In the book I got this question from, the author gives two expressions for the input resistance (\$r_{in}\$, or \$\dfrac{v_{in}}{i_{in}}\$ in the AC model) of this bootstrap bias circuit. The two expressions are below:

1. \$\dfrac{v_{in}}{i_{in}} = \dfrac{R_3}{1-A} \parallel (r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E ))\$

2. \$\dfrac{v_{in}}{i_{in}} = \dfrac{(\beta+1) R_E'R_3 + r_\pi ( R_3 + R_E')}{R_3+r_\pi}\$

Expression 2 is obtained from a thorough analysis of the AC model of the circuit (which I put in the question). Expression 1 uses more simplifying assumptions.

For reference, below are my attempts at finding both expressions for the input resistance.

Solution 1

In this solution, I try to find that \$\dfrac{v_{in}}{i_{in}} = \dfrac{R_3}{1-A} \parallel (r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E ))\$.

Due to the circuit's behavior as an emitter follower (as explained in jonk's answer), the node V has a voltage of approximately \$AV_{in}\$, where A is the gain of the emitter follower (therefore A is very close to 1).

Therefore, the current through the \$R_3\$ branch is about \$\dfrac{v_{in} - Av_{in}}{R_3} = \dfrac{(1-A)v_{in}}{R_3}\$. Since A is very close to 1, \$\dfrac{(1-A)v_{in}}{R_3}\$ is very close to 0.

Now, let's express\$v_{in}\$ in terms of \$i_b\$ (the current through the \$r_\pi\$ branch). Since the current through \$R_3\$ is very small compared to the current through \$ R_2 \parallel R_1 \parallel R_E \$, I will neglect the \$R_3\$ branch for the following calculation, and assume that all of the emitter current (\$(\beta+1)i_b\$) goes through the \$ R_2 \parallel R_1 \parallel R_E \$ combination. Thus, \$v_{in}\$ can be calculated as the voltage across \$r_\pi\$ (which is \$i_br_\pi\$) plus the voltage across \$ R_2 \parallel R_1 \parallel R_E \$ (which is \$(\beta+1)i_b( R_2 \parallel R_1 \parallel R_E )\$):

\$v_{in} = i_br_\pi + (\beta+1)i_b( R_2 \parallel R_1 \parallel R_E )\$

So, the current through \$r_\pi\$ can be expressed as:

\$i_b = \dfrac{v_{in}}{ r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E )}\$

Now, let's calculate \$i_{in}\$. It can be calculated as the sum of the currents through \$R_3\$ and \$r_\pi\$:

\$ i_{in} = \dfrac{(1-A)v_{in}}{R_3} + \dfrac{v_{in}}{ r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E )} \$

Now, let's calculate \$\dfrac{v_{in}}{i_{in}}\$:

\$\dfrac{v_{in}}{i_{in}} = \dfrac{v_{in}}{\dfrac{(1-A)v_{in}}{R_3} + \dfrac{v_{in}}{ r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E )}}\$

\$\dfrac{v_{in}}{i_{in}} = \dfrac{1}{\dfrac{(1-A)}{R_3} + \dfrac{1}{ r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E )}}\$

\$\dfrac{v_{in}}{i_{in}} = \dfrac{1}{\dfrac{1}{\dfrac{R_3}{1-A}} + \dfrac{1}{ r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E )}}\$

\$\dfrac{v_{in}}{i_{in}} = \dfrac{R_3}{1-A} \parallel (r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E ))\$

Solution 2

In this solution, I try to find that \$\dfrac{v_{in}}{i_{in}} = \dfrac{(\beta+1) R_E'R_3 + r_\pi ( R_3 + R_E')}{R_3+r_\pi}\$.

Applying KCL on the node labeled V (the current into this node from the transistor emitter is \$(\beta+1)i_b\$):

\$(\beta+1)i_b = \dfrac{V}{R_1}+\dfrac{V}{R_2}+\dfrac{V}{R_E} + \dfrac{V-v_{in}}{R_3}\$

\$(\beta+1)i_b = V\left ( \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_E} \right ) + \dfrac{V-v_{in}}{R_3}\$

Making \$\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_E} = R_E'\$:

\$(\beta+1)i_b = \dfrac{V}{R_E'} + \dfrac{V-v_{in}}{R_3}\$

Now, expressing \$V\$ in terms of \$v_{in}\$ and \$i_b\$:

\$V=v_{in}-i_br_\pi\$

Making \$V=v_{in}-i_br_\pi\$ in the node equation:

\$(\beta+1)i_b = \dfrac{v_{in}-i_br_\pi}{R_E'} + \dfrac{v_{in}-i_br_\pi-v_{in}}{R_3}\$

\$v_{in} = i_b\left [(\beta+1) R_E' + r_\pi + \dfrac{r_\pi R_E'}{R_3} \right ]\$

Plugging this \$v_{in}\$ expression back into the formula \$V=v_{in}-i_br_\pi\$:

\$V = v_{in} - i_br_\pi = i_b\left [(\beta+1) R_E' + \dfrac{r_\pi R_E'}{R_3} \right ]\$

Now, expressing \$i_{in}\$ as the sum of the currents through \$r_\pi\$ and \$R_3\$:

\$i_{in} = i_b+\dfrac{v_{in}-V}{R_3}\$

Plugging in the expressions found for \$V\$ and \$v_{in}\$ in terms of \$i_b\$:

\$i_{in} = i_b+\dfrac{i_br_\pi}{R_3}=i_b\left ( \dfrac{R_3+r_\pi}{R_3} \right )\$

\$i_{in} = i_b+\dfrac{i_br_\pi}{R_3}=i_b\left ( \dfrac{R_3+r_\pi}{R_3} \right )\$

Finally, calculating the input resistance (\$\dfrac{v_{in}}{i_{in}}\$):

\$\dfrac{v_{in}}{i_{in}} = \dfrac{i_b\left [(\beta+1) R_E' + r_\pi + \dfrac{r_\pi R_E'}{R_3} \right ]}{i_b\left ( \dfrac{R_3+r_\pi}{R_3} \right )}\$

\$\dfrac{v_{in}}{i_{in}} = \left (\dfrac{(\beta+1) R_E'R_3 + r_\pi R_3 + r_\pi R_E'}{R_3} \right )\left ( \dfrac{R_3}{R_3+r_\pi} \right )\$

\$\dfrac{v_{in}}{i_{in}} = \dfrac{(\beta+1) R_E'R_3 + r_\pi ( R_3 + R_E')}{R_3+r_\pi}\$

I'm the OP and below is my own attempt at analysing this circuit (by finding its input resistance).

In the book I got this question from, the author gives two expressions for the input resistance (\$r_{in}\$, or \$\dfrac{v_{in}}{i_{in}}\$ in the AC model) of this bootstrap bias circuit. The two expressions are below:

1. \$\dfrac{v_{in}}{i_{in}} = \dfrac{R_3}{1-A} \parallel (r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E ))\$

2. \$\dfrac{v_{in}}{i_{in}} = \dfrac{(\beta+1) R_E'R_3 + r_\pi ( R_3 + R_E')}{R_3+r_\pi}\$

Expression 2 is obtained from a thorough analysis of the AC model of the circuit (which I put in the question). Expression 1 uses more simplifying assumptions, but it gives more intuition about the circuit's behavior (see Solution 1 below).

For reference, below are my attempts at finding both expressions for the input resistance.

Solution 1

In this solution, I try to find that \$\dfrac{v_{in}}{i_{in}} = \dfrac{R_3}{1-A} \parallel (r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E ))\$.

Due to the circuit's behavior as an emitter follower (as explained in jonk's answer), the node V has a voltage of approximately \$AV_{in}\$, where A is the gain of the emitter follower (therefore A is very close to 1).

Therefore, the current through the \$R_3\$ branch is about \$\dfrac{v_{in} - Av_{in}}{R_3} = \dfrac{(1-A)v_{in}}{R_3}\$. Since A is very close to 1, \$\dfrac{(1-A)v_{in}}{R_3}\$ is very close to 0.

Now, let's express\$v_{in}\$ in terms of \$i_b\$ (the current through the \$r_\pi\$ branch). Since the current through \$R_3\$ is very small compared to the current through \$ R_2 \parallel R_1 \parallel R_E \$, I will neglect the \$R_3\$ branch for the following calculation, and assume that all of the emitter current (\$(\beta+1)i_b\$) goes through the \$ R_2 \parallel R_1 \parallel R_E \$ combination. Thus, \$v_{in}\$ can be calculated as the voltage across \$r_\pi\$ (which is \$i_br_\pi\$) plus the voltage across \$ R_2 \parallel R_1 \parallel R_E \$ (which is \$(\beta+1)i_b( R_2 \parallel R_1 \parallel R_E )\$):

\$v_{in} = i_br_\pi + (\beta+1)i_b( R_2 \parallel R_1 \parallel R_E )\$

So, the current through \$r_\pi\$ can be expressed as:

\$i_b = \dfrac{v_{in}}{ r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E )}\$

Now, let's calculate \$i_{in}\$. It can be calculated as the sum of the currents through \$R_3\$ and \$r_\pi\$:

\$ i_{in} = \dfrac{(1-A)v_{in}}{R_3} + \dfrac{v_{in}}{ r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E )} \$

Now, let's calculate \$\dfrac{v_{in}}{i_{in}}\$:

\$\dfrac{v_{in}}{i_{in}} = \dfrac{v_{in}}{\dfrac{(1-A)v_{in}}{R_3} + \dfrac{v_{in}}{ r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E )}}\$

\$\dfrac{v_{in}}{i_{in}} = \dfrac{1}{\dfrac{(1-A)}{R_3} + \dfrac{1}{ r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E )}}\$

\$\dfrac{v_{in}}{i_{in}} = \dfrac{1}{\dfrac{1}{\dfrac{R_3}{1-A}} + \dfrac{1}{ r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E )}}\$

\$\dfrac{v_{in}}{i_{in}} = \dfrac{R_3}{1-A} \parallel (r_\pi + (\beta+1)( R_2 \parallel R_1 \parallel R_E ))\$

In this approximate expression, we can clearly identify that one of the parallel components, \$\dfrac{R_3}{1-A}\$, is the apparently very large "effective resistance" that the author referred to.

Solution 2

In this solution, I try to find that \$\dfrac{v_{in}}{i_{in}} = \dfrac{(\beta+1) R_E'R_3 + r_\pi ( R_3 + R_E')}{R_3+r_\pi}\$.

Applying KCL on the node labeled V (the current into this node from the transistor emitter is \$(\beta+1)i_b\$):

\$(\beta+1)i_b = \dfrac{V}{R_1}+\dfrac{V}{R_2}+\dfrac{V}{R_E} + \dfrac{V-v_{in}}{R_3}\$

\$(\beta+1)i_b = V\left ( \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_E} \right ) + \dfrac{V-v_{in}}{R_3}\$

Making \$\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_E} = R_E'\$:

\$(\beta+1)i_b = \dfrac{V}{R_E'} + \dfrac{V-v_{in}}{R_3}\$

Now, expressing \$V\$ in terms of \$v_{in}\$ and \$i_b\$:

\$V=v_{in}-i_br_\pi\$

Making \$V=v_{in}-i_br_\pi\$ in the node equation:

\$(\beta+1)i_b = \dfrac{v_{in}-i_br_\pi}{R_E'} + \dfrac{v_{in}-i_br_\pi-v_{in}}{R_3}\$

\$v_{in} = i_b\left [(\beta+1) R_E' + r_\pi + \dfrac{r_\pi R_E'}{R_3} \right ]\$

Plugging this \$v_{in}\$ expression back into the formula \$V=v_{in}-i_br_\pi\$:

\$V = v_{in} - i_br_\pi = i_b\left [(\beta+1) R_E' + \dfrac{r_\pi R_E'}{R_3} \right ]\$

Now, expressing \$i_{in}\$ as the sum of the currents through \$r_\pi\$ and \$R_3\$:

\$i_{in} = i_b+\dfrac{v_{in}-V}{R_3}\$

Plugging in the expressions found for \$V\$ and \$v_{in}\$ in terms of \$i_b\$:

\$i_{in} = i_b+\dfrac{i_br_\pi}{R_3}=i_b\left ( \dfrac{R_3+r_\pi}{R_3} \right )\$

\$i_{in} = i_b+\dfrac{i_br_\pi}{R_3}=i_b\left ( \dfrac{R_3+r_\pi}{R_3} \right )\$

Finally, calculating the input resistance (\$\dfrac{v_{in}}{i_{in}}\$):

\$\dfrac{v_{in}}{i_{in}} = \dfrac{i_b\left [(\beta+1) R_E' + r_\pi + \dfrac{r_\pi R_E'}{R_3} \right ]}{i_b\left ( \dfrac{R_3+r_\pi}{R_3} \right )}\$

\$\dfrac{v_{in}}{i_{in}} = \left (\dfrac{(\beta+1) R_E'R_3 + r_\pi R_3 + r_\pi R_E'}{R_3} \right )\left ( \dfrac{R_3}{R_3+r_\pi} \right )\$

\$\dfrac{v_{in}}{i_{in}} = \dfrac{(\beta+1) R_E'R_3 + r_\pi ( R_3 + R_E')}{R_3+r_\pi}\$