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thank you for your help. I'm new to electronics, and learning as I go. I need help on a circuit that otherwise works great. This is a magnetic power on-off switch, I'm using a relay switch to supply LIPO BAT power to a regulator. When turned on, the line to reg. measures full BAT power as expected, BUT when the switch is turned off the line to the reg. is still measuring around 0.76v, and slowly drains the battery over time. Is anyone able to to help eliminate or reduce drain battery when off?

Answer - my mistake on this, I didn't show the full circuit, there was a resistor down the line causing a problem.

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To turn off the opto relay, the input current must be below 300uA. This may not be fulfilled. Probably Q1 is not completely turned off. Diode D1 doesn't need anything, but a resistor could be useful instead. Exactly replace the diode with a 470 Ohm resistor. This will probably fully turn off the relay.

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  • \$\begingroup\$ How would your suggestion help prevent the battery draining? \$\endgroup\$
    – brhans
    Aug 31 '20 at 0:44
  • \$\begingroup\$ thanks for suggestion, I tried this and it behaves exactly the same though \$\endgroup\$
    – mdrive20
    Aug 31 '20 at 3:45
  • \$\begingroup\$ Then there was not much left. The first step is to measure whether the transistor is closed. That is, there is no voltage at the opto control. I thought of the proposed method because transistor Q1 is also a grounded base here (capacitor on the base). In this case, hundreds of MHz oscillations may become in the PCB or during installation, depending on the collector and emitter wires. This oscillation was stopped by the resistance. The thing was narrowed down to two parts then, transistor Q1 and opto relay. \$\endgroup\$
    – csabahu
    Aug 31 '20 at 7:18

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