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I have a Microchip PIC board (18F46K22) which has a VIN pin to connect to a power source (7vdc + supply).

What happen if I feed 3.3V to the VIN which goes through the 5V linear regulator (ST 78M05)?

The 18F46K22 is able to run on 5v or 3.3v supply. I would like to force the board to run on 3.3V so it will solve my logic voltage level issues easily.

I am asking because I can't find any information regarding going lower voltage than the linear is normally regulating.

Thanks!

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  • \$\begingroup\$ Simple - it won't work. A 78X05 regulator needs about 8 volts in in order to operate. With 3.3 volts in it will not regulate at all, and the board will be dead. \$\endgroup\$ – WhatRoughBeast Oct 22 '16 at 0:39
  • \$\begingroup\$ sparkfun.com/products/526 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 22 '16 at 2:04
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What will happen is that with a light load the 7805 will still drop around 1.5-1.6V and you'll see around 1.6-1.7V at the output, not enough for the board to operate correctly.

You'll need to bypass the regulator. If you jumper directly from input to output on the 7805 (the two outside pins) it should work fine and the 7805 will behave as if it is not there. Better would be to add a new (and incompatible) connector that goes only to ground and the 3.3V rail- and not install the jumper- because if you forget and plug a jumpered board into a higher voltage supply really Bad Things will happen.

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Look for "headroom" or "dropout voltage" in the 7805 (or any other linear regulator) datasheet - that will state the minimum voltage differential across the regulator for correct operation.

You would be very lucky to get 1 volt from a 7805 with a 3.3 volt input.

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