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High-Frequency Amplifier Response

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High-Frequency Amplifier Response Section H5: High-Frequency Amplifier Response Now that we’ve got a high frequency models for the BJT, we can analyze the high frequency response of our basic amplifier configurations. Note: in the circuits that follow, the actual signal source (vS) and its associated source resistance (RS) have been included. As discussed in the low frequency response section of our studies, we always knew that this source and resistance was there but we just started our investigations with the input to the transistor (vin). Remember - the analysis process is the same and the relationship between vS and vin is a voltage divider! Again, in some instances in the following discussion, I will be using slightly different notation and taking a different approach than your author. As usual, if this results in confusion, or you are more comfortable with his technique, let me know and we’ll work it out. High Frequency Response of the CE and ER Amplifier The generic common-emitter amplifier circuit of Section D2 is reproduced to the left below and the small signal circuit using the high frequency BJT model is given below right (based on Figures 10.17a and 10.17b of your text). Note that all external capacitors are assumed to be short circuits at high frequencies and are not present in the high frequency equivalent circuit (since the external capacitors are large when compared to the internal capacitances – recall that Zc=1/jωC gets small as the frequency or capacitance gets large). We can simplify the small signal circuit by making the following observations and approximations: ¾ Cce is very small and may be neglected. ¾ rce >> (RC||RL), so rce may be neglected since rce||RC||RL is dominated by RC||RL. ¾ rb’c is so much larger than all other resistances that it may be considered “open” and removed from the circuit. ¾ rbb’ << rb’e so it may be neglected; i.e., rπ ≅ rb'e . (Equation 10.62) The above simplifications, along with the definition of Rin as Rin = RB || rπ , (Equation 10.63) and the reflection of Cb’c to the input and output circuits using Miller’s theorem C M1 = C b'c (1 − Av ) (input circuit) ⎛ 1 ⎞ , ⎜ ⎟ C M2 = C b'c ⎜1 − ⎟ (output circuit) ⎝ Av ⎠ yields the simplified small signal circuit presented to the right. Note that this is a corrected version of Figure 10.17c in your text. Using the property that capacitors in parallel add, we can define a single capacitor in the input circuit as C in = C b'e + C M1 = C b'e + C b'c (1 − Av ) . (Equation 10.64) Putting all this together gives us the simplified, simplified circuit shown to the left below. Note that, just like we did for the low frequency response, our strategy will be to set all independent sources to zero (which, in this case, also sets the dependent source to zero) as shown in the figure to the right below. We can now calculate the individual time constants using the Method of Open Circuit Time Constants. Just as a sanity check – even if we had not neglected the capacitance Cce in our first simplification, the parallel capacitance CM2 is so much larger that the contribution of Cce would be negligible. The Method of Open Circuit Time Constants is similar to the Method of Short Circuit Time Constants, but now we set all capacitances to zero (open circuit) except the one of interest. Equivalent resistances seen by the capacitance of interest are then derived and the individual RC time constant calculated. For the high frequency response using the poles of the transfer function only, we may approximate the cutoff in radians per second (ω) or Hertz (f) as −1 ⎡ ⎤ 1 ωH ωH ≅ ⎢∑τ i ⎥ = , fH = . ⎣ i ⎦ ∑ C i Reqi 2π i Using the figure above, we have ¾ Cin: with CM2 open, the resistance seen by Cin is equal to RCin = RS || Rin = RS || RB || rπ , since the open current source separates the input and output circuits. ¾ CM2: with Cin open, the resistance seen by CM2 is equal to RCM2 = RC || RL since, once again, the open current source separates the input an output circuits. Putting this all together, we can define the high frequency time constants for the CE circuit as: τ Cin = C in RCin ; τ CM2 = C M2 RCM2 , and an approximate value for the upper corner frequency, in radians per second, is given by 1 1 1 ω = = = H 1 1 τ + τ C R + C R + Cin CM2 in C in M2 C M2 ωPCin ωPCM2 . 