# Transorb, Zener or Resistor for Back EMF Suppression of Relay?

by ĦMR.AWESOME!
Tags: resistor, suppression, transorb, zener
 P: 382 The relay coil is an inductance. When the power source is removed from a relay coil, the current in the relay coil continues. The faster this current in the relay coil is reduced to zero the faster the relay operates. The drive transistor see the coil source voltage plus the voltage across the relay coil. With a diode across the relay coil, the drive transistor see's the coil source voltage plus about 3/4 volt. With a diode in series with a transorb across the relay coil, the drive transistor see's the coil source voltage+ the transorb voltage + 3/4 volt. With no surge suppression across the relay coil, the drive transistor see's the coil source voltage + the voltage across the relay coil. The voltage across the relay coil is limited by arcing (relay contacts) or breakdown (drive transistor), or the energy in the relay coil is stored in stray capacitance and there is a dampened oscillation.
 P: 35 I assume the drive transistor is the PIC microcontroller that I'm using. So what you're saying is that I am worse off with a transorb because it sees the coil voltage plus the transorb voltage plus 3/4 volt. You are also saying that, somehow, arcing across the relay contacts limits the voltage created by the inductance of the coil. Right? Excuse my ignorance, but that seems completely wrong to me. The first three sentences make perfect sense and are congruent with my understanding. The rest confuse me. If you would explain and possibly give some examples, you would be a lot more clear.
P: 4,667

## Transorb, Zener or Resistor for Back EMF Suppression of Relay?

If you decide to use two zener diodes, put them in series (back to back), because a zener is a diode in the forward direction. You could put a resistance in series with the two zeners. The more resistance you put in, the shorter the L/R coil current time constant. But at the same time, a shorter time constant means a higher V = L dI/dt voltage, which could damage an npn collector.

The real value of a shunt across a relay coil is to quickly dissipate the energy stored in the coil inductance; ½LI2. The larger the voltage drop across the shunt (IR drop), the faster the coil discharges.

I recall getting a substantial shock (over 50 years ago!) running a dc relay at the end of a long coax cable on a 6 volt battery. It must have been over 100 volts. Also the high dI/dt couples into other circuits.

Bob S
 P: 382 A relay that has a transorb across it's coil will operate faster than a relay with a diode across it's coil. A relay that has a transorb across it's coil will have a higher voltage across the drive transistor than a relay with a diode across it's coil. WITH NO TRANSIENT PROTECTION: If the current in a relay coil is opened with contacts that have 1000 volt breakdown, then the energy in the relay coil is transferred to stray capacitance and the circuit oscillates. WITH NO TRANSIENT PROTECTION: If the current in a relay coil is opened with a transistor that has a breakdown voltage of 100 volt, then the voltage across the transistor goes to 100 volt. Most likely the transistor will be damaged. Disclaimer: The previous statements are true in theory. In practice there are many things that affect the magnitude of the transient voltage.
 P: 35 Alright. So now I think I get it. I had to do a lot more reading for your responses to even make sense. So I want the current to dissipate quickly to prevent arcing on the relay contacts due to the contacts opening slowly. The quicker the current dissipates, the higher the voltage will be. However, I have to keep the voltage below the maximum of the driving transistor or else the transistor will be destroyed. The transistor I'm using (BD681) has a max collector-emitter voltage of 100V. That means I need to keep the voltage spike below 100V, right? According to Carl, the transistor will see the source voltage+the transorb voltage+0.75V. With a 12V source I would choose a transorb with a Voltage Breakdown below 87.25V, right? I know I would really want want to keep the voltage lower for a safety margin. Now, if I was driving the relay straight from the PIC, what would I do? The pins on the PIC can't take more than about 6 volts. Does that mean I would use a transorb that would keep the voltage under 6 volts? I assume that just isn't done and that a transistor is always used to drive a relay. Please let me know. I have one last question. I have one relay that will be powered by an unknown, 12V power source. I do not know if it will be a transistor or a relay or something else that will switch it. Is there a generalized, safe voltage I should clamp the spike to, or should I use a method other than a transorb? Thanks for your help so far.

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