Chapter 14, Problem 14.33P

(a)

To determine

The value of $I_{S4}$ for the given $C-E$ voltage.

Answer to Problem 14.33P

The value of current $I_{S4}$ is $I_{S_4}=5\times {10}^{-15}\text{A}$ .

Explanation of Solution

Given:

Given bipolar active load diff-amp is,

  

Given parameters are:

  $V^{+}=5\text{V}$

  $V^{-}=-5\text{V}$

The transistor parameters are:

  $V_{AN}=120\text{V}$

  $V_{AP}=80\text{V}$

  $v_{BE\left(npn\right)}=v_{EB(\text{pnp})}=0.6\text{V}$

  $I_{S_1}=I_{S_2}$

  $I_{s3}=5\times {10}^{-15}\text{A}$

  $V_{EC4}=0.6\text{V}$

Calculation:

Let $v_1=v_2$

Since $v_{EB3}=0.6\text{V}=v_{BE1}$ , then for $v_1=v_2$

  $v_{CE1}=V^{+}$

  $v_{CE1}=5\text{V}\dots \dots \dots (1)$

Now

  $v_{EC4}+v_{CE2}=V^{+}+v_{BE2}$

  $v_{CE2}=V^{+}+v_{BE2}-v_{EC4}$

  $v_{CE2}=5+0.6-v_{EC4}$

  $v_{CE2}=5.6-v_{EC4}\dots \dots \dots \dots (2)$

The collector currents, $i_{C1}=i_{C3}$

  $\begin{array}{cc}& i_{C1}=I_{S1}\left(e^{\frac{v_{BE1}}{V_T}}\right)\left(1+\frac{v_{CE1}}{V_{A1}}\right)\dots \dots \dots (3)\\ & i_{C3}=I_{S3}\left(e^{\frac{v_{EB3}}{V_T}}\right)\left(1+\frac{v_{EC3}}{V_{A3}}\right)\dots \dots \dots .(4)\end{array}$

  $\text{And}{i}_{c2}=i_{C4}$

  $i_{C2}=I_{S2}\left(e^{\frac{v_{ma}}{V_r}}\right)\left(1+\frac{v_{CE2}}{V_{A2}}\right)\dots \dots \dots (5)$

  $i_{C4}=I_{S4}\left(e^{\frac{v_{mt}}{V_T}}\right)\left(1+\frac{v_{EC4}}{V_{A4}}\right)\dots \dots \dots \dots (6)$

Assume that $Q_1$ and $Q_2$ are matched, then $I_{S1}=I_{S2}\equiv I_S$ and $V_{A1}=V_{A2}\equiv V_{AN}$

Assume that $Q_3$ and $Q_4$ are slightly mismatched, so that $I_{S3}\ne I_{S4}$ but still assume that

  $V_{A3}=V_{A4}\equiv V_{AP}.$ For $v_1=v_2,$

  $v_{BB1}=v_{BE2};$ also $v_{EB3}=v_{EB4}=v_{EC3}\equiv v_{EB}$

Taking ratio of equations (3), (5) produces,

  $\begin{array}{l}\frac{i_{C1}}{i_{C2}}=\frac{I_{S1}\left(e^{\frac{v_m}{V_T}}\right)\left(1+\frac{v_{CE1}}{V_{A1}}\right)}{I_{S2}\left(e^{\frac{v_{m2}}{V_r}}\right)\left(1+\frac{v_{CE2}}{V_{A2}}\right)}\hfill \\ \frac{i_{C1}}{i_{C2}}=\frac{\left(1+\frac{v_{CB1}}{V_{AN}}\right)}{\left(1+\frac{v_{CE2}}{V_{AN}}\right)}\dots \dots \dots .(7)\hfill \end{array}$

Taking ratio of equations (4). (6) produces,

  $\frac{i_{C3}}{i_{C4}}=\frac{I_{S3}\left(e^{\frac{v_{m2}}{V_r}}\right)\left(1+\frac{v_{EC3}}{V_{A3}}\right)}{I_{S4}\left(e^{\frac{{\gamma }_{E\ast }}{V_T}}\right)\left(1+\frac{v_{EC4}}{V_{A4}}\right)}$

