# How to analize an ideal diode circuit with sinusoidal sources and storage elements (inductors and capacitors)?

In my first Electronics classes, we learned how to analyze a circuit with AC sources and ideal diodes but with only resistive elements. In such cases, the currents and voltages respond instantly to inputs (I mean, there's no lag). What we did was that, during the positive half cycle of the AC voltage source, we assumed the current was exiting from its positive terminal, and from there we imagined how the current would divide in each node as it traveled through the branches. I think this is the common way taught. However as soon as a capacitive or inductive element is added, currents and voltages can have lag, and a transient response exists. Therefore, in the positive half cycle of the AC voltage source, the current doesn't necessarily exits from its positive terminal at all instants during the half positive cycle. The $$\i\$$-$$\v\$$ characteristics of an ideal voltage source is $$\v=v_s(t)\$$ and $$\i=\text{any value}\$$.

My question is how can we analyze/solve a circuit with ideal diodes, AC sources, resistive elements, and energy-storage elements (L, C), in order to analytically obtain the expression for a voltage or current? I wasn't taught about that. My goal is to obtain an exact expression for a voltage or current. While I know using the ideal diode will actually give an approximate value, I'm asking for that model to simplify calculations. As an example, consider the following circuit, where each diode is ideal, $$\R = 1 \text{ } \Omega\$$, $$\L = 1 \text{ H}\$$, and $$\v_s(t) = 12 \sin {2 \pi t} \text{ V}\$$ ($$\f = 1 \text{ Hz}\$$), and the inductor is initially discharged. Let's suppose we want to solve for the inductor's current.

The current through the inductor, choosing the reference direction from node 2 to 4, and the current through D1, are the following according to LTspice.

## Proposed solution

What I'm thinking to do is the following. In BJT circuits, in order to find its operating region, we initially assume the BJT is in active region, therefore we substitute it with its DC equivalent circuit in active region, then solve the circuit and compare if the assumption is correct. If it's correct, the analysis is done; if it's incorrect, we substitute it with the DC equivalent circuit in saturation region or cut-off region. Here I explain a bit more about this method.

So, I suppose that for diode circuits, we can also make an assumption and prove if it's correct or wrong. This answer describes the procedure I have in mind. However, since the source is AC and not DC, I think the solution we obtain will be valid only for half a cycle. So, we can not obtain the complete response for all $$\t\$$, but only up to the cycle we analyze. Is that correct?

• Show the BJT circuit you are talking about. – Andy aka May 27 at 8:45
• Hi Andy. The circuit I'd like to solve is shown above. When I mentioned the BJT, I just meant the way I usually solve circuits containing BJT's. – Alejandro Nava May 27 at 12:56

This is the reason simulators were created, because even with the diode, only, it gets complicated.

Consider the case of a diode as a half-wave rectifier, driving an RL load (i.e. your picture without D2). For a mathematical analysis, the diode should be ideal. This would mean that for the 1st half of the period the diode is shorted out, and the circuit disconnected for the 2nd half. But since there is a reactive element in there, the current will not stop when the input voltage would drop to zero. Then, the inductor current will tend to flow the other way, forward-biasing the diode.

But first, let's analyze the circuit without the diode: a simple series RL. The equations would be:

$$L\frac{\mathrm{d}i(t)}{\mathrm{d}t}+R\,i(t)=0$$

with the solution:

$$i(t)=i(0)e^{-{L\over R}t}\tag{1}$$

To solve for $$\i(t)\$$:

\begin{align} Z&=\sqrt{R^2+\omega^2L^2} \\ L\frac{\mathrm{d}i(t)}{\mathrm{d}t}+R\,i(t)&=V\sin(\omega t) \\ i_{\mathrm{steady}}(t)&={V\over Z}\sin(\omega t-\phi)\tag{2} \\ \phi&=\arctan{{\omega L\over R}} \\ i(0)&={V\over Z}\sin(\phi) \end{align}

So the expression of the total current would be $$\(1)\$$ plus $$\(2)\$$:

$$i(t)={V\over Z}\left[\sin(\omega t-\phi)+\sin{\phi}e^{-{R\over L}t}\right]\tag{3}$$

Plotting, side by side, the calculated current next to a SPICE simulation, they would agree:

If the diode is to be considered in the circuit, then the previous equation would only be valid for the first half of the period, plus the part where the diode is forward-biased by the induced voltage. At that point, the current is zero until the 2nd period starts, when the cycle continues. Shown below are the waveforms for the simple RL and for the RL+D:

So, until now, even with the diode, things look like they could be solved analytically fairly easy. But if you insert the D2 from your picture then the things complicate. Until now, there was only a part when the diode was on, and when it was off, and these were two states which could be separated and "stiched" to give the desired waveform. Now, with two diodes, there are 4 states, made by D1 on/off and D2 on/off. In each of these there are different things happening, each influencing the next:

• in the 1st half of the period, the current rises according to expression $$\(3)\$$.
• then, the induced voltage forward biases D2, not D1, which conducts until the next period, but it starts by continuing where D1 ended.
• D1 again starts to conduct, but this time there are no more zero initial conditions, which means that the waveforms for the two previous states determine the values of the initial conditions for each of the next part.
• etc

What remains, though, is the transient solution, $$\(1)\$$, which is visible in the slowly rising average value of the sum of the two currents, and the solution for the current through D1 (I(R1), green), but only for the first half of the period -- it can be seen that the waveforms coincide in that timespan.

The current through D2 can also be derived (similar to $$\(3)\$$) and calculated, but, as mentioned, the initial conditions are always changing, until steady-state is reached. So each next half-period has distinct solutions with initial conditions based on the previous half-periods.

At this point, even if I could derive the other formula (the same way, but the voltage has a displacement), I'd rather not because I'm hoping you can see why the simulators are used at this point. And, if you think the simulator decomposes the circuit into formulas, I'd have to dissappoint you, because it does nothing but compose a matrix of voltages, currents, resistances, conductances, what have you (simulator specific), then solves this numerically. It has no idea about complex operators, phasors, and the likes. It simply crunches numbers until a convergence is reached, at which point it declares the result satisfactory.

• If you still think it's worth it, and you want to go for transistors, ...good luck. :-) – a concerned citizen May 27 at 12:43
• Thanks, man! I actually tried my proposed solution and it worked. What you've explained is essentially the same thing. Yup, I noted it was getting tedious; I just wanted to know how to solve such a circuit without a simulator. And yes, I've seen LTspice says it uses Newton's numerical method to solve circuits. – Alejandro Nava May 27 at 13:02
• @AlejandroNava There is a shortcut for calculating the initial conditions for each half-period, because both currents through diodes obey the same differential equation solution, A*exp(-R/L*t), but the conclusion remains. If you think this answered your question, mark it down (the check mark), so that others, in the future, searching for similar questions, will know that this question has an accepted answer that they can follow. – a concerned citizen May 27 at 14:06
• Correct, I was going to mark it as accepted answer. But I've heard it's good to way at least a day so others might answer it, too. But your answer is complete, so there's no need for another answer. – Alejandro Nava May 27 at 17:23
• @AlejandroNava You're right to let some time pass, you never know if somebody comes with a better answer. That's why I didn't say "mark it now", only mentioned that you should, and this applies to any other answer. – a concerned citizen May 27 at 18:14