Finding AC equivalent seen between two nodes of a BJT Amplifier

In the textbook, Electronic Devices and Circuit Theory, 11th Edition by Boylestead and Nashelsky, on page 566, the lower cutoff frequency is derived for a voltage divider BJT circuit. The circuit used is shown below.

The lower cutoff frequency of $$\C_E\$$ is to be found. For this, they first find the equivalent AC circuit as seen from $$\C_E\$$.

My question is, how did the obtain this expression?: $$\dfrac{R_1||R_2}{\beta}+r_e$$

I understand that they applied Thevenin's thorem, but I'm stil not sure how this expression comes out. Also, doesn't $$\R_C\$$ play a role here?

• – G36
Commented Mar 22, 2020 at 18:58

Just draw the small signal model of the circuit to calculate the impedance seen at the emitter. Assuming a resistance $$\r_b\$$ is connected to the base, the impedance is calculated as:

simulate this circuit – Schematic created using CircuitLab

Apply KCL at the emitter to give:

$$\beta i_b + i_b +i_x = \frac{v_x}{R_E}$$

Apply KVL in the emitter-base loop:

$$v_x + (\beta + 1)i_br_e + i_br_b = 0$$ $$\implies i_b = -\frac{v_x}{(\beta + 1)r_e + r_b}$$

Putting it in the first equation, $$-(\beta + 1)\frac{v_x}{(\beta + 1)r_e + r_b} + i_x = \frac{v_x}{R_E}$$ $$\implies \frac{i_x}{v_x} = \frac{1}{R_E} + \frac{1}{r_e + \frac{r_b}{\beta + 1}}$$ Thus equivalent impedance is $$\R_E||(r_e+\frac{r_b}{\beta + 1})\$$.
The base resistance $$\r_b\$$ for your case is $$\R_1||R_2\$$ which gives the expression you want ($$\\beta >> 1\$$).
I want to emphasize that the calculated base resistance is valid if the transistor is driven by a current source. In case of voltage driven input, the equivalent base resistance would be zero. Note that BJT is a current controlled current source, so it's input is usually a current source.

Looking from the emitter terminal, collector is a current source whose current cannot be changed. It offers infinite resistance and acts as an open circuit:

$$\R = \dfrac{\Delta V}{\Delta I} = \dfrac{\Delta V}{ 0} = \infty\$$

Thus $$\R_C\$$ doesn't affect the impedance seen from emitter side.

You can obtain the resistance of a block by measuring the voltage and current. Suppose you put a current source $$\i\$$ at the emitter terminal. The voltage developed across the current source will be $$\ir_e + \dfrac{i}{\beta}(R_1\| R_2)\$$. Divide by the current to get the resistance seen at the emitter terminal.

Here $$\r_e\$$ is the resistane of transistor's base emitter junction.

• Comment: The quantity re is NOT the resistance of the base-emitter junction. Instead, it is the inverse of the transconductance re=1/gm=VT/Ic (VT=temperature voltage). More than that, the influence of the ohmic resistor RE was forgotten (it is in parallel to the given expression).
– LvW
Commented Mar 23, 2020 at 14:09
• @LvW oh my textbook defined it as ac resistance of the emitter diode. It wouldn't be surprising for my textbook to be wrong though. It oversimplifies a lot prnt.sc/rl843t Commented Mar 23, 2020 at 14:14
• The input resistance at the emitter involves 1/gm - that is for sure. Please, look at the two links provided by the member G36 directly after the question. by the way - I do not like using the quantity re at all because - in fact - it is NOT a resistance. It is just an expression (unit voltage/current) for 1/gm , but it should not be modelled as a resistance. Here we have a good example how such a view can lead to severe misunderstandings.
– LvW
Commented Mar 23, 2020 at 14:26
• Ah I see your point. So it is not actually any physical resistance of the emitter diode. It is the inverse of transconductance and thus has units of resistnace. Thanks @LvW :) Commented Mar 23, 2020 at 14:37
• To make it clear: The defintion of the quantity to be used for the calculation (as shown in your book as a voltage-current curve, linked by you) is correct. But this quantity should not be called "resistance of the B-E diode". It is nothing else than d(Vbe)/d(Ie)=1/gm (assuming Ie=Ic)
– LvW
Commented Mar 23, 2020 at 14:54

I think that expression should actually be:-

((R1//R2//Rs)/Beta) + re where Rs = signal source resistance

In the ideal case of Rs = 0 the left hand term disappears and the equation reduces to re.

