Relay time without power bridged with capacitor?
Good Morning,
I built a circuit for a sound box that I can use with mains power and batteries!
Now I've installed a relay that switches to mains power when the power cable is plugged in. Otherwise, the box runs on battery power.
Now the problem is that when I pull the power cable out of the socket the box switches to battery power, but there are a few seconds without power and the sound disappears for a moment.
I wanted to fix this "current gap" by placing a capacitor in front of the amplifier. But somehow that doesn't work.
Capacitor is rated at 42 V and has 0.47 Farad
According to my calculations, it should provide power for 2 seconds.
The amplifier runs on 19 V and 5 .
If you can help me, I would be very grateful for your help.
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Your problem is that the relay lasts too long even at lower, decreasing voltage, but the voltage from the power supply is no longer enough to operate the amplifier.
So you have to Converse do (not support the amplifier with capacitor): instead take away the "condenser" from the relay, bring it to waste earlier.
There are several possibilities:
Thank you for your answer.
Got the relay that is for 24v
https://www.amazon.de/Trigger relay module high performance single-use trigger optocoupler separation relay YYG-2low level/dp/B07TXHMF12/ref=asc_df_BO7TWH7FW/?tag=&linkCode=df0&hvadid=38021 5546747
804275395663&psc=1 &mcid=692f0d386891323a 9263eOe60 867c1928&th=1 &ref=&adgrpid=77941887
What did I need for a voltage regulator?
Thank you for your help so far!
Okay, I thought it was a normal (naked) relay like in your sketch.
Now it's such a trigger relay with some electronics before. Unfortunately, I don't know how it behaves exactly electrically. However, since you have and describe the problem, it may be similar to a naked relay if you have already tried. Or even deliberately delayed. You should study the detailed data sheet.
It obviously needs an auxiliary voltage/supply and then the control. Presumably it turns on a specific voltage both and out.
Normal relays have a hysteresis:
They attract, for example, from 80% of the nominal voltage, but only drop below 30% of the nominal voltage.
As written, you have to experimentally find out the waste voltage (run down control voltage slowly while monitoring the NC and NO contacts).
Then select the voltage regulator as closely as possible above this voltage.
In a 24V wheel, the waste voltage could be, for example, 5 or 8V. So take a 6V or 9V controller.
OK, I'll try it out.
Thank you so much.
This is not possible because you switch AC.
No equal current
Then take good gold caps, but watch the tension. Calculate with factor 5 of the capacitance and charge the by no impulse short circuit via a load resistor.
Electrolytic capacitors have an unfavorable unloading curve.
No it is equal
What do you think about 5
And how many Farad does the Goldcap need to understand this by a factor of 5!
But then I need two latencies than for charging must be one before the capacitor and to unload one after the capacitor.
Load resistance in series with gold cap, so that slower charging and ending.
Ah ok and the gold cap I put Dan in front of the amplifier. And the load resistance after the capacitor so that if the capacitor emits too much energy the amplifier does not break, right?
And how do I calculate the Goldcap right now?
Thank you for helping me!
without reckoning I would say your capacitor can never supply the relay with current for so long.
Would you like another solution?
You can't say that without reckoning. The current consumption of the relay depends on the capacitance of the capacitor.
On the relay is 10A 28v But you don't think so?
no, the coil current or resistance is decisive. If you know the coil resistance, the discharge time of the capacitor is approximately t≅5*R*C. Since the relay already drops before the capacitor is completely unloaded, you might need to expect a factor of 10 or 15. If you convert the formula to C, you get
C ≅t / (10*R)