Relay Arc Supression Circuit

richard.bair · 2006-03-14T16:12:48+00:00

I have the following circuit: +5Vdc | +---------------+------------------------+ | | | | | | 330 Ohm Diode | | | | | | 5Vdc Relay Coil | | | LED 5.1V Zener | | | | | | | | | | +---------------+------------------------+ | | +---+---+ To Micro----1kohm--+B C |Transistor | E | +---+---+ | | | GNDI'm concerned about reverse bias on the LED when the realy lets go. Imeasure about 7.5 Vdc across the LED when the relay lets go. I thinkthe zener and the diode act as a good method to handle the coilmagnetic field collapse and ensure fast armature motion but I dovioloate the 5 Vdc LED reverse bias voltage. Any comments orsuggestions for this circuit? I need to maintain the LEDs forvisibility and troubleshooting. I'm hoping this thread discusses somegeneral techniques for realy arc supression and addresses the LED inthis specific example.Thanks!

J

John_H 20 years ago

Series resistor? 10.8V? Your coil is in parallel with a back-to-back diode/zener pair with no series resistor.

Your LED is in parallel with the coil which is charged to give nearly 5V across your coil and across your LED/330ohm combination and discharged to dump the coil energy and give you concerns about LED reverse bias. The activity is all relative to your 5V rail, not ground.

If you see 3V across the 330 ohm resistor in series with the LED, this is the forward biased conducting side of your situation, not the discharge of the coil where the LED reverse bias would happen. When reverse biased, the resistor voltage should be nearly zero (0 relative or the same voltage as your 5V rail).

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richard.bair 20 years ago

Ok, I have this installed

...and have simulated. When the relay releases, I see a spike on the voltage from the cathode of the zener to GND just shy of 16 Vdc. So the transistor needs to accomodate this voltage from collector to emitter while the base is off, right?

Thanks (especially for the simulation)

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David Harmon 20 years ago

On 14 Mar 2006 08:12:48 -0800 in sci.electronics.design, snipped-for-privacy@revco.spx.com wrote,

So put a diode in series with it that can stand the reverse bias. Adjust the resistor.

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John Fields 20 years ago

Right. You can increase the Zener voltage, which will speed up the armature\'s release, but you have to make sure you don\'t exceed the transistor\'s Vce or Vbe. Also, you have to make sure that you don\'t exceed the rated reverse voltage of the diode that\'s in series with the LED.

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John Fields 20 years ago

^^^ Vcb

John Fields Professional Circuit Designer

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ehsjr 20 years ago

Just add a diode. Ed

+5Vdc | +---------------+------------------------+ |a | | Diode | | | |k | 330 Ohm Diode | | | | | | 5Vdc Relay Coil | | | LED 5.1V Zener | | | | | | | | | | +---------------+------------------------+ | | +---+---+ To Micro----1kohm--+B C |Transistor | E | +---+---+ | | | GND

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John_H 20 years ago

If a series diode were used, the reverse bias applied to the diode and the LED would be dependent on the reverse leakage current for each device to determine how much voltage goes to the LED and how much to the "protection" diode. Rather than leaving these numbers to chance, adding that one doide in parallel with the LED should provide better protection from reverse bias.

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Fred Bloggs 20 years ago

Yeah- no way in hell should you be seeing 7.5V across the 5.1V zener+diode combination unless you're blasting the hell out of it. Go to a 1W zener and make the diode a 1N4001 type.

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John Fields 20 years ago

Indeed. Very nice. That\'s how I should have done it on my circuit too.

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ehsjr 20 years ago

He's concerned with the longer dropout of the relay with a diode/resistor path. To avoid that concern, the series blocking diode is better. Leakage current in the series diode is going to be in the 1E-08 range at room temperature. Thus voltage across the 330 is not a worry. Even at 100 C, its still in the low uA range.

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page 3.

But you are right - the number is left to chance and the parallel diode offers better protection. So - add both.

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Spehro Pefhany 20 years ago

I think John Popelish's solution is the most elegant (though for the same LED current, the 330R needs to be reduced to account for Vf of the zener).

LED forward voltage is 5V - Vce -Vf (of zener with LED current only) Typical Vf of a 1N5231B at 10mA/25°C is < 0.8V.

LED reverse voltage (peak) is Vf of diode with relay coil current.

Copied from his post:

+5Vdc | +---------------+------------------------+ | | | | | | 330 Ohm Diode | | | | | +---------+ 5Vdc Relay Coil | | | | LED | 5.1V Zener | | | | | | | | | | | | | +-----+ +------------------------+ | | +---+---+ To Micro----1kohm--+B C |Transistor | E | +---+---+ | | | GND

Best regards, Spehro Pefhany

"it\'s the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com

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Relay Arc Supression Circuit

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