Hello,

I have searched the forums and found much about level shifting but not ultra low power.

I have a digital signal from a microcontroller at 0 and 3.3V. I have a supply of 7V and I want to shift the low-level signal to a high-level signal at 0 and 7V.

But I don't want the circuit to burn more than a micro-amp from either the 3.3V or the 7V supplies in either steady state.

Do you have any suggestions?

By the way, it only has to switch at very low speeds like 5 Hz maximum, and the square-ness of the wave is not the most important thing. It can be a little bit slow if it is really low power.

I figure the circuit cannot really be a pull-up resistor sort of thing, because that must either be very touchy based on a drain-source leakage current versus the size of a resistor... I figure all FETs have at least 1 uA of leakage.

I have a hunch it can be done with some FETs in a couple of CMOS pairs but just not sure how. Any help most appreciated.

Also, if anyone has comments on FET drain leakage and know of special FETs with ultra-low leakages? I did find one that says 0.1 uA at 25Vds with zero gate voltage -- it is BSS138 from OnSemi.

But does anyone have a trick for level shifting?
thanks

 

I may have just found the animal that will do it:
http://www.st.com/stonline/products/literature/ds/2020/hcf40109.pdf

... but that part is going obsolete, and I want to know how they do it inside -- what's the circuit?

 

It looks to me like you have your answer. Just put a 10 megohm resistor from the +7 volts to the drain of the BSS138, ground the source, and connect the 0 to 3.3 volt signal to the gate. That will use 7/10 uamp when the transistor is on and zero when it's off.

If you don't want the results to be inverted, use 20 meg resistors and make 2 inverters. It should switch in less than a millisecond.

 

It looks to me like you have your answer. Just put a 10 megohm resistor from the +7 volts to the drain of the BSS138, ground the source, and connect the 0 to 3.3 volt signal to the gate. That will use 7/10 uamp when the transistor is on and zero when it's off.

If you don't want the results to be inverted, use 20 meg resistors and make 2 inverters. It should switch in less than a millisecond.

Thank you for the suggestion... I agree that it should work, but I shy away from pull-up resistors because the off-state-leakage of the NMOS FET will still pull them a bit below the high rail. Like if I use a 10 MegaOhm resistor, and the FET is leaking 0.1uA, then it will at least pull down 1V from the top rail... if that's driving the gate of a PMOS power FET, it could cause some leakage... I get scared of using impedances larger than 1 MegaOhm due to all sorts of leakages and stray conductances could affect it, even EMI.

I found another idea that could work -- there are some ultra-low-power comparators out there. Here is one from Maxim at 1 uA per gate and up to 10V supply voltage:
http://datasheets.maxim-ic.com/en/ds/MAX406-MAX419.pdf

I have a 2V reference available on my board that I can compare the to uC output.

 

Pardon me for again bumping my thread, but I found an even better low-power comparator for the job:
http://focus.ti.com/lit/ds/symlink/tlv3702-q1.pdf

This one has Iq of 0.5 uA per amplifier, and it can sink and source a couple of mA pretty well on the output. Supply voltage up to 16V. Also cheaper than the Maxim chip.