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Section D2: the Common-Emitter Amplifier

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Section D2: the Common-Emitter Amplifier Section D2: The Common-Emitter Amplifier As stated in the introduction, we are going to concentrate on the transistor circuits that use the linear region of BJT operation, as defined in the conditions assumed for the small-signal model. These single-stage amplifiers are the simplest circuits that allow the BJT to provide voltage and/or current gain. Always keep in mind that the total voltage and current are composed of a dc component and an ac component. For the small-signal model to apply, the transistor must stay in the normal active mode for the entire range of input signals (please review Section C2 for details). The dc voltages and current that characterize the bias conditions must be defined such that the transistor stays in normal active mode for all expected input signals. A generic common-emitter (CE) amplifier circuit is shown to the right (this is a modified representation of Figure 5.1a in your text). If you compare this circuit to Figure 4.12a in Section C.5, the only difference is that the dc source, VEE, has been set to zero. This actually makes things a little simpler, since the dc analysis only involves a single source. The capacitors are open circuits for dc, so the input and output are removed for dc bias calculations, yielding the dc bias circuitry of Figure 4.13. As we discussed earlier, the bypass capacitor (CE) and coupling capacitors (CB and CC) are large enough that they may considered short circuits for the frequency range of input signals. Using this information, the ac small-signal model for the common-emitter circuit is shown to the right (note that this is the same as Figure 5.1b of your text, merely represented as a slightly different graphic.) If you compare this with the hybrid-π model of Figure 4.5, the output resistance rO has been left out. This is often a valid assumption since rO is much greater than RC or RL (recall that equivalent resistances for parallel combinations are dominated by the smaller resistance values), and vout=œβib(RC||RL). However (major life hint), if this is not the case, rO would be placed in parallel with RC and RL and vout would be calculated by œ βib(rO||RC||RL). A powerful tool in the design and/or analysis of BJT amplifiers is gained by realizing that, although the input (base-emitter side) and output (collector- emitter side) are coupled, operations may be performed on the two sides separately. The coupling of the input and output occurs through the dependent current source and is achieved either through the transconductance (gm) or the gain term β, depending on whether you‘re dealing with vbe or ib on the input side. Figure 5.1c, modified and reproduced to the right, illustrates this separation. We‘re going to use this representation extensively in the development of amplifier characteristics. A modified version of Figure 5.2, given to the right, illustrates the small signal amplifier circuit in terms of a two-port network, where everything after the input signal and before the load is shown as a —black-box“ with amplifier characteristics Rin and Rout calculated by looking into the input and output respectively. Note: Although this figure is shown with purely resistive input and output characteristics, occasions may arise where these parameters are complex values. Don‘t let this throw you, the process is the same but life gets a little more complicated and we will be dealing with Zin and Zout. We‘re going to be developing expressions for the input resistance Rin, the output resistance Rout, the current gain Ai, and the voltage gain Av for each of the single-stage amplifier configurations. The procedure is the same for each configuration, so (I‘m sure you will be crushed) any extensive derivations will be limited to this first case as much as possible. Input Resistance, Rin By comparing Figures 5.1c and 5.2 above, we can see that the input resistance is the equivalent resistance of the input side of the circuit —looking in“ after the source. This gives us RB in parallel with rπ, or RB rπ Rin = RB || rπ = . (Equation 5.5) RB + rπ Note that your text derives this equation in a slightly different manner, but ends up the same place. OK, back in section C3 I promised that impedance reflection would come back. To refresh your memory, impedance reflection is a tool that allows us to —see“ an equivalent resistance by looking into the base or