3
\$\begingroup\$

Consider the following problem:

Enter image description here

My attempt:

We have a minimum current of \$I_{zk}=0.5\ \mathrm{mA} \$ so that the Zener diode remains in the breakdown region. The given maximum power dissipated is 18 mW this implies that the current at that resistor is 18 × 10-3 A = 103 x IR2 ==> IR = 4.242 × 10-3 A.

If we perform nodal voltage analysis, we will get: (VS - 5 V)/2x103 Ω = 0.5x10-3 A + 4.242x10-3 A which yields VS = 14.484 V

Now one thing to note is that the Zener voltage of 5 V is parallel with the resistor load that being said means that VO = 5 V and if we execute the formula for the power dissipated we get that the power is 25 mW which exceeds the maximum 18 mW dissipated. Therefore, from here I am unable to continue to determine VS_min from which I can deduce delta VS.

\$\endgroup\$
9
  • 1
    \$\begingroup\$ Are you familiar with the Thevenin method to simplify the circuit? \$\endgroup\$
    – devnull
    Commented Oct 19, 2021 at 20:27
  • 3
    \$\begingroup\$ The 18mW is the power limit for the Zener diode, not the load resistor. \$\endgroup\$ Commented Oct 19, 2021 at 20:28
  • 1
    \$\begingroup\$ Indeed I am, could this be the trick to solve this circuit problem? @devnull \$\endgroup\$
    – UserX
    Commented Oct 19, 2021 at 20:28
  • 1
    \$\begingroup\$ I don't think a Thevenin equivalent will help you in this case. \$\endgroup\$ Commented Oct 19, 2021 at 20:29
  • 1
    \$\begingroup\$ @ElliotAlderson The Zener diode is usually modelled by a resistor in series with a constant voltage drop but in this problem, the resistance is zero thus how could we relate the power dissipated to the zener diode \$\endgroup\$
    – UserX
    Commented Oct 19, 2021 at 20:30

2 Answers 2

4
\$\begingroup\$

The diode Zener voltage is given as \$5.0\;\rm{V}\$.

The minimum diode current is given as \$0.5\;\rm{mA}\$.

The maximum diode current is trivially found from the power limit: \$\frac{18\;\rm{mW}}{5.0\;\rm{V}} = 3.6\;\rm{mA}\$.

Thus \$\Delta I_Z = 3.6\;\rm{mA} - 0.5\;\rm{mA} = 3.1\;\rm{mA}\$

Finally \$\Delta V_s = 2\;\rm{k}\Omega \;\cdot\; 3.1\;\rm{mA} = 6.2\;\rm{V} \$

\$\endgroup\$
2
  • \$\begingroup\$ Thank you for your simplified answer. May I ask, how did you deduce that Delta Vs = R x Delta Iz? \$\endgroup\$
    – UserX
    Commented Oct 19, 2021 at 23:06
  • 1
    \$\begingroup\$ @SofiaYaseveya Since the ideal Zener (rz = 0) always have the same exact voltage, the current through the 1k resistor doesn't change. Hence, every increase in current through the Zener results in the same increased current through the 2k resistor. This allows the Delta Vs calculation directly (Ohm's law), without calculating Vs max and min. \$\endgroup\$
    – devnull
    Commented Oct 20, 2021 at 10:47
1
\$\begingroup\$
  1. convert the voltage source and the resistors to their Thevenin equivalent

  2. from the \$I_{zk}=0.5\$ mA, calculate the voltage across the Thevenin resistor and add it to 5V. This gives you the minimum \$V_{th}\$, which gives you the \$VS_{min}\$

  3. since \$r_z = 0\$, calculate the maximum resistor voltage from the 18 mW and nominal Zener voltage

  1. repeat step 2. and find the maximum \$V_{th}\$, which gives you the \$VS_{max}\$

  2. \$VS_{max} - VS_{min}\$ is indeed 6.2V

\$\endgroup\$
3
  • 3
    \$\begingroup\$ Thevenin method doesn't really help here. \$\endgroup\$
    – Ben Voigt
    Commented Oct 19, 2021 at 22:03
  • \$\begingroup\$ Doesn't Thévenin's theorem only apply to a linear electrical network? We have a non-linear component here, a Zener diode. Is a linear model for it assumed? Can you address that in your answer? (But without "Edit:", "Update:", or similar - the answer should appear as if it was written right now). \$\endgroup\$ Commented Oct 20, 2021 at 10:07
  • \$\begingroup\$ @PeterMortensen: "Doesn't Thévenin's theorem only apply to a linear electrical network?" Yes. And the first step is: convert the voltage source and the resistors to their Thevenin equivalent. \$\endgroup\$
    – devnull
    Commented Oct 20, 2021 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.