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Forum Index : Electronics : Circuit question-PICAXE-based mains timer

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domwild
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Joined: 16/12/2005
Location: Australia
Posts: 873
Posted: 01:12pm 06 Aug 2015
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Hi,

The latest "Silicon Chip" schematic for this timer has me stumped. A 9V Jaycar MM-2017 transformer is feeding 12VDC after diodes to a Jaycar SY-4040 relay. That relay has a 120Ohm coil. The schematic has a 120 Ohm resistor in series in addition and a 470uF 16V cap in parallel to the 12V coil of the relay.

The text reads as follows: "The circuit limits current in the power relay coil, using the charge in 470uF capacitor to actuate the relay and the 120Ohm resistor to hold the relay closed at a reduced current".

Showing my electronic ignorance here I am wondering why it needs a 120Ohm resistor and a 470uF capacitor? Would it burn out the coil without the resistor? And why does it need a stored charge to activate the relay?

dom



Taxation as a means of achieving prosperity is like a man standing inside a bucket trying to lift himself up.

Winston Churchill
 
BobD

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Joined: 07/12/2011
Location: Australia
Posts: 935
Posted: 04:48pm 06 Aug 2015
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I think this is the answer.

The power dissipation in the relay if it was run at full voltage is about 1.2 watts. With the 120 ohm resistor that is down to 0.3 W. The actuals are slightly lower due to voltage drop in the transistor Q1. Maybe this lower current stops the relay coil from getting too hot. Also the power dissipation in Q1 is much lowered. I don't know (and haven't checked) the specs on a BC337 but maybe 100mA was too much for it but 50mA is obviously OK.

With the 120 ohm resistor in circuit, the voltage on the relay would be slightly less than 6 volts. Enough to hold the relay on but not enough to operate it. Thus the need for the capacitor which would supply 12 volts for a short time when the relay is powered and then the voltage on the relay would fall to about 6 volts.Edited by BobD 2015-08-08
 
Downwind

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Joined: 09/09/2009
Location: Australia
Posts: 2333
Posted: 05:12am 07 Aug 2015
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I would say its a well designed circuit for low power use, as the latching current for a relay is far higher than the holding current for a relay.

Try the test on the bench with a variable power supply
Latch a relay at the rated voltage, then dial down the voltage till the relay unlatches, and note the power drop (amps) as you reduce the voltage.

The same for the circuit in question it reduces the current for the latched state, in turn limits the voltage to the coil.

The basics is:-

A coil requires a voltage/current of X to latch it.
Once latched it requires a far lower voltage/current to keep it latched.

You will find it is a very predictable drop out voltage to a relay coil, to the point you can dial it in for circuit use as a low voltage cut out switch.

I use to use this function back in the days of Nicad batteries for a pack discharger, to cycle the pack to 1.2v discharge/per cell.

Pete
Sometimes it just works
 
domwild
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Joined: 16/12/2005
Location: Australia
Posts: 873
Posted: 12:41pm 07 Aug 2015
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Thanks for answers. All the circuits I have seen so far drive relays very directly without this cap/resistor combination.

Perhaps more important in battery-driven applications to reduce the draw, not so much here with a mains situation and the relay coil will last longer, I guess.
Taxation as a means of achieving prosperity is like a man standing inside a bucket trying to lift himself up.

Winston Churchill
 
yahoo2

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Joined: 05/04/2011
Location: Australia
Posts: 1166
Posted: 03:35pm 07 Aug 2015
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it is pretty standard to have a 100uF cap to dampen transient voltages along with the diode across 85 and 86 relay terminals for automotive applications.
I'm confused, no wait... maybe I'm not...
 
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