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Old 2nd November 2014, 07:39 PM   #1
hoihen is offline hoihen  Netherlands
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Default Simple power supply for a Relais

Dear All,

Just developed a very simple power supply to drive a relay, however it is not working as it should. The relay is clicking as being on AC

In my equipment I have 5 Volts AC which I'm using, in the schematic these are "F1" and "F2".

This 5V AC is fed via a Bridge rectifier to a TS2940 regulator, supported by two tantalum capacitors. This to have a very small and simple regulated power supply.

I've attached the circuit I made.

Te components used are:
- Bridge rectifier: B250C1000 (picture of type attached)
- 10 uF Tantalum caps, (both the same, these are according to the datasheet of the regulator)
- TS2940 - 1A Ultra Low Dropout Fixed Positive Voltage Regulator, supply 3.3V
- Zetler AZ822–2C–3DSE Relais.

Can anybody support me in what I'm doing wrong here?

Thanks a lot in advance!

Regards, HJH
Attached Images
File Type: jpg Relais Powersupply.jpg (19.9 KB, 107 views)
File Type: jpg 01470010.jpg (3.4 KB, 101 views)
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Old 2nd November 2014, 08:12 PM   #2
Bill_P is offline Bill_P  United States
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Try adding 470uF in parallel with C1 to reduce power supply ripple. This value is not critical and you can use 1000uF too.
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Old 2nd November 2014, 08:14 PM   #3
Elvee is offline Elvee  Belgium
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C1 needs to be at least a few hundreds µF. Actual value will depend on the relay current draw, but if you use 1,000µF, you'll be safe in most conceivable instances and it won't hurt your head....
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Old 2nd November 2014, 08:20 PM   #4
Symon is offline Symon  United Kingdom
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Just my first thought, but looks like you don't have a smoothing capacitor in
output of the rectifier bridge, C1 is in correct location but too small for this purpose,
and is probably there to maintain regulator stability.

Try placing something like a 470uF electrolytic capacitor across the output
of the bridge this will store more energy and maintain the voltage into the regulator.
The new capacitor is effectively in parallel with C1, but C1 and C2 should be
positioned close the regulator for stability reasons.

Regards, Symon
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Old 2nd November 2014, 08:21 PM   #5
Symon is offline Symon  United Kingdom
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Just my first thought, but looks like you don't have a smoothing capacitor in
output of the rectifier bridge, C1 is in correct location but too small for this purpose,
and is probably there to maintain regulator stability.

Try placing something like a 470uF electrolytic capacitor across the output
of the bridge this will store more energy and maintain the voltage into the regulator.
The new capacitor is effectively in parallel with C1, but C1 and C2 should be
positioned close the regulator for stability reasons.

Regards, Symon
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Old 2nd November 2014, 09:00 PM   #6
hoihen is offline hoihen  Netherlands
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Thanks for the quick responses this really helps!

I understand that I should "smooth" the output of the Bridge before it is entering the 2940.

The reason why I kept the C1 and C2 this small is that I need a fast power off for the relay when I disconnect the AC (which is a transformer output).

Would a larger Capacitor not store so much energy that the relay would be delayed for a few seconds?

The Tantalum capacitors are both within 0.5 cm of the 2940, so this should be Ok.

I attached the datasheet that I'm using, on page 3 is the scheme I'm using, there the 10 uF's are listed. Strange

One other thing I just found is that I see in some circuits a diode across the relay coil, would this be an option?
Attached Files
File Type: pdf 2940 3.3 v regulator.pdf (139.7 KB, 6 views)

Last edited by hoihen; 2nd November 2014 at 09:05 PM.
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Old 2nd November 2014, 09:48 PM   #7
JMFahey is offline JMFahey  Argentina
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Those 10uF caps you see are for stability, but they assume the DC you are`regulating is smooth enough already .
Otherwise you are`feeding it full wave rectfied (but unfiltered) AC; the`regulator can't magically create a DC voltage which is not there between peaks.

As of the turn off time constant, calculate it using the filter capacitance and the`relay coil resistance.

I would not choose that shorter than a second.
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Old 2nd November 2014, 09:54 PM   #8
Bill_P is offline Bill_P  United States
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Larger values of C1 will hold the relay on longer when power is shut off. As you found out 10uF is too small and does not work. Try 470uF and if the drop out time is too long then try 220uF and finally 100uF. Eventually the capacitance reduction will be too much and the relay will not close properly. Hopefully one of the capacitance values turns on the relay reliably while still not holding it on too long when power drops.

Regarding the diode across the relay coil, yes that is usually added to clamp the coil voltage when shutting the relay off.
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Old 3rd November 2014, 08:32 PM   #9
hoihen is offline hoihen  Netherlands
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That indeed worked , I put a 470 uF across C1 and now the relay holds perfectly. The shutdown time is also still short and no problem for what I'm doing.

However, a next question arose.....

The connections I made are F1 and F2 for the AC and the GND. The relay works, but one resistor on the main circuit board gets very hot and smelly....

When disconnecting the GND lead everything still works with only F1 and F2 connected, and the resistor does not get hot.

Somehow I cannot connect both the grounds together? The - (ground) of the Relay power supply and the Ground of the equipment I'm building it into.
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Old 4th November 2014, 12:23 PM   #10
AndrewT is offline AndrewT  Scotland
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For low fi audio I have seen a recommendation of 1000uF (1mF) per ampere of load.
For Higher Fi use 2mF/A
For highest Fi consider 5mF/A

For a relay 1mF/A is perfectly adequate.

What is the current draw of the load?
if the regulator wastes 1mA and the relay uses 45mA then the capacitor required is very approximately {0.001+0.045} * 1 = 46uF
use 47uF or 100uF.

If it's too low the ripple will cause the regulator to drop out during the dips in the input voltage.
The relay won't mind.

If you had an oscilloscope you can check for drop out, particularly when mains voltage is low.

But as I said a relay won't complain.
However a relay supply that uses the minimum smoothing capacitance to ensure reliable operation will release more quickly when mains fails.
This is sometimes referred to as instant off.

Too big a cap and instant off can be many seconds !
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