Actuating a power relay takes more power than HOLDing it?

[cupoftea]

Hi
This relay takes 1.8W to actuate it...but only 0.17W needed to HOLD it ON.

https://www.panasonic-electric-works.com/cps/rde/xbcr/pew_eu_en/ds_hes_en.pdf

...Are there any good modules offtheshelf for activating it with the 1.8W worth of coil current, and then having a little delay, before reducing the current to 0.17W worth of coil current?
...Or do we have to make our own such circuit out of discrete transistors and RC delays?
Do you think its a bit dodgy (in any way) to reduce the current to the HOLDing current level?

 

[KlausST]

Hi,

Be careful not to go to extremes.
When there is too low "holding" currend, then on high mechanical vibration the relay may still drop.
Don´t know how critical the relay works and how critical your application is.

***
To reduce the holding current, just drive it with PWM and about 35% duty cycle.
So basically every microcontroller can do.

There are low side driver ICs that can do ON/OFF/PWM so you don´t need 8 independent microcontroller PWMs for 8 channels.

***
I´ve done this with magnetic valves for air. Installed in a small box they got very hot without power reduction and they were pretty cool with power reduction.

Klaus

 

[cupoftea]

Thanks, may i ask what is the best PWM frequency for this? Unfortunately the relay datasheet does not give the coil inductance?.....i suppose i can buy one and measure it though.
Also, since the relay coil is being turned of and off...do you think the auxiliary coil would make/break in this PWM.....because if it was able to, then it could be used to see if the main contact was welded or not?...(without actually turning the relay properly off)

(the auxiliary contact is ususally opposite of the main contact (ie its on make when mains contact is on break)...but when welded, the aux contact is always on "break".

The relay of the top post dissipates 1.8W in the coil if not PWM'd.....but the datasheet doesnt allure to any corresponding lifetime reduction due to this power and heat.....do you know what the lifetime reduction might be?

 

[KlausST]

Hi,

Frequency: depending on inductance maybe even some 100Hz could work...
The higher the less current ripple but the higher the EMI.

the auxiliary coil

you mean the auxiliary contact? It surely must not open as long as the relay is in "hold" state.

do you know what the lifetime reduction might be?

The lifetime is given in count of contact operations. It does not depend (much) on power reduction.
Aging depends on temperature, like oxidation, or drying out of electrolytic capacitors...thus the lieftime of a whole device may be extended by the decreased temperature.

It´s said that a 10C temperature decrease doubles lifetime.

Klaus

 

[cupoftea]

Thanks, do you know what situations might increase the contact resistance of the relay in the top post?....ie the following relay?....

https://www.panasonic-electric-works.com/cps/rde/xbcr/pew_eu_en/ds_hes_en.pdf

eg humidity,
switching very low current
etc etc?

 

[KlausST]

Hi,

ultrasonic (cleaning), silicone oil around (will find it´s way to the contacts), acid residuals in air, dirt, dust...

Klaus

 

[FvM]

switching very low current

It's a power, not a signal relays. Minimal switching current/voltage is specified as 100mA/5V for the main contacts.

 

[BradtheRad]

It's mechanical springs which must be overcome. Once the contacts close, you can apply reduced voltage to the coil and they remain closed. However the springs are always waiting to pull the contacts open as soon as voltage drops too low.

There is the latching relay, which closes by means of a single pulse to the coil. No current is required to keep it closed. When you wish to open it, send a pulse to the coil in the opposite direction.

 

[crutschow]

Below is the LTspice simulation of a relatively simple PWM hit and hold circuit for a solenoid or relay using one CD4093 chip.
You didn't say what the relay coil voltage is so you will have to adjust the power supply voltage to that value.
(The CD4093 maximum nominal voltage is 15V, so if the relay requires more than that, you will need a separate supply for the CD4993).
Pot U5 adjusts the PWM duty-cycle and thus the hold current you want.

 

[barry]

Waaaaay simpler than PWM is to just power the relay through a resistor with a large cap in parallel. When first powering the relay, the cap looks is a low impedance, and you'll get a big inrush current. Then, as the cap charges, the current into the coil will drop. You just have to size your R and C properly.

 

[KlausST]

Hi,

True, it is simpler but less effective.
Reducing the coil current to 30% with an ideal PWM circuit reduces the overall power (dissipated as heat) to 9%.
Reducing the coil current to 30% with a resistor (diode, transistor...) reduces the overall power to 30% only. This is 3.3 times than with PWM.

Klaus

 

[barry]

Hi,

True, it is simpler but less effective.
Reducing the coil current to 30% with an ideal PWM circuit reduces the overall power (dissipated as heat) to 9%.
Reducing the coil current to 30% with a resistor (diode, transistor...) reduces the overall power to 30% only. This is 3.3 times than with PWM.

Klaus

Less effective? It really depends on what's important to you: power efficiency or cost and reliability.

 

[KlausST]

Hi,

Less effective? It really depends on what's important to you: power efficiency or cost and reliability.

I clearly talked about power efficiency.
And no, "it does not depend what's important for me" --> it depends on what's important for the OP. ;-)

Klaus

 

[dick_freebird]

There used to be peak/hold driver piece parts for
this kind of thing, not sure whether any survive.
The old ones were dissipative (like, bypass a
"hold" current setting resistor briefly, and then
not) so not as efficient as a PWM approach might
be. But much less ancillary components than a
switcher.

 

[c_mitra]

It is good to follow up with a real-life example.

Consider a 5V relay with a coil resistance of 20 Ohms. This is a reasonable value from the datasheet given (5V is more common than 6V and the relay will work reliably at this voltage).

The power dissipated in the relay coil is 5*5/(20)=1.25W; some gain is obtained by reducing the operating voltage (at 6V the coil dissipates 1.8W).

To reduce the holding current, we put a resistor in series (a la post #10); the cap capacity is left for the posterity.

To reduce the holding current to 40% (just to keep some margin), we select a resistor 27 Ohm (you can also select 33); the total resistance now is 47 and the power dissipated is now 0.531W (with the relay current 106 mA and holding power of 0.226 mW; more that the specified power).

Wasted power will be slightly more than 300 mW and the designer must decide whether this is acceptable or more complex design needed to reduce the waste.

This is a power relay and I guess the designer has access to other power sources.

 

[KlausST]

Hi,

What is needed to drive a relay ON and OFF?
A transistor, a free wheeling diode...If a standard BJT then usually additionally a base resistor.

Driven ON/OFF/PWM:
If driven by a microcontroller: no additional components.

When driven with an RC:
additional components: power R, C.
With c_mitra´s example:
* 27 Ohm resistor, continous power dissipation of 0.32W. -> use a 0.5W or better 1W rated one.
* for a tau of 100ms: 20ohm || 27 Ohm = 11.4 Ohm --> C = 0.1 s / 11.4 Ohm = 8.7mF or 8700uF
(check capacitor lifetime)

As mentioned in post#2: There are relay / valve drivers with internal free wheeling diode, SPI interface and the feature to choose each output to be ON, OFF or PWM.
Signals from uC to driver: 3 wires SPI for 8 outputs, one common PWM.

Klaus

 

[crutschow]

Less effective? It really depends on what's important to you: power efficiency or cost and reliability.

The TS mentioned power dissipation, so I assume that is important to him.