Block 08 — Two-port theory and transforms

Here is the solution of the Exercise 1

Here is the solution of the Exercise 2

Using superposition, we create two separate circuits where one source is considered. For these two circuits, we calculate $U_\rm A^{(1)}$ and $U_\rm A^{(2)}$.
To make the calculation simpler, the resistors $R_3$ and $R_{\rm s}$ will be joined to $R_4 =R_3 +R_{\rm s}$.

Now, we can use the requested boundaries:

1. For the minimum input voltage $U_{\rm I}= -15 ~\rm V$, the output voltage shall be $U_{\rm O} = 0 ~\rm V$
2. For the maximum input voltage $U_{\rm I}= +15 ~\rm V$, the output voltage shall be $U_{\rm O} = 3.3 ~\rm V$

This leads to two situations:

So, to get values for the relations, we have to put in the values for the input and output voltage conditions. For $k_{14}$ we get by formula $(2)$: \begin{align*} k_{14} = {{R_1}\over{R_4}} =-{{5 ~\rm V}\over {-15 ~\rm V }} = {{1}\over{3}} \\ \end{align*}

This value $k_{14}$ we can use for formula $(3)$: \begin{align*} {{ R_4 }\over{R_2}} &= {{5 ~\rm V + 15 ~\rm V \cdot {{1}\over{3}} }\over{ 3.3 ~\rm V \cdot {{1}\over{3}} }} - {{1 + {{1}\over{3}} }\over{ {{1}\over{3}} }} \\ &= {{10}\over{1.1}} - 4 \\ k_{42} &\approx 5.09 \end{align*}

We could now - theoretically - arbitrarily choose one of the resistors, e.g., $R_2$, and then calculate the other two.

But we must consider another boundary, a boundary for $R_{\rm S}$. The maximum voltage and the maximum current are given for the sensor. By this, we can calculate $R_{\rm S}$: \begin{align*} R_{\rm S} &= {{ U_{\rm OC} }\over{ I_{\rm SC} }} = {{ U_{\rm S,max} }\over{ I_{\rm S,max} }} = {{ 15 ~\rm V }\over{ 1 ~\rm mA }} \\ &= 15 ~\rm k\Omega \end{align*}

Therefore, $R_4 = R_{\rm S} + R_3$ must be larger than this.

The sensor resistance is \begin{align*} R_S &= 15 {~\rm k\Omega}\\ \end{align*}

We can choose $R_3$ arbitrarily. Here I choose a nice value to get integer values for $R_3$ and $R_1$: \begin{align*} R_3 &= 45 {~\rm k\Omega}\\ R_1 &= {{1}\over{3}} (R_3 + 15 {~\rm k\Omega}) = 20 {~\rm k\Omega} \\ R_2 &= {{1}\over{5.09}}(R_3 + 15 {~\rm k\Omega}) = 11.8 {~\rm k\Omega} \end{align*}

Based on the E24 series, the following values are next to the calculated ones: \begin{align*} R_3^0 &= 43 {~\rm k\Omega}\\ R_1^0 &= {{1}\over{3}} (R_3 + 15 {~\rm k\Omega}) = 20 {~\rm k\Omega} \\ R_2^0 &= {{1}\over{5.09}}(R_3 + 15 {~\rm k\Omega}) = 12 {~\rm k\Omega} \end{align*}

The potential of the node is $U_\rm O$. Therefore the currents are:

1. the current $I_2$ over $R_2$ is flowing to ground: $I_2 = - {{U_\rm O}\over{R_2}} $
2. the current $I_1$ over $R_1$ is coming from the supply voltage $U_{\rm S}$ to the nodal voltage $U_{\rm O}$: $I_1 = {{U_{\rm S} - U_{\rm O}}\over{R_1}}$
3. the current $I_4$ over $R_4$ is coming from the input voltage $U_{\rm I}$ to the nodal voltage $U_{\rm O}$: $I_4 = {{U_{\rm I} - U_{\rm O}}\over{R_4}}$

This led to the formula based on the Kirchhoff's nodal rule:

\begin{align*} \Sigma I = 0 &= I_1 + I_2 + I_3 \\ 0 &= {{U_{\rm S} - U_{\rm O}}\over{R_1}} + {{U_{\rm I} - U_{\rm O}}\over{R_4}} - {{U_\rm O}\over{R_2}} \end{align*}

The formula can be rearranged, with all terms containing $ U_{\rm O}$ on the left side: \begin{align*} {{U_{\rm O}}\over{R_1}} + {{U_{\rm O}}\over{R_2}} + {{U_{\rm O}}\over{R_4}} &= {{U_{\rm S}}\over{R_1}} + {{U_{\rm I}}\over{R_4}} \\ U_{\rm O}\cdot \left( {{1}\over{R_1}} + {{1}\over{R_2}} + {{1}\over{R_4}} \right) &= {{U_{\rm S}}\over{R_1}} + {{U_{\rm I}}\over{R_4}} \\ \end{align*}

Both sides can be multiplied by $\cdot R_1$, $\cdot R_2$, $\cdot R_4$ - in order to get rid of the fractions : \begin{align*} U_{\rm O}\cdot \left( {{R_1 R_2 R_4 }\over{R_1}} + {{R_1 R_2 R_4 }\over{R_2}} + {{R_1 R_2 R_4 }\over{R_4}} \right) &= R_1 R_2 R_4 \cdot {{U_{\rm S}}\over{R_1}} + R_1 R_2 R_4 \cdot {{U_{\rm I}}\over{R_4}} \\ U_{\rm O}\cdot \left( R_2 R_4 + R_1 R_4 + R_1 R_2 \right) &= R_2 R_4 \cdot U_{\rm S} + R_1 R_2 \cdot U_{\rm I} \\ U_{\rm O} &= {{R_2}\over{R_2 R_4 + R_1 R_4 + R_1 R_2 }} \left( R_4 \cdot U_{\rm S} + R_1 \cdot U_{\rm I} \right)\\ \end{align*}

The last formula was just the result we also got by the superposition but by more thinking.
So, sometimes there is an easier way…

• Unluckily, there is no simple way to know before, what way is the easiest.
• Luckily, all ways lead to the correct result.

\begin{align*} R_{\rm in}(R_1, R_2, R_3) &= R_3 + R_1 || R_2 \\ &= R_3 + {{R_1 R_2}\over{R_1 + R_2}} \end{align*}

\begin{align*} R_{\rm in, min} &= {{U_{\rm sense}}\over{I_{\rm sense, max}}} \\ &= \rm {{15 V}\over{1 mA}} \\ &= 15 k\Omega \\ \end{align*}