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[SOLVED] Basic question about common source amplifier with inductor

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palmeiras

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Basic question about commom source amplifier with inductor

Hi guys,


uso do indutor.png

In the figure, the resistor RD is replaced by the inductor LD, RAZAVi says that the new circuit has better frequency response. Could, you please, explain why?
How does the Module of Bode diagram looks with the indutor? Is it a band pass with the center in the resonance frequency . Is that correct?
The resonance frequency is 1/square root (LD*CD)
Now, my doubt:
Doing the AC analysis, LD is in parallel with CD (tank in parallel). The impedance of it (at resonance frequency) is very large. For the other frequencies, the impedance is low. Is that correct?
So at resonance frequency, is like the parallel LC tank is unplugged on node X.
Is my explanation correct? Could you explain better it?
Thanks
 

Re: Basic question about commom source amplifier with inductor

Without knowing for what application the above circuit is intended to be used, it is somewhat difficult to evaluate, but your analysis is correct. For a resonance frequency of the LC tank, the gain would be maximum since the impedance of the resulted reactance of the conjugates XL and XC tends to infinite (resulted admitance of subtraction of both tend to zero), limited by the quality factor of the filter, in other words, of the losses in those components. Note that the value of RD that is needed to bias the transistor M1 in a linear region is a limiting factor for the maximum gain before the RC network start to attenuate the gain. Anyway, all this seems to be rather a circuit analysis within the theoretical scope than a comparison of actual circuits for the same application in real world.
 
Re: Basic question about commom source amplifier with inductor

Hi Andre,

Thanks! How is the Bode diagram for the case using inductor?
 

Re: Basic question about commom source amplifier with inductor

The bode plot for a 2nd order circuit is not so simple as it would be for a 1st order RC circuit, on which the junction of two assimptotes are a reasonable approximation at the cuttoff frequency. For LC circuits, in the case of real world components, there would be an attenuation factor, represented by the losses of the circuit, and this would reduce the crest peak at the resonant frequency. Anyway, you can find a nice tutorial here: https://www.youtube.com/watch?v=HrYo-D3b1RM
 
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