  $\frac{i_{C1}}{i_{C2}}=\frac{I_{S3}\left(1+\frac{v_{FB}}{V_{AP}}\right)}{I_{S4}\left(1+\frac{v_{EC4}}{V_{AP}}\right)}$

  $\left(\because i_{c1}=i_{c3},i_{c2}=i_{c4}\right)\dots \dots \dots \dots (8)$

From equations (7) and (8)

$\frac{\left(1+\frac{v_{CF1}}{V_{AN}}\right)}{\left(1+\frac{v_{CE2}}{V_{AN}}\right)}=\frac{I_{S3}\left(1+\frac{v_{EB}}{V_{AP}}\right)}{I_{S4}\left(1+\frac{v_{EC4}}{V_{AP}}\right)}$

  $I_{S4}=I_{S3}\frac{\left(1+\frac{v_{EB}}{V_{AP}}\right)\left(1+\frac{v_{CE2}}{V_{AN}}\right)}{\left(1+\frac{v_{EC4}}{V_{AP}}\right)\left(1+\frac{v_{CR1}}{V_{AN}}\right)}$

  $I_{S4}=I_{S3}\frac{\left(1+\frac{v_{EB}}{V_{AP}}\right)\left(1+\frac{5.6-v_{EC4}}{V_{AN}}\right)}{\left(1+\frac{v_{EC4}}{V_{AP}}\right)\left(1+\frac{v_{CR1}}{V_{AN}}\right)}$

Substituting equations (1) and (2) in above equation,

  $I_{S4}=5\times {10}^{-15}\frac{\left(1+\frac{0.6}{80}\right)\left(1+\frac{5.6-v_{EC4}}{120}\right)}{\left(1+\frac{v_{EC4}}{80}\right)\left(1+\frac{5}{120}\right)}$

  $I_{S4}=5\times {10}^{-15}\frac{\left(\frac{80+0.6}{80}\right)\left(\frac{120+5.6-v_{EC4}}{120}\right)}{\left(\frac{80+v_{EC4}}{80}\right)\left(\frac{120+5}{120}\right)}$

  $I_{S_4}=5\times {10}^{-15}\frac{(80.6)\left(125.6-v_{EC4}\right)}{\left(80+v_{EC4}\right)(125)}$

  $I_{S4}=3.224\times {10}^{-15}\frac{\left(125.6-v_{ec4}\right)}{\left(80+v_{EC4}\right)}\dots \dots \dots \dots (9)$

Given $v_{EC4}=0.6\text{V}$

Substituting this $v_{EC4}$ value in equation $(9),$

  $I_{S4}=3.224\times {10}^{-15}\frac{(125.6-0.6)}{(80+0.6)}$

  $I_{S4}=3.224\times {10}^{-15}\times \frac{125}{80.6}$

  $\therefore I_{S_4}=5\times {10}^{-15}\text{A}$

(b)

To determine

The value of $I_{S4}$ for the given $C-E$ voltage.

Answer to Problem 14.33P

The value of current $I_{S4}$ is $I_{S_4}=4.939\times {10}^{-15}\text{A}$ .

Explanation of Solution

Given:

Given bipolar active load diff-amp is,

  

Given parameters are:

  $V^{+}=5\text{V}$

  $V^{-}=-5\text{V}$

The transistor parameters are:

  $V_{AN}=120\text{V}$

  $V_{AP}=80\text{V}$

  $v_{BE\left(npn\right)}=v_{EB(\text{pnp})}=0.6\text{V}$

  $I_{S_1}=I_{S_2}$

  $I_{s3}=5\times {10}^{-15}\text{A}$

  $V_{EC4}=1.2\text{V}$

Calculation:

Let $v_1=v_2$

Since $v_{EB3}=0.6\text{V}=v_{BE1}$ , then for $v_1=v_2$

  $v_{CE1}=V^{+}$

  $v_{CE1}=5\text{V}\dots \dots \dots (1)$

Now

  $v_{EC4}+v_{CE2}=V^{+}+v_{BE2}$

  $v_{CE2}=V^{+}+v_{BE2}-v_{EC4}$

  $v_{CE2}=5+0.6-v_{EC4}$

  $v_{CE2}=5.6-v_{EC4}\dots \dots \dots \dots (2)$

The collector currents, $i_{C1}=i_{C3}$

  $\begin{array}{cc}& i_{C1}=I_{S1}\left(e^{\frac{v_{BE1}}{V_T}}\right)\left(1+\frac{v_{CE1}}{V_{A1}}\right)\dots \dots \dots (3)\\ & i_{C3}=I_{S3}\left(e^{\frac{v_{EB3}}{V_T}}\right)\left(1+\frac{v_{EC3}}{V_{A3}}\right)\dots \dots \dots .(4)\end{array}$

