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How does this circuit eliminate Miller Effect?

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iVenky

iVenky

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According to Sedra and Smith book, the voltage at the top of C1 and the voltage at the bottom of C2 are out of phase as their other end is connected to ground- How do you say this?

This seems to be eliminating the feedback as both cancels.

Thanks in advance.
 

It is called a neutralized amp. The output resonant tank is AC ground tapped in between the two ends. This means the AC swing on the far end of the coil (at resonance) is 180 degs output of phase with the AC swing on the collector. The bottom of coil output is fed back to base on transistor to cancel output the coupling internal to the transistors (Ccb).

The capacitance ratio (or tapped coil turns ratio) of output tank determines the amplitude of the far end compared to the collector end. Lets say the far end is 1/3 the voltage swing on the collector end, then the proper feedback cap from far end to base of transistor would be 3 times the Ccb of the transistor to neutralize the amplifier.

Ccb is a function of DC bias on transistor, particularly the DC voltage from base to collector, so you have to keep this voltage relatively constant to avoid Ccb from changing with change in average DC bias.

The main limitation of a neutralized amp is narrow bandwidth, being only that of the operating Q of the output tank circuit.
 

In other words, when looking at the small-signal circuit analysis you will see the one cap at the input of the amp and the other at the output with the inductor as the feedback. Being a CE amplifier, the one cap is 180 degrees out of phase with the other which can create a push-pull effect. Satisfying Barkhuizens criteria.
 

Baas Rietrot said:
In other words, when looking at the small-signal circuit analysis you will see the one cap at the input of the amp and the other at the output with the inductor as the feedback. Being a CE amplifier, the one cap is 180 degrees out of phase with the other which can create a push-pull effect. Satisfying Barkhuizens criteria.
Click to expand...

If you mean "Barkhausen criteria" it is not what is intended for this circuit. The purpose of neutralizaion is to improve stabiltiy of RF amp and prevent it from oscillating. The Barkhausen stability criterion is a mathematical condition to determine when a linear electronic circuit will oscillate.

https://electriciantraining.tpub.com/14180/css/14180_76.htm

When done correctly, the input impedance will be 1/Y11 of the device (at output tank resonance). It is better to inductive couple for amp output. The series matching cap on output will effect ratio of tank feedback (and therefore neutralization) as load impedance changes on series matching cap.
 
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RCinFLA said:
It is called a neutralized amp. The output resonant tank is AC ground tapped in between the two ends. This means the AC swing on the far end of the coil (at resonance) is 180 degs output of phase with the AC swing on the collector. The bottom of coil output is fed back to base on transistor to cancel output the coupling internal to the transistors (Ccb).

The capacitance ratio (or tapped coil turns ratio) of output tank determines the amplitude of the far end compared to the collector end. Lets say the far end is 1/3 the voltage swing on the collector end, then the proper feedback cap from far end to base of transistor would be 3 times the Ccb of the transistor to neutralize the amplifier.

Ccb is a function of DC bias on transistor, particularly the DC voltage from base to collector, so you have to keep this voltage relatively constant to avoid Ccb from changing with change in average DC bias.

The main limitation of a neutralized amp is narrow bandwidth, being only that of the operating Q of the output tank circuit.
Click to expand...


I can understand that the feedback will cancel if there is 180 degree out of phase between the two ends of the coil. What I couldn't understand is how do you say that there is 180 deg phase difference between the two ends of the inductor at resonance?

Thanks in advance.
 

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