# Circuit with zener Diodes

Discussion in 'Homework Help' started by AD633, Jul 6, 2013.

Jun 22, 2013
96
1
In the circuit consider that the current minimum for Z1 and Z2 operate in the zener is 5 mA range V = 0.7V amd PZ2 = 0.8 V

a) Calculate the values ​​of R1 and R2 knowing that the current travels through the diode Z1 and 30 mA, and the current that runs trough Z2 is 15 mA

b) Determine the maximum voltage that is possible to use in Vsupply without damaging the Zener diodes

c) Determine the voltage to the extremes of the resistance of 5000 Ohm when Vsupply = 5 V

a)For this i have done the following meshes:

$
25 V=R1*Iz1-10 V
R1=(35)/(30mA)

R1=1,16 kOhm$

For determing R2

$
25 V=R1*Iz1+(R2//1k)*Iz1+500(Iz1+Iz2)
25 V=1.2kohm*30mA+(R2//1k)*30mA+500(30mA+15mA)$

This is giving me a negative value for the resistor R2,so i must be doing something wrong.

b)I know that Vz1=Vz2=0.8 W

$

Pz1=Vz1*Iz1

0.8 W=10V*Iz1

Iz1max=80mA

Pz2=Vz2*Iz2

0.8 W=5 V*Iz2

Iz2max=160 mA

$

$

Vsupply=R1*Iz1max+R2+(R2//1k)*Iz1max+500(Iz1max+Iz2max)
$

Is this the correct way to determine the maximum value for Vsupply?
Since i don't know R2 i can't determine this

c)When Vsupply=5 V

The zener Z2 will be inversely polarized and so V 500 Ohm=5 V

Is this correct?

Thanks

Last edited: Jul 23, 2013
2. ### WBahn Moderator

Mar 31, 2012
17,736
4,789
What voltage is V?

What voltage is referred to by PZ2?

In the figure, is the resistance of the far right resistor 500Ω? Check your figures to be sure that they contain all of the information that you need to convey.

There isn't enough information to determine this unless we are given either the maximum zener current or the maximum power dissipation of power zener diodes.

What is meant by "to the extremes of the resistance"?

This is both a poor application of KVL and a poor application of Ohm's Law.

In order for R1*Iz1 to be the voltage drop across R1, Iz1 would have to be the current through R1. Ohm's Law relates the resistance of a resistor to the voltage across THAT resistor and the current through THAT resistor.
And so, what? You just go on? What efforts have you made to identify and fix your error?

I can tell you at a glance that it is wrong, as you need to be able to.

CHECK THE UNITS!

On the right hand side you have a voltage added to a resistance added to a voltage added to a voltage (assuming the 500 has units of Ω). Absolutely no point going any further. You KNOW it is wrong is the units don't work out!

Ask if it makes any sense. If the voltage on the top left node is 5V (the supply) and the voltage on the top right node is 5V (your answer), why would any current flow through R1 and/or R2?

Last edited: Jul 7, 2013

Jun 22, 2013
96
1
V is the drop voltage across the diode when is it forward biased.

My mistake ,PZ2 is the maximum power that can be dissipated on Zener 2 and is equal to 0.8W.The zener 1 also has Pz1=0.8 W

Yes the resistance of the far right its a 500 Ohm resistor.

I meant to the terminals of the 500 Ohm resistor.

I didn't understood this.What equation are you talking about.On the rigth side i have a sum of 3 drop voltages.Probably the currents going through the resistors aren't the ones i mentioned.

For question 1)(determing R1 and R2)

For determing R1 , i have done the equation of the mesh which involves Vsupply,R1 and Vz1.I know that the zener 1 operates in the zener zone because its said that the current going through Z1 is 30 mA.It is also said that the mininum current necessary in order for the zener diode to operate in the zener mode is 5 mA.Therefore i know the current which goes through R1,witch looking at the mesh i mentioned is the same as the one that goes through the diode.