  $\text{And}{i}_{c2}=i_{C4}$

  $i_{C2}=I_{S2}\left(e^{\frac{v_{ma}}{V_r}}\right)\left(1+\frac{v_{CE2}}{V_{A2}}\right)\dots \dots \dots (5)$

  $i_{C4}=I_{S4}\left(e^{\frac{v_{mt}}{V_T}}\right)\left(1+\frac{v_{EC4}}{V_{A4}}\right)\dots \dots \dots \dots (6)$

Assume that $Q_1$ and $Q_2$ are matched, then $I_{S1}=I_{S2}\equiv I_S$ and $V_{A1}=V_{A2}\equiv V_{AN}$

Assume that $Q_3$ and $Q_4$ are slightly mismatched, so that $I_{S3}\ne I_{S4}$ but still assume that

  $V_{A3}=V_{A4}\equiv V_{AP}.$ For $v_1=v_2,$

  $v_{BB1}=v_{BE2};$ also $v_{EB3}=v_{EB4}=v_{EC3}\equiv v_{EB}$

Taking ratio of equations (3), (5) produces,

  $\begin{array}{l}\frac{i_{C1}}{i_{C2}}=\frac{I_{S1}\left(e^{\frac{v_m}{V_T}}\right)\left(1+\frac{v_{CE1}}{V_{A1}}\right)}{I_{S2}\left(e^{\frac{v_{m2}}{V_r}}\right)\left(1+\frac{v_{CE2}}{V_{A2}}\right)}\hfill \\ \frac{i_{C1}}{i_{C2}}=\frac{\left(1+\frac{v_{CB1}}{V_{AN}}\right)}{\left(1+\frac{v_{CE2}}{V_{AN}}\right)}\dots \dots \dots .(7)\hfill \end{array}$

Taking ratio of equations (4). (6) produces,

  $\frac{i_{C3}}{i_{C4}}=\frac{I_{S3}\left(e^{\frac{v_{m2}}{V_r}}\right)\left(1+\frac{v_{EC3}}{V_{A3}}\right)}{I_{S4}\left(e^{\frac{{\gamma }_{E\ast }}{V_T}}\right)\left(1+\frac{v_{EC4}}{V_{A4}}\right)}$

  $\frac{i_{C1}}{i_{C2}}=\frac{I_{S3}\left(1+\frac{v_{FB}}{V_{AP}}\right)}{I_{S4}\left(1+\frac{v_{EC4}}{V_{AP}}\right)}$

  $\left(\because i_{c1}=i_{c3},i_{c2}=i_{c4}\right)\dots \dots \dots \dots (8)$

From equations (7) and (8)

$\frac{\left(1+\frac{v_{CF1}}{V_{AN}}\right)}{\left(1+\frac{v_{CE2}}{V_{AN}}\right)}=\frac{I_{S3}\left(1+\frac{v_{EB}}{V_{AP}}\right)}{I_{S4}\left(1+\frac{v_{EC4}}{V_{AP}}\right)}$

  $I_{S4}=I_{S3}\frac{\left(1+\frac{v_{EB}}{V_{AP}}\right)\left(1+\frac{v_{CE2}}{V_{AN}}\right)}{\left(1+\frac{v_{EC4}}{V_{AP}}\right)\left(1+\frac{v_{CR1}}{V_{AN}}\right)}$

  $I_{S4}=I_{S3}\frac{\left(1+\frac{v_{EB}}{V_{AP}}\right)\left(1+\frac{5.6-v_{EC4}}{V_{AN}}\right)}{\left(1+\frac{v_{EC4}}{V_{AP}}\right)\left(1+\frac{v_{CR1}}{V_{AN}}\right)}$

