Given the circuit diagram below, indicate the on/off state of each diode and calculate the output voltage \$V_o\$ in each of the two half-cycles of source \$V_1\$, which applies a square wave with an amplitude of \$24\ \mathrm{V}\$. The threshold voltage of each diode is \$0.6\ \mathrm{V}\$. Resistor units are in \$\Omega\$.
I wish to understand the teacher's solution for the negative period of the voltage source; it consists of a couple of lines:
The entire bottom of the source is shorted, while the mesh on the right has \$D_1\$ in reverse \$\implies\$ only the left mesh remains \$\implies\$ \$I= 24\ \mathrm{V} / 6\ \mathrm{kΩ} = 4\ \mathrm{mA}\$ and \$V_o = - 6\ \mathrm{kΩ} · 4\ \mathrm{mA} = 24\ \mathrm{V}\$.
Why is the bottom half (everything below the source, apparantely) shortcircuited?
Why is \$D_1\$ in reverse?
The current \$I\$ computed is that which goes through the \$6\ \mathrm{k\Omega}\$ resistor. How exactly is that current used in computing \$V_o\$? That is, where does the equation \$V_o = -6\ \mathrm{k\Omega}\cdot 4\ \mathrm{mA}\$ come from?
My attempt (not needed to answer the post): I label the currents as follows:
Applying Kirchoff's voltage law to each quadrant gives us four equations:
$$\varepsilon_0: V_1 = -I_0\cdot 6k\Omega$$ $$\varepsilon_1: V_1-0.6V = -2I_1\cdot k\Omega - I_2k\Omega$$ $$\varepsilon_2: 0.6V = I_1\cdot k\Omega -3I_2\cdot k\Omega +I_3\cdot k\Omega$$ $$\varepsilon_3: 0.6V = I_2\cdot k\Omega - I_3\cdot k\Omega$$
Equation \$\varepsilon_0\$ always applies as is, while the other equations are subject to the fact that if diode \$D_k\$ is off, current \$I_k=0\$.
The problem is divided into two depending on whether \$V_1\$ is positive or negative. Assume the latter for now. My strategy was then to subdivide the problem as follows: for each of the eight possible configuration of the diodes (e.g. \$D_1\$ on, \$D_2\$ off, \$D_3\$ on) I solve the equations \$\varepsilon_1\$, \$\varepsilon_2\$, and \$\varepsilon_3\$. For example, our equations now are:
$$\varepsilon_1: -24.6V = -2I_1\cdot k\Omega$$ $$\varepsilon_2: 0.6V = I_1\cdot k\Omega +I_3\cdot k\Omega$$ $$\varepsilon_3: 0.6V = - I_3\cdot k\Omega$$
giving \$I_1=12.3\ \mathrm{mA}\$ and \$I_3 = -0.2\ \mathrm{mA}\$. As the latter current should be positive, we have reached a contradiction, and I discard the current configuration.
As another example, assume all diodes are on. Our equations are $$\varepsilon_1: -24.6V = -2I_1\cdot k\Omega - I_2k\Omega$$ $$\varepsilon_2: 0.6V = I_1\cdot k\Omega -3I_2\cdot k\Omega +I_3\cdot k\Omega$$ $$\varepsilon_3: 0.6V = I_2\cdot k\Omega - I_3\cdot k\Omega$$
which may be written in matrix form as
$$ \begin{pmatrix} -2 & -1 & 0\\ 1 & -3 & 1\\ 0 & 1 & -1 \end{pmatrix} \begin{pmatrix} I_1\\ I_2\\ I_3 \end{pmatrix} = \begin{pmatrix} -24.6\\ 0.6\\ 0.6 \end{pmatrix} \implies \begin{pmatrix} I_1\\ I_2\\ I_3 \end{pmatrix} = \frac{1}{5} \begin{pmatrix} -2 & 1 & 1\\ -1 & -2 & -2\\ -1 & -2 & -7 \end{pmatrix} \begin{pmatrix} -24.6\\ 0.6\\ 0.6 \end{pmatrix} = \frac{1}{5} \begin{pmatrix} 50.4\\ 22.2\\ 19.2 \end{pmatrix} $$
As no contradiction is found, I posit the configuration to be one of the many 'possible' ones and continue to compute \$V_o\$:
$$V_o = -24\ \mathrm{V} - I_2\cdot\ \mathrm{k\Omega} = -24\ \mathrm{V} - 4.44\ \mathrm{mA}\cdot\ \mathrm{k\Omega} = -28.44\ \mathrm{V}.$$
I continue this procedure for the remaining configurations.
