Electronic – solving minimum voltage to turn on a diode

You can't use Ohm's law to calculate the current through a diode.

However, the resistor is in series with the diode and, as you've learned, series connected circuit elements have identical currents.

Thus, you can use Ohm's Law to find the current through the resistor and diode but, to do that, you need the voltage across the resistor.

When you subtract the diode voltage from the source voltage, by KVL, what remains is the voltage across the resistor \$V_R\$.

Now, if you know the desired current, e.g., \$20mA\$, you can calculate the resistance required for that series current.

Thus, by Ohm's Law, the desired resistor value is:

$$R = \dfrac{V_R}{20mA} = \dfrac{V_S - V_D}{20mA} = \dfrac{9V - 1.7V}{20mA}$$

KCL for the junction is $$I_1=I_2+I_3$$ The voltage on the junction is \$V_1\$ Then $$I_1=(4.7-V_1)/40$$ $$I_2=V_1/50$$ $$I_3 = (V_1-0.6)/20$$

Solving: $$ (4.7-V_1)/40 = V_1/50 + (V_1-0.6)/20 $$ giving us $$V_1=1.55V$$ Hence $$I_3=(1.55-0.6)/20=0.0475A=47.5mA$$ Close enough to \$45.3\$..`.