Substituting equations (1) and (2) in above equation,

  $I_{S4}=5\times {10}^{-15}\frac{\left(1+\frac{0.6}{80}\right)\left(1+\frac{5.6-v_{EC4}}{120}\right)}{\left(1+\frac{v_{EC4}}{80}\right)\left(1+\frac{5}{120}\right)}$

  $I_{S4}=5\times {10}^{-15}\frac{\left(\frac{80+0.6}{80}\right)\left(\frac{120+5.6-v_{EC4}}{120}\right)}{\left(\frac{80+v_{EC4}}{80}\right)\left(\frac{120+5}{120}\right)}$

  $I_{S_4}=5\times {10}^{-15}\frac{(80.6)\left(125.6-v_{EC4}\right)}{\left(80+v_{EC4}\right)(125)}$

  $I_{S4}=3.224\times {10}^{-15}\frac{\left(125.6-v_{ec4}\right)}{\left(80+v_{EC4}\right)}\dots \dots \dots \dots (9)$

Given $v_{EC4}=1.2\text{V}$

Substituting this $v_{EC4}$ value in equation $(9),$

  $I_{S4}=3.224\times {10}^{-15}\frac{(125.6-1.2)}{(80+1.2)}$

  $I_{S4}=3.224\times {10}^{-15}\times \frac{124.4}{81.2}$

  $\therefore I_{S_4}=4.939\times {10}^{-15}\text{A}$

(c)

To determine

The value of $I_{S4}$ for the given has a $C-E$ voltage.

Answer to Problem 14.33P

The value of current $I_{S4}$ is $I_{S_4}=4.811\times {10}^{-15}\text{A}$

Explanation of Solution

Given:

Given bipolar active load diff-amp is,

  

Given parameters are:

  $V^{+}=5\text{V}$

  $V^{-}=-5\text{V}$

The transistor parameters are:

  $V_{AN}=120\text{V}$

  $V_{AP}=80\text{V}$

  $v_{BE\left(npn\right)}=v_{EB(\text{pnp})}=0.6\text{V}$

  $I_{S_1}=I_{S_2}$

  $I_{s3}=5\times {10}^{-15}\text{A}$

  $V_{EC4}=2.5\text{V}$

Calculation:

Let $v_1=v_2$

Since $v_{EB3}=0.6\text{V}=v_{BE1}$ , then for $v_1=v_2$

  $v_{CE1}=V^{+}$

  $v_{CE1}=5\text{V}\dots \dots \dots (1)$

Now

  $v_{EC4}+v_{CE2}=V^{+}+v_{BE2}$

  $v_{CE2}=V^{+}+v_{BE2}-v_{EC4}$

  $v_{CE2}=5+0.6-v_{EC4}$

  $v_{CE2}=5.6-v_{EC4}\dots \dots \dots \dots (2)$

The collector currents, $i_{C1}=i_{C3}$

  $\begin{array}{cc}& i_{C1}=I_{S1}\left(e^{\frac{v_{BE1}}{V_T}}\right)\left(1+\frac{v_{CE1}}{V_{A1}}\right)\dots \dots \dots (3)\\ & i_{C3}=I_{S3}\left(e^{\frac{v_{EB3}}{V_T}}\right)\left(1+\frac{v_{EC3}}{V_{A3}}\right)\dots \dots \dots .(4)\end{array}$

  $\text{And}{i}_{c2}=i_{C4}$

  $i_{C2}=I_{S2}\left(e^{\frac{v_{ma}}{V_r}}\right)\left(1+\frac{v_{CE2}}{V_{A2}}\right)\dots \dots \dots (5)$

  $i_{C4}=I_{S4}\left(e^{\frac{v_{mt}}{V_T}}\right)\left(1+\frac{v_{EC4}}{V_{A4}}\right)\dots \dots \dots \dots (6)$

Assume that $Q_1$ and $Q_2$ are matched, then $I_{S1}=I_{S2}\equiv I_S$ and $V_{A1}=V_{A2}\equiv V_{AN}$

Assume that $Q_3$ and $Q_4$ are slightly mismatched, so that $I_{S3}\ne I_{S4}$ but still assume that

