Here is my answer.

1) Remove the diode from the circuit. Calculate the voltage across R2 as per voltage divider formula.

Knowing this voltage, you can obtain an estimate of the expected voltage when the diode is inserted (whether forward or reverse biased). Since current is expected to flow through the diode one would expect the voltage across the diode and R2 to be lower than that calculated in (1).

The voltage across the resistor is equal to the voltage across the diode. Period. End of argument.

2) There are three ways to determine the voltage across the diode.

#1 - You can apply the diode equation and solve the equation for diode current and voltage.
#2 - You can apply a graphical solution (which relies on the diode equation or some estimate of the diode I-V characteristic curve).
#3 - You can make an assumption that the diode voltage has reached the diode forward turn on voltage and assume an approximate forward voltage of about 0.7V in which case the voltage across R2 is 0.7V.

That is not the question, no one claimed they are not equal, it was rather misplaced question of where did the rest of the supply voltage go and that is R1.

 

From what I must admit a skim read of this posting,
It looks like the answers have all been given.

The circuit is a resistor divider, with the lower resistor in parallel with a diode.
As has been said,
first thing is to see what voltage would be across lower resistor without diode,
then from the diode data sheet, it can be seen if the diode is conducting or not,
and if it is conducting, then the current through the diode is going to dominate.

As the current through the top resistor would increase if diode was conducting,
which would drop the voltage seen by the diode,

As diode voltage drop is determined by its current,
A bit of a repeat calculation needed to find the steady state.

But in reality ,t he diode is going to have most of the current flowing through it, so the resistor in parallel to it is in practice immaterial, the bottom resistor will have a fairly constant current through it.

 

What prompted me to this was analysis of slayer exciter circuit. It is said purpose of the diode you see is to prevent base-emitter junction from excessive reverse bias during negative half-cycle. And we are talking kilovolts here, at very low current tho.

View attachment 226472

re this rathe different circuit,

The diode is across the secondary of the transformer,
thus when the lower end of the secondary goes below zero volts, the diode will conduct,
preventing a reverse voltage Base Emitter greater than what the transistor can take.

 

re this rathe different circuit,

The diode is across the secondary of the transformer,
thus when the lower end of the secondary goes below zero volts, the diode will conduct,
preventing a reverse voltage Base Emitter greater than what the transistor can take.

Yes, it's different but diode is used as voltage regulator, i understand it's purpose and operation. EDIT: i must add however i read of slayers without this diode working normally. I guess transistors don't get burned cause current is tiny.

Also, i agree with before said, this is common approach, calculate voltage for voltage divider, see if diode is forward biased, if so diode dominates.

 

Ok. All the explanations are in.

Can we close this thread now?

 

Sure