Given the circuits below with ideal diodes 1 and 2 on the right and the left, respectively.
.Show that
$\ln\left(\frac{i_1}{i_2}\right) = \ln\left(\frac{I_{S1}}{I_{S2}}\right) + \left(\frac{r_2i_2 - r_1i_1}{V_T}\right)$ (Eq. 1)
Give expression for V1-V 2 in terms of $r_1$, $r_2$, V1-V 2
Now suppose $V_T = 25$ mV, $I_{S1} = 4 I_{S2}$.
$r_1 = 40 \Omega$, $r_2 = 10 \Omega$ and $I = 20$ mA
Determine value for current in each branch
using simple iterative method.
To do this,
Step 1: first assume $i_1 = 10$ mA then
determine values for $i_2$. Using these
values determine values for Left Hand Side
(LHS) and Right Hand Side (RHS)
of equation (1).
Step 2:If they are not equal, repeat Step 1
with a different value for $i_1$
Continue step 2 by trying yet a different
value for $i_1$ until you obtain $0.002 < E < 0.004$
$N = (LHS) - (RHS)$
where $|E| = \frac{N}{D} = \frac{(LHS)-(RHS)}{(LHS)+(RHS)}$
Make a Table with different rows for
different values of $i_1$ and different columns
for (LHS), (RHS), N, D, N/D, and E values
for each row
