Timeline for answer to Using two regulators to create new ground: How do I calculate needed power? by russ_hensel
Current License: CC BY-SA 3.0
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| when toggle format | what | by | license | comment | |
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| Feb 5, 2012 at 10:24 | vote | accept | AndrejaKo | ||
| Jan 30, 2012 at 21:17 | comment | added | supercat | ...pull the ground down to a point that's "at least" 5 volts below the positive rail. Note that the 7905 would not prevent the "ground" from going more than 5 volts below the positive rail; if more current would be sinked into "-5" than sourced from "+5" when ground is midway between the rails, then "ground" would be pulled to a point more than five volts below the "+5" supply. | |
| Jan 30, 2012 at 21:13 | comment | added | supercat | @AndrejaKo: Conceptually, it's pretty simple. If one were to eliminate the 7905 and simply let the "ground" be floating relative to the source supply rails, the ground would float to a point where the current sourced into "ground" from the "+5" rail was equal to the current sunk from the "-5" rail. It's not terribly likely, however, that that point would be halfway between the two rails. If one can be assured that the point would be above the midway point, one can use a chip like the 7905 to... | |
| Jan 30, 2012 at 16:38 | comment | added | AndrejaKo | @supercat Can you write a bit more on that or post some links? Looks pretty interesting. | |
| Jan 30, 2012 at 16:27 | comment | added | supercat | If the current drawn from the +5 will always exceed that drawn from the -5 supply, one might be able to rework the circuit using a 7905. There are some real scenarios where the -5 draw will never be very big, and the minimum draw from the +5 supply will exceed that. Even in such scenarios, it might not be a bad idea to add a 5.1-volt or 5.4-volt zener on the +5 supply to ensure that even if the expected load isn't present, the "5-volt" supply voltage will not get so high as to cause damage. | |
| Jan 30, 2012 at 15:45 | history | answered | russ_hensel | CC BY-SA 3.0 |