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PLEASE, PLEASE! My question is about power consumption of the CPU during EEPROM write NOT about the power supply! PLEASE! Thank you!

I have a device (ATmega48PA) powered through a 150k resistor directly from the line (with regulation of course!):

enter image description here

In more details (however this is FAR from the question and I don't know why so many people discussing the power supply schematic and NOT the question I asked):

enter image description here

I've estimated the consumption of the CPU itself about 300-400 uA which looked realistic with 2.5V VCC and 1M frequency.

However after I tried to write some data to CPU's EEPROM VCC goes dramatically down (generating BOD reset). After some investigation I realized that the current goes to 7 mA (and the current growing dramatically further if VCC higher).

I tried to find any information in the datasheet, but the only thing I found was this chart (on page 538):

enter image description here

Which is not really comply with my experience (and it looks like FLASH programming, not EEPROM).

So what is the reality of the current consumption of CPU during EEPROM write? And is it possible to reduce it?

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    \$\begingroup\$ Is there a reason you're powering the device through a large resistor in the first place? \$\endgroup\$ – Nick Johnson Nov 12 '15 at 16:11
  • \$\begingroup\$ It is actually a terrible idea. \$\endgroup\$ – Eugene Sh. Nov 12 '15 at 16:20
  • \$\begingroup\$ @NickJohnson of course! If the resistor will be smaller (in Ohms) it becomes pretty hot! And the device can become a energy inefficient despite the fact it is a micropower. Why I don't use any SMPS solutions: device should be extremely cheap and small. \$\endgroup\$ – Roman Matveev Nov 12 '15 at 16:22
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    \$\begingroup\$ People are asking about your power regulation because the question you asked 'smells' like your problem isn't what you think it is - in this case, the issue only arises because you're unable to draw more than a tiny amount of power without your rail browning out. \$\endgroup\$ – Nick Johnson Nov 12 '15 at 16:53
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    \$\begingroup\$ Repeat: You DO NOT have an LDO, even with the additional schematic. You have a 7805 (L78L05) which has a dropout of 1.7V, which doesn't qualify as low. You have a linear regulator, and are following the really bizarre habit people have gotten into of calling all linear regulators LDO (Low Drop Out) even when (as in the case of the 7805) they cannot be considered low dropout at all. \$\endgroup\$ – JRE Nov 12 '15 at 17:13
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The complete datasheet for the ATMega48 has more details on power consumption, but nothing specific to EEPROM write. It's not unusual that writing to EEPROM would take a lot more current than regular operation, however.

Instead of powering your AVR via a large resistor, you should use a linear regulator so its input voltage doesn't depend directly on the current it consumes.

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  • \$\begingroup\$ Please see my edited original question for clarification regarding the power supply schematic. \$\endgroup\$ – Roman Matveev Nov 12 '15 at 16:36
  • \$\begingroup\$ I've done my best to answer your original question given what little information the manufacturer provides. You're not likely to find more information from official sources; if you want exact figures you'll have to characterize the device yourself. \$\endgroup\$ – Nick Johnson Nov 12 '15 at 16:54

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