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I'm designing a PCB that receives 3 power inputs 3.3v, 5v and 12v.
I need simple way to switch all of them of in one go.

Here is a schematic of my PCB PCB schematic

My 12V line is for running a ws2815 led strip and may go up to 10A.
The 5V line is just for the level shifter.
The 3.3V is for running a NodeMCU ESP-12E board.

I'm using EVGA SuperNOVA 650 G1+ to drive everything .
I can switch the outside power source but it should run a couple of this boards and I want to be able to Independently shut them off.
I need to run a lot of those and I'm already running 7A on the 12v line so I can't just keep drawing power form only the 12v line there is the matter of the cable and how much amp it can run.
I can use what ever components I want I just need to keep it small.

Using help form an answer on a previous post of mine see here I came up with this switching solution:

CircuitLab Schematic 9u2s5v58b9hb

Is this a correct use of BJT?
Do I need to add fly-back diode or a resistor somewhere?

This is just a test circuit to see if it even works everything at default and the diode is there just a place holder.

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  • \$\begingroup\$ The point of a flyback diode is to protect devices switching inductive loads. There's nothing inductive here except tiny parasitics, so you certainly don't need one of those. \$\endgroup\$ – Hearth Mar 9 at 17:34
  • \$\begingroup\$ You must NOT use this design as it switches a voltage source to a diode without any control of current limit. \$\endgroup\$ – Sunnyskyguy EE75 Mar 9 at 17:42
  • \$\begingroup\$ Why bother with the MOSFETs in that simple example? Are the LEDs representative of something else? The BJT alone would be all you need to sink current thru an LED with proper current limit resistor. Are the values just the default values that the schematic tools brings up? \$\endgroup\$ – CrossRoads Mar 9 at 17:49
  • \$\begingroup\$ @CrossRoads yes the led are representative of something else. Yes everything at the default value as it's just a test to see if it even works. I need the mosfet because I have 3 power source that I need to be able to switch off at the same time using one switch see here: electronics.stackexchange.com/questions/424810/… \$\endgroup\$ – Rxzlion Mar 9 at 18:10
  • \$\begingroup\$ Edits to clarify what it's going to be used for with details. \$\endgroup\$ – Rxzlion Mar 9 at 18:36
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I agree that you do not need the bipolar transistors, but you cannot simply tie the three MOSFET gates together. The three gate pull-up (turn-off) resistors go to three different voltages. Without some isolation, the pull up resistors form voltage dividers. The 3.3 V source could turn on the 12 V FET.

Replace each 3906 with a 1N914 or 1N4148 diode. Anodes to the individual gates, cathodes tied together to the switch. Low Vgs might be an issue for the FET switching the 3.3 V source. in that case ...

Is the unswitched 12 V source available all the time? If yes, then the 12 V can be used to turn off all three FET gates. Now you don't need the isolation diodes.

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  • \$\begingroup\$ What an idiot i'm the 12v is indeed available all the time just using it removes the need for all of this complication. That's why you always need a fresh set of eyes . Thank you for the simple and esay soultion \$\endgroup\$ – Rxzlion Mar 9 at 19:09
  • \$\begingroup\$ Glad it works for you. Upvote - ? \$\endgroup\$ – AnalogKid Mar 9 at 19:40
  • \$\begingroup\$ I did but I have less the 15 rep. \$\endgroup\$ – Rxzlion Mar 9 at 19:52
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It's not ideal. The bases of the transistors will be pulled up to +11.3V by Q2 E-B junction so Q1 will see a reverse bias of 8.7V on the B-E junction, which violates the 5V absolute maximum value and is at the edge of where the B-E junction typically breaks down.

You would be better to use NPN transistors rather than PNP with 3 base resistors. That will eliminate the problem mentioned above, and M12 will get more gate voltage when "on". It may not be adequate even with that but I can't analyze it because there is yet another problem- the LED currents are uncontrolled because you didn't add the required series resistors to set the currents. As it is, it will likely destroy the LEDs.

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  • \$\begingroup\$ I edited the question to clarify this is a test circuit with default values and that the led is just a place holder. \$\endgroup\$ – Rxzlion Mar 9 at 18:20
  • \$\begingroup\$ More edits to clarify what it's going to be used for with details. \$\endgroup\$ – Rxzlion Mar 9 at 18:35
  • \$\begingroup\$ Pretty much the same answer after your edit, except I think your 3.3V high side switch M12 is inadequate for the peak current. \$\endgroup\$ – Spehro Pefhany Mar 9 at 18:40
  • \$\begingroup\$ did you read all the edit I clearly stat that everything on default i'll pick the right component after I know the concept is even correct. \$\endgroup\$ – Rxzlion Mar 9 at 18:44
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I think you could simplify it to this. You'll have to find MOSFETs that will work at the voltages you had, especially some that will turn on/off with just 3.3V.

schematic

simulate this circuit – Schematic created using CircuitLab

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