Series diode and a low power device

[TheComet]

I'm designing a device that consumes 450nA in idle (up to 2mA peak) and the maximum allowed voltage is 3.3V. The power is supplied by a battery with a voltage of 3.6V.

One of the problems I've run into is: All LDOs I could find have a quiescent current consumption (Iq) greater than 450nA. The one I ended up choosing is the TPS7830 with an Iq of 500nA, but I was wondering whether it be possible to instead use a series diode to drop the 3.6V down to ~3.0V and save those 500nA.

Of course, the "0.7V rule" for diode forward voltages does not apply with these kinds of currents. It got me wondering: If I were to use a diode in this situation, would it end up damaging the ICs that have an absolute maximum rating of Vcc=3.3V? Or would the overvoltage cause an increase in current through the diode, enough to increase the diode's forward voltage and drop Vcc to a safe level? Is there a way to calculate this current?

It seems strange that an almost "static" overvoltage could cause damange to an IC, because as soon as you start drawing any amount of current necessary to cause permanent damage, the voltage would drop to a safe level.

Are there perhaps other options besides LDOs and series diodes that consume less current?

 

[donpacino]

I'm designing a device that consumes 450nA in idle (up to 2mA peak) and the maximum allowed voltage is 3.3V. The power is supplied by a battery with a voltage of 3.6V.

One of the problems I've run into is: All LDOs I could find have a quiescent current consumption (Iq) greater than 450nA. The one I ended up choosing is the TPS7830 with an Iq of 500nA, but I was wondering whether it be possible to instead use a series diode to drop the 3.6V down to ~3.0V and save those 500nA.

Of course, the "0.7V rule" for diode forward voltages does not apply with these kinds of currents. It got me wondering: If I were to use a diode in this situation, would it end up damaging the ICs that have an absolute maximum rating of Vcc=3.3V? Or would the overvoltage cause an increase in current through the diode, enough to increase the diode's forward voltage and drop Vcc to a safe level? Is there a way to calculate this current?

It seems strange that an almost "static" overvoltage could cause damange to an IC, because as soon as you start drawing any amount of current necessary to cause permanent damage, the voltage would drop to a safe level.

Are there perhaps other options besides LDOs and series diodes that consume less current?

you would likely run into similar issues using both a series and shunt diode as you would the ldo.

You can use a set load (parallel resistor) to drive force the ldo to drive above the idle current

also you can use a part similar to this to get close, only need a load resistor to pull a small amount
http://www.ti.com/lit/ds/symlink/tps783.pdf

 

[TheComet]

also you can use a part similar to this to get close, only need a load resistor to pull a small amount
http://www.ti.com/lit/ds/symlink/tps783.pdf

You linked the exact device I said I was using in my first post

 

[davenn]

It seems strange that an almost "static" overvoltage could cause damange to an IC, because as soon as you start drawing any amount of current necessary to cause permanent damage, the voltage would drop to a safe level.

no, as the voltage increases, so does the current and that is the problem/danger of overvoltage for a device

 

[TheComet]

no, as the voltage increases, so does the current and that is the problem/danger of overvoltage for a device

...I'm not sure how you drew this conclusion. I clearly stated that this is battery powered and there is a series diode with very little current, and the voltage the diode would drop over the diode is 0.3V, meaning the diode will be nearly non-conducting.

Maybe I didn't explain the problem properly. I'll try to answer my own question because I think I understand it.

The question I want answered:
Can the device be damaged if exposed to a (conceptually initial) voltage that is higher than the maximum ratings (see picture).

The anwer:
No it will not. In theory, as Vcc is increased over 3.3V the device will start consuming a lot of current (which would lead to permanent damage if you did this for a longer period of time). We can model the worst-case scenario by making the device a 3.3V voltage source:

Because the drop is only 0.3V (3.6 to 3.3), the operating point of the diode will be here:

In other words, the diode is in a non-conducting state (a simulation with the 1N4148 shows 2uA) and therefore the device is not harmed.

 

[donpacino]

No it will not.

there are dangers other than overheating the device. A high voltage can cause damage to silicon devices permanently changing their functionality.

 

[berkeman]

The power is supplied by a battery with a voltage of 3.6V.

Can you just choose a different battery chemistry? Is this for a rechargeable application, or non-rechargeable?

https://learn.sparkfun.com/tutorials/battery-technologies

Do the IC Application Notes suggest a particular battery technology that is compatible with driving the IC directly? Can you say more about your application? There are a number of important circuit design tricks to use in very low power applications, especially if it involves any radio/RF electronics.

 

[berkeman]

there are dangers other than overheating the device. A high voltage can cause damage to silicon devices permanently changing their functionality.

Yeah, oxide punch-through comes to mind...

 

[davenn]

...I'm not sure how you drew this conclusion.

I accurately answered you specific question, which again was ...

It seems strange that an almost "static" overvoltage could cause damange to an IC, because as soon as you start drawing any amount of current necessary to cause permanent damage, the voltage would drop to a safe level.

it doesn't matter what the supply or device is
... overvoltage produces extra current and damage is the likely result to the device

 

[davenn]

The anwer:
No it will not. In theory, as Vcc is increased over 3.3V the device will start consuming a lot of current (which would lead to permanent damage if you did this for a longer period of time).

you just contradicted yourself

 

[Baluncore]

Of course, the "0.7V rule" for diode forward voltages does not apply with these kinds of currents. It got me wondering: If I were to use a diode in this situation, would it end up damaging the ICs that have an absolute maximum rating of Vcc=3.3V? Or would the overvoltage cause an increase in current through the diode, enough to increase the diode's forward voltage and drop Vcc to a safe level? Is there a way to calculate this current?

The first crude silicon diode rule says 0.7V at 10 mA. The second crude Si diode rule says 0.1V per decade of current change.
10 mA = 0.7V, 1 mA = 0.6V, 100uA = 0.5V, 10uA = 0.4V, 1uA = 0.3V. Everything is dependent on diode area.

For a zero (<1 nA) increase in quiescent current I would consider an old N-channel depletion mode FET wired like a simple constant current sink, but with the source resistor replaced by the voltage sensitive device. You need a pinch-off Vgs of maybe 3.1 volt.
Select the FET to have Vgs = 3.0V with a 6M8 resistor in place of the device. ( 3.0V / 450nA = 6.66Meg ohm ). The FET will carry the 2mA Imax without a problem. I am not sure at those low currents but would look at things like the 2N5484, 2N5485 and MPF102 or similar. Look at the data sheet to find the pinch-off voltage and the operating point over the current range expected.