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I need to create a negative voltage rail (from positive supply rail), say -5V to keep a P-MOS transistor in ON-state even when the voltage on its [S] source is 0V. So I found this schematic in an answer of another question here, but does this actually work?

enter image description here

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    \$\begingroup\$ C1 needs to be between pins 1 and 3 instead (I'm assuming you're using CUI's P7805). \$\endgroup\$ – joribama Sep 1 '19 at 2:20
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Contrary to popular belief this does work, but the P7805 is a DC to DC converter and it can sink current. A regular 7805 voltage regulator requires a negative supply, this does not. Here is a excerpt from the P7805 datasheet:

enter image description here
Source: https://www.cui.com/product/resource/p7805-s.pdf

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    \$\begingroup\$ Uhhhh, let marketing decide the name of a component is harmful… +1 for pointing this out! \$\endgroup\$ – Janka Aug 26 '19 at 22:12
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    \$\begingroup\$ Well, they wanted it to be drop in comparable with a 7805... I agree bad idea \$\endgroup\$ – Voltage Spike Aug 26 '19 at 22:13
  • \$\begingroup\$ Wow, this is entirely different thing than one may think of...bad naming 100%. Thank you! \$\endgroup\$ – velizarw Aug 27 '19 at 9:32
  • \$\begingroup\$ For anyone else not spotting it immediately: the 2 and 3 pins stealthily switch places. \$\endgroup\$ – jpa Aug 27 '19 at 12:24
  • \$\begingroup\$ Also, note that in this circuit C1 is between pin 1 and 3 of the converter, while in OP's version it's between pins 1 and 2. \$\endgroup\$ – Nyos Aug 27 '19 at 12:36

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