Assume that the diode has a constant voltage drop in forward mode of and that

Assume that the diode has a constant voltage drop in

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Assume that the diode has a constant voltage drop in forward mode of , and that all transistor base-emitter junctions have a forward voltage of | | . Calculate the value of and the collector current of Q3 and Q4 (which is the primary bias current for this circuit). Class Notes Additional BJT Problems
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237 Solution On the surface, this looks like a very complex circuit with many components. However, when we break down the circuit, we can see that it is nothing more than small building-block circuits which we have already studied. To begin, since we are solving the circuit at DC, we know that the capacitors will be open-circuit. Thus, is not connected, and the capacitor between the bases of Q1 and Q2 has no effect: Let us try to use iteration to produce the answer to this question it would be quite difficult to create and solve a direct set of equations with so many transistors. To begin this process, we need an initial guess for , , and , our goal values. Working backwards, we know that for Q3/4 will depend on base current, which in turn depends on the collector currents of Q1/Q2, which in turn depend on the base currents of Q1/Q2. We do not have an easy, direct method for finding these base currents yet. So, as in previous examples, we can assume that the base current will be very small, effectively zero for the first iteration: This is typically a safe assumption if we expect the transistor to operate in active mode, and/or if the transistor has a high value for . Since this is identified as an amplifier circuit, and , we have both conditions, meaning this is a relatively safe starting assumption. We can now calculate the values of and directly by first finding the current, , in the series-connected string of R2 D1 R3 R4: ( ) ( ) ( )( )
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238 Note: Be careful with the number of significant digits which you carry for ; it actually has a large effect in this problem. If you carried only three, then , which would not be sufficient for the BJT to be in a conducting mode. We will try to avoid introducing this issue on exam problems, but this is not always possible when in doubt, carry a few extra digits, and apply basic sanity checks to your solutions (i.e., identifying in the above that 0.699 would put the BJT in cut-off mode). ( )( ) ( )( ) ( ) We now know the emitter voltages of Q1 and Q2, assuming that both are in either conducting mode with | | : Note that we are now working with PNP transistors. They use all the same equations as the NPN transistors, except the polarities of currents and voltages are inverted: PNP Transistor Notice that current flows into the emitter, and out of the collector and base. The fundamental equation of the BJT still holds: . Given the above directionality of currents, the equations of the PNP transistor in active mode also hold: . To check the conditions, we use the same conditions for the NPN transistor with reverse polarity for all voltages. In other words, what was before becomes , and so on. This will become clearer when we check the conditions for this problem.
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