6.2.1. Initial Guess and Iteration

Yes, if we consider the same numerical problem, namely if I consider the V here and then we do have the resistance of 10 k, and then we do have the diode here, and then if we observe the corresponding output, by considering one initial guess. And, this initial guess it is not just arbitrary, typically we know that if it is silicon diode and if the diode is on the drop across this diode is roughly 0.6 V.

In solving electronic circuit problems, especially those involving non-linear components like diodes, an initial guess is required. This guess isn't arbitrary; often, for silicon diodes, we know that when they are conducting ('on'), they have a voltage drop of approximately 0.6 volts. So, we start our calculations with that assumption, which simplifies our analysis and helps us approach the correct values quicker.

Imagine tuning a radio. You don't just randomly turn the knob; you know that certain stations are found around certain frequencies. By starting close to those known frequencies, you can zero in on your desired station much faster.

And, with this guess, if I consider V = 0.6. The value of this I you will be obtaining it d R is . So, what you are getting here it is 0.94 mA. Now, if you compare our previous numerical value, which it was close to if, I recall correctly 0.94023 something like this mA.

Once we set our initial guess for the voltage at 0.6V, we can calculate the corresponding current through the circuit using Ohm's Law. With a resistance of 10 kΩ, this gives us a calculated current of approximately 0.94 mA. Comparing this with a more precise value obtained through iterative methods reveals that our initial guess was quite accurate, with less than 0.03% error. This shows the effectiveness of making a good initial assumption.

Think of baking a cake. You might start with a recipe that tells you to use a certain amount of flour. If you start with that quantity, your cake is likely to rise well, but if you tried a wildly different amount, you might end up with a flat cake or a mess. Starting with a good guess sets you up for success.

So then just by one step itself we can find the solution. This is of course, one indication that how we are trying to get a practical method by the virtue of guess and proceed by one iteration, but then some people may say that no I do have a diode, it may be silicon diode, but I know that it is diode drop it is roughly 0.7. And, if we use a instead of 0.6 if you use 0.7, then also if you see the corresponding value of this I is .

After making our initial guess, if we adjust our guess based on varying criteria, like the specific type of diode we are working with, we can still attain reasonable results. For example, if we guessed a 0.7V drop instead of 0.6V for the silicon diode, our calculations would yield a slightly different current value. This illustrates the importance of understanding component behavior and characteristics for practical applications.

It's like adjusting the temperature for a recipe based on the oven type you’re using. If you're using an oven that runs cooler than usual, you adjust your baking temperature accordingly, leading to a better end product.

So, we can say that this gives us one indication that probably we may have some practical method to replace this diode by a something call some model. So, what is that model? We may consider if the diode is on drop across this diode it is may be around 0.6 or 0.7 and let you call this voltage is Vγ. And, but then if depending on the current level, the voltage drop across this resistance diode it may not be remaining same.

To simplify our calculations, engineers often replace the diode with a model that reflects its behavior more accurately in various circumstances. For instance, we can define a 'cut in' voltage (Vγ) representing the voltage at which the diode turns on and behaves predictably. However, we also note that as current varies, the voltage drop across the diode can change, so we need to account for that variation in our calculations.

Consider how car engines and their efficiency can vary by conditions. Just as you wouldn't assume that the engine will perform the same under all conditions, you account for changes in performance over different speeds and loads. Similarly, we model diodes to account for variability in their operation.

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