how to increase input impedance via bootstraping?

Question

According to theory in order to increase the input impedance bootstrap capacitor is employed…which couples the same output voltage to point B ..and as the circuit is a voltage follower Vin = Vout ..thus the voltage difference across R1 is very small ..which means that no or very little current will flow through R1 ..which can be also visualised as a very high resistance present..

Analysis:

Without bootstrap capacitor Cb:

Zin = R1; figure 1..

With bootstrap capacitor:

Zin = (1+Aol)R1 ....where Aol = open loop gain of opamp.

As its clear from above equation that input impedance increase by a factor of (1+Aol)..

since from figure 1. we have

i1 = V1/R1 = 500mv / 68K = 7.35 mA.

from figure 2 ..we have

Zin = (999.996 - 999.998)mV / 7.35mA (forget Zin = (1+Aol) R1 for the time being) = 2.72* 10^-4 k

But the value of Zin should be very 2 large ..but is not agreeing with the theory.. why?? please justify..

Answer 1

I think you're mixing up the circuit's behaviour for DC (constant over time) currents and voltages and AC (varying over time like a sinewave) currents and voltages.

If someone talks about "small signal" then that's AC !

You have to "bias" your circuit so that it can work, that is DC. Let's see if your circuit is biased properly. For DC you can pretend all the Capacitors are gone, because Capacitors cannot conduct DC current. This would leave the + input of the opamp connected to ground via 2 resistors. You know no current flowes into the input of an opamp so the + input is at 0 V DC Since Cb is out, the - input is directly connected to the output of the opamp so we have a unity gain buffer. This means the output voltage will copy the voltage at the + input, so it will be... 0V DC ! Does it matter at what DC voltage Vin is ? No, because the Capacitor will block the DC whatever it is !

Now we will discuss AC :-) Let's assume the input signal is a 1 V sinewave (AC) at the input with a frequency that is high enough such that the capacitors have a very low impedance. Let's also assume the opamp is ideal and that it can easily handle this high frequency.

Since the Cap at the input has a very low impedance (for our assumed signal) the signal will also appear at the + input, so 1 V AC at the + input of the opamp. The opamp is still connected as a unity gain buffer isn't it ? So at the opamp's output we will also have the same (a copy of) the 1 V sinewave. Capacitor Cb makes sure this 1 V AC will also appear at the node between the 2 68 k ohm resistors. Now what will be the AC voltage across R1 ?

There's your bootstrap ! If you didn't have the bootstrapping in place the input impedance (for AC) would be 2 x 68 k ohms = 136 komhs (or 1 x 68 komhs in the top circuit). But because of the bootstrapping the AC voltage across R1 is zero meaning that no AC current will flow meaning infinitely high input impedance for AC ! So in theory, even if R1 and R2 had a very low value, the input impedance would be very high ! So is the magic of bootstrapping ;-)

Answer 2

Here's your 2nd picture with the error marked in red: -

How on earth you get 999.998mV at point B baffles me. Because the 0.5uF capacitor is assumed to be much lower impedance at AC than either R1 or R2, the voltage at point B is 999.994mV leaving an AC voltage across R1 of 2uV.

This implies a current thru R1 of 2uV / 68k = 29 pA.

This current is sourced from the top of R1 i.e. 0.999996V therefore the input impedance is approximately 1 V / 29 pA = 34 Gohms.

Now clearly the real input impedance will be lower because the op-amp input will have some relevance to the story but, theoretically, with an infinite op-amp impedance the bootstrapping yields many G ohms input impedance.