# DC voltmeter yields a value with 21% error

In the following circuit, the dc voltmeter yields -2.36V while the analytical value is -3V which means there is an error of 21%. The error is rather huge. I've searched through the documentation but with no useful information. Is there any setting need to be changed regarding this issue? With the assumption that the diode is an ideal diode. • It's not clear to me, but I think you are showing the output of a simulator and not the result of actually trying this? Could you confirm that point? (Proteus, in short?) – jonk Oct 21 '16 at 23:41
• Place a voltmeter across the diode and check – JIm Dearden Oct 22 '16 at 11:33

The diode is not ideal. If you right click the diode and select "Display Model Help", it will show you the model parameters used for simulating the diode: When a string of voltages does not add up to what you expect, you check each individual part to understand why.

Vf for small Silicon diodes and Vbe near 1mA is around 0.65V.

Note that just by rearranging the same loop it may be easier to understand the R1 controlled loop current.

If we assume they are ideal 3V batteries and a real diode, I would expect smilar value of what you show, except the opposite polarity. simulate this circuit – Schematic created using CircuitLab

Note your meter polarity is reversed.

It's pretty clear that the diode is accounting for (I assume) a simulated voltage reading in Proteus. I can't speak to how "ideal diode" applies to your simulator, as I don't have or use it. But broadly speaking the loop equation looks about like:

\begin{align*} 0 &= +3\:\textrm{V} - I\cdot R_1 - \frac{n k T}{q}\cdot \textrm{ln}\left(\frac{I}{I_s}\right) + 3\:\textrm{V} \\ \\ 6\:\textrm{V} &= I\cdot R_1 + \frac{n k T}{q}\cdot \textrm{ln}\left(\frac{I}{I_s}\right) \end{align*}

Solving this in closed form would use a product-log function. But we can avoid that. Just ignore the diode for a moment and find a first estimate for $I=600\:\mu\textrm{A}$. I can't speak for your "ideal diode" in Proteus, but I can use rough values I know for real diodes like the 1N4148. In this case, $n\approx 1.75$ and $I_s\approx 1\:\textrm{nA}$, which suggests that the diode's voltage drop is about $600\:\textrm{mV}$.

A second iteration through the equation then estimates $I=540\:\mu\textrm{A}$ and plugging that back in still gives me about the same $600\:\textrm{mV}$ estimate for the diode drop.

That's really what it must be, if this is about the simulator. Regardless of protestations otherwise about ideal diodes, the simulator must be using some given set of basic Silicon-based diode model parameters here. It simply explains your observation to a T. Anything else fails, I think.

Feel free to check out the current through the resistor, $R_1$, for example. I expect to see it less than $600\:\mu\textrm{A}$.

• Thanks for the informative answer. The current according to KVL is correct in your case however, the voltage drop at the diode should be assumed to be zero in the ideal case in which the diode works as a short circuit. I've searched in the documentation but I couldn't find data regarding to the ideal assumption. – CroCo Oct 22 '16 at 0:00
• @CroCo Well, that's the problem then, isn't it? You need to find out. I think the simulator is already telling you almost everything you need to know, though. It might be nice to see it in writing. But I'm pretty sure this accounts for the simulation results, regardless of what you feel it should be. I just don't have Proteus and can't do much more than I have about it. I apologize for my lack on that score. – jonk Oct 22 '16 at 0:04
• It seems the simulator utilizes the equation you post it. According to the documentation, $I_s=1e-14$ and $n=1$. Also, I can change these properties to model an ideal diode with zero voltage drop so that the diode works as a switch. Thank you. – CroCo Oct 22 '16 at 3:02
• @CroCo Well, you have your own answer then. I figured it would be in the docs somewhere. It's modeling a diode that is close to the behavior of the BE junction of a 2N3904. – jonk Oct 22 '16 at 4:05

The difference seems to be close to a ~ 0.7 V drop across the diode. Could that be the source of the error?

Ideal diodes are almost always modeled as having a constant 0.7 V drop regardless of the current. That is the ideal part. This seems to be exactly what is happening in your circuit. The voltage drop seems to be modeled as 0.64 V.

A non-ideal model would use the full diode equation to calculate voltage and current.

• This is an ideal diode. – CroCo Oct 21 '16 at 23:26
• Contrary to your belief, an ideal diode has a forward voltage which is described by the Shockley diode equation. See en.wikipedia.org/wiki/Shockley_diode_equation – Janka Oct 21 '16 at 23:35
• This depends on your definition of an ideal diode. I think of an Ideal Diode as one that has no forward voltage drop. Your simulator apparently believes that an Ideal Diode follows the Schockley diode equation. – Peter Bennett Oct 21 '16 at 23:52
• I didn't invent that definition. Ideal Si diode (as the schematic symbol suggests) means Shockley diode. It doesn't make much sense to use a simulation tool with components that don't exist. – Janka Oct 22 '16 at 0:10
• Almost always Ideal diodes are modeled as having around a 0.7 V drop. – rtclark Oct 22 '16 at 0:20