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I am designing a SIM card with RF radio included. The radio with its CPU needs minimum 2.7V to run. The current consumption is my problem: - for a period of 50msec (during transmission), it requires 100mA or more - for the same period afterwards, there is an idle period where it needs less than 10mA

The SIM card is plugged into any cell phone, which supplies on the SIM connector 3V/50mA guaranteed.

My idea is to take advantage of my radio idle period, charge something up, and add that charge to my high demand period.

In short: - supply to SIM card from the phone is 3V / 50mA (no more current allowed) - the SIM circuit requires 2.7V minimum, but the consumption needs to alternate between 10/100mA (50mA supply is not sufficient)

Sound like 'just add a capacitor'... but here is my last obstacle: SIM card is TINY, we have a total of about 8 x 10mm

We have tried to use a charge-pump to hold the 2.7V for as long as possible, but unless we are wrong, the capacitance needed is 1mF which we do not see how to do in the small size we have available.

Anybody has any smart idea how to solve this?

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  • \$\begingroup\$ I don't really understand what product that is.Are you're making a converter? \$\endgroup\$ – Daniel Tork Feb 17 '17 at 20:09
  • \$\begingroup\$ Are you making the SIM card? What exactly are you making? Please be more specific. \$\endgroup\$ – Daniel Tork Feb 17 '17 at 20:12
  • \$\begingroup\$ I updated the text. Should I write there more? \$\endgroup\$ – EmbeddedGuy Feb 17 '17 at 20:13
  • \$\begingroup\$ Well let's see. You say something about a converter(in the title).Then, you specify "a product". You state that "you need more current" so that make me think that this circuit has a converter on it, but it can't output more current? There's a mess in my head. Help me get rid of it by specifying: \$\endgroup\$ – Daniel Tork Feb 17 '17 at 20:18
  • \$\begingroup\$ Do you have a converter for the product? Is the voltage ok?More details about product like the entire setup: SIM card with wi fi adapter and bluetooth module for example. \$\endgroup\$ – Daniel Tork Feb 17 '17 at 20:20
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I found this battery which should have enough capacitance and ESR is ≤ 650 mΩ http://www.powerstream.com/thin-lithium-ion.htm

2x12x12.5 mm is too big but there is hope you may find one that fits your package.

But it may have to be current limited for charging so as not to pull down your 50mA supply voltage.

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  • \$\begingroup\$ I wrote them an email to request information on tiny size. Thank you. \$\endgroup\$ – EmbeddedGuy Feb 18 '17 at 5:03
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It looks like you have a circuit that is powered by 2.7V and draws 100mA for 50% of the time and 10mA for the other 50% of the time. That is 2.7V x 0.1A = 0.27W for half of the time or 0.135W average. Then it draws 0.01A from the 2.7V or 2.7 x 0.01 = 0.027W for the other half of the time or 0.0135W average. That's 0.135W + 0.0135W = 0.1485W average. You say you have 3V @ 50mA available or 3 x 0.05 = 0.15W. So you have 0.15 - 0.1485 = 0.0015W to spare. If you have a converter that is 99% efficient, there you are. On top of that, you have 8mm x 10mm = 80 sq mm of space available for this circuitry. I found one solution may be close to achieving your goal. TI makes a DC to DC converted TPS622316. It has a 2.7V fixed output in a 1mm x 1.5mm package. It requires minimal components. They should fit in under 20 sq mm. The efficiency is around 90%. This might work. But then consider you are powering a radio and you're putting a switching regulator in close proximity to it. You might consider shielding and input and output filtering. It's still close. Good luck.

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