Diode: reverse and forward bias current and voltage

"Assume that diodes have voltage of 0.7 V and resistance 4Ω. Find the current through the forward and reverse bias of diodes. Draw the characteristic of diode through the given values." My question is that how do I approach these type of questions and what "diodes have voltage of 0.7 V" means. Does that mean "the voltage drop" of each diode in the forward bias(I presume not in reverse bias)? Also, why do I need to know the resistance of the diodes since we know the voltage drops, and aren't the 4 ohms negligible compared to 1.5 K ohm?

In addition, can I thread the two diodes as one diode in reverse bias?

My, that's a lot of questions.

Presumably the diode model you instructor wants you to use is an ideal diode with 0.7V drop in series with a 4 ohm resistor. So, when the diode is conducting, it behaves like a voltage source with a resistor in series. When the forward bias is less than 0.7V it does not conduct.

Is the 4 ohms per diode negligible in comparison to 1.5K? Well, it's more than 0.5% for two diodes so it will drop tens of mV. That might be negligible or it might not be, depending on the application. Since the instructor gives you the value, I suggest it might not be negligible in terms of getting the correct answer.

Two such diodes in series behave like one 1.4V diode with 8 ohms in series.

• To liken a conducting diode with a voltage source is quite misleading... When conducting, the diode is sooner a "constant-voltage non-linear resistor"... a kind of a dynamic resistor that changes its instant resistance so that to keep an almost constant voltage drop across it... Feb 23 '15 at 21:15
• @Circuitfantasist I fail to see any practical difference (other than that a newbie may find it hard to analyze such a nonlinear magic resistor construct). Feb 23 '15 at 21:33

@Spehro Pefhany, I just wanted to show that it is misleading to mimic a resistor by a source... A nonlinear resistor should be mimicked by a dynamic (a kind of a "self-varying") resistor. I illustrated this "trick" many years ago in the sketch below. 