I am confused about this circuit found in Electronic Devices and Circuit Theory By Robert Boylestad 10th Ed.

Electronic Devices and Circuit Theory By Robert Boylestad. Fig 2.211.

I don't understand why D1 (left diode) has a Vd1 = 0v in equivalent circuit:

Equivalent circuit

When I simulated on Proteus ISIS the polarization voltage of D1 is 0.43V:

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Even when I made the physical circuit (2 diodes 1N41004, 1 resistor of 73Kohms "Note: I dont have a value shown in the problem but I use this as reference").

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D1 shows a polarization voltage of around -0.5V.

enter image description here Could someone explain to me which analysis is correct?

  • \$\begingroup\$ Can you adjust the range of the ammeter? Are you sure the simulator isn't modeling the diodes with a tiny (nanoamps or microamps) reverse leakage current? \$\endgroup\$
    – The Photon
    Commented Sep 16, 2020 at 5:33
  • \$\begingroup\$ What diodes did you use in the physical circuit? \$\endgroup\$ Commented Sep 16, 2020 at 7:49
  • \$\begingroup\$ I used 1N4004 diodes... I added new images and information. \$\endgroup\$
    – rm1234
    Commented Sep 16, 2020 at 20:17
  • \$\begingroup\$ can I ask which exercise is this one? I'm studying the same argument in the same book \$\endgroup\$
    – Carlo
    Commented Sep 17, 2020 at 16:08
  • \$\begingroup\$ Example 2.8 page 69 \$\endgroup\$
    – rm1234
    Commented Sep 18, 2020 at 2:27

2 Answers 2


The total current through the real circuit will be dominated by the leakage of the reverse-biased diode, so a few nA. The forward voltage of a similar silicon diode at that current will be tens of mV so close enough to zero.

Your measuring devices may be causing enough current to flow to get more forward voltage, but that would require more like 100uA than nA for a silicon diode to drop ~500mV, so not sure where your error is coming from.

Default resistance of Proteus virtual voltmeters is apparently 100M, so ~0.2uA should be flowing unless you've reduced it. Just to be sure you can repeat the simulation with the RH voltmeter removed. Also, pick a diode type with a known model.

Basically, I agree with the textbook (within < 0.1V), not the other two numbers for D1 voltage.

If I do an LTspice simulation with 1N4148s (and no faux meters) I get 33mV across D1 and 14uV across the resistor, with current of 2.54nA, which matches my expectations.

  • \$\begingroup\$ I will try it on LTspice... Thanks \$\endgroup\$
    – rm1234
    Commented Sep 16, 2020 at 20:19

What your simulation and actual real life measurement showed are in line with what I would expect. In order for a diode to start conducting, you need a forward voltage sufficient enough to overcome the depletion region. This is usually around 0.3V-0.7V, depending on the diode.

Now, this translates in your circuit as this:

  • 20 V see the anode of D1, and since is grater that 0.4V, which is the minimum voltage required in your situation for the diode, it conducts.
  • There is the expected voltage drop, so to "other side" of diode, the cathode, the voltage is now 20V - 0.4V = 19.6V.
  • Since it hits the cathode of the second diode, the latter cannot conduct because it is reverse biased, so it creates and open circuit.

The reason why you measure 0.4V difference is because the voltage cannot "make it" to the other side without "giving" 0.4V to the diode, ever. Think about it, if the cathode had a voltage of 20V, which is what your figure actually represents (20V-20V = 0V), then it would have found no "resistance" from the diode.

Just because it is an open circuit on the other side, this does not mean that the 0.4V drop would not exist, like if a simple resistor was there.

  • \$\begingroup\$ Yes I think D1 should get a polization voltage as shown in Proteus and real circuit; however in real circuit the polarization voltage is negative... meanwhile the voltage on D2 is positive... \$\endgroup\$
    – rm1234
    Commented Sep 16, 2020 at 20:18

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