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I am driving a row of LEDs using an P channel logic level MOSFET (SI2323DS-T1-E3CT-ND) to switch power from a bench power supply. The gate is connected via a current limiting resistor to an output of a 74HC165 8 bit shift register. The shift register VCC is being powered from an Arduino 5V supply, I have connected the grounds between the Arduino power supply and the bench power supply.

schematic

simulate this circuit – Schematic created using CircuitLab

The gate of the Si2323DS has a pull up resistor, the intention being to turn off the leds until Q1 pulls it low, everything works as expected until I turn off the power to the Arduino, at which time the MOSFET turns on.
It seems that in the "powered off" state the output from the 74HC595 acts like a low resistance pathway to ground and the pull up resistor is unable to raise the gate to a logic 1?

It's important that the MOSFET is off under normal conditions because if all the rows are enabled by default in the final design, an over current condition will occur.

Given this configuration, is there any way to ensure the MOSFET is turned off when the 74HC595 is powered off?

I thought it might be possible to invert the logic using a transistor such that a logic high form the 74HC595 would be required to turn on the MOSFET, but if the problem can be solved with fewer components that would be optimal.

I am only using 2 power supplies during the prototype phase, ultimately the Arduino and the LEDs will be powered from the same PSU. Thanks for any help/suggestions you can offer.

I should also point out this was just a bench test for using the MOSFET, ultimately I am going to use current sinking high power shift registers (MIC5821) to sink the columns of my LED matrix similar to this design.

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  • \$\begingroup\$ You are totally on the right track. If it is allowable to use low side switching, you could use a power N-channel MOSFET instead of the P-channel you are currently using. Put the N-channel between the low-side of the LED's and GND, and drive the gate of that MOSFET directly from the 595. But inverting will also work. You could just add a single N-channel signal MOSFET to drive the gate of the current P-MOS. That is only one extra part, and it would preserve the single GND arrangement you have now, which may be required, depending on what you are doing. \$\endgroup\$
    – user57037
    Commented Oct 1, 2017 at 16:36
  • \$\begingroup\$ Oh, wait, your mosfet, the si2323, is a P-channel mosfet. I just noticed that you refer to it as N-channel in the text. But it is P-channel. \$\endgroup\$
    – user57037
    Commented Oct 1, 2017 at 16:43
  • \$\begingroup\$ Also, I think your schematic is wrong. The bench supply in your schematic does not share a GND with the arduino or HC595 as you say in your text. \$\endgroup\$
    – user57037
    Commented Oct 1, 2017 at 16:46
  • \$\begingroup\$ I corrected the mistakes in my diagram, be gentle this is the first time I have created one :-) \$\endgroup\$
    – mikee
    Commented Oct 1, 2017 at 18:55
  • \$\begingroup\$ It looks like the bench supply negative terminal is still not connected to GND. \$\endgroup\$
    – user57037
    Commented Oct 1, 2017 at 19:05

1 Answer 1

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Its just as you said, you can use an N fet connected to the diode resistors and to ground, that has 4 advantages:

1: Your problem gets instantly solved, and

2: N channel MOSFETs tend to have lower resistance than a P channel ones of the same size

3: You don't need a pull up resistor which consumes power when the MOSFET is off (not that it is a lot of energy tough).

4: if you decide to use many diodes in series hence increasing the needed led PSU voltage the circuit remains unchanged (unless the Vds of the chosen fet is too low).

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