I'm learning electronic circuits on my own, and I got stuck. Please have a look at the book's question and answer below. I cannot understand why V2 = V1 when V1 ≤ 0.6 V. Should not the voltage drop across the resistor R be included in this equation?

My understanding is that an ideal diode has a voltage drop of 0. And if two parallel branches have different voltages, the current will favour the branch with a smaller voltage drop.

I'll have no trouble plotting the voltage transfer ratio once I understand what's going on.

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  • \$\begingroup\$ Think about what the voltage drop across the resistor would be: V=IR, what is I in this case? \$\endgroup\$
    – BeB00
    Nov 27, 2022 at 8:18
  • \$\begingroup\$ when V1 ≤ 0.6 V, I think that current going through the resistor would be equal to the current coming from the plus side of V2...from V2 the voltage would drop by IR, and then by V1 to reach 0. \$\endgroup\$
    – dboko
    Nov 27, 2022 at 8:52
  • 1
    \$\begingroup\$ @dboko How will current enter from V2? Think of the + and - as the two probes of a multimeter that is used to measure the voltage in the arm containing the diode. \$\endgroup\$ Nov 27, 2022 at 8:56
  • \$\begingroup\$ Got it, thanks. In my head I was imagining that this circuit would be connected to something else. \$\endgroup\$
    – dboko
    Nov 27, 2022 at 9:00
  • \$\begingroup\$ The ideal diode and voltage source look like what you're actually working with is a slightly less ideal model of a diode, where the diode is modelled as a voltage source (of somewhere between 0.55 and 0.7 V) when forward-biased and an open circuit when reverse-biased, rather than the ultra-simplified short-circuit/open-circuit model you may have been using previously. \$\endgroup\$
    – Hearth
    Nov 27, 2022 at 19:46

3 Answers 3


When V1 is less than or equal to 0.6 V, no current will enter the branch containing the diode. Since no other closed loop exists, the current flowing in your circuit will be 0. Now, apply KVL in the only loop within in your circuit. It will be: $$V_2 - V_1 +0*(R)=0 \Rightarrow V_2=V_1$$


The current across a diode (taken positive for current flowing in p-side of diode) in series with a resistor R can be written as I=max(0, Vdiode/R) [i.e. mathematical way of saying current cannot be negative and as long as it is positive diode is like non-existing/short], here Vdiode = (V2-.6); now you can start by making equations

V2=V1-I*R ----(1)

I=max(0, (V2-.6)/R) ----(2)

If you substitute back V2=V1-max(0, (V2-.6)/R)*R You can substitute V2<.6 in the above equation and verify the transfer characteristics.



Parallel voltage sources...

You are right - really, there are two voltage sources which at some point are connected in parallel. One of them is imperfect with varying input voltage V1 and internal resistance R; the other is perfect with constant reference voltage 0.6 V and zero internal resistance. The input source is always connected to the output (load, not shown here). The diode acts as a switch that controls the reference source.


If V1 < 0.6 V, the diode switch is off and the reference source is disconnected. The input voltage is applied through the resistor to the output. There is no problem if there is no load connected (open circuit) - the full input voltage appears at the output. But if a significant load RL is connected, a voltage divider R-RL is formed and the output voltage decreases.

If V1 > 0.6 V, the diode switch is on and the perfect reference source is connected in parallel to the imperfect input source. The perfect source dominates over the imperfect one and its voltage appears at the output. There is no problem if a significant load is connected since the internal source resistance is low.

General idea

From these observations we can formulate a rule for parallel connection of voltage sources:

If we connect in parallel two voltage sources - perfect and imperfect, the voltage across this network is equal to the voltage of the perfect source.

The advantage of this clever trick is that two voltage sources can be switched to the output by a simple 2-terminal switch (SPST) instead by the more sophisticated 3-terminal switch (SPDT). SPST can be implemented by a diode.

What is the role of R?

At first glance, the resistor R is redundant and only worsens the circuit operation... but in fact its role is extremely important. By it the input source is "deliberately worsen" to prevent the conflict between two perfect voltage sources.


With the same success, we can swap the source's properties making the input source perfect and the reference source imperfect. This means to swap the resistor and diode. Thus we will obtain another kind of diode limiters.

In total, there are eight variations of this device known as a "diode limiter".


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