the transistor input. More often, Cb is chosen to have a capacitance of several microfarads or tens of microfarads. For this reason, Cb in low-frequen­cy circuits is usually a small-sized electrolytic capacitor.

Fig. 4.16. Base biasing circuit and temperature compensation of a transistor

 

The scheme shown in Fig. 4.16 be called the emitter temperature-stabilized scheme. The resistor RE sets up negative feedback in terms of direct current. Should a rise in temperature cause the currents in the transistor to rise, the increase in IE0 will raise the emitter voltage UE, and lower the base bias voltage UBE in proportion, thus leading to a fall in the currents. The resistor RE is placed in shunt with a sufficiently high-valued capacitor CE so as to prevent it from setting up negative feedback in terms of alternating current.

Neglecting UBE in comparison with the other voltages, the resistance values for the emitter compensation scheme can be calculated, using the following approximate equations:

R1 ≈ (ЕC – UE)/(IB + I1); R2 ≈ UE/I1; RE = UE/IE .

A further requirement is to choose the value of UE in view of the likely increase in EC. The divider current is usually anywhere between three to five times IB0.

 


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