Solution To begin you might notice that we are missing an important piece of

Solution to begin you might notice that we are

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Solution To begin, you might notice that we are missing an important piece of information. Since the base- emitter forward voltage drop is not given, we must assume a value. In general, if is not given, assume that it is 0.7V when forward biased. We must also assume an operating mode for the two transistors. This particular arrangement is called a darlington arrangement; it is commonly used with BJTs to greatly increase the effective . Basically, the two transistors together form a single NPN transistor with a total gain of , where and are the gains of the individual transistors. This is very useful when attempting to construct single-stage, high-gain amplifiers. However, let us assume that we did not know this fact, and wanted to analyze the circuit at DC. We would begin by assuming a mode. Like the MOSFET, we will assume active mode, as the majority of circuits we will analyze (amplifiers) will use this mode. Remember that saturation mode for the BJT (like the triode mode for MOSFET) is primarily used as a switching or ‘ON/OFF’ mode. We can begin by identifying a few of the voltage identities caused by the connection of Q1 and Q2: (1)
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232 It is intuitively simple to see how this circuit will operate. The emitter current of Q1 will become the base current of Q2, so if ( ) , then ( ) ( )( ) . This roughly gives us the darlington equation from earlier, assume the two values are large enough that the ‘+1’ does not matter. At this point, we have a choice. The original procedure calls for us to potentially iterate; this is a circuit where this method will make the calculations quite simple if we iterate. However, it can (like all circuits) be solved directly through equations and linear solution methods. Either is allowable on a test, though the iterative method is strongly recommended for complex circuits; the systems of equations can get very large quite quickly. Iterative Solution The basis of an iterative solution (assuming active mode for all transistors) is to obtain an initial guess at the emitter and collector currents. This guess is used to calculate an accurate base current, which then drives the next iteration. To get our initial guess, we typically assume that if is large, then should be small we can initially assume it to be zero, in fact: (1) We now have our initial guess for the emitter current of Q2. If we desired, we could also find the collector current of Q2. Instead, we will work backward to find the base current of Q2, and eventually the base current of Q1. This more accurate estimate will then drive our next iteration. (2)
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233 We can see that the base current is indeed quite small, meaning we will probably not need too many iterations to arrive at a suitable solution. The last step is to close the loop to start the next iteration; given this value of current , we calculate a new (recall this was our starting point): (3) Iterating, we can watch the value of to see that our solution has converged: ( ) We can see that after just one iteration, we would arrive at a converged solution. In general, it is
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