As a way to power-on/-off a circuit load, a MCU is used to drive a relay through two MOSFETs. The problem is, when the MCU GPIO output is measured 0-volt, the relay could stay connected.

Can anybody demonstrate how this could have possibly happened?

MOSFET Controled Relay Circuit

More details that may or may not be relevant to the problem:

This happens very rarely. We have not got enough chance to troubleshoot the issue.

Usually the process to operate the circuit is: MCU set GPIO 0; human switch the input off; the load circuit is changed; human switch the the input on; MCU set GPIO 1. When the input switch is turned off, the voltage will drain slowly as the schematic can show there are big decoupling capacitors.

We suspected that the relay is bad. Changing to another piece of relay only marginally changed the chance of running into the problem. Multiple boards are showing the same problem, thus it seems to rule out a damaged relay or MOSFET or any other single component.

  • \$\begingroup\$ As always, thank you very much for any comments or hints! \$\endgroup\$ – minghua May 7 '19 at 4:50
  • \$\begingroup\$ What current does the relay switch? (inrush current/switch off current) \$\endgroup\$ – Huisman May 7 '19 at 6:31
  • \$\begingroup\$ The datasheet doesn't give \$R_{gate}\$, but I'd suggest to increase R219 to a few ohms. There's no need for a super fast charging of the gate of Q4. Now, it only contributes to EMI. \$\endgroup\$ – Huisman May 7 '19 at 6:58
  • \$\begingroup\$ The current switched is typical 0.15A at 12V, Max 0.3A. The relay is rated DC 1A, max 3A, same as ones used on dfr0144. \$\endgroup\$ – minghua May 7 '19 at 15:02
  • \$\begingroup\$ As @huisman reminded in the answer section, the relay has been changed to a different model rated DC 220V 2A. \$\endgroup\$ – minghua May 7 '19 at 15:20

If the contacts stick, then this could be a problem of a welding of the metal. Yes, this can happen. Please check the contact rating in the datasheet. One possible cause: If the driving voltage /current through the coil is not strong enough to quickly turn on the relay, the armature may move rather slowly and thus a small electrical arc could develop. The arc in turn melts the surface of the contacts and they may weld.

Two ideas to remedy this: 1) Use a higher voltage to drive the coil (or a relay which is rated for a lower coil voltage) 2) Put a ceramic capacitor across the contact side of the relay.

  • 1
    \$\begingroup\$ Regarding the coil operating voltage: according the datasheet it should be at least 9V. It is currently 6V, which is too low. \$\endgroup\$ – Huisman May 7 '19 at 6:48
  • \$\begingroup\$ The same can also happen by wear & tear, if the relays exceed the specified amount of operations. \$\endgroup\$ – Lundin May 7 '19 at 7:26
  • \$\begingroup\$ Thanks for sharing the insights! If I understand it, two causes: [1] low operating voltage that does not drive the relay on. [2] welding/arc causing the contacts stuck on. Sounds [2] is in accordance with what we are experiencing. Isn't [1] the opposite effect? \$\endgroup\$ – minghua May 7 '19 at 15:07
  • \$\begingroup\$ Ok, I see [1] actual affects indirectly. Excellent explanation! \$\endgroup\$ – minghua May 7 '19 at 15:22
  • \$\begingroup\$ We measured the voltage on R217, it is actually 12V when the relay is driven on. Thus the low-voltage assumption does not really be a direct factor. The cause must be welding contacts. \$\endgroup\$ – minghua May 8 '19 at 4:19

I suspect the load has some capacitance so that the circuit resembles that of a capacitive discharge welder. The charging of the load capacitance (effectively through the 100uF or so you show) causes a very high peak current which pits the contact and can weld them together. In such a case, physically tapping the relay sharply (when de-energized) may allow the contacts to open.

The solution depends a bit on exactly what your load is, but other than using a massive contactor or replacing the relay with a MOSFET it's not necessarily straightforward if you need the capacitance on the switched side. You might be able to move your capacitors to the switched side and add an inductor and flyback diode, for example, to lower the peak current.

  • \$\begingroup\$ The load is a circuit board product being tested, yes lot of capacitance on its input side. You explanation makes a lot of sense for me to quickly understand and grasp the idea. Thanks a lot! \$\endgroup\$ – minghua May 7 '19 at 15:17

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