Relay latching and getting stuck - Page 1
Forum section: electronics / projects

author: dansku
author_rank: Regular Contributor
timestamp: July 30, 2016, 09:17:26 am
content: Hey I am using a relay to turn ON/OFF a DC signal that run on 55 volts (low current).The problem that I am having is that after toggling a few times, it is getting stuck and not releasing anymore.When I touch the two ends of the cable by hand there's a small spark due the high voltage. I am wondering if that could the problem, or in-rush current or something else.I am using this relay:http://www.mouser.com/ds/2/307/G6C_0911-14342.pdfThis is the circuit:I was wondering if anyone has any experience with this, could adding a inductor in series with the circuit do any good?Or if I should use a Solid State Relay, would it be better? Any advices there?Thank you

author: MosherIV
author_rank: Super Contributor
timestamp: July 30, 2016, 09:33:06 am
content: The relay is not rated for the DC voltage that has been applied and the contacts have welded together.Use a relay that is rated for the DC voltage.

author: johnkenyon
author_rank: Regular Contributor
timestamp: July 30, 2016, 09:35:28 am
content: Assuming that the problem is caused by the relay contacts welding together...If the relay is driving an inductive load, then you'll need higher rated contactsA snubber across the relay contacts - resistor and capacitor in series may help.

author: dansku
author_rank: Regular Contributor
timestamp: July 30, 2016, 09:44:33 am
content: Thank you, I think I misread the datasheet?Max switching voltage to 125VDC, where am I checking it wrong?Thanks

author: tatus1969
author_rank: Super Contributor
timestamp: July 30, 2016, 09:45:37 am
content: also a capacitive load can cause contact welding. Can you tell us more about the load that you switch here?

author: dansku
author_rank: Regular Contributor
timestamp: July 30, 2016, 09:49:37 am
content: Quote from: tatus1969 on July 30, 2016, 09:45:37 amYes it is a capacitive load! I am turning on an electric vehicle controller, which has some big capacitors in it.

author: Gyro
author_rank: Super Contributor
timestamp: July 30, 2016, 09:53:24 am
content: You could maybe try a low value resistor (<1 ohm) or NTC thermistor in series.

author: johnkenyon
author_rank: Regular Contributor
timestamp: July 30, 2016, 10:21:46 am
content: Quote from: dansku on July 30, 2016, 09:49:37 amSo a large inrush current... Higher current rating for the relay or a resistor or NTC as suggested by Gyro.

author: Ian.M
author_rank: Super Contributor
timestamp: July 30, 2016, 10:22:18 am
content: In addition to using an appropriately rated relay, to minimise the risk of the contacts welding due to arcing during the period of contact bounce on open or close, if the load is inductive, you should add a RC snubber across the contacts, sized to divert the max load current for a time-constant of whichever is longer of the contact bounce time or the opening time, ad for the max voltage across the contacts at opening not to exceed the relay's max voltage rating. If the load is capacitive, you may need a NTC thermistor in series with the contacts to keep the inrush current within the relay's ratings, but this is much more problematic if the off time is short., and may require another time-delayed relay to short it out so it can cool down.Also the simple anti-parallel diode to clamp the back EMF slows down the decay of the magnetic field holding in the armature. If you add a 12V Zener in series with the diode, (cathode to cathode) it will cause the field to decay a lot faster, which results in 'snappier' contact opening. The down-side is that the transistor will see a peak Vce of double the supply voltage.In the case of your original relay, and a 1N4148 diode, it will take 3ms for the coil current to decay by 90% to the minimum dropout current. Adding a BZX84C12L reduces that to 0.9us. As the typical release time is 2ms, this makes a significant difference.

author: tatus1969
author_rank: Super Contributor
timestamp: July 30, 2016, 10:31:13 am
content: Quote from: dansku on July 30, 2016, 09:49:37 amImagine that your power source has a reasonable output capacitor that is fully charged, and your EV controller has an empty input capacitor. As the relay effectively shorts them out, the current is only limited by wiring, ESR and alike, and can be hundreds of amps. If your load current is not too high, then as mentioned: resitor. Otherwise NTC.Long time ago I did a two-step approach to switch a large transformer: one relay connects the load via a NTC and "pre-charges" the load, after a short delay a second relay shorts out the NTC.