1 = C in (RS || RB || rπ ) + C M2 (RC || RL ) Since the capacitance on the input side is much larger than in the output circuit, the high frequency cutoff is determined by dominant pole formed due to the input resistance and capacitance. Expressing the high frequency cutoff as a single dominant pole, we have 1 ωH 1 ωH = ; fH = = . (Equation 10.65) Cin(RS || RB || rπ ) 2π 2πCin(RS || RB || rπ ) Note that, if the source resistance, RS, becomes very small, the dominant pole will switch from the input circuit to the output circuit. This results in a higher cutoff frequency and a larger available bandwidth. Since the expressions of Equations 10.64 and 10.65 are formulated in terms of Rin and Av, the same equations may be used to calculate the high frequency cutoff of the emitter resistor amplifier. The difference between the common-emitter and emitter-resistor configurations is that Av is smaller and Rin is larger for the ER, the cumulative effect of which results in a larger bandwidth (higher cutoff frequency). High Frequency Response of the CB Amplifier The common base amplifier circuit is shown below and left, while the high frequency small signal equivalent is given below and to the right (based on Figures 10.21a and 10.21b of your text). Note that, unlike the CE/ER amplifier, no internal feedback capacitance exists for the common base configuration. This is the most important characteristic of the CB stage and means that there is no Miller effect, which means that the capacitances are smaller, which means that the cutoff frequency will be higher. Cool, huh? Using the method of open circuit time constants, we can define equivalent resistances for each of the remaining capacitances. With vS=0 and, therefore, gmvbe=0, we have ¾ Cb’e: with Cb’c set to zero, the resistance seen by Cb’e is equal to r R = π || R || R = r || R || R , Cb'e β E S e E S since we have to reflect rπ to the emitter circuit and assuming β >>1 so that β+1 ≈ β. ¾ Cb’c: with Cb’e set to zero, the resistance seen by Cb’c is equal to RCb'c = RC || RL . Putting this all together, we can define the high frequency time constants for the CB circuit as: τ Cb'e = C b'e RCb'e ; τ Cb'c = C b'c RCb'c , and an approximate value for the upper corner frequency, in radians per second, is given by 1 1 1 ω = = = H 1 1 τ + τ C R + C R + Cb'e Cb'c b'e C b'e b'c C b'c ωPCb'e ωPCb'c . 1 = C b'e (re || RE || RS ) + C b'c (RC || RL ) If the source resistance is not extremely small, the dominant pole will normally be at the input circuit, so that the high frequency cutoff is 1 ωH 1 ωH = ; fH = = . (Equation 10.72) Cb'e(re || RE || RS ) 2π 2πCb'e(re || RE || RS ) Note that the frequency defined in Equation 10.72 is quite high since Cb’e and re are quite small. At such high frequencies, it is often necessary to take into account effects that may generally be considered negligible. To more accurately determine the high frequency cutoff of a CB configuration, a more elaborate transistor model is necessary and it is generally a good idea to verify design using a computer simulation package such as Spice. High Frequency Response of the EF (CC) Amplifier The generic circuit for the emitter follower (common collector) amplifier is given to the left below and the high frequency small signal circuit is shown below and to the right (Figures 10.22a and 10.22b of your text). Your author states that the load for an EF amplifier is small and often capacitive, so he has included the load capacitance, CL. Also, since the voltage gain for an EF stage is approximately equal to positive one, the Miller capacitances are both close to zero. The equivalent circuit using Miller’s theorem is given in Figure 10.22c in your text and is reproduced to the right, where C M1 = C b'e (1 − Av ) ⎛ 1 ⎞ and R = R || [r + β(R || R )]. ⎜ ⎟ in B π E L C M2 = C b'e ⎜1 − ⎟ ⎝ Av ⎠ Note: It looks like your author is making β(RE||RL) >> rπ so that Rin=RB||β(RE||RL). This may be true, and probably will be true, but I’m more comfortable leaving rπ in for now. So…as has happened so many times before…some of my equations won’t exactly match the text. We can again define an equivalent capacitance for the input side and an equivalent capacitance for the output side as C = C + C ≅ C in b'c M1 b'c , C out = C L + C M2 ≅ C L and, by the method of open circuit time constants, the resistances seen by each capacitor as ¾ Cin: with Cout open, the resistance seen by Cin is equal to RCin = RS || Rin .
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