Fastest way to switch off DC relay or valve

Question

Switching on a relay is simple: just power it. Switching it off can be another matter.

What is the fastest way? (Yes, relays are slow by electronic standards, but we can still try to make them as fast as possible.)

In most cases I just insert a free wheeling diode:

^{simulate this circuit – Schematic created using CircuitLab}

This gets rid of the energy in the coil over time. Because the diode only drops about 0.7V, I mostly have the resistance of the coil to dissipate the energy. This is not ideal since the current is slowly reduced which will make it somewhat difficult to predict when the relays will really switch off (and in the worst case might even lead to more abrasion the the contacts of the switch if they are under load.)

Now I could put a resistor in series to the diode:

^{simulate this circuit}

I'll have to select the resistor so that at the rated current of the relay coil Vcc + +VD1 VR2 is smaller than the breakdown voltage of Q1.

Something that seems to be included in dedicated driver circuits. Some sort of Zener diode or other voltage limiting circuit.

^{simulate this circuit}

What's the fastest clean way to switch of a relay or other magnetic? What are the benefits and downsides like EMI, energy consumption, etc.?

Answer 1

Here's how the zener alone works. I modeled the relay as an inductor of 100 mH with a series resistance of 240 Ω so that the steady-state current is 100 mA.

^{simulate this circuit – Schematic created using CircuitLab}

If you run the simulation, you'll see that the current drops to zero in about 50 µs. However, note that when switching on, it takes about 680 µs for the current to rise to 80% of its peak value.

What if you also care about how quickly the relay turns on? Here's a thought I had: Why not store the energy in a capacitor and use it to speed up the current rise at switch-on time. It took a bit of work to come up with a workable circuit — I had to add a transistor that switches on only when its emitter is driven below ground by the capacitor. When the driver transistor switches on, the charge on the capacitor drives Vc strongly negative, increasing the rate at which the current in the coil rises.

^{simulate this circuit}

You can choose the capacitor value based on the energy stored in the coil and how high you want the voltage to rise at Q1's collector.¹ This particular value gives the same 200 V peak as the zener above.

It does work, at least in simulation (I haven't tried to build it). The current fall time is a little longer at 83 µs, but the current rise time (to 80%) is decreased to just 60 µs. And note that the coil's stored energy isn't dissipated, eliminating that source of waste heat.

Of course, in a real application, the effectiveness of this circuit would depend on the length of time that the relay is off. If it's too long, the charge on C1 will leak away before it can be used.

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¹ It's a simple matter of energy transfer. The energy in the coil is \$\frac{1}{2}L I^2\$ while the energy in the capacitor will be \$\frac{1}{2}C V^2\$. We can set these equal to each other and solve for C, giving:

$$\frac{1}{2}C V^2 = \frac{1}{2}L I^2$$

$$C = L \frac{I^2}{V^2}$$

Plugging in our known values (L = 100 mH, I = 100 mA, V = 200 V) gives us the value 25 nF.

This works for an ideal inductance, but a real relay will require some experimentation, since it is non-ideal in several significant ways.

Answer 2

You can keep npn bjt but use a higher volt part like say 300V.Then a 200 Volt zener and no Freewheel diode will help speed.