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I have a voltage source that will be either 3.7v, 5v or 12v and would like to use a 6v 300mA rated switch to turn the power on and off by using the on/off pin on a MC33375D regulator. Would appreciate if someone could tell me if my layman's understanding is correct in the below schematic.

schematic for turning ldo on off

The 4.7k 6.8k res are to reduce the voltage to 5v if the 12v source is in use. The 10k is to make the pin set to off by default. Of course, my overall assumption is that the schottky diodes attached to each source will result in only one of them being used even if all are attached. Is this correct?

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    \$\begingroup\$ The wire linking D2 to D3 shorts the 4k7 resistor. \$\endgroup\$ – brhans Sep 8 '18 at 20:03
  • \$\begingroup\$ Using these as a non-load carrying switch, I'd say you could use it without the voltage divider. It's only carrying single digit milliamps if that. \$\endgroup\$ – Passerby Sep 8 '18 at 23:13
  • \$\begingroup\$ @Passerby the switch is rated at 6v. Wouldn't this make it unsuitable for instances when the 12v is connected? \$\endgroup\$ – Philip Lemon Sep 9 '18 at 19:58
  • \$\begingroup\$ Depends. You can derate, or use it at a different spec then provided, by considering how it is going to be used. Here, you use it not as a load carrying component, but as a small signal switch. 12V at 1 mA or less is a very insignificant power compared to 6V 300 mA. It would be very very very unlikely to degrade the switch or cause it to fail. You could reach out to the manufacturer and ask if it's okay to use in that situation. You can also consider, are you using this in a small project run or a large quantity production. It's okay in a small project to make due \$\endgroup\$ – Passerby Sep 9 '18 at 20:45
  • \$\begingroup\$ But better to replace the switch in a large commercial product due to liability or cost of replacement if failed x how many units, etc. It's always a cost benefit analysis. \$\endgroup\$ – Passerby Sep 9 '18 at 20:46
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I think you mean the MC33375D, not the MCC33375D (which doesnt exist). Your circuit wont work because of the shorting of the 4.7k top resistor mentioned by brhans.

The ON/OFF input requires at least 2.4V to turn on. You would get that with the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

It's a bit unweildy though, it would be much better to just get a higher rated switch.

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  • \$\begingroup\$ thanks. Are the R1 and R3 resistor values changed from my original specs because of the voltage drop from the diodes, or did I just have them wrong in the first place? When selecting the Schottkys for D1 - D6, is it ok to just use the same rated one for each? The 12V source is 2A so I thought i'd just use 20V 2A rated schottkys all round. Thanks. \$\endgroup\$ – Philip Lemon Sep 9 '18 at 20:03
  • \$\begingroup\$ The resistors are different because R4 affects the voltage output from the potential dividers. You can use the same diodes for everything, but you dont need schottkys for D4-6, they can be normal cheap low power diodes \$\endgroup\$ – BeB00 Sep 9 '18 at 22:20

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