Mosfet relay output safety flyback shunt/crowbar circuit design

Passing the loudspeaker current through a MOSFET might introduce distortion, it will certainly cause some losses. Also consider that the diode symbol in the MOSFET is effectively a real diode and it will conduct if there is a discharge from the loudspeaker or its filter components so the problem may not be as bad as you predict.

Regardless, the far simpler method is just to wire two diodes, one from ground to output and one from output to supply, both orientated so they are not normally conducting. Any voltage outside of ground or supply will be clamped to ground or supply so over-voltage damage should not occur.

Brian.

Hi,

I see this less critical .. regarding "destroying the speaker".

True is:
* If you OPEN the circuit of a coil: The current after opening is not higher that before. So if the speaker withstood the current before it also will stand the current during/after opening.
* The voltage across a coil can/will increase when yu suddenly open the circuit. But as before the current does not increase. Also the "open path" now is high ohmic, high voltage across the opening contacts, maybe arc ... this also means the energy is dissipated via the arc (very fast decay) and only a tiny part becomes dissipated by the speaker.
* "short circuiting" is counter productive: It prevents fast decay and thus more energy needs to be disspated by the speaker.

An optimized circuit could be to keep the voltage high (for fast decay) but limit the voltage form going too high (arc building, overvoltage on capacitors...). Maybe zeners to limit the voltage while dissipating most of the stored energy.

Klaus

KlausST said:

Hi,

I see this less critical .. regarding "destroying the speaker".

True is:
* If you OPEN the circuit of a coil: The current after opening is not higher that before. So if the speaker withstood the current before it also will stand the current during/after opening.
* The voltage across a coil can/will increase when yu suddenly open the circuit. But as before the current does not increase. Also the "open path" now is high ohmic, high voltage across the opening contacts, maybe arc ... this also means the energy is dissipated via the arc (very fast decay) and only a tiny part becomes dissipated by the speaker.
* "short circuiting" is counter productive: It prevents fast decay and thus more energy needs to be disspated by the speaker.

An optimized circuit could be to keep the voltage high (for fast decay) but limit the voltage form going too high (arc building, overvoltage on capacitors...). Maybe zeners to limit the voltage while dissipating most of the stored energy.

Klaus

Click to expand...

Maybe I wasn't clear, I am not worried about current and current generally cannot increase much (otherwise free energy...) but voltage can and does rise whne a current is suddenly interrupted. Such voltage spikes might not do anything to the low and mid range speakers but they can harm the tweeter.

betwixt said:

Passing the loudspeaker current through a MOSFET might introduce distortion, it will certainly cause some losses. Also consider that the diode symbol in the MOSFET is effectively a real diode and it will conduct if there is a discharge from the loudspeaker or its filter components so the problem may not be as bad as you predict.

Regardless, the far simpler method is just to wire two diodes, one from ground to output and one from output to supply, both orientated so they are not normally conducting. Any voltage outside of ground or supply will be clamped to ground or supply so over-voltage damage should not occur.

Brian.

Click to expand...

Where would the mosfet diode conduct to? given they are put in series back to back (source to source together) both body diodes are now in opposing polarity so that when the mosfets are off they don't conduct in any direction. Isn't that the whole purpose of a safety relay ?

Well given mosfet is a bidirectional device in theory I would need just one but due to the built in diode I need two in series so that when they are OFF they don't conduct. In theory I could use just one mosfet and one diode in series but a simple diode has a larger voltage drop than the mosfet channel so two mosfets in series would still make a lower drop.
As for distortion I'm not sure , modern mosfets have ON resistance in the few miliohm range IIRC, so two mosfets in series would make the equivalent resistance of a longer piece of wire I think.

How exactly did you think wiring the diodes?
I have seen something similar in power amplifiers where the output is tied to both + rail and - rail with diodes put in reverse.

Hi,

Salvador12 said:

Maybe I wasn't clear, I am not worried about current and current generally cannot increase much (otherwise free energy...) but voltage can and does rise whne a current is suddenly interrupted. Such voltage spikes might not do anything to the low and mid range speakers but they can harm the tweeter.

Click to expand...

I understand what you say,
But the tweeter has an impedance. And if the current through the tweeter is limited, then the voltage also is limited according it´s impedance. (Worst case if no crossover exists and all of the energy theoretically could go to the tweeter)

And to physically destroy something you need some amount of energy. The energy stored in the circuit is limited, and as already explaind.. if most of this energy is dissipated in the "open switch" circuit, then it can not harm that much to the tweeter.

Klaus

KlausST said:

Hi,

I understand what you say,
But the tweeter has an impedance. And if the current through the tweeter is limited, then the voltage also is limited according it´s impedance. (Worst case if no crossover exists and all of the energy theoretically could go to the tweeter)

And to physically destroy something you need some amount of energy. The energy stored in the circuit is limited, and as already explaind.. if most of this energy is dissipated in the "open switch" circuit, then it can not harm that much to the tweeter.

Klaus

Click to expand...

In a mosfet relay, where would this "open switch" circuit be that can dissipate the energy you talk about?
In a mechanical relay it would be the arc between the contacts , but in mosfets there is no such thing,

Hi,

I already told you to use "something" across the switch to limit the voltage while dissipating power. Maybe use zeners.

Last paragraph of post#3

Klaus

Cable inductance is possible source of overvoltage, a bidirectional TVS diode should be sufficient to protect the MOSFETs. I don't see a risk to damage speakers by switching transients, their energy is far too low.