Given D1 is a 10 V Zener diode, R1 = 1000 Ω and the following circuit diagram

Figure 1

Find the current \$I_D\$ through the diode (D1).

\$I_D = \frac{11 - 10}{1000} = 1\$ mA, is the maximum current through the Zener diode.

Find the current \$I_R\$ through the resistor (R1).

\$I_R = I_D\$, same current because they are in series.

Find the voltage \$V_D\$ across the diode (D1).

\$V_D = 11 - (0.001 \times 1000) = 10\$ V, I feel like 10 V is given to me but I need to show calculations...

Find the voltage \$V_R\$ across the resistor (R1).

\$V_R = 11 - V_D = 1\$ V

Despite how basic the problem seems, I am used to solving a Zener diode with \$V_Z = V_{Z0} + I_Z r_Z\$ and feel very caught off guard. Is the question missing information or are the calculations I have done valid? I also reconstructed the circuit in a simulator and the values it spits out are close but not exact to mine. This only adds to my lack confidence in my reasoning.

  • \$\begingroup\$ Well, computing the diode voltage drop can be done by subtracting the resistor voltage drop from the source voltage if you have to do this, indirectly. Your equation for VR is labeled wrong, though. That VR equation is really the VD equation since you are subtracting the resistor voltage drop (1 mA times 1 k Ohm) from 11. \$\endgroup\$
    – jonk
    Oct 27, 2021 at 18:50
  • \$\begingroup\$ Whoops, sorry I do see that mistake thank you. I have edited the problem to fix that. \$\endgroup\$
    – TBD
    Oct 27, 2021 at 18:57
  • \$\begingroup\$ Why is it that if the simulator produces nearby, but not exactly the same values, that this causes you to feel a lack of confidence? It shouldn't. \$\endgroup\$
    – jonk
    Oct 27, 2021 at 19:00
  • \$\begingroup\$ The simulator returned 1.077 V for the voltage drop across the resistor. The difference of 77 mV feels substantial compared to what I calculated (1 V). In addition I was compelled to build the circuit and measured an even more substantial change +490 mV or 1.49 V total. I consulted the data sheet and it specified that the voltage tolerance is 5%, so now I feel foolish because that falls within range. \$\endgroup\$
    – TBD
    Oct 27, 2021 at 19:15
  • \$\begingroup\$ The zener datasheet will specify the current through the zener that yields the nameplate voltage, too. This varies by zener voltage. So 1 mA may not be the "ideal" current, too. There are lots of factors. But yes, niggling over a few tens of millivolts is probably over-thinking things. Simulators also take into account details that a simplistic view cannot easily handle. So there are a number of sources for minor differences in results. \$\endgroup\$
    – jonk
    Oct 27, 2021 at 19:38

1 Answer 1


Is the question missing information?

Yes and no. When no information about the Zener diode is given (other than its Zener voltage), you can go two ways:

  1. Assume it's an idealized Zener with \$r_Z=0\$. That's what you did. Such a Zener is modeled by an ideal voltage source set for the Zener voltage, in series with an ideal diode. Of course you have to use said Zener's 10V voltage rating, otherwise you can't solve this circuit. You should redraw the circuit with a voltage source and a diode in place of the Zener, to show what it is you're actually analyzing! And then you'll quickly realize that there are two voltage sources in series, you can reduce them to a single 1V voltage source :)

  2. Go to Digi-Key. Search for 10V Zener diodes. Sort by stock quantity, descending. Take the diode that has the highest stock level - it'll be highly likely to be a popular product, representative of typical signal/analog applications. Take the \$r_Z\$ from that diode's characteristic curves in the datasheet and use it in your calculations. Justify your choice for \$r_Z\$.

The benefit of approach #2 is that what you are doing is extremely likely (>99%) to be representative of anyone actually having an application where they use the Zener as a high impedance voltage reference, and not as a shunt power supply (say with >=50mW of load on the stabilized voltage).


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