The relay switches both the line and neutral, and I need to detect if one of the contacts are welded. I've made this circuit, but if a load is connected, a small current flows through the diodes and the output is always on.
How should I solve this problem?
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\$\begingroup\$ Do you have (1) the ground connection and/or (2) any limits on passing the voltage/current to the load when the relay is off? \$\endgroup\$– Eugene RyabtsevCommented Aug 9 at 5:36
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\$\begingroup\$ (1) Yes, the ground connection is available. (2) No voltage is allowed on the outputs. The output ends in a socket. \$\endgroup\$– germitronCommented Aug 9 at 11:26
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\$\begingroup\$ Do note that any variant of this will cause the output to never be truly disconnected. The 470 kohm resistors pass enough current to be felt and to fry electronics, even through it is not dangerous to a person. Some power supplies may also end up trying to start up repeatedly as their input capacitors charge, leading to extra wear. \$\endgroup\$– jpaCommented Aug 9 at 11:47
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\$\begingroup\$ Yeh, I'm start concerning about this. I get this idea from a TI ref design, but they are using a digital isolator. Do you have a better way to solve this problem? \$\endgroup\$– germitronCommented Aug 9 at 12:17
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1\$\begingroup\$ Use a relay with guided contacts. These are used in safety systems for this very reason. If one contact welds, the others will be held closed. Use another set of contacts to determine open/closed. Or maybe a contactor if the load is sufficient to justify the size/cost. \$\endgroup\$– KartmanCommented Aug 10 at 11:53
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3 Answers
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This should do it! Use unidirectional optocouplers so you won't get AC current passed through them. Since optocoupler diodes aren't made for extreme reverse voltages I put a series diode (use a generic rectifier high voltage low current one like 1n400X). The parallel resistor is placed to make sure the optocoupler diode does not compete in terms of reverse bias current with the diode. The other side of the optocoupler will receive a pulsating current and I put an RC filter to filter it out.
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2\$\begingroup\$ Please use the built in editor to draw schematics. \$\endgroup\$– Voltage Spike ♦Commented Aug 8 at 19:33
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1\$\begingroup\$ If you add the diodes as depicted, the optocouplers can well be left bi-directional, meaning it's no difference, the OP can use whatever ones are more convenient to procure. \$\endgroup\$ Commented Aug 9 at 5:39
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There is a current transfer ratio pretty much always on the LTV-8141, even down to currents of 0.1mA and maybe even lower than what the graph displays
Source: LTV-8141
You could try and increase R1 and R2 to a higher value, maybe 2MegΩ IDK what will work, even at 2Meg you'll still get 0.1mA .
This is to be expected with a darlington pair optotransistor, it would be better to select a optocoupler with a regular transistor IMO.
answered Aug 8 at 17:48
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Why did you cross-connect the optos on the input side of the contactor?
It would be more useful to connect each opto directly across one of the sets of contacts.
- When the contactor is energized, both will be shorted out and inactive.
- When the contactor is deenergized, they will be in series across the line (with the load also in the middle) — both active.
- If one of the contacts welds closed, only one of the optos will be active when the contactor is deenergized, and your control logic can detect that condition.
See this recent question for an example of what that logic might look like: A simple logic control yet it does not work
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\$\begingroup\$ Because it should work when the load is not connected (sorry for not indicating this on the schematic). \$\endgroup\$ Commented Aug 9 at 11:07
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\$\begingroup\$ Really? Why would the contacts weld if the load is disconnected? In that case, Mahdi's answer makes sense. \$\endgroup\$ Commented Aug 9 at 11:15
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\$\begingroup\$ Yes, because in any situation (reset, power cycle) the system should detect the welded contacts. \$\endgroup\$ Commented Aug 9 at 12:06
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1\$\begingroup\$ @germitron You can just add a large resistor in parallel with the load connections to make this work without external load. \$\endgroup\$– jpaCommented Aug 9 at 17:43