Here is my task:
Here is solution:
Although solution is there, I'm stuck. Firstly, how did they find transfer function? How did they treat diodes?
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1\$\begingroup\$ For the transfer function, assume the circuit to be linear; that's what they seem to do. Then use the usual method: represent in s-domain and solve via your favorite circuit analysis method, e.g. node method. \$\endgroup\$– got trolled too much this weekCommented Nov 21, 2015 at 20:55
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\$\begingroup\$ So to calculate transfer function, I analyse this circuit oi65.tinypic.com/14d0mu0.jpg , right? \$\endgroup\$– hariCommented Nov 21, 2015 at 21:00
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1\$\begingroup\$ Yes, that's what they did: imgur.com/PSb5SHO \$\endgroup\$– got trolled too much this weekCommented Nov 21, 2015 at 21:44
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1 Answer
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Hari - here is a much simpler method for deriving the bandpass transfer function:
Without R1 and C the rest of the circuit (at node 1) is nothing else that the classical active inductor circuit based on an GIC block (Generalized Impedance Converter). It is easy to show that the inductive input impedance (node 1) is Zi=sCR²=sL with L=R²C.
Hence, together with the parallel capacitor C you can treat the whole circuit as an ideal tank circuit (L||C) with an input impedance Zt=sL/(1+s²LC). Now, the tank circuit forms - together with R1 - a dynamic voltage divider with a bandpass response (at node 1):
H(s)=Zt/(Zt+R1)=....=sL/[R1(1+s²LC)+sL].
Introducing the given expression for L you arrive at the desired transfer function (at node 1). More than that, it is easy to show that - for ideal opamps - the output voltage at the opamp outputs is twice the voltage at node 1 (for equal resistors R within the GIC block). Thus, the task for deriving the filters transfer function is solved.
The circuit can be transferred to an oscillator by closing the loop using another resistor R between opamp output and filter input (resp the finite filter input resistance, which is identical to R1 at the oscillation frequency).
Hence, we have a voltage division ratio (at w=wo): R1/(R1+R). This ratio must be chosen to be slightly larger than "1/2". In this case (because of the filter gain of "2") we have a loop gain slightly larger than unity at w=wo (safe start of oscillation). For rising amplitudes the diodes will reduce the loop gain to unity - thereby fulfilling the condition for steady-state oscillations.
