I'm using this circuit to turn on and off the system.


The power supply is 12V. I use an LM2576 to decrease to 5V. Here is the circuit:


Firstly I build on-off circuit also I added an LED at the 12V and on-off circuit works well. Then I added the voltage regulator circuit and the on-off circuit stopped working. It just turns on the system at the first push then it always stays turned on. When I remove the voltage regulator capacitors the on-off circuit continues to work.

I would like to know what is the reason of this problem and how can I fix it?

edit: This is my circuit. enter image description here

  • \$\begingroup\$ Could you show how and where you attach the LED when it is working (including a resistor if there is one), and the same with the regulator. Don't just give us separate circuits and describe in text what you do with them. SHOW us in your drawing or a photo how exactly do you connect them. The devil is in the details. \$\endgroup\$ Apr 17 '20 at 7:26
  • \$\begingroup\$ Additionally, do you connect the voltage regulator with a load connected to it, or just by itself? \$\endgroup\$ Apr 17 '20 at 7:40
  • \$\begingroup\$ @EdinFifić Hi, i add my circuit. I just use a LED as a load. \$\endgroup\$
    – km-74
    Apr 17 '20 at 8:08

I have 3 possible solutions:

  1. I think that the reason it doesn't work with the regulator is because this switch requires a resistive/linear load to work, more specifically less than 0.6V across its load output in order to turn off when you push the button to turn it off.
    This is because in order for the PNP transistor (T1) to turn off, its base needs to be less than -0.6V compared to its emitter, or nearly at the same potential as the emitter.
    Although this doesn't explain how would an LED work if connected as a load, since it is non-linear and has at least 1.5V drop across it, but I suspect the LM2576 also acts like a non-linear load which requires a minimum voltage (like maybe 2V) to start conducting more than a few microamps.
    I have located the circuit author's page and he is obviously using a linear/resistive load (incandescent light bulb):
    - My first suggestion would be to try placing a resistor (maybe 10K) across the switch's load output to bring voltage across its leads to 0 when it's switched off.

  2. Increase the C1 value to 220nF or just another 100nF in parallel with the existing one. This may be needed to swamp out the MOSFET capacitance, as the author suggests on his page.

  3. You could use the LM2576 digital switch function (on pin 5) together with your switch circuit by first placing a 10k resistor on your switch output (+ and -), disconnecting the LM2576 pin 5 from ground and connecting it to the negative (-) output of your switch. The +Vin (pin 1) and the GND (pin 3) of the LM2576 would go directly to your 12V source.
    The only "problem" you would have with this arrangement is that the regulator would be turned on as soon as you connect the 12V power supply, but you should be able to turn it off and on after that.
    This solution would be better in terms of wasted power when turned on, because the MOSFET would drop about 0.27V at 1A and would dissipate about 0.27W if all of the output power was connected to it, and there would be almost no wasted power if using the switch pin of the LM2576.
    ADDITIONAL reason why this solution would be better is that you would have a common ground for your power supply and the regulator's output. Otherwise, the output ground would be floating and would bypass your switch if it touched the supply ground.


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