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Design >> Analog Design >> Bandwidth of Oscillator
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Message started by Brandon on Nov 16^th, 2011, 1:37pm

Title: Bandwidth of Oscillator
Post by Brandon on Nov 16^th, 2011, 1:37pm

Attached is schematic of oscillator whose output frequency depends on the LC tank circuit. The inductance by which this oscillator seems to be 'stable' is 1nH - 10mH. I need to know what is the purpose of the negative feedback with a bypass capacitor to ground and the purpose of the resistor network on top. Is there an effective way to modify the circuit to handle higher inductor values?

thanks

Title: Re: Bandwidth of Oscillator
Post by raja.cedt on Nov 16^th, 2011, 1:58pm

hello,
1.What is your Frequency of interst? Why you want to maximize the L, you dont have smaller inductors?

2. -ve feedback(at least DC) through is essential because you need -ve resistance to compensate finite Q of the tank. Top resisters are for biasing the opamp i guess.

3. Try to post small schematics so that it will easier to view.

Thanks,
Raj.

Title: Re: Bandwidth of Oscillator
Post by Brandon on Nov 16^th, 2011, 2:25pm

This oscillator will be used to determine the value of inductance or capacitance by measuring the output frequency from a microcontroller.
That is why I need to increase the bandwidth. The circuit is attached again.

Title: Re: Bandwidth of Oscillator
Post by aaron_do on Nov 16^th, 2011, 11:14pm

Hi Brandon,

I think the negative feedback is used to provide a DC bias to the negative input pin of the op-amp. If you want the circuit to work for higher L values, maybe you can increase the size of the two large capacitors. You should simulate it to verify.

regards,
Aaron

Title: Re: Bandwidth of Oscillator
Post by buddypoor on Nov 16^th, 2011, 11:59pm

Yes, of course the negative R-C feedback is necessary to ensure a bias point at approx. half of the supply voltage.
This is the clasisical bias concept for single supply circuits:
DC bias at the pos. input and negative feedback for dc unity gain of the opamp.