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Simple relay driver circuit - Will this be reliable in the long run?

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Please see attached schematic

K1 will be energized almost continuously dropping out perhaps 25 times a day for 2 seconds coil current is ~130mA

Q1 is rated 45V, 500mA, and will have a fair bit of copper soldered to pin 2
D2 is rated 100V, 300mA
Ambient temperature is typically 70F, could reach 100 on a hot day.

Will this circuit prove reliable in the long run 10-20 years?
Are Q1 and D2 adequately rated or should I use parts with higher current rating?
Relay drive.png
 

Is your relay rated for 182,500 operations (20 years)?
What's the current rating of the contacts, and how much do they carry?

I wouldn't worry about the diode or transistor, I'd worry about the relay. You should be able to calculate temperature rise of the diode and transistor and determine if they need heatsinking.
 

barry said:
Is your relay rated for 182,500 operations (20 years)?
What's the current rating of the contacts, and how much do they carry?

I wouldn't worry about the diode or transistor, I'd worry about the relay. You should be able to calculate temperature rise of the diode and transistor and determine if they need heatsinking.
Click to expand...

The relay is rated for > 200,000 operations at rated current and this will be used at a fraction of that, the focus of my question is the drive circuit.

data sheet for Q1 https://www.mouser.com/ds/2/302/BC817_BC817W_BC337-840614.pdf
data sheet for D2 https://www.mouser.com/ds/2/115/ds30086-60872.pdf
 

Hi,

You don't give relay coil current.
-->
* bjt continous current rating should be larger than relay current
* bjt power dissipation should be low, i recommend a temperature rise smaller than 20K
* diode peak current rating should be larger than relay current

R2 is useless.

I see no problem in your circuit.

Some hints for improvements. Not necessary, but maybe you find it usefull:
* use a relay with NC contacts to save power
* you could use a "digital bjt" with resistor inside to reduce part count
* or you could use a small mosfet to reduce transistor power dissipation. (Not overall power dissipation)

To reduce power with your configuration (2s OFF):
* switch the relay ON with 1s and 100% duty cycle, then reduce duty cycle to 50% (check datasheet).
This reduces power dissipation and heating of the relay coil.
No need for high frequency. Everything above some 100Hz is OK.

Klaus
 

The current supplied by the out pin should be sufficient to switch on the transistor.
(approx 0.5ma to 1mA) It is not shown whether the out pin is from a microcntroller or else.

The sourcing capability of the source "out" should be checked.
In the continiuous rating the heating effect of the controller is to be considered.

If you have the choice, make a "0" to turn ON instead of "1" to do that.
 

KlausST said:
Hi,

You don't give relay coil current.
-->
* bjt continous current rating should be larger than relay current
* bjt power dissipation should be low, i recommend a temperature rise smaller than 20K

* diode peak current rating should be larger than relay current

R2 is useless.

I see no problem in your circuit.

Some hints for improvements. Not necessary, but maybe you find it usefull:
* use a relay with NC contacts to save power
* you could use a "digital bjt" with resistor inside to reduce part count
* or you could use a small mosfet to reduce transistor power dissipation. (Not overall power dissipation)

To reduce power with your configuration (2s OFF):
* switch the relay ON with 1s and 100% duty cycle, then reduce duty cycle to 50% (check datasheet).
This reduces power dissipation and heating of the relay coil.
No need for high frequency. Everything above some 100Hz is OK.

Klaus
Click to expand...

Thanks to all

"coil current is ~130mA"

I calculate 4 degree rise
Q1:
VCE sat = 0.05V, so Pd = 0.13 * 0.05 = 6.5mW
Rth = 625 deg/W, so temp rise = 0.0065 * 625 = 4 degrees

It just seems like a really small transistor to drive a big relay

Peak diode current rating is 1A
 

brushhead said:
Just a word of caution using NC contacts don't present a 'fail safe' state.
Click to expand...
Also NC contacts are more sensitive to vibration because of limited contact pressure.
 

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