The equation for the mesh:
$
Vsupply=R1*Iz1-VZ1

25 V=R1*30mA-10V

R1=1,16 kOhm$

For determing R2

I tried to the equation(KVL) that involves Vsupply,R1,R2,and the 500 Ohm resistor.The current that goes trough R1 now,has to be equal to Iz1.The current that goes through R2 is the current that goes through R1,but part of it goes through the 1kOhm resistor which is im pararel with R2.
The current that goes through 500 ohm is equal to IZ2 since Z2 is also in the zener mode since the current that goes through it is 15 mA>8mA

From this it results the equation that i made.I know this is wrong because with the current i said and the values of resistance i found the drop voltage across the mesh is superior to Vsupply which is not possible.(This is the reason i am getting a negative value for R2)

So i must have calculated badly R1 or i am not thinking well in terms of the currents that go through the resistors.

Jun 22, 2013
96
1

Vsupply=R1*Iz1max+R2+(R2//1kOhm)*Iz1max+500(Iz1max+Iz2max)

I mistaped the equation the R2 wasn't supposed to be there

Vsupply=R1*Iz1max+(R2//1kOhm)*Iz1max+500(Iz1max+Iz2max)

5. ### LDC3 Active Member

Apr 27, 2013
920
160
How can the current that goes through the zener go through the resistor when they are not connected in series?
I think you need to label all the currents and show what currents add together to get the other currents.

Jun 22, 2013
96
1
The current that goes through 500 Ohm resistor , is equal to current that goes through R2 minus the current that goes through the zener 2.

$

Vsupply=R1(I1)+(R2//1kOhm)*I2+I500 Ohm* R
Vsupply=R1(IR1+Iz1)+(R2//1kOhm)*I2+500 Ohm(I2-Iz2)

$

This way i get to may variables..Do i have to do a system of equations to solve this?

Thanks

7. ### LDC3 Active Member

Apr 27, 2013
920
160
You still don't have the right equation.
The currents you need to use are:
Current through 500Ω, let's call it I500.
Current through zener 2 = IZ2
Current through R2 = IR2 or I500 + IZ2
Current through 1K = I1k
Current through wire between R1 and R2 connections = Iw or IR2 + I1k
Current through zener 1 = IZ1
Current through R1 is IR1 or Iw + IZ1

Now, you know that the voltage across R500 is the same as zener 2 (because they are in parallel). Calculate the current, I500.
There is a start, see if you can find IZ2, then IR2, ...

Jun 22, 2013
96
1

I 500 Ohm

$
I500 Ohm=(5V)/(500 Ohm)=10 mA
$

About Iz2 its said that it is 15 mA.

$
IR2=I500 Ohm +Iz2=10mA+15mA=25mA
$

Now for calcuting $Iw=IR2 + I1k$

The 1kOhm resistor is in parallel with zener 1.Since zener 1 operates in the zener zone.The voltage across 1kOhm resistor will be 10 V

Therefore the current $I1kOhm=(10V)/(1kOhm)=10mA$

Therefore $Iw=I1kOhm+IR2=10mA+25mA=35mA$

Iz1 it is said to be 30 mA.

$IR1=Iz1+Iw=30mA+35mA=65mA$

Is this correct?

Now for determing R1

$
25 V=R1*IR1-10V

R1=(538,5 Ohm)$

For determining R2

$
25 V=R1*IR1+R2*IR2+500Ohm*I500Ohm

25 V=538,5*65mA+R2*35mA+500Ohm*10mA

$

It is not correct yet..

Thanks

Last edited: Jul 7, 2013
9. ### LDC3 Active Member

Apr 27, 2013
920
160
You have all the current correct, but the voltage across R2 is not right.
What is the voltage on the left of R2?
What is the voltage on the right of R2?
Now, with the current through R2, what is the resistance?

10. ### WBahn Moderator

Mar 31, 2012
17,736
4,789
Take a step back and annotate your diagram and then catalog what you know.

From the given information, you know:

Va = 25V
Vb = 10V
Vc = 5V

Iz1 = 30mA
Iz2 = 15mA

Notice that I have done nothing but organize the information given in your original post. But now it becomes clear that we know ALL of the node voltages.

Now look at each resistor in turn. If we know any two of V, I, or R for that resistor, we can find the third. For the two resistors with known values, we also know the voltages across them so we can find the current through them with Ohm's Law.

Which currents remain unknown? One of these can be found immediately by KCL and now the value of the corresponding resistor can be found. Now work back toward the other unknown current and do the same thing.

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Jun 22, 2013
96
1
The voltage on the left of R2 will be imposed by the zener 1

$VLeft R2-VRight R2=R2*IR2
10V-5V=R2*(25mA)
R2=200 Ohm$

Is this correct?

