Need help with a relay switching circuit....

[navienavnav]

I have created a relay based switching circuit which switches between a battery source and the mains source for my Router. Everything is working as expected and the switching is taking place the way it was meant to. However, the "Switch" is not fast enough and thus between switches the router resets. Is it possible somehow to avoid this? I had the idea of using a RC (resistor and capacitor in series) circuit in parallel with the final output for the router which (could) store energy and supply it for that tiny moment when the switch is taking place. This is just an idea which came to my mind and I haven't actually implemented it, yet. I thought I'd rather first ask over here than to waste my energy. So is it theoretically (and maybe possibly practically) possible? If this RC method won't work then is there any other way?

Thanks.

 

[Pjdd]

I have some ideas that will probably solve your problem, but to make sure I understand what it is you're trying to do, please clarify these first:
Is the purpose to use the battery as a backup during a mains failure?
Is the relay coil operated by the normal router supply so that it de-energises when the mains fails and the contact switches over to the battery?
Is the router power supply a regulated switched-mode type?
What is the voltage of the power supply?
What battery - voltage and type - are you using?

 

[wa1kij]

Put a capacitor across the output from the relay. You'll have to do some trial-and-error to find the right value. 1000 microfarads might be enough, but then again it might not. Just make sure the capacitor voltage ("WVDC") is greater than the power supply voltage. Keep in mind that those little power supplies with the plug built in are unregulated, and often put out a higher voltage than they say they do.

 

[navienavnav]

@Pjdd

Yes, it's meant to be a backup power supply.

The relay coil is powered by an independent full wave center tapped rectifier outputting 22.8 volts of DC which takes its power from an independent mains socket.

Router is powered through the ac adapter that came with it which I am guessing is a smoothened full wave/bridge rectifier type (just a guess).

Voltage of the power supply for the router is ~12 V and the ac adapter for the router outputs ~11.68 V.

I am using 8 Size "C" Type batteries in series and this combination successfully runs the router. Each battery is of 1.5 V.

---------- Post added at 09:15 ---------- Previous post was at 09:10 ----------

@wa1kij


I tried doing that but I guess the capacitance wasn't enough haha as I used only a 100 microfarad capacitor in series with a 50 K-ohm resistor.

Should I even use a [resistor in series with the capacitor] which is in parallel to the output of the relay? Or should I just use a single capacitor in parallel to the same output?

The capacitor has 63 V printed on it...is that the "WVDC" you are talking about?

One more thing...I don't have any capacitor with that high a capacitance. Can I use several capacitors connected in parallel (C=C1+C2+...+Cn) to achieve the same result?

Thanks.

 

[s_guria]

@Pjdd

Yes, it's meant to be a backup power supply.

The relay coil is powered by an independent full wave center tapped rectifier outputting 22.8 volts of DC which takes its power from an independent mains socket.

I think the problem is with the relay power supply. I guess filter capacitor in the relay power supply prevents relay from de-energizing quickly.

I would suggest you the following idea illustrated in the diagram below, which eliminates the need of extra power supply for just relay.


Use existing router power supply for driving 12v relay (you mentioned it has 12v output).
Nowadays almost all router comes with SMPS based power adopter, with enough current capacity to drive extra relay.
Don't worry about diode voltage drop, Internally router works on 3.3v or 5v DC, with linear regulator inside.
Capacitor near the router will take care of power supply glitches during switchover. Use around 1000uF.

goodluck

 

[Pjdd]

First, it's a refreshing change, compared to some other threads, to see an OP who supplies clear and precise answers to requests for more information. However, your statement below brings up another question:

Voltage of the power supply for the router is ~12 V and the ac adapter for the router outputs ~11.68 V.

By the symbol "~", did you intend it to mean "approximate" or "AC"? Is that the way it's printed on the product? If it's printed like that, it means that the output of the power supply is AC and a capacitor cannot be used to store energy during the switchover period. In fact, a capacitor placed across it is likely to explode and it's only your use of a series resistor that has prevented it from already doing so.

Another thing is that, if the output is AC, s_guria's scheme is not suitable because 1) the relay is for DC and 2) the diode provides half-wave rectification which will produce an unsymmetrical load for the transformer and is not desireable for anything but very light loads.

So, the crucial question is whether the output of the AC adapter is AC or DC.

 

[navienavnav]

I have managed to gather 640 micro-farad of capacitance by attaching two capacitors in parallel. I'll try and see if that solves my problem. I'll report back soon. :-o

@Pjdd

by '~' I meant approximation.

 

[sreepss]

Relay is a mechanical device, which having some mechanical inertia. so when u try to turn on the relay, it will take about 20-30 milliseconds to switch on the contacts.. for the electronic circuit, this much delay was not tolerated.. so one opinion was use a mosfet switch instead of mechanical relays..

 

[wa1kij]

A bipolar transistor could be used for the purpose. Put a diode in series with the power supply's positive output. Connect the emitter of a PNP transistor to the battery's positive terminal. Connect the collector to the load side of the diode. Connect the base of the transistor (through a resistor) to the power-supply side of the diode. When the power supply output is 0.6 volts or more lower than the battery voltage, the transistor will pass voltage from the battery to the router.