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I built this latching switch circuit: https://www.edn.com/design/power-management/4427218/Latching-power-switch-uses-momentary-pushbutton

When I connected it to my project (~140 LEDs) I found that I could switch the circuit on but (often) not off. A more careful reading of the article told me that this is expected because "If the load impedance contains active devices such as LEDs, the load voltage at the instant Q2 turns off may be large enough to bias Q1 on via R4, thereby preventing the circuit from turning off properly. The presence of R5 pulls the OUT (+) terminal down to 0V when Q2 turns off, thus ensuring that Q1 turns off rapidly, and allowing the circuit to revert to its unlatched state in a proper manner."

I swapped out the 10K on R5 for 5K and the switch works reliably with my project, but I don't understand this explanation. What was happening when I pressed the button but the circuit didn't turn off?

enter image description here

  • Vs = approx. 4.5V from three alkaline AA batteries
  • Q1 = BC847C NPN transistor
  • Q2 = TSM2307CX P-channel mosfet
  • load = several strings of LEDs, in parallel. Strings are one colour of either white, blue or purple. They are different lengths (20-50 LEDs). I added a 15R to each of the longer strings because that's what I found in the original battery boxes that I removed, and a 56R to each of the shorter ones so they'd have roughly similar brightness.
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    \$\begingroup\$ Switching back and forth between your question and the schematic is cumbersome. Also include a picture of the schematic in your question. \$\endgroup\$ Jun 9, 2018 at 14:28
  • \$\begingroup\$ What actually are Vs, Q1, Q2 and the load? \$\endgroup\$
    – HarryH
    Jun 12, 2018 at 15:18
  • \$\begingroup\$ This explanation makes little sense to me. To begin with, since when are LED active components? \$\endgroup\$ Jun 14, 2018 at 6:38
  • \$\begingroup\$ This circuit was designed for a higher voltage. I think you should put it in a simulator and see what you need to tweak to get it to work. \$\endgroup\$
    – τεκ
    Oct 30, 2018 at 2:21

3 Answers 3

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An easy to guess explanation: High load causes substantial voltage drop in the mosfet, capacitor C1 is charged in ON state to far lower voltage than Vs. Connecting C1 to the gate cannot turn the mosfet non-conductive. A fix: Use higher operating voltage to get enough Vgs drive to the mosfet to keep voltage drop Vds low.

Another possibility is that Vs drops too much due the loading. Trying to turn OFF reduces the loading of Vs, so it jumps higher and what happens to be in C1 isn't enough to cause full turn off. A fix: Use stiffer source of Vs.

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When you push the switch in order to toggle the circuit off, Q2 is closed as long as the switch is pressed and C1 is charged, whether Q1 is 'biased' or not.
I guess your LEDs behave capacitively?

That would explain that Vout doesn't 'immediately' return to zero, but that R5 takes some time to discharge the capacitance in the load, which however at the same time will charge C1 through R2.
Your reduction of R5 speeds up the discharge, but increases the power dissipation in the on-state.

Maybe an increased (doubled) value of C1 would give the original R5 more time to discharge the load's capacitance.

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  • \$\begingroup\$ I measured the capacitance of the load: 4.9nF. On the original (R5=10K) circuit, increasing C1 to 660nF had no effect on behaviour. The LEDs dim momentarily when I press the button. \$\endgroup\$ Jun 9, 2018 at 18:36
  • \$\begingroup\$ If the LEDs don's switch off, your capacitor may be too small. \$\endgroup\$
    – HarryH
    Jun 10, 2018 at 12:09
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I think the issue might be that the combined forward voltage of the leds is causing the load to no longer conduct once the voltage drops to say.. 3v. The load needs to pull the output low for the circuit to switch off, and 3v might not be low enough for that.

Does it work with just the 10k, and no LEDs?

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