Bootstrapping and miller's theorem

[ShreyasR]

I am trying to understand how the input impedance of a bootstrapped darlington emitter follower is analysed. They make use of Miller's theorem. But honestly, I have not understood this theorem as well... All I know is bootstapping increases input impedance but i have no clue as to how? And why is a capacitor connected between the input and output terminals? Can someone link me to some place where a bootstrapped darlington circuit is analysed? I googled it but did not find any helpful links...

 

[berkeman]

I am trying to understand how the input impedance of a bootstrapped darlington emitter follower is analysed. They make use of Miller's theorem. But honestly, I have not understood this theorem as well... All I know is bootstapping increases input impedance but i have no clue as to how? And why is a capacitor connected between the input and output terminals? Can someone link me to some place where a bootstrapped darlington circuit is analysed? I googled it but did not find any helpful links...

IIRC, the Miller capacitance is between base and collector, and provides negative feedback (slowing down frequency response). Bootstrapping is positive unity feedback from output to input (for any amplifier) to increase input impedance, so it would be between emitter and base in this case.

The most common bootstrapping I'm familiar with is bootstrapping the shield capacitance on an input to an amplifier. If the input coax cable (or triax cable) has enough capacitance to cause a LPF rolloff of the input signal, you can drive the shield with a buffered version of the signal to eliminate the capacitance to the adjacent shield.

http://www.ti.com/general/docs/lit/getliterature.tsp?literatureNumber=snoa664&fileType=pdf

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[Jony130]

Hi, Miller's theorem is better know as a Miller effect

http://en.wikipedia.org/wiki/Miller_effect
http://web.mit.edu/klund/www/papers/jmiller.pdf

We simply have a ideal voltage amplifier with gain equal to A = -10V/V

Next we connect a resistor R = 10Ω between the input and the output of the amplifier.

Now let as try to find a input resistance.

Rin = Vin/Iin

In = (Vin - Vout)/R = (Vin - A*Vin)/R = Vin * (1 - A)/R

Rin = Vin/Iin = R/(1 - A)

Iin = (1V - (-10V))/10Ω = 1.1A

So Rin = 1V/1.1A = 0.909Ω

So as you can see our Rin resistance is (1 - A) smaller then R if we have inverting amplifier .
And this is what we call a Miller effect

Now let us consider different situation. We replace our amplifier with "voltage follower" amplifier.
But now the gain is equal to A = 0.5V/V

So if Vin = 1V we get 0.5V at the output. So the input current is equal to:

Iin = (Vin - Vout)/R = 0.5V/10Ω = 50mA and therefore

Rin = 1V/50mA = 20Ω

If we increase the gain to 0.9V/V we have

Iin = (Vin - Vout)/R = 0.1V/10Ω = 10mA

So the input resistance is equal to

Rin = 1V/10mA = 100Ω

Also we can use Miller's theorem to find Rin.

Rin = R/(1 - A) = 10Ω/(1 - 0.5) = 20Ω

Rin = R/(1 - A) = 10Ω/(1 - 0.9) = 100Ω

But this time we call this bootstrap
http://electronics.stackexchange.co...analysis-of-a-emitter-follower-with-bootstrap

Any questions ?