• Just for clarification: re=1/gm=VT/IC (VT: temperature voltage VT)
– LvW
Commented Mar 23, 2020 at 14:33

Well, we have the following circuit:

simulate this circuit – Schematic created using CircuitLab

When analyzing a transistor we need to use the following relations:

• $$\text{I}_\text{E}=\text{I}_\text{B}+\text{I}_\text{C}\tag1$$
• Transistor gain $$\\beta\$$: $$\beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}\tag2$$
• Emitter voltage: $$\text{V}_\text{E}=\text{V}_1-\text{V}_2\tag3$$

When we use and apply KCL, we can write the following set of equations:

$$\begin{cases} \text{I}_2=\text{I}_1+\text{I}_9\\ \\ 0=\text{I}_8+\text{I}_9+\text{I}_{10}\\ \\ \text{I}_{10}=\text{I}_6+\text{I}_{11}\\ \\ \text{I}_7=\text{I}_5+\text{I}_{11}\\ \\ \text{I}_\text{E}=\text{I}_6+\text{I}_8\\ \\ \text{I}_1+\text{I}_3=\text{I}_\text{B}+\text{I}_2\\ \\ \text{I}_4=\text{I}_\text{C}+\text{I}_7\\ \\ \text{I}_5=\text{I}_3+\text{I}_4 \end{cases}\tag4$$

When we use and apply Ohm's law, we can write the following set of equations:

$$\begin{cases} \text{I}_1=\frac{\text{V}_\text{x}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_\text{y}-\text{V}_1}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_\text{y}-\text{V}_3}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_3-\text{V}_4}{\text{R}_5}\\ \\ \text{I}_6=\frac{\text{V}_2}{\text{R}_6}\\ \\ \text{I}_7=\frac{\text{V}_4}{\text{R}_7}\\ \\ \text{I}_8=\frac{\text{V}_2}{\text{R}_8} \end{cases}\tag5$$

Solving these systems of equations is possible, I used Mathematica:

In[1]:=FullSimplify[
Solve[{VE == V1 - V2, β == IC/IB, IE == IB + IC, I2 == I1 + I9,
0 == I8 + I9 + I10, I10 == I6 + I11, I7 == I5 + I11,
IE == I6 + I8, I1 + I3 == IB + I2, I4 == IC + I7, I5 == I3 + I4,
I1 == (Vx - V1)/R1, I2 == (V1)/R2, I3 == (Vy - V1)/R3,
I4 == (Vy - V3)/R4, I5 == (V3 - V4)/R5, I6 == (V2)/R6,
I7 == (V4)/R7, I8 == (V2)/R8}, {IB, IC, IE, I1, I2, I3, I4, I5, I6,
I7, I8, I9, I10, I11, V1, V2, V3, V4}]]

