# Explanation about op-amp bias resistor and bootstrapping  I am reading a book on op-amp and its applications in linear-integrated circuits and came across this topic of creating an op-amp circuit with very high input impedance. I believe that I understand the part about the dc-coupled voltage follower (correct me if I am wrong) where the author advises the reader to use an op-amp with a bias current in the range of nA (for a voltage follower) so that the voltage drop across the source resistance $$\R_{in}\$$ is minimum. This leads to minimum difference between the input voltage at the non-inverting terminal and the output voltage $$\V_o\$$ and hence the performance of the voltage follow is satisfactory.

Now from the part about the ac-coupled voltage follower, I understand that the coupling capacitor eliminates the dc offset (if any) in the input signal. From this document, I understood that

For the op-amp to operate correctly, these inputs must be DC biased. That is, the DC bias currents ($$\I_{B+}\$$ and $$\I_{B-}\$$), must be able to flow into or out of the input terminals.

and hence we need to add a bias resistor $$\R_1\$$ that connects the non-inverting terminal to the ground and creates a path for $$\I_{B+}\$$. Here the book mentions that:

this bias resistor drastically reduces the input resistance of the follower circuit. In fact, the input resistance is equal to the bias resistance.

Here I want to understand how the bias resistor has reduced the input resistance and how, specifically the input resistance is now equal to the bias resistance.

Then the author explains that we can fix this by bootstrapping the bias resistance. This is done by adding a resistance $$\R_2\$$ and connecting the output to the node in between $$\R_1\$$ and $$\R_2\$$ via a capacitor $$\C_2\$$ (as shown in fig 6-22(b)). As the gain of the voltage follower is 1, drop across $$\R_1\$$ is now almost zero (see text) and the input resistance is now high again.

I want to understand this part, please explain bootstrapping and how it is useful here and why do we need the second capacitor when we have already eliminated the dc offset using the first capacitor. Also by reducing the drop across $$\R_1\$$ we are now restricting the bias current so doesn't this defeat the original purpose?

• Could you please tell me what book is that? I'm looking into the design of very high impedance buffer circuits and I'm struggling to find a good ammount of didatic materials Sep 30 at 21:31
• @RodrigoAnjosdeSouza The name of the book is Op-Amps and Linear Integrated Circuits. But I am not sure if this book will serve your purpose as it is a general book on op-amps and these images along with a few more pages are the only parts dedicated to this topic. Anyways, you can find a pdf for this book online and check if the content helps. Oct 7 at 20:13

Without $$\C_2\$$ the input resistance seen by the signal source is equal to $$\R_1 + R_2 = 200kΩ\$$

But for the signal frequency, the capacitor will act just like a short circuit ($$\X_{C2} \approx 0\$$). So the situation will look like this: simulate this circuit – Schematic created using CircuitLab

And finally, we have: simulate this circuit

As you can see now, for the AC signal our $$\R_1\$$ resistor is connected between the input and the output of the amplifier. And the voltage across $$\R_1\$$ resistor is equal to the difference between $$\V_{IN}\$$ and $$\V_{OUT}\$$.

Now we can try to find the input resistance. The input resistance is equal to $$\R_{IN}=\frac{V_{IN}}{I_{IN}}\$$ additional as you know the opamp input bias current is very low we can see that $$\I_{IN} = I_{R1}\$$. Of course, the current that will flow through $$\R_1\$$ is determined by the voltage that appears across the resistor. Because your amplifier is a voltage follower which means that the output is equal to the input voltage there will be almost no voltage drop across $$\R_1\$$ resistor. All this means that the input resistance will be very high.

But let us assume that the voltage gain of our voltage follower is equal to $$\A\approx 1\$$

$$R_{IN}=\frac{V_{IN}}{I_{IN}}$$

$$I_{IN} = \frac{V_{IN}- V_{OUT}}{R_1} = \frac{V_{IN} - A V_{IN}}{R_1} = V_{IN} \times \frac{1 - A}{R_1}$$

$$R_{IN} = \frac{V_{IN}}{I_{IN}} = \frac{R_1}{1 - A}$$

Notice that for $$\A = 1\$$ we will have $$\R_{IN} = oo\$$ but for $$\A = 0.98\$$ (for Vin = 1V---> Vout = 0.98V)

$$\R_{IN} =\frac{R_1}{1 - A} = \frac{100k\Omega}{1 - 0.98} = 5M \Omega\$$.

• Thanks for the explanation. Based on what you and others have posted I have written a note as a pdf here. I can't post the entire note as it is quite long so please take a look at it and check if I understood it right. Sep 19 at 19:58
• @rustyelectron It looks like you got it. As a side note, that if you have inverting amplifier you no longer have a bootstrap (positive feedback) but a Miller effect (negative feedback) instead electronics.stackexchange.com/questions/234349/…
– G36
Sep 19 at 20:10
• Thanks for the heads up. I will look into it. Sep 19 at 20:16

A bit of a hand-wavy answer but:

The input capacitor should be chosen to appear as a short to the signal of interest and an open circuit to DC. After the capacitor, you have the parallel combination of the opamp input impedance and the bias resistance. Since the opamp input impedance is assumed to be very high in comparison to the bias resistance, the bias resistance value dominates.

The bootstrapping circuit with the feedback capacitor is neat in that it turns R1 into an essentially open circuit for the AC input signal, thus restoring the high input impedance but keeping the impedance "low" for the DC bias current which, recall, is coming out of the input which is already at Vin. The trick in the analysis is to remember that the the bias current is DC.

• Your short explanation also helped me write this note on what I understand so please take a look into this and check my interpretation. Sep 19 at 20:18

Capacitor C2 is necessary to separate the DC bias that is coming from the ground and the copy of the AC input signal that is from the output of the amplifier.

If C2 was not present (ie replaced by a short-circuit) the ground reference would be overridden by the DC at the output of the amplifier and the boas would not be stable.

In cases where the amplifier does not itself have zero offset (a simple emitter or source follower), the output may be at a slightly different voltage than ground anyway.