A potential divider connects a 48V(worst-case) battery supply and a micro-controller for voltage sensing. Is it enough if I take care of the resistor values of the potential divider so as to limit the current,within the microcontroller's limit or should I opt for an isolation circuit? I can understand the reason behind the use of isolation in power supply circuits, for they are provided as a safety measure to counteract the unexpected behavior of the AC mains. Can DC act as unexpected as the AC mains? What factors decide the necessity if an isolation in a circuit?

The circuit looks like this: enter image description here

  • \$\begingroup\$ Why would you use a divider there? \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 7 '14 at 7:01
  • \$\begingroup\$ Voltage sensing? \$\endgroup\$ – Adithya Nov 7 '14 at 7:01
  • \$\begingroup\$ Might want to add that important fact to the question. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 7 '14 at 7:09
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    \$\begingroup\$ There will be almost 1A current (I = U/R) and about 40W heat on your voltage divider when you connect 47+4.7ohm to 45V (if 45V source can provide such current). Are you sure you want to connect ~52ohm divider there? \$\endgroup\$ – Kamil Nov 7 '14 at 12:00
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    \$\begingroup\$ Extremely sorry! It is a kohm! Will edit the circuit \$\endgroup\$ – Adithya Nov 7 '14 at 12:17

Your question is not totally clear, but I assume that your uc must measure the 48V, not just know whether it is (within some wide margin) present at all.

First: Is it a problem if the 48V and the uc share the ground? If so, you will need isolation and you will have an interesting design challenge.

If a common ground is no problem, I see no need for isolation (unless there are considerations that you did not tell us about).

You need an input voltage range that your uc can handle, with an impedance that is not too high for the A/D input. You can use a resistor divider: choose the lower resistor equal to the highest impedance that your A/D allows; choose the upper resistor to divide the maximum of your range-of-interest (for instance 30 .. 60V) down to what your A/D input can measure.

If the 48V can be higher than the highest value you considered you can either take that value into account when calculating the upper resistor (which reduces your accuracy), or add some overvoltage protection on your A/D input (like a zener clamp, or skottky diode clamp to Vcc if it can handle the extra current).

If you or your safety guys are worried about the consequences of a failure of the upper resistor you can split it into two equal values, and design the clamp to handle the current when one of the two shorts.

Do note that the accuracy of your measurement will be lower for lower '48'V voltages, and it will depend on the accuracy of your reference voltage (often the Vcc of the uc).

  • \$\begingroup\$ I've added the circuit for a clearer understanding of the problem. 45V is the maximum voltage the battery can attain when its fully charged.So, the worst case voltage that can be sensed by the A/D converter would be: 45 * 4.7/(47+4.7)=4.1V(approx) which is well within the reference voltage of the ADC which is fixed to 5V as can be seen from the circuit diagram. The chosen resistor values limit the worst-case current to a 0.8mA which is also well within the current-limits that the microcontroller can handle. \$\endgroup\$ – Adithya Nov 7 '14 at 12:15
  • \$\begingroup\$ (continued..) Thanks for pointing out the impedance consideration for the ADC i/p s even though it didn't matter when I chose the resistors and implemented using an arduino. The voltage measurement was quite accurate. Now,I have to do the same using a stand-alone uc and I am not so sure about the isolation stuff which I carelessly forgot to include in the Arduino implementation. Any light on when a common ground would become a problem? (with/without reference to my circuit) \$\endgroup\$ – Adithya Nov 7 '14 at 12:15
  • \$\begingroup\$ Common ground: probably no problem for the circuit itself, but is it allowed from a safety point? \$\endgroup\$ – Wouter van Ooijen Nov 7 '14 at 13:14
  • \$\begingroup\$ I think I get the picture now. If I get a DC-DC converter with an inbuilt isolation circuit,my problem would be solved, wouldn't it? \$\endgroup\$ – Adithya Nov 9 '14 at 11:03
  • \$\begingroup\$ What problem? And would it make any difference for the common ground? \$\endgroup\$ – Wouter van Ooijen Nov 9 '14 at 19:31

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