Determine I through load resistor in circuit with diodes in series

diodes

I'm trying to find an expression for the current through the resistor \$R\$ when the AC-source is at max voltage of 5 volts.

\$U_T\$ represents the voltage drop across each diode.

\$R_D\$ represents the the internal resistance of each diode.

As you can see on the picture, I'm wondering how the expression for the current \$I\$ through the resistor is derived?

Now I got the wrong answer if I said that after all the voltage drops the voltage is zero, therefore I'm trying to understand how the formula on the picture has been derived.

2 diodes in serires with a resistor

Best Answer

If you rearrange your formula, you get $$ (U_m - 2U_T) = I_m (2R_d + R) $$ which looks like $$ V = IR $$ where \$R\$ is the sum of all of the resistors in the loop and \$V\$ is the voltage across that summed resistor.

Kirchoff tells you that the voltage across this resistor is the same as the voltage looking at the rest of the circuit, which is \$U_m - 2U_T\$, because there must be a voltage drop across each of the diodes for current to flow.