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Suppose I have a device with a number of ICs (microcontroller, logic gates, MEMs sensors, etc) which can operate from a supply voltage in the range of, say, 1.7V to 3.6V. Will the current draw be different if I power the devices at the low-end (1.7V) vs the high-end (3.6V) of the operating range? Does the answer vary depending on the specific IC or is there a general rule-of-thumb?

The datasheets I've looked at don't seem to go to this level of detail, and instead only specify current draw at a single supply voltage. I'm interested because my device is battery powered and I'm considering whether to run direct from the battery or to use a LDO regulator.

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    \$\begingroup\$ Yes it will change in general. Yes it will depend on the IC. The rule of thumb is that it's the power which is taken, not the current unless the IC has some internal LDO or similar. \$\endgroup\$ – Eugene Sh. Nov 2 '17 at 14:16
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    \$\begingroup\$ There is no general answer, how much supply current is drawn depends on the IC and if it varies a lot then it should be mentioned in the datasheet. I could generalize a bit and mention that most analog ICs like opamps draw a nearly constant current independent of the supply voltage as their internal circuit often depend on a constant biasing current. However, if you load the same opamp with a resistor to ground and pull the output high then the current will depend on the supply voltage. If your circuits can work on a lower voltage I'd recommend using an LDO anyway. \$\endgroup\$ – Bimpelrekkie Nov 2 '17 at 14:18
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    \$\begingroup\$ Indeed what Eugene says, even if the supply current remains constant then an increased supply voltage will still result in more power consumption. If your battery supplies 3.6 V and more but your circuits need only 1.7 V then consider using a buck regulator as that could make up to 2x difference in battery life. \$\endgroup\$ – Bimpelrekkie Nov 2 '17 at 14:21
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    \$\begingroup\$ Here's another question that is related to this question. \$\endgroup\$ – Harry Svensson Nov 2 '17 at 14:35
  • \$\begingroup\$ @Harry Svensson Thanks for the link. There's some good info there relating to CMOS devices. \$\endgroup\$ – knick Nov 2 '17 at 23:52
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It depends on the IC.

For ordinary digital chips, the dominant current draw is charging and discharging capacitances to full rail voltages, so current draw is likely to be roughly proportional to voltage.

For analog chips the current is also likely to be positively correlated to the voltage, but the correlation may be weaker than with digital parts as analog designs are often based around (roughly) constant currents derived from some reference.

And very occasionally you will come across a device with a built-in switched mode power supply so the current drops as the voltage increases.

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  • \$\begingroup\$ Good point on charging/discharging capacitances. Thanks for the comprehensive answer. \$\endgroup\$ – knick Nov 2 '17 at 23:47
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In general, yes, working at the lower voltage range will drop the current pretty much directly proportional to the supply voltage.

However, since you are dealing with non-linear elements, there are many exceptions to this rule.

However, dropping the voltage to save current adds a couple of issues.

  1. Your supply needs to be more accurate to ensure any ripple or variation does not drop the supply below the minimum required by your devices.

  2. Dropping signal levels makes you more susceptible to noise. When running at the lower end of the supply, more care needs to be taken to reduce system noise.

For your specific case, which seems to have a fair amount of complexity, you will be adding a lot of heartache trying to get it to run at the lower end. If the minimum is 1.7V and you try to run it at 1.7V that means you need a perfect power source and zero system noise.

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  • \$\begingroup\$ working at the lower range will drop the current - Hm. Not the opposite? \$\endgroup\$ – Eugene Sh. Nov 2 '17 at 14:22
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    \$\begingroup\$ @EugeneSh. last I checked Ohm's law.. \$\endgroup\$ – Trevor_G Nov 2 '17 at 14:23
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    \$\begingroup\$ OK, so we are understanding "IC" differently, I guess. I am thinking of some complex IC requiring specific power to function, so increasing the voltage will drop the current.. \$\endgroup\$ – Eugene Sh. Nov 2 '17 at 14:25
  • \$\begingroup\$ @EugeneSh. yes indeed there are exceptions galore. \$\endgroup\$ – Trevor_G Nov 2 '17 at 14:26
  • \$\begingroup\$ Thanks for the tips on operating at the low-end regarding system noise. I'll be mindful of this. \$\endgroup\$ – knick Nov 2 '17 at 23:49

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