Analysis method of PNP transistor

Many textbooks introduce the analysis method of NPN transistor, but for the PNP type transistor, the analytical method is similar to that of the former. How a similar law, but not detailed description. The following is an introduction to the PNP type three-tube analysis method:

We know that NPN transistors have three working areas (cutoff area, amplification area, saturation zone), and the same PNP transistor also has three working areas. The following figure is an input-output characteristic curve for two types of transistors:

Fig. 1 characteristic curve of NPN type transistor

Fig. 2 Characteristic curve of PnP type transistor

It is observed that the input and output characteristic curves of NPN transistor are in the first quadrant, while the PNP type transistor is in the third quadrant. It is indicated that the current and voltage of PNP transistor are opposite to that of NPN transistor, so the voltage and current can be taken absolute value when analyzing the working state of the PNP transistor. The subsequent analytical method is completely similar to the NPN tube. The following examples illustrate:

Example: A switch circuit known to be composed of PNP tubes is shown in Figure 3. If the conduction voltage Ube = -0.1v, saturation uces= 0.1V, ask: The UI is 0V, -2V and -5v, the working state of the pipe, the corresponding uo of how many volts.

Figure 3

Solution: (1) when u1 = 0V, the transistor is in a cut-off state (the pressure drop on RB is not considered during analysis).

(2) When u1 = -2v, transistor conduction, transistor may work in the amplification area, may also work in the saturation zone, at this time assuming that the transistor works in the amplification area.

| Ib | = (| u1|-| ube|) /RB = 0.0475mA

| ic| = Beta*ib = 4,75ma

The voltage drop at both ends of the RC is:

URc = IC*RC = 4.75V

Uo = uRc-10 = -5.25v

UBC = Ub-uc = Ub-uo = 5.15V

So transistor sets the inverse bias, assuming that the transistor works in the amplification area.

(3) When U1 = -5v, transistor conduction, the same (2) transistor may also have two operating states, assuming that it works in the amplification area.

| Ib | = (| u1|-| ube|) /RB = 0.1125mA

| ic| = beta*| ib| = 11.25mA

URc = IC*RC = 11.25V

Uo = uRc-10 = 1.25V

UBC = Ub-uc = Ub-uo =-1.15

Transistor collector Junction is biased, assuming that the transistor is operating in the saturation zone.

Because uces = 0.1V

So the output circuit can be equivalent to the following image

Uo = Uce = -0.1v

Summary: The value of the voltage and current of the PNP tube is absolute, just to correspond with the analytical method of the NPN tube. The actual analysis, can also not take absolute value, according to the general analysis method. That is to determine whether the transistor's transmit junction is conduction, if there is no conduction, then the transistor must work in the region. If the transmit junction is on, the transistor may work in the amplification area, or it may work in the saturation zone, which requires specific analysis. When the junction is inverted, the transistor works in the amplification area, and vice versa, working in the saturation zone. In the analysis, if the crystal is not determined to work in that state, it is possible to assume a state (amplification or saturation) and then determine its working state by calculating the deflection state of the set junction according to the known data. In addition, the analysis of the input and output characteristic curve is very important when analyzing the working state of the transistor, and it will be more effective with the analysis of the characteristic graph.