Question

The circuit below uses AD620AN instrumentation amp and allows the use of a single power supply: a 9V battery.

Vcc Vcc Black 47K 47K 1K Ou 680K Red Green 75K 33K Vcc Vcc Spld Power Followve 47K

The gain of the AD620AN chip is set by the resistor across the pins G- and G+, and is given by G = 1+ 49.4kΩ / R_G so we have set the gain to be approximately 50 by setting R_G as 1K. This, coupled with the back end op-amp gain of 24 will result in a combined gain of 1,200. This will amplify the mV level ECG signal to the level of a few volts.

A)Find the transfer function of the filter network. To do this properly requires a network analysis. I encourage you to do that, but at least, make a sketch of the LogLog transfer function showing the slope of the low, middle and high-frequency regions. You may also use a model using the online circuit analysis tool PartSim.

You should find that you have a bandpass filter that includes the relevant frequencies in the ECG. Find the low and high-frequency 3dB voltage cutoffs. This is where the voltage has decreased by 50%.
B) What do you think the 0.1μF capacitors across the power supply battery is for?

C) Why does the voltage follower providing circuit ground have a 47W resistor in series with its output?

Answer 1

Consider the op-amp circuit Vcc Vcc Black 47K 4.7K w Vout 0.1MF 68OK Red Green 75K $3.3K Draw the circuit in s-domain and mark the node voltages Vcc Vcc Black 4.7KV 47K Veutl Va Vond 101 106 Red Green _75K 23.3K Determine the output voltage of the second op-amp 3.3 k out 3.3 k75 k 3.3 k75 k V_ V. out 3.3 k 23.73V out Apply KCL at the node V1 1-и 106 4-4 0 47,000 И + outl 1 1 + 4700 2.34x1010 10s)V-2.128 x10^V, =2.128 x103V outl -5 2.128 x10 2.34 x101062+Vou) 21.3 1 2 outl 4+ 4) s+11

Apply KCL at the node V2. V, -и V -V 0 47,000 10 S S -2.128x10 (2.128 x103 +1.1x10s)V, =10^sV 21.3 (V, +V) for V,. Substitute ut s+11 2.128x10 21.3 ( -2.128 x10 +(P4+4) 1.1x10s |V = 10 sV S+11 2.128 x10 4.53x10 =10 sV 4.53 x10 + 4 s+11 autl S+11 1.1x10s 4.53 x10 4.53 x 10 V, + - =10sV S+11 s+11 1.1x10 (s +19.34) | 4.53 x10 1.1x10 (s2 30.34s 212.74) 4.53x10 V 10sV + 4 S 11 utl s+11 4.53 x 10 V, + s30.34s624.56 -Va10sV 1.1x10 s11 s+11 s230.34s+624.56 1.1 453 V2SV out1 S+11 s+11 s(s+11) 1.1(30.34s+624.56) 453 V 1.1(52 ourl +30.34s624.56 0.909s(s11 411.82 V, = s230.34s624.56) V. 1.1(s 30.34s +624.56) Apply KCL at the V+ node 1- и 0= = 100 680,000 S (1.47X1010SV = 10 sV (s1.47)V sV

Substitute 0.909s(s11 411.82 for V2 -V 1.1s30.34s 624.56 outl (s2 s 30.34s624.56 0.909s(s1 V 624.56 + +30.34s S (s +1.47)V = 411.82 V out1 1.1(52 30.34s624.56) 0.909s(s11 374.38 (s +1.47)+ (C out s30.34s+624.56 +30.34s624.56 (s' .1)V 28.9s2688.25s +918 =-374.38V out1 The voltage ratio of the first op-amp is 1200 V 1200V out (s° 374.38 X1200V +28.9s688.25s +918 + 449256 V -V 28.9s688.25s +918.1 2 + S According to virtual ground concept, V V V. 23.73V out V. -23.73x 449256 out 28.9s688.25s +918.1 V 10.66 x106 V f V. out 328.9s2688.25s +918.1

Sketch the magnitude plot of the transfer function num 10.66E6; den=[1 28.9 688.25 918.11; sys-tf (num, den); bodemag (sys) Bode Diagram 100 80 60 40 20 0 -20 -40 10 10 10 10 10 10 Frequency (rad/s) (B) The capacitive reactance is, 1 Х. For low frequencies, X^ is very high. Hence, the capacitor acts as an open-circuit For high frequencies, Xc is low and acts as a short-circuit So, all high-frequency components are by-passed through the 0.1 MFcараcitor. Thus, 0.1 /HF capacitor across the power supply battery is to filter all high-frequency signals, and allow only low frequency signals to the op-amp circuit. The RC component acts as a low pass filter. (C) The 47 2 series resistor is connected at the output of the voltage follower circuit to limit the load current. Magnitude (dB)