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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.
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}\$.
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.
