A new sleep mode using 225 nA

[perky]

The is generally true I think, but no guarantee you won't get a batch that are near their maximums so a bit dodgy for volume stuff. Did you test this at high temperature? There's a non-linear logarithmic relationship with leakage and temperature, that 60M resistor is only going to need 16nA to drop a volt and possibly turn the top FET on.
Mark.

[WhiteHare]

I  see that Joe has already proven out the REG102 concept, except using a buck regulator instead:  https://lowpowerlab.com/forum/low-power-techniques/making-a-lower-power-moteino/msg13181/#msg13181

I suppose even if the buck regulator circuit had a relatively high quiescent current when turned off, it might still come out ahead of a REG102 simply because the power conversion is more efficient, whereas the REG102 would be burning off as heat all voltage greater than 3.3v.  If there is a buck regulator that can be turned on/off like the REG102, and with a quiescent current when OFF as low as the REG102, it would be pretty cool.

[perky]

Well here's the thing. If you look at the voltage Vin against Vout curve you'll see it simply passes the voltage through until it regulates, and the spec says it acts like a 0.6R resistor while doing it, so you could just use 2 cells. If it was at 2.2V for Vin it's output would be 2.2V with a 0.6R resistor in series. Similarly for 3V, the output would be 3V with 0.6R in series. Vin doesn't need to be above 3.3V. This is how LDOs usually work, the MPC1703 is not actually typical in that respect (although the MPC1703A now has similar Vin v. Vout curves but the quiescent current is a little high). 

[WhiteHare]

I'm afraid the leakage current of the FETs might scupper this idea. The  SI2356DS for example has 1uA max zero gate voltage drain current (Idss) at 25 deg C (admittedly at 40V), and this rises to 10uA at 40V for 55 deg C. It'll be less for lower voltages of course but it's still significant, and your 60M pull-up might not actually pull it up leaving the load switch FET permanently on. You are going to have to find ultra low leakage FETs I think.
Mark.

@perky Just a rough idea, but how about substituting the  SI2356DS with TI's TPS22860 instead?  http://www.ti.com/lit/ds/symlink/tps22860.pdf 
Ultra-Low Leakage Current
– VIN Leakage Current = 2 nA Device Information(1)
– V BIAS Leakage Current at 5.5 V = 10 nA
Would that overcome your concerns?

[perky]

Mmm. Interesting part, but unfortunately Figure 1 (Ron v. Vout) gives pretty poor output impedance at lower bias voltages. Unless you have a higher Vbias available this would mean a large voltage drop as batteries deplete.

The killer here is always the voltage drop due to resistance when using batteries down to depletion levels, not only have you got higher internal resistance of the batteries themselves to deal with but any load switches or input circuitry resistance as well. It's almost like you really need supercaps to hold enough energy to do a full TX/RX cycle and only gradually top that up between cycles using much lower currents.

Mark.

[WhiteHare]

It's almost like you really need supercaps to hold enough energy to do a full TX/RX cycle and only gradually top that up between cycles using much lower currents.

I was leaning that direction anyway--a cap and no battery--just to get a feel for the whole energy harvesting thing by using only ambient light and the power scavanged from a cheap solar cell (canabalized from a $1 garden light).  How much Farad, when fully charged to 3.3V, do you calculate would be needed to support a full transmission?  Order of magnitude is good enough.  I was going to determine an answer experimentally by just plugging in different value caps and seeing which would support a full transmission, but if you have a ballpark on the cap value, it might save a few steps of testing.

[perky]

OK, back of a fag packet calculation:

Energy in a cap is C(V*V)/2. Let's assume you could have an allowable drop from 3.3V to 2V for a full transmission, which lasts 10ms and takes a constant current of 100mA.

We'll give a very worst case energy used during transmission and say that's at constant 3.3V (voltage is dropping from 3.3V to 2V so this is worst case). Energy is power times time, so that's 3.3*100e-3*10e-3, i.e. 3.3mJ. That has to be equal to the loss of energy in the caps, so:

C*(3.3*3.3)/2 - C*(2*2)/2 = 3.3e-3.

C = (2*3.3e-3)/(3.3*3.3 - 2*2), which is roughtly 958uF. So you're looking at 1mF at least as a ball-park.
Mark.

Edit: Also remember that to charge this cap up you'd ideally want to to do it in an energy efficient manner. Just using a resistor will waste power in that resistor due to the I^2R losses. So some buck-boost regulator might be needed, but one that uses very low current (possibly even a switched capacitor regulator, although I have no real idea how efficient that might be). Unless of course you simply keep it charged up continually by the source and use a load switch to isolate it while transmitting..

[WhiteHare]

OK, back of a fag packet calculation:

Energy in a cap is C(V*V)/2. Let's assume you could have an allowable drop from 3.3V to 2V for a full transmission, which lasts 10ms and takes a constant current of 100mA.

We'll give a very worst case energy used during transmission and say that's at constant 3.3V (voltage is dropping from 3.3V to 2V so this is worst case). Energy is power times time, so that's 3.3*100e-3*10e-3, i.e. 3.3mJ. That has to be equal to the loss of energy in the caps, so:

C*(3.3*3.3)/2 - C*(2*2)/2 = 3.3e-3.

C = (2*3.3e-3)/(3.3*3.3 - 2*2), which is roughtly 958uF. So you're looking at 1mF at least as a ball-park.
Mark.

Edit: Also remember that to charge this cap up you'd ideally want to to do it in an energy efficient manner. Just using a resistor will waste power in that resistor due to the I^2R losses. So some buck-boost regulator might be needed, but one that uses very low current (possibly even a switched capacitor regulator, although I have no real idea how efficient that might be). Unless of course you simply keep it charged up continually by the source and use a load switch to isolate it while transmitting..

Thanks, Perky.   

[WhiteHare]

Regarding the original post, I notice that a chip like the LTC2935 (datasheet: http://cds.linear.com/docs/en/datasheet/2935fa.pdf) might also be used to turn the circuit on again.  That way you wouldn't have to wait for the voltage to fall all the way down to 1.3v, but could instead resume at 1.8v (or higher) if desired to stay within spec.

While the quiescent current is low (500na is "typical"), it's not as low as the OP circuit.  Also, it's not as minimalist or as inexpensive as the OP circuit either, and so those are possibly trade-offs.

More simply, though, I wonder if a diode might be used to drop voltage to a pin, say D3, so that the atmega328p can be awakened using pin change interrupt to restart the cycle, yet the atmega328p itself would still be powered at >= 1.8v (so as to stay within Atmel's spec)?