Is a wetted relay a better choice for low level AC?
I would argue that, as much as switching 'noise' maybe induced by the coil discharge and PSU disturbance - the signal passing through the relay should have absolutely zero DC current/bias.
That way you can be sure of the source of EMI.
I would argue that, as much as switching 'noise' maybe induced by the coil discharge and PSU disturbance - the signal passing through the relay should have absolutely zero DC current/bias.
That way you can be sure of the source of EMI.
They make shielded relays...
https://www.te.com/usa-en/product-CAT-AX41-H519A.html
EDIT: I know these are for "RF" but they would work fine for AF - they are rated down to DC.
The only annoyance IMHO is the AC coil.
https://www.te.com/usa-en/product-CAT-AX41-H519A.html
EDIT: I know these are for "RF" but they would work fine for AF - they are rated down to DC.
The only annoyance IMHO is the AC coil.
Turning off (or on) a DC relay coil typically forces a step current load change on the supply rail for the coil. Some equipment may just use a convenient supply rail, which may allow the step load to induce a disturbance in to the audio chain. That can be alleviated by either appropriate distribution and filtering of that step load current, or as Elvee points out by controlling the coil switching device - they have nothing per se to do with the quality or cost of the relay, but have a lot to do with the nous of the circuit designer.
Maybe a stupid question, but can't you put a diode in series with the coil to prevent back EMF instead of reverse shorting it with a diode?
The only reason I can see not to is the forward biased diode will draw a small power and drop ½ a volt or so, the reversed diode won't.
The only reason I can see not to is the forward biased diode will draw a small power and drop ½ a volt or so, the reversed diode won't.
Putting diode in series with coil will just reduce coil voltage and emf voltage by diode drop voltage. Polarity of emf is reversed to polarity supplied to coil ,so voltage just sums ,like connecting batteries in series. A better for me looks to use RC filter , which will also reduce voltage on coil . As we know , relay release voltage is smaller than needed to turn on . Practically 12v relay will turn on at 10v ,maybe even less ,and turn off at 5-7v . RC filter like 100ohms 100uf would reduce voltage ,when relay is turned on , but will give needed voltage and current to start relay .Emf will be reduced too. And for small signal relay, there is no contact welding and such problems,as with power relays , so relay release time is nonsense , you can use just properly rated diode , also may add capacitor in parallel to coil , to slow down things a bit .
When the power source to a coil is opened the magnetic field that has been created is of course power. It has to go somewhere. What happens is the magnetic energy tries to dissipate as a voltage into a load. Without a load the voltage will continue to rise. For even a simple 12 volt relay this can go to hundreds of volts.
For the classic demonstration use a 9 volt battery to power a relay coil. As you disconnect the battery just hold your fingers across the coil contacts. After all how can just 9 volts affect you?
Right, I get that, but that hundreds of volts would be blocked from damaging the transistor by the diode and dissipated as heat somewhere, no?
Those hundreds of volts are there because no load , and its duration is short. But if placing diode in parallel to coil , then yes , returning from coil current will be dissipated to heat by Vf drop on diode . But pulse is short and in practice there is no heat .
Because that doesn't prevent back EMF at all, it does precisely nothing. The current only ever flows in one direction through the inductor, its the rate of change of current that matters. Such a series diode will always be forwards biased.
When the switching device turns off the rate of change of current is very large and negative, leading to the inductor responding with a large positive voltage, forcing avalanche breakdown of the switching device and probably destroying it instantly. Having a conducting diode in that current path is not going to change that.
Having an antiparallel diode to clamp the inductor's voltage to supply + 0.7V prevents the voltage transient completely.. The current is diverted, not stopped.
The problem isn’t the diode - it’s everything else. Don’t corrupt the power or ground for the audio with the current drawn by the relay. Return the ground (emitter of the driving transistor, I assume) back to the power supply reservoir through a path NOT shared by any of the audio ground - including the local bypass caps on op amps. Minimize the base current so it’s transient is small too. Keep the traces at right angles to the signal. Do not use the regulated voltage supply to power the relay! Take its power from the raw side. If you need to drop voltage, use a resistor or a separate voltage reg if one is used for fans, lights, or digital controllers.
No you just built a very small spark gap transmitter! Although some diodes will break down from the voltage rise.
Any diode will break down from the voltage rise. I’ll bet it will get high enough to zener a 1N4007. Enough to shock the **** out of you if your fingers are across it. A 4000 series diode is more likely to handle the energy than a 2N3904 driver transistor, but the voltage spike will be larger for sure.
I use TIP100’s for relay drivers because I have several hundred I got for free. Absolute overkill, but again, free. They will handle the energy in a typical coil in an ice cube relay without a fly back diode - but why push one’s luck?. Or have the ~150 volt spike that the transistor allows before breaking down (harmlessly).
I use TIP100’s for relay drivers because I have several hundred I got for free. Absolute overkill, but again, free. They will handle the energy in a typical coil in an ice cube relay without a fly back diode - but why push one’s luck?. Or have the ~150 volt spike that the transistor allows before breaking down (harmlessly).
MPSA42 actually breaking down at 600 V may be about right. Use of a beefier transistor that breaks down at a lower voltage is likely safer.
It was 600V into a scope lead. Put the DCR of the average fingers in parallel to it and see if it gets that high... 🙂
If I use a switch instead of a transistor, where does the energy go when I open the switch?