Analog modeling of a diode clipper (2): Discretization

Let’s start with the two equations we got from the last post and see what we can do with usual/academic tools to solve them (I will tackle nodal and ZDF tools later in this series).

Euler and trapezoidal approximation

The usual tools start with a specific form: $$ \dot{y} = f(y) $

I’ll work with the second clipper whose equation is of this form:

\frac{dV_o}{dt} = \frac{V_i - V_o}{R_1 C_1} - \frac{2 I_s}{C_1} sinh(\frac{V_o}{nV_t}) = f(V_o)

Forward Euler

The simplest way of computing the derivative term is to use the following rule, with h, the inverse of the sampling frequency:

V_{on+1} = V_{on} + h f(V_{on})

The nice thing about this rule is that it is easy to compute. The main drawback is that the result may not be accurate and stable enough (let’s keep this for the next post, with actual derivations).

Backward Euler

Instead of using the past to compute the new sample, we can use the future, which leads to

V_{on+1} = V_{on} + h f(V_{on+1})

As the result is present on both sides, solving the problem is not simple. In this case, we can even say that the equation has no close form solution (due to the sinh term), and thus no analytical solution. The only way is to use numerical methods like Brent or Newton Raphson to solve the equation.

Trapezoidal approximation

Another solution is to combine both solution to have a better approximation of the derivative term:

V_{on+1} = V_{on} + \frac{h}{2}(f(V_{on}) + f(V_{on+1}))

We still need the numerical methods to solve the clipper equation with this method, but like the Backward Euler method, this one is said to be A-stable, which is a mandatory condition when solving stiff systems (or systems that have a bad condition number). For a one variable system, the condition number is of course 1…

Other approximations

There are different other ways of approximating this derivative term. The most used one is the trapezoidal methods, but there are others like all the linear multistep methods (that actually encompass the first three).

Numerical methods

Let’s try to analyse a few numerical methods. If we used trapezoidal approximation, then the following function needs to be considered:

g(V_{on+1}) = V_{on+1} - V_{on} - \frac{h}{2}(f(V_{on}) + f(V_{on+1}))

The goal is to find a value where the function is zero, called a root of the function.

Bisection method

This method is simple to implement, from two starting points, on either side of the root. Then, we take the middle of the interval, check the sign of it and keep the original point that has a different sign, and keep on until we get close enough of the root.

What is interesting with this method is that it can be vectorized easily by checking several values in the interval instead of just one.


This numerical method requires the derivative function of g(). Then we can start from the original starting point and iterate this series:

x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)}

For those who are used to optimize cost function and know about the Newton method, it is exactly the same as this one. To optimize a cost function, we need to find a zero of the derivative function. So if g() is this derivative function, then we end up on the Newton method to minimize (or maximize) a cost function.

That being said, if the rate of convergence is quadratic for the Newton-Raphson method, it may not converge. The only way to achieve convergence is to be close enough to the root we are looking for and have some conditions that are usually quite complex to check (see the wikipedia page).

Starting point

The are several ways of starting the methods.

The first one is any enough: just use the result of the last optimization.

The second one is a little bit more complex: approximate all the complex functions (like sinh) by their tangent and solve the resulting polynomial.

The third one is derived from the second one and called pivotal method/mystran method. Instead of using the tangent, we use the linear function that crosses the origin and the last point. The idea here is that it can be more stable that the tangent method (consider doing this for the hyperbolic tangent, the resulting result could be quite far).


Of course, there are other numerical methods that I haven’t spoken about. Any can be tried and used. Please do so and report your results!

Let’s see how the ones I’ve shown behave in the next post.

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