author: danadak
author_rank: Super Contributor
timestamp: July 30, 2016, 12:28:07 pm
content: You also might not be saturating Q1. Rule of thumb is Ib = Ic / 10to saturate it. If you are coming from a UP there will not be enoughdrive to saturate the transistor.So you can use a darlington transistor or a MOSFET.Additionally the R from B-E on the transistor should be ~ 10 x theR from UP to transistor base, to maximize drive to bipolar transistor.Its primary function is to absorb leakage and prevent transistor turnonwhen the UP output is tristated.Regards, Dana.

author: Ian.M
author_rank: Super Contributor
timestamp: July 30, 2016, 12:38:25 pm
content: The 12V coil version of the O.P's original relay nominally draws 16.7mA so even if the MCU has 3.3V I/Os that can only source 1mA, any transistor with HFE>100 would saturate with plenty of margin, even without increasing R4.

author: danadak
author_rank: Super Contributor
timestamp: July 30, 2016, 04:07:19 pm
content: You should be fine with the 12 V coil. I had thought worst case you were on 3.3 coil, reasonfor my prior post.Regards, Dana.

author: Ian.M
author_rank: Super Contributor
timestamp: July 30, 2016, 04:47:08 pm
content: No, its still worth mentioning, as the O.P. has to upgrade the relay, and one with a higher DC rating is likely to draw more coil current due to the need for more contact travel to break the arc effectively.

author: MosherIV
author_rank: Super Contributor
timestamp: July 30, 2016, 06:13:15 pm
content: QuoteSorry for the delay, busy during the day.The spec says the relay contacts are only rated for 30VDC, you are applyjng 50VDCYes, the relay may be able to withstand 125V surge during switching (probably for inductive load back emf)not 50VDC continuous.

author: calexanian
author_rank: Super Contributor
timestamp: July 30, 2016, 07:38:51 pm
content: What is the exact part number of that relay. I see it is available in a latching coil version. If you have that, it would explain what is happening. They are designed to "Latch" On and require either a reverse current pulse to release, or a pulse on a second release coil depending on the type of part, and this particular part comes in both versions. A little more info is required. As for as capacitive loads are concerned on the contacts, that has to be determined in your particular application. Generally it is not a big problem. Obviously if you have many thousands of microfarads worth of capacitance you are going to need to address it, but for just general purpose use, saw switching on a large piece of equipment, you should have no problem. Inductive kickback and arcing during the opening of the relay is typically a larger problem for relays.

author: tatus1969
author_rank: Super Contributor
timestamp: July 30, 2016, 07:51:57 pm
content: Quote from: MosherIV on July 30, 2016, 06:13:15 pm30VDC is the limit for breaking 10A because of arcing. He wrote that the load is light, so the relay is eore than appropriate. Only thing to be solved is capacitive load handling.

author: Zero999
author_rank: Super Contributor
timestamp: July 30, 2016, 09:11:51 pm
content: Why use a relay?If isolation isn't needed, then a BJT or MOSFET could do the job, without any fear of arcing.If it does need to be isolated, then use a solid state relay or make your own with an opto-coupler and MOSFET.

author: tatus1969
author_rank: Super Contributor
timestamp: July 30, 2016, 10:48:09 pm
content: Quote from: Hero999 on July 30, 2016, 09:11:51 pmYou still need to care about inrush current for capacitive loading - look for single-pulse avalanche energy, if the transistor's die is too small then it instantly overheats during the switching process. The switching process is normally fast enough that heat transfer to case/heatsink cannot even start, and the die stores all thermal energy that is created during the switch. If wiring is low impedance and both supply and input capacitors have low ESR then your transistor vaporizes itself.

author: Zero999
author_rank: Super Contributor
timestamp: July 31, 2016, 03:26:56 pm
content: Quote from: tatus1969 on July 30, 2016, 10:48:09 pmYes you're right but there are things you can do with a transistor to limit the current which can't be done with a relay.

