Hartley Oscillator Op-amp Circuit

[akbarza]

hi
the attached pic is from https://www.electronics-tutorials.ws/oscillator/hartley.html
how is the oscillator frequency calculated?
in the above address is said that " the gain of the circuit must be equal too or slightly greater than the ratio of L1/L2. If the two inductive coils are wound onto a common core and mutual inductance M exists then the ratio becomes (L1+M)/(L2+M)."

please explain how is obtained that the gain must be larger the ratio L1/L2 ?also explain about (L1+M)/(L2+M).
thanks

 

[BradtheRad]

By making L1 L2 into a coupled inductor (or perhaps a tapped inductor), the current in the windings reinforce each other. A resonant loop is set up (including the capacitor), so that oscillations are more stable and more easily sustained. So less gain is needed.

 

[FvM]

The OP circuit is no functional Hartley oscillator, although similar circuits are often shown in literatur. Hartley oscillator requires a high impedance (> characteristic resonator impedance) at amplifier in and output. This is the case with the usual transistor and tube circuits.

In the present circuit, the low impedance OP output is shorting the resonator and you never achieve required 180 degree phase shift to fulfill the oscillation condition.

 

[BigBoss]

I have done a simple math to find Transfer Function of this circuit and I couldn't see L2 in TF.

 

[BradtheRad]

Here's my simulation with an NPN transistor rather than an op amp. (Running in Falstad's animated interactive simulator.)

The coupled inductor acts in a manner to reinforce oscillations. (In the simulator it's a transformer.)

The yellow dots represent current bundles moving through wires. The dots move up in unison through both sides of the transformer. Then they move downward in unison.

Oscillations weaken if I reduce the coupling coefficient.

 

[LvW]

I have done a simple math to find Transfer Function of this circuit and I couldn't see L2 in TF.

Yes, L2 acts as a simple load to the opamp without being part of the feedback circuit.
Reason: The circuit was transferred from a transistor circuit without considering the fact that a BJT works as a current source - but the opamp is a voltage source.
Hence, use an additional resistor between opamp output and L2.