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I'd like to ask you a question about my circuit to automatically turn off battery charger. It works now fine except when voltage is close to the switch of treshold, coil of relay sounds noisy so I know that I need some histeresis but have now idea how to implement it there.

Thank you![Circuit]1

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  • \$\begingroup\$ R6 value is 10kOhm and D1 is 5V1 Zener \$\endgroup\$
    – pablos91
    Commented May 18, 2020 at 11:46
  • \$\begingroup\$ Is this for educational purposes? Otherwise you would be better off buying an IC that does the job for you. \$\endgroup\$
    – DimP
    Commented May 18, 2020 at 11:50
  • \$\begingroup\$ absolutely just for amateur needs \$\endgroup\$
    – pablos91
    Commented May 18, 2020 at 11:55
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    \$\begingroup\$ Try a feedback resistor from LM358 output pin to non-inverting input - try 100 kohm initially - if too much hysteresis then try a higher value. \$\endgroup\$
    – Andy aka
    Commented May 18, 2020 at 12:19

1 Answer 1

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Hystersis is positive feedback. You may know you can get negative feedback by connecting the amplifier output with a resistor to the inverting input. So you can get positive feedback by connecting a resistor to the positive input. Sometimes a small capacitor across the feedback resistor, perhaps 100 pF or so, is useful to speed up the transition.

If the signal coming into the +ve input was from a voltage source, then this wouldn't work. However, the input is from an effective resistance of a little below 1 kΩ to a couple of kΩ depending on the position of the potentiometer. That variability will make the size of the hysteresis window dependent on the pot position.

Andy's suggestion of 100 kΩ is a good place to start for the feedback resistor, and will give you a hysteresis of one percent or two of the opamp output swing.

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  • \$\begingroup\$ I can get rid of potentiometer or lock it when I choose specific voltage like 14V (turn of chargin above this value). But I don't it will work. I think that difference between turning off and on relay might be about 1V. \$\endgroup\$
    – pablos91
    Commented May 18, 2020 at 13:11
  • \$\begingroup\$ Will it help when I swap potentiometer output with voltage source ? \$\endgroup\$
    – pablos91
    Commented May 18, 2020 at 13:14

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