Circuits Ci it II EE221 Unit 6 Instructor: Kevin D D. Donohue Active Filters, Connections of Filters, and Midterm Project Load Effects If the output of the filter is not buffered, then for different loads, the frequency characteristics of the filter will change. Example: p Both low-pass p filters below have fc = 1 kHz and GDC = 1 (or 0 dB). 1 k 1 k + vi(t) 1 2 F + vo(t) - vi(t) 1 2 - F 10 k + vo(t) - Find: fc and GDDC when a 100 load is p placed across the output p vo(t). Result: fc = 11 kHz and GDC = 1/11 (or -21 dB) for the passive circuit, while fc and GDC remain unchanged for the active circuit. Filter Order The order of a filter is the order of its transfer function (highest power of s in the denominator). If the filter order is increased, a sharper transition between the stopband b d and d passband b d of f the h fil filter iis possible. ibl Example: For the two low-pass filters, determine circuit parameters such that fc is 100 Hz, and GDC is 3 2 1.586 Plot transfer function magnitudes to observe the transition near fc. C R R + vi(t) C - Rf + vo(t) - R + vi((t)) C - (K 1)Rf (K-1)R R1 Rf + vo(t) - Transfer Function Results First Order 1 Second Order Rf R1 Hˆ ( s ) s 1 1 RC GDC = 1+Rf / R1 fc = 1 / (2RC) Hˆ ( s ) K s s2 1 (3 K ) 1 1 2 RC RC For GDC = K = 3 2 1.586 , fc = 1 / (2RC) Formula not valid for any other value of K. Value of K was contrived so the cutoff would come out this way. For a general K value fc = / (2RC) , where 7 6K K 4 2 2 7 6K K 2 2 Plot Comparison of Filter Order Effect first order (-), second order (---) 1.4 1.2 3d cut-off 3dB ff gain 1 0.8 0.6 0.4 0.2 w = [0:1024]*2*pi; s = j*w; j* 0 100 200 300 400 500 600 700 800 Hz h1 = (3-sqrt(2)) ./ (1 + (s / (2*pi*100))); h2 = (3-sqrt(2)) ./ (1 + (sqrt(2)*s / (2*pi*100)) + ( s / (2*pi*100)) .^2); plot(w/(2*pi),abs(h1),'b-', plot(w/(2 pi),abs(h1), b , w/(2 w/(2*pi), pi), abs(h2), 'b:') b: ) title('first order (-), second order (---)'); xlabel('Hz'); ylabel('gain') 900 1000 SPICE Transfer Function Analysis The simulation option for “.AC Frequency Sweep” with plot transfer function for simulated circuit circuit. The frequency range (in Hz) must be selected along with plot parameters such as log or linear scales. Both the phase and magnitude can be plotted if requested. The input can be a voltage or current source with amplitude of 1 and phase 0. Selection of the frequency range is critical. critical If a range is selected in an asymptotic region (missing the dynamic details of the transfer function) the plot will be misleading. The range can be determined by looking at large scale plots and making adjustments. Selecting a large range on a log scale may make it easier to identify the frequency range where g is happening, pp g and then a smaller range g can be selected change to better show the details of the plot. Spice Example: Design the second order LPF so that it has a cutoff at 3.5 3 5 kHz with a gain of 3 2 1.586 Verify V if th the d design si with ith a SPICE R simulation. C R + The h key k design d equations become: K 3 2 3500 (2 ) vi(t) 1 RC So let C = 0.01F and Rf = 10k. This implies p R = 4.55k and (K-1)Rf = 5.86k C - (K 1)Rf (K-1)R Rf + vo(t) - SPICE Results (Plot range 10 to 10 MHz) The magnitude plot in dB. Crosshair marker near 3 dB cutoff (What is happening near 100kHz)? TFLPF2ex-Small Signal AC-6-Graph 10.000 100.000 1.000k 10.000k 100.000k 0.0 -20 20.000 000 -40.000 FREQ 3.447k D(PH DEG(v(IVm1))) D(PH_DEG(v(IVm1))) 87 325 87.325 DB(v(IVm1)) 1.116 D(FREQ) 0.0 Frequency (Hz) 1.000Meg 10.000Me SPICE Results (Plot range 10 to 10 MHz) The phase plot in degrees. Crosshair marker near 3 dB pp g near 100kHz))? cutoff (What is happening Frequency (Hz) TFLPF2ex-Small Signal AC-6-Graph 10.000 100.000 1.000k 10.000k 100.000k 1.000Meg -100.000 -200.000 -300.000 FREQ Q 3.548k D(DB(v(IVm1))) -1.000 PH_DEG(v(IVm1)) ( ( )) -91.852 D(FREQ) ( Q) 0.0 10.000Me