emitter. Specifically, if a resistance is in the emitter, it —looks“ β times larger to the base. Alternatively, a resistance in the base, —looks“ β times smaller to the emitter. β r = βr = π e g m (Equation 4.14: Modified) r r = π = g r e β m π Heads up… the relationships given in the equation above are assuming that β >> 1, so that β+1 is approximately equal to β. If this is not the case, the technique still works but β must be replaced with β+1. Using the relationship between re and rπ, we may express the input resistance in terms of the emitter resistance: R β r R r R = B e = B e . (Equation 5.6) in R RB + β re B + r β e Output Resistance, Rout 9e use the same strategy to derive an expression for the output resistance, but now we‘re looking into the output side before the load resistor, as illustrated in Figure 5.2 above. Referring back to Figure 5.1a, we can see that the output resistance for this particular configuration is simply RC (it‘s the only thing left on the output side), so Rout = RC . (Equation 5.14: Modified) There are a couple of points that need noting however: As stated by your author, we are formally finding the Thevenin equivalent resistance. However, if the independent source is set to zero, the dependent source is also zero (if vin=0, ib=0, and βib=0), so the simplified strategy above is valid. Be very careful to be sure that rO can be neglected! If rO is on the same order of magnitude as RC, it must be included in the small-signal model and the output resistance will be Rout = rO || RC . (Equation 5.14: Modified) Current Gain, Ai By convention, gain terms are denoted by the symbol ”A‘ with a subscript that defines the particular gain of interest. Specifically, subscripts ”i‘, ”v‘, and ”p‘ indicate current, voltage and power, respectively. Also, gain refers to the ratio of the relevant output quantity over the relevant input quantity. Using this information, the current gain is defined as the output current (the load current in the figures above) over the input current, or iL Ai = . (Equation 5.7) iin The load current is a portion of the collector current (ic=βib) and may be expressed in terms of circuit components using current division: − RC β ib iL = (Equation 5.9) RL + RC Note that the negative sign is introduced in Equation 5.9 because of the defined current directions for iL and βib. Similarly, the base current may be expressed as a portion of the input current using current division: RBiin ib = . (Equation 5.8) RB + rπ which allows us to get an expression for iin in terms of ib: (RB + rπ )ib iin = . RB Using the expressions for iL and iin in Equation 5.7: ≈ ’ − RC βib ∆ ÷ i « R + R ◊ − R R β A = L = L C = B C , i ≈ ’ iin (RB + r )ib (RB + rπ )(RL + RC ) ∆ π « RB or, by factoring β out of the numerator and rearranging, − R R − R R A = B C = B C , (Equation 5.10: Modified) i ≈ R r ’ ≈ R ’ ∆ B + π ÷(R + R ) ∆ B + r ÷(R + R ) « β β ◊ L C « β e ◊ L C where, once again, if β is not much larger than one, all β terms become β+1. 5oltage Gain, Av The voltage gain is defined as the ratio of the output voltage to the input voltage. By using Ohm‘s law and recognizing the current gain ratio, Av becomes: ≈ ’≈ ’ vout iL RL ∆ iL ÷∆ RL ÷ RL Av = = = ∆ ÷∆ ÷ = Ai (Equations 5.11 and 5.12) vin iin Rin « iin ◊« Rin ◊ Rin Using the relationships defined for Rin (Equation 5.6) and Ai (Equation 5.10), the voltage gain expression may be derived as follows: ≈ ’ ∆ ÷ ∆ ÷ ≈ ’ ≈ R ’ ∆ ÷∆ ÷ − R R R ∆ B + r ÷ − R R ∆ R ÷ B C L « β e ◊ A = ∆ B C ÷ L = . v ∆ ≈ R ’ ÷∆ ≈ ’ ÷ ≈ R ’ ∆ B r ÷(R R ) ∆ ÷ ∆ B r (R R )(R r ) ∆ + e L + C ÷ ∆ R r ÷ + e L + C B e « « β ◊ ◊∆ ∆ B e ÷ ÷ « β R ∆ ∆ B + r ÷ ÷ « « β e ◊ ◊ Whew! Still looks pretty ugly, huh? Well, after canceling out a whole bunch of stuff, recognizing a parallel resistance relationship, and recalling that 1/re=gm, we get − RC RL − RC || RL Av = = = −g m (RL || RC ) . (Equation 5.13) (RL + RC )re re So, the voltage gain does not exhibit any β dependencies and is a straightforward relationship depending upon circuit components and the transconductance of the BJT. A word of caution: These are the standard techniques for developing expressions for current and voltage gain based on Figure 5.2, however, you must be careful to include all relevant components in the expressions for Rin, iin and iL. Finally, your author develops the gain impedance formula in this portion of your text. This relationship allows us to express the voltage gain (or current gain) in terms of the load and amplifier input characteristics.
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