author: Seekonk
author_rank: Super Contributor
timestamp: July 31, 2016, 06:28:38 pm
content: If you are going to use a relay on DC outside of its intended voltage, you certainly don't want to use one with Ag contacts. Tungsten will give you a fighting chance. I've seen some use multiple contacts in series.

author: tatus1969
author_rank: Super Contributor
timestamp: August 01, 2016, 08:05:53 am
content: Quote from: Hero999 on July 31, 2016, 03:26:56 pmAgree, I wanted to put this as a remark in general, I would also use a transistor here. Let me share my experience in more detail:Application was 25V / 15A, with a Lipo as the source and ~500uF (low ESR) at the load. Transistor was soft-switching in ~ 1ms. The transistor (SO8 package, I dont remember the rating but I surely had chosen 40V / > 50A) that only switched a few times before breaking (shorting D-S). Then realized that the switch process is overheating the transistor. Then, by simulations, I realized that I cannot solve this by extending the soft switching time (unless I go into the range of > 100ms to allow heat transfer to case/ambient start developing, which was also impractical because it was a motor driver that would already have started spinning the motor under load, then I would again have blown the transistor). I realized that the amount of energy that is dumped into the transistor does not depend a lot on the soft start duration. The only solution was to use a transistor with a larger die size and thus thermal capacity. The thermal capacity is mostly given as the single avalanche pulse energy, so this is the number to look at. I ended up with a (more expensive) transistor that could handle ~ 500mJ of energy.Attached are two simulations that illustrate why extending the soft start period does not help (green: voltage at load, magenta: power at transistor).240us soft start --> 120mJ1.5ms soft start --> 150mJEnergy is even increasing with time, because in my example the transistor also has to supply the load while still in its linear region.

author: Zero999
author_rank: Super Contributor
timestamp: August 04, 2016, 07:58:53 am
content: Quote from: tatus1969 on August 01, 2016, 08:05:53 amDid you try implementing current limiting, using a sense resistor and negative feedback to reduce the gate voltage if the current is too high?

author: tatus1969
author_rank: Super Contributor
timestamp: August 04, 2016, 09:23:25 am
content: Quote from: Hero999 on August 04, 2016, 07:58:53 amMy attached circuits are just doing that, by raising the gate voltage slowly using high ohmic gate resistors. (They are of course over simplified but show the principle.) As you see from the graphs, this is already quite close to an ideal constant current circuit.My original problem (motor inverter as the load) does not allow reducing the charging current too much, as the load will start drawing significant current after booting. I think this is common for many applications (but not for the one that dansku originally posted, he said that his load is not drawing lots of current so soft starting would be appropriate).What I wanted to point out is to be careful with soft switching large capacitive loads: increasing the soft switching time doesnotreduce the stress to the pass transistor - as long as you stay in the time region where heat flow to base/heatsink does not play a role. We could probably derive a model based on the thermal impedance and avalanche characteristics of a particular transistor, my first guess would be that that time region ends somewhere between 10ms and 100ms.

author: dansku
author_rank: Regular Contributor
timestamp: August 04, 2016, 10:22:44 am
content: Thank you all for the answers, my circuit needs to be isolated. In the end I went with the RC SnubberWith a 10 ohm resistor and a 330uF Capacitor with a time constant of 3ms. I am wondering if I should make the time smaller and also what should be the power of the resistor.The max power on the resistor is 250W, but its for a super small time.Any suggestion on this part? I am doing some readings online about it.Thank you!!!

author: dansku
author_rank: Regular Contributor
timestamp: August 04, 2016, 11:00:50 am
content: Check if my ideas are correctI found this resistor for example:http://www.mouser.de/search/ProductDetail.aspx?R=0virtualkey0virtualkeyAS25J10R0ETIt's a 1.5W / 10ohm and the pulseWhat do you think?

