Relay coil energise by smoothing capacitor

Hi,

I am trying to implement a circuit that will energise the relay coil.

Attached is the screen shot of the LT spice simulation.

instead of relay coil I have used a inductor with 390uH and 18 ohm resistance, 1A current capability.

instead of using a load resistor with 18ohm i have used a inductor.

The spec of relay coil is 9 - 36V and 18ohm resistance.

my questions are

1. I am using a 24V to energise the coil and does it means 24V/18 ohm = 1.33A. So my input power should be able to deliver 1.5 A at 24V?

2. in the lt spice the blue trace is only 3.6V when simulating because if the 18ohm load on the inductor coil. So what does the 3.6V represent? will there be voltage drop of 20V and 4 volt approx will be still available in the capacitor when the coil is energised since the 18 ohm of the coil act as a load resistance ?

Don't see the purpose of diode, capacitor, zener diode, parallel resistor. What do you want to achieve with the complex circuit?

Presume the "1A current capability" is your arbitrary choice and not substantiated by the relay spec?

By applying ohms law, you'll understand why the relay voltage can't achieve the specified minimum of 9V and thus the relay won't be ever activated.

You may use a series resistor to reduce the coil voltage and current and respectively power dissipation, but not below 9 V, and probably not below 9V + a certain margin, e.g. 11 or 12V. In so far, an 18 ohm series resistor might be used under circumstances.

- - - Updated - - -

To energize a coil by a capacitor, there must be a switching element and some kind of sequential action like firstly charging the capacitor, then connecting it to the coil. You also need to know the minimal pulse width required for activating the relay coil. What's the intended action after the capacitor has discharged to the coil?

Hi,

I assume the relay specification is wrong. 24V 18 Ohms means 32W of power (heat) disssipatin. This is more than my soldering iron needs.
Are you sure this is a DC relay?

What´s the idea behind the capacitor? Can you give a description about what you want to achieve?

Klaus

Hi,

See the attached image. it's some kind of the brake circuit implementing for a 3 phase motor.

yes I am sure about the relay as it can handle 310A of current and it s latching type relay.

Why not mention latching relays type in the first place?

The final circuit is implementing the sequential action assumed in my previous post, if you want to simulate it's operation, change your Ltspice circuit respectively. You should be aware of pulse width requirements, check the relays specification.

FvM said:

Why not mention latching relays type in the first place?

The final circuit is implementing the sequential action assumed in my previous post, if you want to simulate it's operation, change your Ltspice circuit respectively. You should be aware of pulse width requirements, check the relays specification.

Click to expand...

Hi,

in my simulation when the relay coil is not connected I see a voltage of 24V measuring at the capacitor positive.

When the coil is connected i see a voltage of 3.5V which I hope is because of the 18ohm resistance of the relay acting as a load. am i correct ?

So is that not enough to activate the relay ?

I have got no info about the PWM of the relay.

I have updated my circuit switches as per the attached picture.

Hi,

Now with the new circuit and the "latching relay" information ...it makes a lot more sense.

When the coil is connected i see a voltage of 3.5V which I hope is because of the 18ohm resistance of the relay acting as a load. am i correct ?

Click to expand...

It´s a simple voltage divider (for DC voltages).
24V - 100R - 18R GND
V(18R) = 24V * 18 Ohms / (100 Ohms + 18 Ohms) = 3.66V

This is DC state condition that never is true for the original circuit.
It´s only true for your ltspice circuit.

Klaus

Hi,

It a latching relay type as shown in the attached image earlier.

can any one please explain me the working principle.

from my understanding the relay gets energised by the capacitor and K2 will be open.

when the k2 is closed the relay click and moves from NC to K3.

Now what happens ? does the relay come back to the NC position again. how is it possible with out the switch ?

Hi,

impossible to answer without knowing K2 and K3 functions.

During normal operation C1 becomes energized

I assume on power_loss K3 becomes de-energized and falls back to normal state.
Now the energy - previously stored in C1 - makes K1 to become activated.

Klaus

K2 is like a tactile switch

K3 is another relay contact. Initially the coli is connected to NC of K3.

In your hand drawn schematic you show k3 contacts as normally closed. Are they actually normally closed. Or normally closed because the module is keeping the coil energized.

K3 contact is normally closed as the module is keeping the coil energised.

The coil is energised when the tactile switch K2 is closed.

Unless you link all the component specs and circuit parameters, it is guess work.

Hi

The K1 relay and R1 resistor is mentioned in the hand drawn sheet.

The K1 coil voltage that we are planning to energise with is 24v.

The coil resistance is 18 ohm.

The aim is to store sufficient energy to allow brake contactors latching coil to be activated when the power is lost.

provide control to allow latching brake contactor coil to be reset. (connect reverse polarity).

provide control a feedback interface to main system PLC.

Rest of the components I need to select by myself considering in mind that coil voltage is 24V

insufficient data... lost power for how long? what about brown out? Can you compute Energy Required 1/2LI^2 then maybe CV^2 with SCR trigger

vikash23 said:

Hi,

It a latching relay type as shown in the attached image earlier.

can any one please explain me the working principle.

Click to expand...

From my experiments with latching relays...

The contacts close when you send a brief DC pulse in one direction. Nothing happens if you send more pulses in the same direction.

The contacts open when you send a brief pulse in the opposite direction.

To operate a latching relay, your circuit must be designed to send current in either direction through its coil.

The latching relay uses a permanent maget in each state to hold the contacts.

A coil with a magnetic force and pulsed current is capable of pulling the armature contacts away from the permanent magnet with much higher strength to accelerate with sufficient velocity that the transition time bewteen states to minimize contact chatter or perhaps the arc time for inductive currents.
en-g5rl_uk.pdf G5RL-K1A-E-DC12 Omron Electronics Inc-EMC Div | Relays | DigiKey

Drivers
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you have a choice between dual coils or single coil which requires dual low side drivers or a dual half bridge or full bridge.

The flux in the solenoid action must change direction. This can be a single coil with a center tap common to the V+ supply with either winding shunted to ground to force the momentary acceleration ( hopefully in the opposite direction of the current state. THese are usful for low side dual drivers using common emitters a.k.a. open collector or open drain switches to ground. ( or smart "low side switches" with thermal protection.

The switched current starts at zero for the inductive load then rises quickly with L/R time constant as it accelerates to the opposite state, where at maximum velocity counter EMF is produced and thus current is miniimum until it slams again the end stop and then current ramps up to the V+/DCR resistance of the coil again.

Since these types of relays have greater magnetic force needs, the resistance of the coil and thus current is a much higher percentage of the maximum contact thermal ( resistive ) current ratings.

It is common for DC relays to have 12V coils and 240V contact ratings with a coil current / contact current ratio of 1:2000 compared to a transistor this may seem like a pretty high current gain and it is, but not for a FET switch. However for Latching ratios, this RATIO is much lower due to the forces required and thus the time for the operation must follow the device specs but consider the size of the part , self heating applied/rated voltage ratio and applied/required time to complete the transition.

This transfer can be defined in terms of energy so that sufficient stored energy is kept with a AC charged storage cap to dump the energy to transfer the power to the relay coil.

I dont know the exact formula but the basics are integrated RMS energy min. E= V(t)*I(t)*t vs stored inductive energy + losses Pd= 1/2LI^2 + I*DCR*t and similar for the Cap. Pc = 1/2C(Vi^2-Vf^2) + I(t)^2*ESR where final voltage results in no force against opposing permanent magnet.