Electronics - Diodes and Applications - Discussion

If the zener is opened, we have only R1 and R2 resister are in series.Therefore now we have a battery of 24V , R1 , R2 are all in series.

Apply voltage divider rule and find the voltage across the R2.It will come nearly 4.8v by doing calculation

i.e ., Voltage at R2=Vcc*R2/(R1+R2)
=24*2.5*10^3/(10K+2.5K)
=4.8v

This 4.8v is nearly indicated as 4.790 in the voltmeter shown in the diagram across the 2.5kohm resistor.

This is possible only if the zener is made as open.

simple:-)

Parallel= voltage is same current different.
Series =voltage is different, current same.

But here r1 and r2 are parallel voltages are different .
So open the diode,r1 and r2 are series so different voltages

i=v/r; r=r1+r2;
i=24/12.5
i=1.92mA
v1=1.92mA*10k ohm=19.2v
v2=1.92mA*2.5K ohm=4.608v (4.7v)

v=v1+v2 = 19.2v+4.608v

=23.8v (24v)

Here the breakdown voltage of diode is 15V so, it may get conducted by the source of 24V. So, if it is in conduction then the reading of voltmeter(4.79V) will become wrong. So, To get that amount of voltage we have To keep diode open.

Here the zener diode is reverse biased, and in reverse bias condition the resistance across it is very high, for this reason diode should be open.

If the diode is close then the voltage across the diode is Vcc-Vd = 9v.

If a zener is open then the current is not open so the answer is 2.

In case diode is close, then what is the voltage in voltmeter.

Thank you for your explanation @Shalini.

Thats cool. Thank you for explaination.

I don't know this. Please teach me.