Example CircuitMidband AnalysisHigh-frequency analysisThe dominant pole approximationLow-frequency analysisThe dominant pole approximationFinding the other poleECE 304: Open- and Short-circuit Time Constants See S&S, pp. 575-578, 497-503 Example Circuit +{C_1}0PARAMETERS:R_1 = {R_L}R_2 = {R_H}R_3 = {R_L}R_4 = {R_H}C_1 = {C_LF}C_2 = {C_LF}C_3 = {C_HF}C_4 = {C_HF}C_LF = 1uFC_HF = 1pFpi = 3.1415926R_L = {10/(2*pi)}R_H = {1E3/(2*pi)}+{R_3}+{C_3}+{R_1}+{R_4}+{R_2}VDB+{C_4}Sweep+-ACV11V+{C_2} OUTFIGURE 1 An RC-circuit that is a midband filter. The first two sections in Figure 1 are high-pass filters and the last two are low-pass filters. The factor 2π is introduced to make the frequencies easier to estimate. Frequency100Hz 10KHz 1.0MHz 100MHz 10GHz 1.0THzVDB(OUT) dB( 9.804E+08/Frequency) dB(9.804E+08/1.02E+11)+2*dB( 1.02E+11/Frequency)dB( Frequency/1.0200E+05) 2*dB( Frequency/980.3921569)+dB(980.3921569/1.0200E+05)-100-500(102.000K,-2.9688)(980.390,-43.313)(980.400M,-2.9686)(102.000G,-43.312)(10.000M,-921.072u) FIGURE 2 Bode magnitude plot for circuit of Figure 1. Midband region lies between the lower 3dB-corner frequency of 102 kHz and the upper 3dB-corner frequency of 980 MHz; midband gain is approximately zero dB Frequency100Hz 10KHz 1.0MHz 100MHz 10GHz 1.0THzP(V(OUT))-190d0d180d(102.000G,-134.722)(980.400M,-45.270)(102.000K,45.272)(980.392,134.724)(Midband,10.000M,0.000) FIGURE 3 Bode phase plot with phases marked at the pole frequencies for each section. The phases are slightly affected by the nearness (proximity) of other poles. For isolated poles, the phases at the pole frequencies taking midband as 0° are 135°, 45°, −45°, and −135° Unpublished work © 3/7/2005 J R Brews Page 1 3/7/2005Midband Analysis At midband, all low-frequency capacitors are short-circuits and all high-frequency capacitors are open circuits. That is, C1 and C2 are shorts, and C3, C4 are opens, making the gain VOUT/VIN = 1 V/V or 0 dB. High-frequency analysis At high frequencies, near or above the upper corner frequency of 980MHz, we can approximate the two low-frequency capacitors C1 and C2 by short circuits. The result is shown in Figure 4: the high-frequency gain of the two circuits is the same, but the low-frequency behavior is not, because C1 and C2 are not present in the approximate circuit. Frequency100Hz 10KHz 1.0MHz 100MHz 10GHz 1.0THzDB(V(OUT)) DB(V(HF))-100-500Full circuitLF caps short-circuited FIGURE 4 Comparison of high-frequency approximation to circuit (capacitors C1 and C2 replaced by short circuits) and the original circuit of Figure 1 0+{C_3}+{R_4}Sweep+-ACV31VVDB+{R_3}+{C_4} HFFIGURE 5 The approximate circuit valid at high-frequency Circuit analysis of Figure 5 shows the gain to be EQ. 1 []4343233434SOUTRRCC)j(RC)RR(Cj11VVω+++ω+=. The dominant pole approximation The denominator of EQ. 1 is quadratic in ω, which means there are two poles in the gain function and two high-frequency break points. In Figure 3, these occur at 980 MHs and 102GHz, for this example. In the case that these two poles are well separated (the dominant pole case) the linear term in ω determines the corner frequency that marks the end of the midband region. If the term linear in ω is examined more carefully, it is composed of two RC time constants, one related to each of the two high-frequency capacitors. For example, the term C4(R3+R4) uses the resistance seen between the nodes where C4 is attached if the applied signal source is shorted out. This resistance can be found by putting a test current or voltage in place of C4, shorting Vsh and open-circuiting C3 (see Figure 6) Unpublished work © 3/7/2005 J R Brews Page 2 3/7/2005160.746V+{R_4}1A1.00000A+{R_3}+{R_2}0+{R_1} FIGURE 6 Circuit for finding the Thevenin resistance (R3+R4) = 160.7 Ω seen by C4 In Figure 6 a test current was used so we can read the resistance seen by C4 simply as the voltage at the top of the test source. Likewise, the term C3R3 is found using Figure 7. 0+{R_1}+{R_4}1.59155V0A0A1A1.00000A+{R_2}+{R_3} FIGURE 7 Circuit for finding the Thevenin resistance R3 = 1.59 Ω seen by C3 In Figure 7 a zero-current source is added at the end of R4 because PSPICE does not allow dangling resistors. The zero-amp current source tells PSPICE that there is no current in R4, as is appropriate when the circuit ends with an open circuit because C4 has been replaced by an open circuit. The general approach to finding the high-frequency corner is then as follows: 1. Draw the small-signal circuit 2. Short all the low-frequency capacitors 3. Short all independent AC voltage sources and open all independent AC current sources 4. Select a particular high -frequency capacitor. Replace all the others by open circuits. 5. Put a test voltage at the site of the selected capacitor and find the resistance seen at this site. 6. Multiply this resistance by the value of the selected capacitance → τ1, say 7. Do the same for all the other high-frequency capacitances. 8. The high-frequency corner frequency is fC = 1/(2π) × 1/(τ1 + τ2 + τ3 + …) Low-frequency analysis +{R_1}Sweep+-ACV21V0+{C_2}+{R_2}+{C_1} LFFIGURE 8 The approximate low-frequency circuit Unpublished work © 3/7/2005 J R Brews Page 3 3/7/2005Frequency100Hz 10KHz 1.0MHz 100MHz 10GHz 1.0THzDB(V(OUT)) DB(V(LF))-100-500Full circuitHF caps open-circuited FIGURE 9 Bode gain plots comparing the low-frequency circuit of Figure 8 with the complete circuit of Figure 1 Figure 9 shows the low-frequency circuit and the original circuit agree at low frequencies, but not at high frequencies, which is the reverse to Figure 4. Circuit analysis of Figure 8 shows the voltage gain is given by EQ. 2 ω++ω+=2121222211SOUTRRCC1)j(1RC1)R//R(C1j111VV The dominant pole approximation In complete analogy with the high-frequency case, the denominator of EQ. 1 is quadratic in ω, which means there are two poles in the gain function and two low-frequency break points. In Figure 3, these occur at 980 Hz and 102 kHz, for this example. In the case that these two poles are well separated (the dominant pole case) the linear term in 1/jω determines the corner frequency that marks the start of the midband region. Looking at the linear term, it is composed of two RC time constants, one related to each of the two low-frequency capacitors. For example, the term C1(R1//R2) uses the resistance seen between the nodes where C1 is attached if the applied