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REVIEW: Bipolar transistors are so named because the controlled current must go through two types of semiconductor material: P and N. The current consists of both electron and hole flow, in different parts of the transistor. Bipolar transistors consist of either a P-N-P or an N-P-N semiconductor “sandwich” structure. The three leads of a bipolar transistor are called the Emitter, Base, and Collector. Transistors function as current regulators by allowing a small current to control a larger current. The amount of current allowed between collector and emitter is primarily determined by the amount of current moving between base and emitter. In order for a transistor to properly function as a current regulator, the controlling (base) current and the controlled (collector) currents must be going in the proper directions: meshing additively at the emitter and going against the emitter arrow symbol.

REVIEW: Transistors may be used as switching elements to control DC power to a load. The switched (controlled) current goes between emitter and collector; the controlling current goes between emitter and base. When a transistor has zero current through it, it is said to be in a state of cutoff (fully nonconducting). When a transistor has maximum current through it, it is said to be in a state of saturation (fully conducting).

How to Solve BJT Circuits: The state of a BJT is not known before we solve the circuit, so we do not know which model to use: cut-o, active-linear, or saturation. To solve BJT circuits, we need assume that BJT is in a particular state, use BJT model for that state to solve the circuit and check the validity of our assumptions by checking the inequalities in the model for that state. A formal procedure will be: 1) Write down a KVL including the BE junction (call it BE-KVL). 2) Write down a KVL including CE terminals (call it CE-KVL). 3) Assume BJT is in cut-o (this is the simplest case). Set iB = 0. Calculate vBE from BE-KVL. 3a) If vBE , then BJT is in cut-o, iB = 0 and vBE is what you just calculated. Set iC = iE = 0, and calculate vCE from CE-KVL. You are done. 3b) If vBE > v , then BJT is not in cut-o. Set vBE = v  . Solve above KVL to nd iB. You should get iB > 0. 4) Assume that BJT is in active linear region. Let iE iC = iB. Calculate vCE from CE-KVL. 4a) If vCE > v  , then BJT is in active-linear region. You are done. 4b) If vCE  , then BJT is not in active-linear region. It is in saturation. Let vCE = vsat and compute iC from CE-KVL. You should nd that iC iB. You are done.

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