So I'm still reading The Art of Electronics Third Edition. On P.39, when dealing with inductive "kick", I found this sentence:

For the fastest decay with a given maximum voltage, a zener with series diode (or other voltage-clamping device) can be used instead, giving a linear-like ramp-down of current rather than an exponential decay (see discussion in Chapter 1x)


simulate this circuit – Schematic created using CircuitLab

Which circuit does the author mean? In fact, neither of them seems to work to me:

  • In circuit A, when the switch is closed, D2 is actually used as a zener, but it isn't supposed to do this job. Also, after the switch is open, it can reduces the voltage across D1 by its forward voltage drop, but can't clamp the voltage.

  • In circuit B, when the switch is closed, D1 is forward biased, which may attract the current through L2. When the switch is open, the maximum voltage across L2 would be the forward voltage drop of D2, then why even use a zener? Also, I don't think this structure can be called "a zener with series diode".

So, what do you think the author wants to express by saying "a zener with series diode"?

  • 4
    \$\begingroup\$ I don't see any sense using circuit B. If the switch is ope or closed the coil is always energized. \$\endgroup\$ Commented Jul 9, 2016 at 8:07
  • 1
    \$\begingroup\$ @sunqingyao: We recommend waiting a day or two before accepting an answer to allow others time to post. If you've really got what you want that's fine but accepting too quickly may dissuade others from answering and you might miss out. \$\endgroup\$
    – Transistor
    Commented Jul 9, 2016 at 9:23
  • \$\begingroup\$ @transistor Got it. \$\endgroup\$
    – nalzok
    Commented Jul 9, 2016 at 9:24

2 Answers 2


The left circuit (A) is what he means.

The Zener diode limits the voltage across the inductor to its nominal value. The other diode is there just to block the Zener diode in the forward direction.

Concerning the decay time, this is better than just having the normal diode, because a normal diode just yields a voltage of ca. 0.7 V across the diode. For fast decay, we want the maximum tolerable voltage (e.g. as much as the switching transistor can handle) in order to have high power which means dissipating energy in less time.

More detailed explanation:
Let's consider the time after the switch opens (after the coil has been energized):
The current through the inductor drops suddenly which results in a reverse voltage across the inductor ("reverse" means that now the bottom terminal is more positive than the top terminal). That voltage can damage the transistor (acting as the switch) or create unwanted sparks etc.
In order to avoid or limit that voltage it can be clamped by a normal diode (circuit A without Z-diode). What actually happens is that the total energy, \$E = \frac{1}{2}LI^2\$, stored in the inductor must be dissipated. Energy is the integral of power over time \$E = \int P dt\$, i.e. it takes some time depending on the dissipating power and the time is shorter if the power is higher. The power on the other hand is the product of voltage squared over resistance (inductor and diode resistance) \$P = \frac{V^2}{R}\$, i.e. we have a higher dissipating power, i.e. higher energy decay rate, if the voltage is kept higher (circuit A with a Zener diode).

Here is an LTspice simulation of the situation

  • without a Zener diode (left sub-circuit) and
  • with a 10 V Zener diode (right sub-circuit)

Enter image description here

The top graph shows the power dissipation in both inductors. The area under both power curves must be the same (= energy stored in the inductor).

The bottom graph shows the switching signal (red) and the voltages at the bottom terminals of the inductors.