Out[1]={{IB -> -(((R6 + R8) (R1 (R2 + R3) VE + R2 R3 (VE - Vx) -
R1 R2 Vy))/(R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β)))),
IC -> -(((R6 + R8) (R1 (R2 + R3) VE + R2 R3 (VE - Vx) -
R1 R2 Vy) β)/(R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β)))),
IE -> -(((R6 + R8) (R1 (R2 + R3) VE + R2 R3 (VE - Vx) -
R1 R2 Vy) (1 + β))/(R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β)))),
I1 -> (-R2 R3 (R6 + R8) (VE - Vx) + R3 R6 R8 Vx (1 + β) +
R2 R6 R8 (Vx - Vy) (1 + β))/(R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β))),
I2 -> (R1 R3 (R6 + R8) VE + R3 R6 R8 Vx (1 + β) +
R1 R6 R8 Vy (1 + β))/(R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β))),
I3 -> (-R1 R2 (R6 + R8) (VE - Vy) -
R2 R6 R8 (Vx - Vy) (1 + β) +
R1 R6 R8 Vy (1 + β))/(R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β))),
I4 -> (R2 R6 R8 (R5 (Vx - Vy) + R3 Vy) +
R2 (-R3 R7 (R6 + R8) (VE - Vx) + R5 R6 R8 (Vx - Vy) +
R3 R6 R8 Vy) β +
R1 ((R3 -
R5) R6 R8 Vy - (R3 R7 (R6 + R8) VE + (-R3 +
R5) R6 R8 Vy) β +
R2 (R5 (R6 + R8) (VE - Vy) + R3 R6 Vy + (R3 + R6) R8 Vy -
R7 (R6 + R8) VE β +
R6 R7 Vy β + (R6 + R7) R8 Vy β)))/((R4 + R5 +
R7) (R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β)))),
I5 -> (R2 R6 R8 (-(R4 + R7) Vx + (R3 + R4 + R7) Vy) +
R2 (-R3 R7 (R6 + R8) (VE - Vx) + R3 R6 R8 Vy +
R6 (R4 + R7) R8 (-Vx + Vy)) β +
R1 (R6 (R3 + R4 + R7) R8 Vy - R3 R7 (R6 + R8) VE β +
R6 (R3 + R4 + R7) R8 Vy β +
R2 (-R4 (R6 + R8) (VE - Vy) + R3 R6 Vy + R3 R8 Vy -
R7 R8 (VE - Vy) (1 + β) -
R6 (R7 VE - (R7 + R8) Vy) (1 + β))))/((R4 + R5 +
R7) (R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β)))),
I6 -> -((R8 (R1 (R2 + R3) VE + R2 R3 (VE - Vx) -
R1 R2 Vy) (1 + β))/(R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β)))),
I7 -> (R2 R6 R8 (R5 (Vx - Vy) + R3 Vy) +
R2 (R3 (R4 + R5) (R6 + R8) (VE - Vx) + R5 R6 R8 (Vx - Vy) +
R3 R6 R8 Vy) β +
R1 (R3 R6 R8 Vy +
R3 ((R4 + R5) (R6 + R8) VE + R6 R8 Vy) β -
R5 R6 R8 Vy (1 + β) +
R2 (R6 R8 Vy + R3 (R6 + R8) Vy +
R4 (R6 + R8) (VE - Vy) β + R6 R8 Vy β +
R5 (R6 + R8) (VE - Vy) (1 + β))))/((R4 + R5 +
R7) (R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β)))),
I8 -> -((R6 (R1 (R2 + R3) VE + R2 R3 (VE - Vx) -
R1 R2 Vy) (1 + β))/(R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β)))),
I9 -> (R1 R3 (R6 + R8) VE + R2 R3 (R6 + R8) (VE - Vx) -
R2 R6 R8 (Vx - Vy) (1 + β) +
R1 R6 R8 Vy (1 + β))/(R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β))),
I10 -> (R1 R2 R6 (VE - Vy) (1 + β) +
R2 R6 R8 (Vx - Vy) (1 + β) - R1 R6 R8 Vy (1 + β) -
R2 R3 (VE - Vx) (R8 - R6 β) +
R1 R3 VE (-R8 + R6 β))/(R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β))),
I11 -> (R2 R6 R8 (Vx - Vy) +
R2 (R3 (R6 + R8) (VE - Vx) + R6 R8 (Vx - Vy)) β +
R1 (R3 R6 VE β + R3 R8 VE β +
R2 (R6 + R8) (VE - Vy) (1 + β) -
R6 R8 Vy (1 + β)))/(R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β))),
V1 -> (R2 (R1 R3 (R6 + R8) VE + R3 R6 R8 Vx (1 + β) +
R1 R6 R8 Vy (1 + β)))/(R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β))),
V2 -> -((R6 R8 (R1 (R2 + R3) VE + R2 R3 (VE - Vx) -
R1 R2 Vy) (1 + β))/(R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β)))),
V3 -> (R2 (R3 R4 R7 (R6 + R8) (VE - Vx) β -
R4 R5 R6 R8 (Vx - Vy) (1 + β) +
R3 R6 (R5 + R7) R8 Vy (1 + β)) +
R1 (R3 R6 (R5 + R7) R8 Vy + R3 R4 R7 (R6 + R8) VE β +
R3 R6 (R5 + R7) R8 Vy β +
R4 R5 R6 R8 Vy (1 + β) -
R2 R4 (R6 + R8) (VE - Vy) (R5 - R7 β) +
R2 (R5 + R7) Vy (R3 (R6 + R8) + R6 R8 (1 + β))))/((R4 +
R5 + R7) (R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β)))),
V4 -> (R7 (R2 R6 R8 (R5 (Vx - Vy) + R3 Vy) +
R2 (R3 (R4 + R5) (R6 + R8) (VE - Vx) + R5 R6 R8 (Vx - Vy) +
R3 R6 R8 Vy) β +
R1 (R3 R6 R8 Vy +
R3 ((R4 + R5) (R6 + R8) VE + R6 R8 Vy) β -
R5 R6 R8 Vy (1 + β) +
R2 (R6 R8 Vy + R3 (R6 + R8) Vy +
R4 (R6 + R8) (VE - Vy) β + R6 R8 Vy β +
R5 (R6 + R8) (VE - Vy) (1 + β)))))/((R4 + R5 +
R7) (R2 R3 R6 R8 (1 + β) +
R1 (R2 R3 (R6 + R8) + R2 R6 R8 (1 + β) +
R3 R6 R8 (1 + β))))}}


Now, I've a question for you: what do you mean by 'The lower cutoff frequency of CE is to be found'? The lower cutoff frequency of a component is a weird question.

• Yes - you are right. But to me it was clear that the cut-off frequency CAUSED by CE and the surrounding resistive elements is wanted.
– LvW
Commented Mar 23, 2020 at 14:28
• @LvW So if I understand correctly, we want to find the -3dB point (3dB lower than the maximum voltage possible for a given frequency) of the voltage V2 in my circuit? Commented Mar 23, 2020 at 14:29
• No - your circuit does not represent the original circuit. I do not see any capacitor.
– LvW
Commented Mar 23, 2020 at 14:32
• @LvW Yes I understand that, but I can set it to be one using R1=1/(sC1). If I do so is that than what you want to find that -3dB point? Commented Mar 23, 2020 at 14:37
• For a first-order RC section the -3dB cutoff is always the inverse of the corresponding time constant RC. If we feed a test current into the emitter node and measure the corresponding voltage at this node, the -3dB frequency is defined where this voltage is 3dB lower than the maximum (for very low frequencies)
– LvW
Commented Mar 23, 2020 at 14:46