I'm designing a computer vision system and my 3 main components are

  1. microcontroller (VDD = [1.8V, 3.6V])
  2. microSD card (4 bit mode, VDD = [2.7V, 3.6V] according to SD 2.0 standard)
  3. cheap cell phone camera (using the 8 bit parallel DVP interface, VDD = [1.8V, 3.0V], Vanalog = [2.6V, 3.0V])

I suppose using lower voltages will reduce power, but I'm starting to doubt this because I found all the components have an internal switched capacitor regulator that change the voltage further (1.3V for the microcontroller, 1.5V for the camera, and probably a high voltage for the microSD). So, using a lower voltage will just result in a higher current draw. The only exception is for the camera, whos datasheet says it uses a linear regulator to drop the external DOVDD (I/O voltage) to the 1.5V digital core. In that case, I can save (2.7V - 1.8V) * 0.1A = 90mW by using 1.8V instead of 2.7V. This is about a 15% power saving for the whole system.

Currently, I can get away with using 2.7V for all my components. I'm only working on this as a hobby and only have a 1 layer PCB, so I'm hesitant about using multiple voltages.

So my questions are:

  1. How much core/internal power does using a lower external voltage save? None?

  2. How much I/O power does using a lower voltage save? I'm aware from Bill Dally's presentations that data transfer often consumes more power than computation, but I'm not operating at GHz speeds. According to the camera data sheet, table 8-3, when you don't use the internal regulator for generating the 1.5V core voltage, the I/O current, Idd-do is only 9-12 mA, so using a lower swing wouldn't save much power.

  3. How do you use the 1.8V mode on SD cards? Initially, you need to operate at 2.7 - 3.6V. After sending a CMD5 requesting 1.8V, then you can drop to 1.8V. But this seems hopelessly complicated to me? That means I need to drop the voltage to the micro controller too, right? and do it gradually. Are there any regulators that allow smooth transitioning from 2.7V to 1.8V?

  • \$\begingroup\$ Rule #1 about voltage vs. power saving: P=E²/R \$\endgroup\$ Oct 4, 2015 at 1:54
  • \$\begingroup\$ It's not so much the voltage a device is operated at, but the clock frequency. Microcontrollers are often rated at so many µA's (or mA's) per MHz. This is why they have so many sleep modes. I assume SD cards would be the same, they can be run off of an SPI bus anywhere from several hundred kHz to 40 or so MHz, a 100:1 ratio. I would guess there would be at least an order of magnitude power savings over that range. \$\endgroup\$
    – tcrosley
    Oct 4, 2015 at 3:15
  • \$\begingroup\$ @Ignacio, silicon isn't an ohmic material, so E^2/R doesn't apply, only P = V * I. I'm expecting a much more in depth answer that considers the efficiency of the switching regulators used. Yesterday, I found that voltages closer to the internal core voltage of a device are more efficient for capacitative regulators (charge pumps) because it means less energy is wasted in charging/discharging the capacitor. \$\endgroup\$
    – Yale Zhang
    Oct 6, 2015 at 3:21
  • \$\begingroup\$ @tcrosley - I will optimize the clock speed, but for now, let's assume it's already minimized. \$\endgroup\$
    – Yale Zhang
    Oct 6, 2015 at 3:21

1 Answer 1


Circuits that have a switch mode power supply (either a DC/DC converter, or a switched capacitor regulator) will not change the total power consumption significantly as the input voltage changes.

Circuits that don't include a pre regulator will generally consume less power when powered by a lower voltage. When you reduce the voltage, most digital circuits will reduce their current proportionately, so the overall power will decrease as V^2 (lower V and lower current will both reduce power). Analog circuits are generally biased differently and while the power will reduce, the current may remain the same. That is still lower power.


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