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I have the following problem:

Consider the circuit below:

enter image description here

The component values are: \$R_1 =5000 \Omega\$, \$R_2 =1000 \Omega\$, \$V_z = 5 \text{V} \$.

When \$i_1\$ jumps from \$12 \text{mA} \$ to \$0 \text{mA}\$, what is the time constant for discharging the inductor immediately after?

  1. \$ \tau = L/R_1\$

  2. \$ \tau = L/R_2\$

  3. \$ \tau = L/(R_1+R_2)\$

  4. \$ \tau = L/(R_1||R_2)\$

Here are my thoughts.

Since current through an inductor can't change momentarily, \$i_L=12 \text{mA}\$. The current will run through the zener-diode, when \$V_z=5 \text{V}\$.

The zener-diode allows a current of \$i = \frac{5 \text{V}}{5000 \Omega}=1 \text{mA}\$ to run through \$R_1\$, while there flows \$11 \text{mA}\$ through the zener-diode. These two currents meet at the middle node, and causes \$ 12 \text{mA}\$ to flow through \$R_2\$.

So as far as I see it, the current flows through bots \$R_1\$ and \$R_2\$ back into the inductor. \$R_1\$ and \$R_2\$ are in parallel, so my guess it that the time constant is expressed as: \$ \tau = L/(R_1||R_2)\$.

HOWEVER, it turns out the correct answer is \$ \tau = L/R_2\$ which I don't quite understand.

Can someone explain to me why this is the case?

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4 Answers 4

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Analysing the voltage loop during the discharge time you have:

$$-L\dfrac{di_Lt}{dt}=5V+i_L(t) R_2$$

Taking the laplace transform gives:

$$L(sI_L(s) - I_L(0))=-5V-I_L(s) R_2$$

Rearranging:

$$I(s)=\dfrac{I(0)L-5V}{Ls+R_2} = \dfrac{I(0)-\dfrac{5V}{L}}{s+\dfrac{R_2}{L}}$$

Taking the inverse laplace:

$$I(t)=(I(0)-\dfrac{5V}{L})\cdot e^{-t\dfrac{R_2}{L}}$$

$$t\dfrac{R_2}{L}= - \ln(\dfrac{LI(t)}{LI(0)-5V})$$

$$t= - \dfrac{L}{R_2} \ln(\dfrac{LI(t)}{LI(0)-5V})$$

Meaning, that the discharge time is proportional to \$\dfrac{L}{R_2}\$

NOTE The negative sine of the voltage across the inductor in the first equation is due to the fact that it is discharging at the point of interest.

Calculation must be double checked!

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First of all, when the Zener diode would not be there or would not be functional, R1 and R2 are in series, not parallel.

The ideal Zener behaves like a voltage source, which can pass through any current.
The Zener diode clamps the voltage across it to 5V. It doesn't matter what the value of R1 is, the current through R1 will be \$V_Z / R1 \$, and the remaining current caused by \$I_L\$ flows through the Zener.
Therefore, R1 does not affect the discharge current and is the time constant only determined by R2 (and L1) as long as the Zener diode is clamping.

Below a simulation in LTspice to show what happens.

enter image description here

As the simulation shows, node clamp is clamped to 5V and I(R1) = 1 mA, until about t=0.1us where the Zener stops clamping. Then, R1 also starts playing a role in the discharge time.

Obviously, the voltage across L1 eventually become 0V. Therefore, at some time, the voltage across the Zener will be less than 5V at some time, and so, there always will be a tail where the (remaining) discharge time is determined by \$L/(R_1+R_2)\$.

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The natural time constant of a network is obtained when all stimuli are turned off: a voltage source is replaced by a short circuit while a current source is open-circuited. Then, temporarily remove the energy-storing element and "look" through its connecting terminals what resistance \$R\$ do you see. Then if you deal with a capacitor, \$\tau=RC\$ and with an inductor, \$\tau=\frac{L}{R}\$ with a dimension in [s].

In your circuit, there is a Zener diode which can be modeled by a voltage source associated with a dynamic resistance \$r_d\$ whose value depends on the Zener diode operating point. An equivalent circuit could thus be the below one:

enter image description here

If we consider a small \$r_d\$, then the lower-side resistance \$R_1\$ is shorted and what remains is \$R_2\$. Therefore, the time constant is \$\tau=\frac{L}{R_2}\$. However, this expression pertains to a linear system but as the inductor depletes, the operating point of the Zener diode will change and so will the time constant considering \$r_d\$ also changing. Unless it is not considered here in this simple exercise.

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Vz dominates over R1 so R1 is being neglected. Certainly, R1 is neither in parallel nor in series with R2.

So as far as I see it, the current flows through bots R1 and R2 back into the inductor. R1 and R2 are in parallel, so my guess it that the time constant is expressed as: τ=L/(R1||R2).

This is wrong in more ways than one. I don't know how you arrived with parallel from this logic and not series. But even if you arrived at series it is also wrong.

Parallel = same voltage, series = same current. R1 and R2 share neither.

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