given : (r2/r1 ) = (r4/r3 )

c = 1nf , F= 2 Mhz

Required : output voltage and oscillation frequency at 3 cases

a- (r2/r1 )= (r4/r3 ) b- (r2/r1 )> (r4/r3 )

c-(r2/r1 )< (r4/r3 )


intial conditions : I (coil )= Io (i.e. doesnot equal 0 )

Vcap = 0

My attempts so far :

1- i deduced the L that satisfies parallel resonance between the parallel R-L-C , such that L= 6.33uH and at resonance the Z ( total ) = L /C

2-Since the RLC are at resonance , then the intial current Io travels in the parallel RC giving a rise to the cap voltage.

3- since the cap voltage equals the V+ then V+ increases

what i failed to understand is that :

from the voltage division
V- = Vo ( r1 / r1+ r2 )
V+ = Vo ( Z / Z+ r4 ) where z is the parallel RLC ; Z= L/C

so when Vc increases V+ increases and so does V- , this doesnt seem logical and i can't figure out what im missing out here

Thx in advance for your help

 

Why do you think it is not logical?

 

hi mik3

because its supposed to be an oscillator , Vo should change .

If V+ and V- and Vo increase together , Vo would reach Vcc and then stabilzes

 

When Vo will be at its peak value, the inductor will be discharged and the capacitor charged. Then the capacitor will discharge and V+ will reduce again. The capacitor will discharge, V+ will be at its lower value and the charged inductor will discharge again chatging the capacitor and risisng V+. This happens continuously.

 

What you have here is a "negative impedance converter" (NIC).

http://en.wikipedia.org/wiki/Negative_impedance_converter

The impedance looking into the plus input of the opamp needs to be slightly smaller in magnitude (and negative in sign) so that the parallel combination of the impedance presented by the NIC and R3 is a net negative impedance. When it's sufficient to overcome the losses of the inductor (and capacitor), the circuit will oscillate.

If the inductor is fairly good quailty, it won't take much to sustain oscillations. It would be reasonable to assume that when the negative input impedance of the NIC is equal to R3, that will be the boundary between oscillations and no oscillations.

If the negative impedance of the NIC overwhelms the positive impedance of R3 by a large margin, the oscillations will be quite non-sinusoidal; the opamp output will be driven to the rails.