NerdKits - electronics education for a digital generation

You are not logged in. [log in]

NEW: Learning electronics? Ask your questions on the new Electronics Questions & Answers site hosted by CircuitLab.

Support Forum » NK component current requirements

January 27, 2010
by mjswan
mjswan's Avatar

It's easy to drain a 9V battery with even the simplest NK project, for example, the LM34-based thermometer. In an attempt to understand why there is so much current draw, I tried to find the draw of most of the major components.

  • The ATmega168 in Active Mode says it draws 250uA at 1.8V and 1MHz. Since NK is driving it at 5V, I think that means 0.6mA at 1 MHz. Also, the NK clock runs at 14MHz. I don't think this means a 14x increase in current but a 10x increase is probably safe so my estimate would be about a 6mA current draw.

  • From other threads on the Forum, I gather the 4-line LCD uses about 180mA.

  • The LM34 draws 90uA.

  • I read the L7805C (voltage regulator) datasheet but am lost with respect to determining its current draw. The datasheet does show a "Quiescent current" value of 8mA so I'm guessing this is the regulator's draw although I don't believe the regulator is drawing 33% greater current than the MCU.

Do the values above appear correct and, if not, what are the corrections? If these calculations are correct, then the LCD is by far the biggest current draw, at 30x of the MCU.

Finally, my electronics background is more theoretical than practical so please feel free to correct my terminology or conceptual errors. :-)

January 28, 2010
by mrobbins
(NerdKits Staff)

mrobbins's Avatar

Hi mjswan,

A few notes:

For the ATmega168 itself, look at page 317 of the datasheet, and specifically Figure 29-2. At 5.0 volts and 14.7456 MHz, the current draw is around 9mA. (However, I do suggest you look at mikedoug's thread about sleep modes, as well as a earlier thread about understanding battery life.)

For CMOS-style logic, which the ATmega168 is, it is very typical that the current consumption is directly proportional to the clock frequency. That's because each gate is like a tiny capacitor that is charged and discharged, and the more often you do that, the more power is consumed. That's why lots of modern computers try to reduce power consumption by frequency scaling (lowering the number of cycles per second), or by shutting off unused parts of the chip (lowering the number of gates that have to be switched in the first place).

The LCD itself uses only a few milliamps of current. I don't have a number, but you should be able to measure it. The backlight is what could possibly come close to 100-200mA. If you don't use the backlight, then you're much better off for battery life. (The same is true with laptops and cell phones!)

Quiescent current is the number you're looking for. This current is used to set up an absolute voltage reference and to power the feedback loop. If power consuption is a critical element of your system, then you'll definitely want to look for a voltage regulator with lower quiescent current. The tradeoffs will be increased cost, and perhaps greater noise / looser voltage regulation tolerances.

So in summary: it's probably closer to 20-25mA total (without the LCD backlight). But you should definitely look at the forum threads I linked to above. Hope that helps!


February 01, 2010
by mjswan
mjswan's Avatar

Hi Mike,

Thanks for the detailed answers and the links. I'd previously seen the first link but not the second. I'm glad to know I was understanding most of the concepts.

You wrote, "The LCD itself uses only a few milliamps of current. I don't have a number, but you should be able to measure it." As a practical matter, how would I go about measuring that current? Would I use a DVM to measure the resistance from the +5V input (pin 2) to the Ground (pin 1) while the LCD is powered up then apply Ohm's Law?


February 05, 2010
by mrobbins
(NerdKits Staff)

mrobbins's Avatar

Hi Michael,

Does your DVM have a current measuring mode (maybe labeled "mA" for milliamps)? If so, you could remove the wire that connects LCD pin #2 to +5V, and instead use the DVM between those two nodes. That essentially forces the current to flow through the multimeter, which should let you measure the current.

(Note that a multimeter, when used to measure voltage, measures "across" an element, and the multimeter is designed to have very high resistance. But when used to measure current, it measures "through" itself, and the multimeter is designed to have very low resistance like a wire.)


February 10, 2010
by BobaMosfet
BobaMosfet's Avatar

There is another way, using a plain old voltmeter to measure your current draw. It's not 100% accurate unless you have 1% resistors, but it will give you a very good idea of how much current your circuit has flowing through it:

On pin 2 of the LM7805 regulator, put a 1% 1 Ohm resistor in place of the ground wire to the ground rail. This is called a "shunt". Use the PIE formula to calculate the size (watts) of the resistor so you don't fry it. A 1/2 Watt resistor should work, as I recall.

Use your voltmeter and measure the voltage drop across this resistor when your circuit is on. Because of the relationship of EIR and the rating of the resistor, the reading may be interpreted as Amps, not volts.

For me, the LCD project (with some additions and other sensors) returned a reading of 97mA.

Enjoy. BM

February 10, 2010
by mrobbins
(NerdKits Staff)

mrobbins's Avatar

Hi BobaMosfet,

Does your LCD have a backlight connected? If not, 97mA sounds quite high -- but then again I'm not sure what other sensors you might have connected.

I just measured a my DEMOfall '09 badge circuit, which has the LCD (no backlight), ATmega168 scrolling text, plus it's running the LCD version of the piezoelectric sound meter code. For the 7805 + running ATmega168 + LCD + the analog amplifier, the total current drawn from the battery was 17.5 mA -- much more of what I expected!

Out of 17.5 mA, I separately measured:

  • LCD = 1.3 mA
  • 7805 = 3.3 mA (with nothing connected to its +5V output)

By design, I know my analog circuit will draw about 0.2mA. Making some assumptions (particularly that the 7805 doesn't suddenly start wasting more current), this leaves about 17.5-1.3-3.3-0.2 = 12.7mA for the microcontroller running at 14.7456 MHz. (The microcontroller and analog amplifier were hard to measure directly because each had multiple power/gnd pins.)

LCDs use a surprisingly tiny amount of current. Most of that 1.3mA is probably for the digital controller logic, and not in the display elements! (Think of a digital watch, where the LCD runs for months or years off a tiny battery.)


Post a Reply

Please log in to post a reply.

Did you know that the printf format string "%.3f" will show three digits after the decimal point? Learn more...