High Frequency Response Using Miller's Approximation

At high frequencies, the voltage gain of amplifiers usually decreases. To demonstrate this consider the circuit in Fig.4.7.

  

Figure 4.7: CE Amp with Miller Capacitor

As we saw in the preceding sections, capacitors C₁, C₂ and C_E tend to give rise to increasing gain with high frequency. Here, on the other hand, we have another capacitor C₄, which tends to cause the gain to decrease as frequency increases. The effect of this capacitor can be explained qualitatively by the following discussion. Recall that the signal at the collector is approximately 180^o out of phase with the input signal at the base. The capacitor C₄ provides a path for some of that signal at the collector to be fed back to the base. Since collector signal is out of phase with the base signal, this feedback has the effect of partially canceling the input signal at the base. Since the total signal going into the base is thus reduced, the overall gain decreases. It can be shown that the effect of the capacitor C₄ can be approximated by the circuit in Fig. 4.8, which is known as Miller's Approximation.

  

Figure 4.8: CF Amp using Miller's Approximation

The circuit shows that the effect of C₄ can be modeled by replacing it by C_M to ground, where C_M is given by

C_M=(|A|+1)C₄

Where A_m is the midband gain of the amplifier stage. For example, if the circuit in Fig. 4.7 has a midband gain of -g_mR_C||R_L, then C_M=(g_mR_C||R_L+1)C₄. As implied by Fig. 4.8, the base signal then sees a low pass filter consisting of R_S, C_M and R_in, thereby attenuating the high frequency response of the circuits. For example, the gain of the amplifier in Fig. 4.7, assuming that the frequency is high enough so that C₁, C₂, C_E can be considered short circuits, is given by  

$$\frac{v_{out}}{v_{s}}=\frac{-R_{in}\vert\vert\frac{1}{sC_M}} {R_{in}\vert\vert\frac{1}{sC_M}+R_S}g_mR_C\vert\vert R_L$$ (82)

It is interesting to note that Equation (4.17) can be arranged into the following:  

$$\frac{v_{out}}{v_{s}}=\left (\frac{-R_{in}}{R_{in}+R_{S}}g_m... ... ) \left( \frac{1}{1+\frac{sC_MR_{in}R_S}{R_{in}+R_S}} \right )$$ (83)

By recalling the results from Chp. 2, we can see that the expression in the first set of parentheses is the midband gain A_m. Furthermore, the expression in the second parentheses is just that of a first order low pass filter, with the pole $p= -\frac{R_{in}+R_S}{C_MR_{in}R_S}$.Thus, we can express the gain of the circuit at high frequencies as a midband gain A_m multiplied by a low pass filter response.  

$$A(s)=A_m\frac{1}{1-\frac{s}{p}}$$ (84)

The Bode plot for the high frequency response given by equation (4.18) can quickly be drawn starting with a horizontal line of magnitude $20log\frac{-R_{in}}{R_{in}+R_{S}}g_mR_C\vert\vert R_L$,and then having the gain fall off with a slope of 20dB/dec at frequencies $\omega \geq \frac{R_{in}+R_S}{C_MR_{in}R_S}$