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I have to programmatically switch a 50 W load (20 V, 2.5 A) and I'm using a Bosch DC relay to do it (exact part is pictured; see page 6 of the datasheet).

Bosch relay with diode

When I apply 12 VDC the relay clicks, so that works. The issue is that the coil and contacts get very hot (can't touch with a finger) after 5-10 minutes of operation. I need this to operate 24/7, unless I power it off eventually and power on again.

Measured coil resistance is 90 Ω, so according to some formula I've found it's V2/R = 2 W of power dissipation.

What am I doing wrong? Is it a faulty part? Should I put a resistor in series to increase resistance? I better replace it with another relay but I would prefer going with a dedicated part, not a module (for my application space is a big concern).

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    \$\begingroup\$ I havent found a datasheet (if you have one you should post it), but what I have found suggests that the relay is rated for 12V so that should be fine. You should check the polarity that you are driving it at though, since there is a diode that will conduct if you connect it the wrong way round. What is the voltage/current that you are switching? \$\endgroup\$
    – BeB00
    Oct 28, 2022 at 23:05
  • \$\begingroup\$ @BeB00 there's a link to datasheet now but it doesn't say much \$\endgroup\$
    – user37741
    Oct 28, 2022 at 23:28
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    \$\begingroup\$ A brief look at a similar relay from another manufacturer shows a rating of 1.2 to 1.5 W coil power, so it looks like your 2 W observation isn't that unusual. \$\endgroup\$
    – Hearth
    Oct 28, 2022 at 23:40
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    \$\begingroup\$ I wonder if it's designed to sink heat through the terminals. Seems doubtful, it would most likely be stuck in a relay/fuse box surrounded by still more plastic -- but stranger things have happened. Also if it's designed for engine bay service, it could simply be rated much hotter than fingers can go, and self-heating isn't a problem. (But just a possibility: @Hearth's similar datasheet gives 85C, fairly low as electronics are concerned.) \$\endgroup\$ Oct 29, 2022 at 0:20
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    \$\begingroup\$ @user37741 - Hi, I don't know if this is related to your question - hence it's a comment: I noticed that you are switching a 20 V, 2.5 A load (you didn't mention the type of load e.g. resistive, motor etc. and that does affect the required rating). That 20 V might be a concern. These relays are typically for automotive use e.g. 12 V coil & 12 V load. Look at the first page of the datasheet you linked - it says: "Rated voltage (load and excitation circuit) [...] 12 V" so your 12 V relay isn't rated for a 20 V load. \$\endgroup\$
    – SamGibson
    Oct 29, 2022 at 17:44

4 Answers 4

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Should be good for 85°C ambient and 16VDC on the coil => 78% more power into a an ambient 60°C hotter compared to 25°C and 12VDC (in reality it won't be that bad for long since the coil resistance will increase with temperature by ~0.4%/K).

"Can't hold a finger" is typically >60°C so that would imply a surface temperature in perhaps the 110°C range worst-case. I suppose that's plausible, internal coil temperature would require high temperature insulation, maybe class F.

You could consider pulling the relay in with 100% and then backing off to a PWM-controlled output after perhaps 100 milliseconds. That might not require any additional components at all.

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    \$\begingroup\$ I went with PWM in the end. I put a transistor in the cable for the PWM signal and now relay is holding at 35% duty cycle while consuming only 100mW! \$\endgroup\$
    – user37741
    Nov 1, 2022 at 14:18
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You should get 1.6 W at 12 V and 90 Ω.
Sometimes a series resistor helps if you add a big capacitor in parallel to the resistor. This way the coil initially gets the full 12 V and the voltage drops while the capacitor charges up. Try 47 Ω and 220 μF for a start.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ wouldn't capacitor charge and discharge all the time? if I got you correctly \$\endgroup\$
    – user37741
    Oct 29, 2022 at 10:41
  • \$\begingroup\$ @user37741 The capacitor discharges when the relay is turned off and charges up to around 4 V after relay turn on. This is a solution for one or less switching cycles per second, because the capacitor must fully discharge during relay off periods. \$\endgroup\$
    – Jens
    Oct 29, 2022 at 16:44
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It shouldn't get extremely hot at 2W, it will get hot though. Make sure you are not driving the diode (the relay has polarity). The relay will get very hot if you drive the diode.

Sometimes you can do whats called a 'hit and hold', the drawback is you need two switching circuits. Usually the coils will take -say- 12V to actuate and then something like 8V to hold. There is nothing in the datasheet about this so YMMV. You can experiment with this if you have a bench supply and turn on the relay and then lower the voltage until it switches off, if you go 1V above this it should remain on. If you do that you could try hit and hold circuits to lower the power.

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  • \$\begingroup\$ 2W is quite a lot of heat for a small space. \$\endgroup\$ Oct 29, 2022 at 0:30
  • \$\begingroup\$ What you call a "hit and hold circuit" is something I've generally seen called a "relay economizer", if you want another search term. \$\endgroup\$
    – Hearth
    Oct 29, 2022 at 3:21
  • \$\begingroup\$ I've connected +12V to the 1st contact (cathode side), as per datasheet/diagram (where + sign is). relay economizer might be a solution indeed, but that's probably gonna be a PCB I don't really have space for \$\endgroup\$
    – user37741
    Oct 29, 2022 at 10:42
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I agree with the others that if the relay says it's rated for 12V, then it will probably last, even though it gets hot.

That being said, according to the datasheet it will operate at 8V, so why not try it at a lower voltage? At 8V it will dissipate only 0.7W.

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