  $V_{A3}=V_{A4}\equiv V_{AP}.$ For $v_1=v_2,$

  $v_{BB1}=v_{BE2};$ also $v_{EB3}=v_{EB4}=v_{EC3}\equiv v_{EB}$

Taking ratio of equations (3), (5) produces,

  $\begin{array}{l}\frac{i_{C1}}{i_{C2}}=\frac{I_{S1}\left(e^{\frac{v_m}{V_T}}\right)\left(1+\frac{v_{CE1}}{V_{A1}}\right)}{I_{S2}\left(e^{\frac{v_{m2}}{V_r}}\right)\left(1+\frac{v_{CE2}}{V_{A2}}\right)}\hfill \\ \frac{i_{C1}}{i_{C2}}=\frac{\left(1+\frac{v_{CB1}}{V_{AN}}\right)}{\left(1+\frac{v_{CE2}}{V_{AN}}\right)}\dots \dots \dots .(7)\hfill \end{array}$

Taking ratio of equations (4). (6) produces,

  $\frac{i_{C3}}{i_{C4}}=\frac{I_{S3}\left(e^{\frac{v_{m2}}{V_r}}\right)\left(1+\frac{v_{EC3}}{V_{A3}}\right)}{I_{S4}\left(e^{\frac{{\gamma }_{E\ast }}{V_T}}\right)\left(1+\frac{v_{EC4}}{V_{A4}}\right)}$

  $\frac{i_{C1}}{i_{C2}}=\frac{I_{S3}\left(1+\frac{v_{FB}}{V_{AP}}\right)}{I_{S4}\left(1+\frac{v_{EC4}}{V_{AP}}\right)}$

  $\left(\because i_{c1}=i_{c3},i_{c2}=i_{c4}\right)\dots \dots \dots \dots (8)$

From equations (7) and (8)

$\frac{\left(1+\frac{v_{CF1}}{V_{AN}}\right)}{\left(1+\frac{v_{CE2}}{V_{AN}}\right)}=\frac{I_{S3}\left(1+\frac{v_{EB}}{V_{AP}}\right)}{I_{S4}\left(1+\frac{v_{EC4}}{V_{AP}}\right)}$

  $I_{S4}=I_{S3}\frac{\left(1+\frac{v_{EB}}{V_{AP}}\right)\left(1+\frac{v_{CE2}}{V_{AN}}\right)}{\left(1+\frac{v_{EC4}}{V_{AP}}\right)\left(1+\frac{v_{CR1}}{V_{AN}}\right)}$

  $I_{S4}=I_{S3}\frac{\left(1+\frac{v_{EB}}{V_{AP}}\right)\left(1+\frac{5.6-v_{EC4}}{V_{AN}}\right)}{\left(1+\frac{v_{EC4}}{V_{AP}}\right)\left(1+\frac{v_{CR1}}{V_{AN}}\right)}$

Substituting equations (1) and (2) in above equation,

  $I_{S4}=5\times {10}^{-15}\frac{\left(1+\frac{0.6}{80}\right)\left(1+\frac{5.6-v_{EC4}}{120}\right)}{\left(1+\frac{v_{EC4}}{80}\right)\left(1+\frac{5}{120}\right)}$

  $I_{S4}=5\times {10}^{-15}\frac{\left(\frac{80+0.6}{80}\right)\left(\frac{120+5.6-v_{EC4}}{120}\right)}{\left(\frac{80+v_{EC4}}{80}\right)\left(\frac{120+5}{120}\right)}$

  $I_{S_4}=5\times {10}^{-15}\frac{(80.6)\left(125.6-v_{EC4}\right)}{\left(80+v_{EC4}\right)(125)}$

  $I_{S4}=3.224\times {10}^{-15}\frac{\left(125.6-v_{ec4}\right)}{\left(80+v_{EC4}\right)}\dots \dots \dots \dots (9)$

Given $v_{EC4}=1.2\text{V}$

Substituting this $v_{EC4}$ value in equation $(9),$

  $I_{S4}=3.224\times {10}^{-15}\frac{(125.6-2.5)}{(80+2.5)}$

  $I_{S4}=3.224\times {10}^{-15}\times \frac{123.1}{82.5}$

  $\therefore I_{S_4}=4.811\times {10}^{-15}\text{A}$