Does the voltage drop off entirely after diode voltage has been overcome?

Regard the following circuit. Let $$\U_e\$$ be $$\-5\text{ V} \leq U_e \leq 5\text{ V}\$$. Also, let the diode be an ideal diode with the current being 0 at 0.7 V and 100% when over 0.7 V.

Is my assumption correct that the entire voltage $$\U_e\$$ is dropped off at the diode if $$\U_e\$$ is over 0.7 V? Would therefore be $$\U_a=0\$$ when this is the case? I am not sure if I understood diodes correctly, at least in this context.

• If $U_a$ is 0 V, is the diode forward biased according to your model? Commented Feb 7, 2022 at 23:03

You’re close. If Ue > 0.7V, then the voltage across the diode (Ua) would be 0.7V and the rest of the voltage dropped across the resistor.

Having said that, real diodes aren’t quite as perfect as we’d like. The actual voltage drop across the diode depends on the type of diode, current and temperature. As well, the transition from non conducting to conducting is not sharp - it is a curve. If you put your circuit into a simulator and Ue is an AC signal and plot Ua you’ll see what I mean.

Real diodes also have many non-ideal properties like leakage, capacitance, switching time etc. There are also different types of diodes that have different characteristics to suit certain applications.

Remember that the 0.7V drop is a gross simplification that is adequate for a general understanding of a silicon diode, but it is not the whole story.

• Okay I think I begin to understand diodes a little bit better. Is it true that we would measure no current if Ue is under 0.7V then? (In this fictional, ideal situation at least) Commented Feb 8, 2022 at 9:53

$$\U_a\$$ is just the voltage across the diode. By KVL and Ohm's Law the current $$\i\$$ through the diode (from top to bottom) is $$i = \frac{U_e - U_a}{R}$$ (where $$\R\$$ is the resistor of unknown value). $$\i\$$ cannot be negative since a diode only allows current to flow in one direction. Rearranging we have $$U_a = U_e - iR$$

According to your diode model, $$\i = 0\$$ and therefore $$\U_a = U_e\$$ if $$\U_a \leq 0.7\text{ V}\$$. This occurs if $$\U_e \leq 0.7\text{ V}\$$.

$$\U_a = 0.7\text{ V}\$$ if $$\i > 0\$$, in which case $$0.7\text{ V} = U_e - iR \to U_e = 0.7\text{ V} + iR$$

We don't know the value of $$\R\$$ so we can't compute $$\i\$$, but the voltage drop across the resistor is the difference between $$\U_e\$$ and $$\U_a = 0.7\text{ V}\$$.

Your assumption that the entire voltage $$\U_e\$$ is dropped off at the diode if $$\U_e\$$ is over 0.7 V (and that $$\U_a=0\$$ when this is the case) is not correct.

Your assumption is wrong except for the case of an ideal diode with no voltage drop. Then Ua = 0. Otherwise use a model for a diode and load line analysis to find the operating point.

https://web02.gonzaga.edu/faculty/talarico/EE303/HO/ch4_diodemodels.pptx.pdf

Replace the diode with a model such as shown below:

Or with a diode curve model as shown below:

Then the operating point is determined by load line analysis as shown: