Voltage drop - Diode reverse biased

In each circuit, calculate the current flowing through the 10kΩ resistor and calculate the voltage drop across the germanium diode. Please assume that a germanium diode has a forward-biased junction potential of 0.3V. Circuit (a)

Supply Voltage = 5V, R= 10kΩ, Junction potential = 0.3V, I =? VDrop=?

VD = (5V – 0.3V) = 4.7V
I = VD/R = 4.7V / 10kΩ = 0.47mA

Circuit (b)

Supply Voltage = 10V, R= 10kΩ, Junction potential = 0.3V, I =? VDrop =?

VD = (10V – 0.3V) = 9.7V
I = VD/R = 9.7V / 10kΩ = 0.97mA

Can you please review my solution?

I am not sure about the voltage drop, which in my calculation is indicated as VD. Also the second circuit, which is reverse biased, I'm not sure about the calculation I made.

• Look more closely at the two drawings. And remember what the basic function of a diode is. – Hearth Jan 29 at 19:14
• I had a science teacher back in grade school with the last name Dibenedetto. But back to your question. It's also important to note that you should consider drawing an equivalent circuit of a diode as well. – KingDuken Jan 29 at 19:23