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DC level filter

Hi, I need to determine the transfer function Vout/Vin. A traditional differential amplifier transfer function without the 2.2uF on the input side would be G1 = (Z2/Z1)*Vin, where Z2 = (2.2uF // 150kOhms) and Z1 = 450kOhms + 150kOhms . I am having some trouble to model de 2.2uF input capacitor. The output filter stage (Rout = 10k and Cout = 2.2uF) transfer function is G2 = 1/(sRC+1). The total transfer function without the input capacitor is G1*G2. How do I deal with the input capacitor filter stage (2.2uF connecting both inputs after the 450k resistor)?

EDIT: enter image description here

I believe this is the correct transfer function for the input filter, it gives the 1/4 gain with some low pass dynamics Vout/Vin = G(s) = (1/4)*2.02/(s + 2.02)

Second stage considering only the op amp ( (2.2uF//150kOhms)/(150kOhms) ratio ): G1(s) = 3.03/(s + 3.03)

Output stage considering the RC filter: G2 = 1/(sRC+1) = 45.45/(s + 45.45)

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It is still a function of impedance ratios but a passive LPF loaded by an active filter raises the breakpoint and also attenuates to 1/4 or -12dB at DC with these 4:1 R ratios.

-3dB BW 242 mHz, ... 469mHz without C1

DC gain -12 dB ........ same
2 Hz -40dB ............ -24 dB
2kHz -80dB ............ -44 dB
-40 dB/dec ............ -20dB/dec

What do you really want? 2nd order filter with 0dB DC gain at 0.33 Hz?

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  • \$\begingroup\$ Hint : my “?” Has a sim for improved design , maybe but may want to answer me. \$\endgroup\$ – Sunnyskyguy EE75 Aug 15 '18 at 3:04
  • \$\begingroup\$ Hi Tony thanks for your answer, I would like to know how to obtain the transfer function equation when I have this kind of input filter on the op amp. I got the frequency response running anac sweep on the circuit but I would like to know how to obtain the equations. I need this transfer function because I will design a control loop for a variable that is measured by this circuit (already implemented on hardware). \$\endgroup\$ – Schwalb Aug 15 '18 at 3:38
  • \$\begingroup\$ What phase margin and slew rate is needed? \$\endgroup\$ – Sunnyskyguy EE75 Aug 15 '18 at 3:51
  • \$\begingroup\$ You have to short the Op Amp inputs to get the voltage on C1. \$\endgroup\$ – Sunnyskyguy EE75 Aug 15 '18 at 5:39
  • \$\begingroup\$ Thank you for your answer, I believe I got the correct transfer functions (I edited the post including the answers I got.). This circuit is being used to extract the DC value from an AC voltage (127Vrms/60Hz). I tuned a PI control loop to render a phase margin of 60 degrees and cutoff frequency of 0.3Hz to eliminate the DC value from a power inverter (LC filter drived by a half-bridge). \$\endgroup\$ – Schwalb Aug 15 '18 at 21:24

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