Thanks

12. ### LDC3 Active Member

Apr 27, 2013
920
160
Yes, it's correct.
Now, check to see if the numbers make sense. Place the currents beside the places they occur and the resistances on the resistors.

Jun 22, 2013
96
1

$

Vsupply=R1*IR1+R2*IR2+500Ohm*I500 Ohm

$

This doesn't make much sense since the supply voltage is 25 V.Looking just at the drop voltage on R1 it will be $538,5Ohm*65mA=35 V$ which is bigger than the supply voltage.Is the value of R1 correct?

$25 V=538.5 Ohm*65mA+200 Ohm*25mA+500 Ohm*10mA￼$

For question 2) I know that

$Vz1=Vz2=0.8 W

PZ1=VZ1*IZ1
IZ1max=80mA
PZ2=VZ2*IZ2
IZ2max=160mA
$

I have to find the values of all currents again:

$

I 500 Ohm=10 mA

IR2=10mA+160 mA=170 mA

Iw=10mA*170mA=180 mA

IR1=180mA+80 mA=260 mA
$

And then i can Vsupplymax through this equation,right?

$

Vsupply=R1*IR1+R2*IR2+500Ohm*I500 Ohm

$

14. ### LDC3 Active Member

Apr 27, 2013
920
160
Why do you think this is correct?
According to the calculations:
The voltage drop on R1 is 15V.
The voltage drop across the zener is 10V.
Even though there is more to the circuit, it does not influence the supply.
If you replace the zener with a resistor, you will find that you cannot have both the current and the voltage match, so the zener does something special in the circuit.

15. ### LDC3 Active Member

Apr 27, 2013
920
160
Sorry, I didn't notice this earlier.
Why are you subtracting the zener voltage?
What is the voltage on either side of R1?

16. ### WBahn Moderator

Mar 31, 2012
17,736
4,789
This was pointed out in the very first response. That is what makes it a bad application of KVL.

Jun 22, 2013
96
1
On the left side its 25 V and on the right side its 10 V.So the drop voltage across it its 15 V.

$
25 V=15V+5V+5V$
,it makes sense now.

Jun 22, 2013
96
1
I suppose that the value of R1 that i had found was incorrect..It should be

$25V-15V=R1*65mA
R1=230,1 Ohm$

For question 2) I know that

$Vz1=Vz2=0.8 W

PZ1=VZ1*IZ1
IZ1max=80mA
PZ2=VZ2*IZ2
IZ2max=160mA
$

I have to find the values of all currents again:

$

I 500 Ohm=10 mA

IR2=10mA+160 mA=170 mA

Iw=10mA*170mA=180 mA

IR1=180mA+80 mA=260 mA
$

And then i can Vsupplymax through this equation,right?

$

Vsupply=R1*IR1+R2*IR2+500Ohm*I500 Ohm
Vsuppymax=(230,1 Ohm)(260mA)+200Ohm*170mA+500Ohm*10mA
Vussplymax=98,826V

$

Is this correct?

Thanks

Last edited: Jul 8, 2013
19. ### LDC3 Active Member

Apr 27, 2013
920
160
It looks good.

Jun 22, 2013
96
1
c) Determine the voltage to the extremes of the resistance of 500 Ohm when Vsupply = 5 V

If Vsupply=5V then the zener 2 will be directly polarized and therefore i can apply KVL do determine the current that goes through R1 since its the same as the one which passes through the zener rigth?

$
5V=R1*IR1+0.7 V

5V=230,1 Ohm*IR1+0.7 V

IR1=18,68 mA$

$

VRightR1=5-230.1 Ohm*18,68mA=0,7V

I1kOhm=700 uA

Iw=18,68 mA+700 uA

VR2=IR2*200 Ohm

0.7V=IR2*200 Ohm

IR2=3,5mA+18,68mA+700 uA

$

The calculations don't seem correct.I would say that the voltage at the terminals of the 500 Ohm resistor is 0,7 V ,since the diode is directly polarized (if vsupply is equal to 5 V considering the voltages drop across the circuit it is impossible for the diode z2 to be in zener mode) and therefore it has a 0,7 V voltage drop at the terminals.Since 500 Ohm resistor is in parallel with the zener it also has 0,7 V at its terminals.

Thanks