bjt-4 - Copy - Figure 2.25(a shows a common-emitter circuit...

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Figure 2.25(a) shows a common-emitter circuit with a pnp bipolar transistor, and Figure 2.25 (b) shows the dc equivalent circuit. In this circuit, the emitter is at ground potential, which means that the polarities of the V BB and V CC power supplies must be reversed compared to those in the npn circuit. The analysis proceeds exactly as before, and we can write ----- (2.30) -----(2.31) and -----(2.32) We can see that Equations (2.30), (2.31), and (2.32) for the pnp bipolar transistor in the common-emitter configuration are exactly the same as Equations (2.26), (2.27), and (2.28(b)) for the npn bipolar transistor in a similar circuit, if we properly define the current directions and voltage polarities. Figure 2.25 (a) Common-emitter circuit with pnp transistor and (b) dc equivalent circuit. Transistor equivalent circuit is shown within the dotted lines with piecewise linear transistor parameters. In many cases, the pnp bipolar transistor will be reconfigured in a circuit so that positive voltage sources, rather than negative ones, can be used. We see this in the following example.
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Example Analyze the common-emitter circuit with a pnp transistor. For the circuit shown in Figure 2.26(a), the parameters are: V BB = 1 . 5V, R B = 580k , V + = 5 V, V EB (on) = 0 . 6 V, and β = 100. Find I B , I C , I E , and R C such that V EC = ( ) V + . Figure 2.26 Circuit for Example 5.4; (a) circuit and (b) circuit showing current and voltage values Solution: Writing a Kirchhoff voltage law equation around the E B loop, we find the base current to be Comment: In this case, the difference between V + and VBB is greater than the transistor turn-on voltage, or (V + V BB ) > V EB ( on ) . Also, because V EC > V EB (on), the pnp bipolar transistor is biased in the forward-active mode.
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Discussion: In this example, we used an emitter-base turn-on voltage of V EB (on) = 0 . 6V, whereas previously we used a value of 0.7V. We must keep in mind that the turn-on voltage is an approximation and the actual base emitter voltage will depend on the type of transistor used and the current level. In most situations, choosing a value of 0.6V or 0.7V will make only minor differences. However, most people tend to use the value of 0.7V. Exercise Problem The circuit elements in Figure 2.26(a) are V + = 3 . 3V, V BB = 1 . 2V, R B = 400k , and R C = 5 . 25k . The transistor parameters are β = 80 and V EB ( on ) = 0 . 7V. Determine I B , I C , and V EC . (Ans. I B = 3 . 5 μ A, I C = 0 . 28 mA, V EC = 1 . 83V) Load Line and Modes of Operation The load line can help us visualize the characteristics of a transistor circuit. For the common-emitter circuit in Figure 2.24(a), we can use a graphical technique for both the B E and C E portions of the circuit.
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  • Spring '16
  • Transistor, Bipolar junction transistor, npn Transistor Circuit

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