transistor-2 - IV. Transistors (Biasing & Small-Signal...

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Unformatted text preview: IV. Transistors (Biasing & Small-Signal Model) 4.1 Introduction Amplifiers are the main component of any analog circuit. Not only they can amplify the signal, they can be configured into may other useful circuits with a proper “feedback” (you will see this in 100 for OpAmps). In this course, we focus on simple transistor amplifiers. Transistor amplifiers utilizing BJT or MOSFET are similar in design and analysis. As such, we discuss them together. However, first we need to review some concepts which are essential in the design and analysis of the amplifier circuits. Consider the circuit below with a NPN BJT. The operating point of the BJT is shown in the i C v CE space. V BB R C V CC I B V BE V CE I C B R +-- +- +- + Let us add a sinusoidal source with an amplitude of ∆ V BB in series with V BB . In response to this additional source, the base current will become I B + ∆ i B leading to a collector current of I C + ∆ i C and a CE voltage of V CE + ∆ v CE . V BB B R R C V BB V CC i B I B v BE V BE i C v CE V CE I C-- + +- + ∆ ~ +∆ +∆ +∆ +∆- +- + Assume that without the sinusoidal source the base current is 150 μ A, I C = 22 mA, and V CE = 7 V (the Q point). If the amplitude of ∆ i B is 40 μ A, then with the addition of the sinusoidal source I B + ∆ i B = 150 + 40 cos( ωt ) μ A and i B varies from 110 to 190 μ A. As the BJT operating point should remain on the load line, the collector current and CE voltage change with changing base current while remaining on the load line. For example when base current is 190 μ A, the collector current is 28.6 mA and CE voltage is about ECE65 Lecture Notes (F. Najmabadi), Spring 2010 4-1 4.5 V. As can be seen from the figure above, the collector current will approximately be I C + ∆ i C = 22 + 6 . 6 cos( ωt ) mA and CE voltage is V CE + ∆ v CE = 7 − 2 . 5 cos( ωt ) V. This example shows that the signal from the sinusoidal source ∆ V BB is greatly amplified and appears as signals in collector current and CE voltage. It is also clear from the figure that this happens as long as the BJT stays in the active state. As the amplitude of ∆ i B is increased, the swings of BJT operating point along the load line become larger and larger. At some value of ∆ i B , BJT will enter either the cut-off (when i B + ∆ i B ≤ 0) or saturation state and the output signals will not be a sinusoidal function. The above circuit, however, has several major issues: 1) The input signal, ∆ V BB , is in series with the V BB DC voltage. As typically the input signal is the output from another two-port network, this DC voltage will also appear in the output of the previous two-port network, making two-port networks dependent on the next and system design difficult. Similarly, the output signal is usually taken either across R C as R C × i C or as v CE . These output voltages have DC components which is of no interest and can cause problems in the design of the next two-port network. Basically, we have twoand can cause problems in the design of the next two-port network....
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This note was uploaded on 01/19/2011 for the course ECE 65 taught by Professor Coles during the Spring '08 term at UCSD.

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transistor-2 - IV. Transistors (Biasing & Small-Signal...

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