Calculating time-constant with z-Diodes

Question

I'm a student in Computer Sciences and am currently preparing myself for an Exam in Electronics. I came across this Exercise:

(U_0(t) = 10V), u(0) = 0V)

I don't have any solutions for the exercise, which is why i'm asking here:

The exercise is to calculate the time constants as well as the end values for the capacitor-voltage ( u(t) ) in case of a charging or discharging the capacitor. It seems to be an easy task in case of the discharging: The capacitor can only apply it's voltage to R_2 because of the Diode, so tau_1 = R_2 * C. Also the end-value for the charging-procedure (for me) seems to be equal to the z-voltage.

But what about the time-constant for the charging-procedure? We're assuming, that the diode and z-diode are absolutely ideal here. So modeling the z-diode with a voltage-source and a resistor r_z of 0 Ohm would lead to a short-circuit when calculating the pre-resistor for the capacitor!

What am i missing here?

This is my first post here and i would consider myself not really experienced, so take it easy on me!

Answer 1

The actual "r_z of the Zener" in datasheets is called Zzt or the impedance Z of the zener z at the test current e.g. 5mA.

Since this is an "incremental" or "knee" slope on the curve above some threshold voltage, Vth which is just below the rated Vz voltage.

It includes the knee voltage in the equation Vzt= Vth + Izt*Zzt

Your assumption is incorrect as an ideal voltage source because here it is an incremental resistance only when operating in Zener mode current and not at 0V.

It would be the same as an ideal battery except only during Charge mode.

When the input=0V , and (Vcap = +), D is no longer forward biased and blocks the zener with open circuit

Answer 2

So i figured it out myself:

When the capacitor in the given example is charged, current is applied to both of the present resistors R_1 and R_2. To calculate the time-constant tau (tau = R * C), we have to first calculate the pre-resistor in front of the capacitor which in this case (not taking the diodes or voltage-sources into account) is

$$R_1||R_2 = \frac{R_1\cdot R_2}{R_1+R_2} = \frac{33k\cdot 100k}{33k+100k} = 24.812k$$

So tau would be:

$$\tau = R \cdot C \approx 25k\Omega \cdot 8nF = 0.0002S$$

When the voltage source is detached, the capacitor itself will become a voltage source, which discharges in the direction of the Diode. No positive current will flow through the Diode, so the capacitor will apply current only to R_2. So

$$\tau_2 = R_2 \cdot C = 100k\Omega \cdot 8nF = 0.0008S$$