Determine the range of values ​​of the voltage VE ensures a constant value of
Vo = 9.1V.

I think that for the diode to operate in the zener current must be between [100 micro A, 10 milli A]

Writing the equations of the mesh for both cases:

\( VE=Rs*Izmin+Rl*Izmin
VE=(1 kOhm)*(100 micro A)+(4 kOhm)*(100 micro A)=0,5 V

VE=Rs*Izmax + Rl*Izmax

VE=(1 kOhm)*(10 mili A)+(4 k Ohm)*(10 mili A)=50 V \)

Is this correct?

Thanks

 

No, how Ve = 0.5V can be smaller than the Vz voltage?

 

No, how Ve = 0.5V can be smaller than the Vz voltage?

It can't.So is the equation wrong?Have i missed something?

 

Yes you equations are wrong. This is not the right answer. You simply forget about Zener voltage and Kiechhoff''s law. Because as we can see on the diagram the Zener diode is connected in parallel with RL resistor. So voltage across RL is equal to Vz voltage.

 

Hello,

Have a look at this page of the eBook about zener didoes:
http://www.allaboutcircuits.com/vol_3/chpt_3/11.html

Bertus

 

Yes you equations are wrong. This is not the right answer. You simply forget about Zener voltage and Kiechhoff''s law. Because as we can see on the diagram the Zener diode is connected in parallel with RL resistor. So voltage across RL is equal to Vz voltage.

Yes you are right the voltage across RL will be equal to 9,1 V if the diode is in the zener mode.I did not remember how the zener diode worked.

So VE will have to be between [9,2;19,1] V for the diode to operate in the zener region.What if i used VE greater than 19,1 V.This would damage the zener because the current across it would be to large right ?

Thaks

 

Why you skip RL resistance in you calculations? Also if Iz > 10mA the power dissipation will finally destroyed the diode.

 

Why you skip RL resistance in you calculations? Also if Iz > 10mA the power dissipation will finally destroyed the diode.

Because i already know that voltage drop across RL will be 9,1V since it is in parallel with the zener diode.Since i am trying to determine VE ,what i want to calculate is the voltage drop across the resistors in the circuit.

 

But if we have RL in the circuit Ve voltage can be greater than 19.1V. Why? Simply because RL "steals" some current from the Zener diode.
Is = Iz + IL

 

Writing the equations of the mesh for both cases:

\( VE=Rs*Izmin+Rl*Izmin
VE=(1 kOhm)*(100 micro A)+(4 kOhm)*(100 micro A)=0,5 V

VE=Rs*Izmax + Rl*Izmax

VE=(1 kOhm)*(10 mili A)+(4 k Ohm)*(10 mili A)=50 V \)

Is this correct?

No.
Why are you multiplying the series resistance by the current flowing through the diode instead of the current flowing through the series resistance? Again the load resistance by the current flowing through the diode instead of the current flowing through the load resistance?