enter image description hereAccording 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..


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..

enter image description here

  • \$\begingroup\$ Justify 7.35k and 7.35 mA first. \$\endgroup\$ – Andy aka Sep 4 '15 at 6:55
  • \$\begingroup\$ oh sorry its a typo... \$\endgroup\$ – partykid Sep 4 '15 at 7:24

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 ;-)

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  • \$\begingroup\$ @FakeMoustache..if i have a multimeter adjusted to the ohm setting..and i measure resistance across R1..then what reading will it show ..as it is being bootsrapped..will it show something higher value..or the plain value of 68k. \$\endgroup\$ – partykid Sep 4 '15 at 11:52
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    \$\begingroup\$ Again you fall in the AC vs DC trap ! You cannot measure that AC properly with a ohmmeter. The bootstrapping is an AC effect, the ohmmeter a DC only device. So the ohmmeter would measure 68 kohms. While the circuit is working with an input signal you could use an oscilloscope and see that there would be no voltage across R1. \$\endgroup\$ – Bimpelrekkie Sep 4 '15 at 13:14

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

enter image description here

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.

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  • \$\begingroup\$ @Andy..since bootstrapping makes Zin looks much higher..and as Zin=R1..that technically means R1 will appear to be at a very high value..You have considered the voltage to be 1V...which brings R2 into consideration...which seems against my generalization...since moreover we are considered with R1..so its quite obvious for me to involve the voltage drop across R1 into calculation...which is (999.996-999.994)m...But if I do it then according to ohms law it wont yield a high input impedance ..and this is exactly my problem... \$\endgroup\$ – partykid Sep 4 '15 at 11:48
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    \$\begingroup\$ Zin does not equal R1. The volt drop across R1 is tiny due to bootstrapping ie 2uV therefore the current is in the pA range. This current flows from Vin therefore the bootstrapped impedance is Vin / 29 pA. \$\endgroup\$ – Andy aka Sep 4 '15 at 12:14
  • \$\begingroup\$ thanks i got it..i mistakenly thought that Zin depends upon R1 in bootstrap configuration .... Zin=Vin/i1 ..so silly of me \$\endgroup\$ – partykid Sep 4 '15 at 13:44
  • \$\begingroup\$ @partykid are you done with this question now? If so, please formally accept one of the answer else raise a new comment for further clarification. \$\endgroup\$ – Andy aka May 6 at 13:20

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