Enter image description here

  • \$\begingroup\$ So the non-zener diode can't do voltage clamping? \$\endgroup\$
    – nalzok
    Commented Jul 9, 2016 at 8:11
  • 1
    \$\begingroup\$ It clamps to just 0.7V which is quite low. If you clamp to e.g. 60V by Z-diode you get higher dissipating power, i.e. more energy per time, i.e. less time. \$\endgroup\$
    – Curd
    Commented Jul 9, 2016 at 8:13
  • \$\begingroup\$ Please elaborate on it a little more: the normal diode clamp what to 0.7V when the switch is off? It doesn't seem like the voltage across L1. \$\endgroup\$
    – nalzok
    Commented Jul 9, 2016 at 8:23
  • \$\begingroup\$ @sunqingyao when a diode is forward biased, the voltage across it is 0.7V. Since the D is in parrallel with L, the voltage across the inductor must also be 0.7V. This is the clamp. The inductor voltage cannot exceed this because the diode will restrict the voltage across the inductor to 0.7V. \$\endgroup\$
    – efox29
    Commented Jul 9, 2016 at 8:24
  • \$\begingroup\$ @sunqingyao: if there was only the normal diode it would clamp the voltage across the inductor to only 0.7V (in the situation after the switch is opened). For more details see edited answer above. \$\endgroup\$
    – Curd
    Commented Jul 9, 2016 at 8:38


simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) The standard snubber diode configuration. (b) The equivalent circuit after SW1 opens. (c) The zener snubber. (c) The zener circuit equivalent after SW2 opens. \$ R_L \$ is the inductor's coil resistance.

The problem with the circuit shown in Figure 1a is that it maximises the drop-out delay of the relay. This can be a problem when a rapid response is required but also tends to open the contacts slowly and this may cause arcing.

Imagine that the relay coil was purely inductive and had no resistance and that D1 was an ideal diode with no voltage drop. Then when SW1 opened the inductance would keep the current flowing around the loop forever. In any practical circuit the coil resistance, \$ R_L \$ will burn up the energy and the current will decay.

If we were to adding extra resistance in the loop by adding a resistor in series with D1 (see \$ R_{SNUB} \$ in Figure 2) we can improve the speed of drop-out by burning up the energy more quickly. There are two things to be aware of:

  1. There will be a voltage drop across the additional resistor and the switch needs to be able to cope with this.
  2. The current decay will be exponential as in a standard LR circuit. i.e., The energy loss will be mostly through Rs and will decay exponentially with the current.

An improvement can be made by adding Zener diode D3 as shown in Figure 1c:

  • D2 prevents forward current through D3 shorting out the relay coil.
  • When SW2 opens current flows as shown in the equivalent circuit Figure 1d.
  • D2 is forward biased and D3 is reverse biased.
  • The inductance will cause the voltage to rise until D3 breaks down in reverse mode.
  • From the formula \$ V = -L \frac {di}{dt} \$ we can deduce that since L and V are constant then \$ \frac {di}{dt} \$ the rate of change of current will be constant too. i.e., the current will fall linearly until there isn't enough energy left to exceed the Zener breakdown voltage.

Figure 1c will drop the relay out faster than 1a.

Your circuit B is wrong. Both diodes will conduct and short the supply to GND (with two diode forward voltage drops).

Clarification on additional resistance


simulate this circuit

Figure 2. (a) Addition of snubber resistor to speed relay drop-out. (b) Equivalent circuit when SW1 is opened.

In the circuit of Figure 2 the inductance stored energy is dissipated in its own internal resistance, \$ R_L \$ and the external snubber resistor \$ R_{SNUB} \$.

The potential problem with this is that the current will generate a voltage across \$ R_{SNUB} \$ of voltage IR. If this voltage exceeds the rating of SW1 then damage may occur. (SW1 could be a mechanical or semiconductor switch.)

  • \$\begingroup\$ But the winding resistance of the inductor doesn't change, and the current through it would be decreasing. When the switch is opened, voltage across the inductor will decrease from some positive value(V+ on your diagrams) to 0, which raises the voltage across the opened switch from 0 to that value. So I don't think the switch should be able to cope with the voltage drop across the resistor, instead, it need to be able to cope with the absence of that voltage drop. \$\endgroup\$
    – nalzok
    Commented Jul 9, 2016 at 9:50
  • 1
    \$\begingroup\$ I've clarified this, I hope, in the edit. \$\endgroup\$
    – Transistor
    Commented Jul 9, 2016 at 9:51

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