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'[EE]: Latching relay'
2001\12\19@185247 by Tal Bejerano - AMC

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Hi to all

does anyone can send me the formula for latching relay?
I want to hold the a 12v relay for certain time after power-off (about 20
seconds) the relay consume 37mA/12v (I think it's coil resistance is 30-32
Ohm) what is the size of the capacitor I should use?

p.s: across the coil there is a 1n4001 diode does it important for
calculation?

Thanks

Tal

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2001\12\20@062530 by Russell McMahon

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> does anyone can send me the formula for latching relay?
> I want to hold the a 12v relay for certain time after power-off (about 20
> seconds) the relay consume 37mA/12v (I think it's coil resistance is 30-32
> Ohm) what is the size of the capacitor I should use?
>
> p.s: across the coil there is a 1n4001 diode does it important for
> calculation?


At 27 mA and, say 6v drop for a 20 second delay you would need about 4 FARAD
= 4,000,000 microfarad.
This is far too large a load to supply for a long period directly from a
capacitor.
What you need to do is to use some electronics that need much less drive
current to drive the relay.
Then apply the delay to the drive circuit.


If you drive the relay with a MOSFET then by placing a 1 mohm resistor from
gate to ground and a 10 uF from gate to ground you will get a circuit that
will stay on for ABOUT 20 seconds after power is removed from the gate. This
is not an ideal circuit but is a starting point.
A better solution would be a digital one as timing delays ca  be more
accurately established and altered. A very cheap PIC would do this well or
you could use a counter IC and some extra logic. The capacitor will need to
be a low leakage type. A solid aluminium would be OK. A Tantalum cap would
be useable but select a cap voltage 3 or more times higher than the supply
voltage.

The reverse biased diode should not affect operation.


     Russell McMahon
_____________________________

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2001\12\20@091933 by Vasile Surducan

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from a few miliseconds up to 2 minutes circuit delay:

one cmos 4093 nand gate, two rc and one transistor:
input signal active low on one input, the gate output through a
polarised capacitor ( 1uF...to 4.7uF ) connected to the transistor base;
1M resistor and 1n4148 protection diode against reverse voltage on BE
junction, emiter to ground, colector to VDD through a 10K...250K; from
colector to ground a polarised capacitor ( 1uF to 22uF ), from the same
collector point to the other gate input a negative feedback ( one wire )
Output from gate out. For a relay coil load another transistor is
required. Suplementary protection diode from both gate inputs
to Vdd ( cathode to VDD ). Optional a derivating RC on the input so the
lenght of the input pulse will not count ( the trigger will run on hi-low
input pulse transition and at power up )
It's a stable circuit up to 2 minutes. No PIC inside.

Vasile



On Fri, 21 Dec 2001, Russell McMahon wrote:

{Quote hidden}

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2001\12\21@020624 by Anand Dhuru

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How about a small PCB mountable ni-cad?

Anand Dhuru


----- Original Message -----
From: "Russell McMahon" <spam_OUTapptechTakeThisOuTspamCLEAR.NET.NZ>
To: <.....PICLISTKILLspamspam@spam@MITVMA.MIT.EDU>
Sent: Thursday, December 20, 2001 04:56 PM
Subject: Re: [EE]: Latching relay


> > does anyone can send me the formula for latching relay?
> > I want to hold the a 12v relay for certain time after power-off (about
20
> > seconds) the relay consume 37mA/12v (I think it's coil resistance is
30-32
> > Ohm) what is the size of the capacitor I should use?
> >
> > p.s: across the coil there is a 1n4001 diode does it important for
> > calculation?
>
>
> At 27 mA and, say 6v drop for a 20 second delay you would need about 4
FARAD
> = 4,000,000 microfarad.
> This is far too large a load to supply for a long period directly from a
> capacitor.
> What you need to do is to use some electronics that need much less drive
> current to drive the relay.
> Then apply the delay to the drive circuit.
>
>
> If you drive the relay with a MOSFET then by placing a 1 mohm resistor
from
> gate to ground and a 10 uF from gate to ground you will get a circuit that
> will stay on for ABOUT 20 seconds after power is removed from the gate.
This
> is not an ideal circuit but is a starting point.
> A better solution would be a digital one as timing delays ca  be more
> accurately established and altered. A very cheap PIC would do this well or
> you could use a counter IC and some extra logic. The capacitor will need
to
{Quote hidden}

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'[EE]: Latching relay'
2002\02\02@233308 by Richard Graziano
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If the Relay is a DC relay, then a capacitor accross the coil will hold it
in after power is removed because the capacitor will discharge through the
coil.  The larger the capacitor, the longer the hold time.  The picking time
will also be delayed if that is imporant. Also, you will need to remove the
diode or make sure that it is reversed biased with respect to the capacitor
polarity.
{Original Message removed}

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