RBB IB

VBE

RC

RE

IE

IC

VCC

VCE

M

VBB

Figure4

RB1

RB2

IB

VBE

RC

RE

IE

IC

VCE

M

Figure3

Static study of the transistorExercise 1: Calculation of the polarization of a transistor using the characteristics

We consider a transistor NPN germanium. The plot characteristics in the laboratory in the common emitter configuration has given the output characteristics, which are represented by the network of Figure 1 and the input characteristic, shown in Figure 2. The operating point is selected on the network output; he has coordinates (5V and 3mA).

1) Preliminary study: Starting from the given characteristics to determine the numerical values:

1-1) of the coordinates of the point of operation on the input characteristic.1-2) The input resistance, the output resistance and the current gain of the transistor

at the operating point selected and the parameter of transistor.1-3) knowing that the limit of the power transistor is 20 mW; draw the hyperbolic

power after fulfilling Table 1 (word doc: complement the work-home)2) Study of the polarization

The transistor is biased by a bridge according to the scheme of Figure 3. The resistors RB1 and RB2 are the bridge base bias. This bridge will be chosen such that the current through RB1 equals 10IB (base current). The RC resistance allows fixing the current IC and the RE resistance allows a better thermal stability. The bias voltage is VCC.2-1) Respecting the rule of half power draw the load line and write its numerical equation.2-2) Neglecting the current IB, express the slope of the load line as components of the mounting.2-3) Show that the circuit in Figure 3 can be brought back to Figure 4. Establish then the expressions of RB and VBB.2-4) Knowing that the slope of line of attack is -12,5.103, Determining the resistors RB1, RB2, VBB, RC, RE, VCC.

RB

IB

VBE

RC

IE

IC

VCC

VCE

MFigure5

Exercice 2 : Calcul de la polarisation sans utilisant les caractéristiques

1) operating modes study Consider network of Fig. 5 wherein BJT is

silicon NPN and for witch: =100, all reverse saturation currents are to be neglected and

saturation voltage limit is equal to 0.3V.The bias power supply is VCC = 10V

and the collector resistance RC= 2K.1-1) How about BJT operating mode?1-2) Assume that if BJT operates in: - Active region then VBE=0.7V,- Saturation region then VCESAT=0.2V and

VBESAT=0.7V.It is desired to determine the current IB, IC and VCE for the following cases:

1-2-1) RB = 300K.1-2-2) RB = 150K.

1-3) How about polarizations in both cases above? Justify by necessary calculations if any.

2) Voltage-Divider Bias: consider now the network of Fig. 3 with = 150. In the active region VBE = 0.7V

and all reverse saturation currents are to be neglected. The saturation voltage limit is equal to 0.3V and VCC = 12V, RB1 = 110K, RB2 = 11K, RC = 1.8 K, RE = 1,2K.

2-1) Determine the Thevenin equivalent base circuit.2-2) Determine IC and VCE

2-3) Determine RC for which the transistor is in saturation limit, the active region current gain is to be used.

3) operating point determination Fig. 3 network parameters are now: VCC = 28V, RB1 = 90K, RB2 = 10K, RC = 6,8K,

RE = 1,2K.3-1) Determine the operating point for the following cases:

3-1-1) b = 603-1-2) b = 1503-2) Compare the results 3-1-1) and 3-1-1) by computing /, IB/IB, IC / IC, VCE / VCE and conclude.

20µA

70µA

10µA

IC(mA)

8

0

1

2

3

4

5

6

7

Vce(V)5 10

30µA

40µA

50µA

60µA

0µA

Figure1

VBE(V)

0

0,1

0,2

IB(µA)50 100

Figure2

VCE en V

Ic= en mA