author: tatus1969
author_rank: Super Contributor
timestamp: August 04, 2016, 12:13:16 pm
content: Quote from: dansku on August 04, 2016, 11:00:50 amAny resistor that has enough wattage to deliver the current to your load should be suitable. In order to calculate this, we need to have a rough figure of the current consumption of your load. The bulk capacitance that it has would also be important to thermally estimate the switch process.For example: I assume that your load has a maximum current consumption of 100mA at 55V. You want to limit the inrush current to a value that is below the relay's contact rating. As good engineering practice, I would not go to the limits but use half of that rating instead, especially because you are also operating it above its voltage rating.Your relay has a rated current of 10A for DC, so lets limit the current to 5A.The resistance value would be 55V / 5A = 11 ohms, your choice of 10 ohms should be good then.The power dissipation would be 10ohms * 100mA * 100mA = 0.1W. Double this again to be on the safe side. So you could safely choose a 1/4W leaded resistor here. But you should know your maximum load current well here. If you are in doubt, use an NTC instead, because it is protecting itself just in case. But be aware that an NTC can also act as a "resettable fuse". If it "trips" (heats up too much) during the switch, or during overcurrent, then it can stay hot because of enough static current draw of your load, and you need to completely disconnect the load in order to "reset" the NTC.I also assume that you have 1000uF of capacitance at your load. LTspice (attached) tells me that the resistor has to "digest" 1.55 joules of energy during the switch transition. A quick googling for a rough figure of the thermal capacity of a 1/4 watt resistor gave me this:http://electronics.stackexchange.com/questions/131808/whats-the-thermal-time-constant-capacitance-of-a-1-4w-resistor. They calculate 18mJ / K for that part, the math looks good. If I take that here (although I think that the number is too optimistic because the metal film will heat up more than the ceramic substrate), I get an instantaneous temperature rise of 1.55J / (18 mJ/K) = 86 K. Add max ambient temperature to this, lets say 50°C, then you get the max temp of the resistor = 50°C + 86K = 136°C. That should be okay for any 1/4W part. You may choose a 1/2W resistor again to be on the safe side (needs some more research go get a good number for the heat capacity).Remember, all these equations base on my guess of 100mA and 1000uF.Edit: regarding your chosen resistor: we can derive the thermal capacity from its datasheet. The single-pulse graph shows that the device can handle 4000W for 100us. That is probably valid for ambient = 25°C, although they missed to state the conditions. The max operating temp is 155°C, equivalent to a single-pulse temperature rise of 130K. From this, we get the heat capacity = (4000 * 100e-6) / 130 = 3mJ/K. Divide our 1.55J by this number, and you get a temperature rise of 503K, which is by far too much. I would expect that this device will just vaporizeEdit2: I recommend to choose a wire-wound resistor instead of a metal film type, because these simply have a lot more copper to take the pulsed heat. For example: this onehttp://www.nikkohm.com/nikkohm_e_pdf/rnp50s_e20140701.pdfis a very large device, and still has only 34mJ/K of heat capacitance (!!) according to a Nikkohm appnote (http://www.wdi.ag/files/manufacturer/nikkohm/Nikkohm_Capability-Brochure-Power-Resistors.pdf). The IRV100 (http://www.nikkohm.com/nikkohm_e_pdf/irvirh_e20150101.pdf) has the same rated power (comes with builtin heatsink) but is wire-wound and has a heat capacity of 2.2J/K. That is 65 times better!The heat capacity calculated above for the 1/4W metal film resistor is most probablywayoff. And a 1/2W metal film resistor will neither do it.Edit3: found some information at Vishay:http://images.vishay.com/books/VSE-DB0007-0805_Leaded%20Fixed%20Film%20Resistors_INTERACTIVE.pdf. On page 32 they specify pulse ratings of a 1/4W resistor (MBA/SMA 0204). Doing the same calculation as for your part, I get (9W * 10ms) / (155°C - 25°C) = 0.07mJ/K. With our 1.55J of energy that one would heat up by 1.55 / 0.07m = 2239K (!!). So definitely not suitable, and also 1/2W and 1W metal film resistors will be inadequate.You see that this topic involves quite some math and research beforehand. If my guess with 1000uF and 100mA is close to your reality, I suggest to use a wirewound resistor and check its datasheet for the heat capacity. Should be more than 15mJ/K to stay on the safe side.