Back driving voltage regulator behaviour

[sebmadgwick]

I am designing a battery powered device which has the option to also be powered directly from a 3.3V supply (when battery disconnected).  The battery would otherwise power the circuit through a 3.3 V regulator (MIC5319, see attached)

My question is: What is the expected voltage of the regulator input when no battery is connected and regulator output is at 3.3V from an separate supply?

This matters because a uC measures the battery voltage on start up and will immediately shutdown the circuit if the battery voltage is too low.  I do not want the circuit to shutdown if no battery is connected and the circuit powered from a separate 3.3 V source.  I did some quick tests and the back-driven regulator seems to be at around 3.2 V and is able to drive a fair bit of current.

[orin]

Check out Figure 33 in the LM1086 datasheet and Figure 7 of Linear Technology AN53 for ideas.

From the LM1086 datasheet, it looks like a series diode might be a good idea.

I just built the LTC1154 circuit to replace the diode battery/line switching in my Fluke 731B voltage reference.  The Si9956DY MOSFET pairs can be found on ebay and there are equivalents, though I forget the part number.

[Jon Chandler]

Most regulators don't like having voltage on the output with nothing on the input.  The easy solution is to put a reverse-biased diode from output to input.  With the normal higher voltage on the input, the diode won't conduct.  If the circuit is powered from a different regulated source, the diode conducts, keeping Vin approximately the same as Vout.

[sebmadgwick]

I used a scope to plot (Vout - Vin) when the regulator (MIC5319) is back-driven (no power source on Vin side) and the voltages were equal.  The voltages remained equal even when I add a variable load resistor on the Vin side.

I was expecting to see some relationship between the Vin and Vout voltage; e.g. voltage drop due to diode or series resistance.

The MIC5319 block diagram (attached) shows a MOSFET linking Vin to Vout which in itself does not explain the observed back-driven behaviour.  Perhaps the MOSFET is coupled with a diode, but I didn't observe a froward voltage drop, so perhaps I made a mistake...

Any comments welcome.

[Clear as mud]

My guess is that it has a small internal diode from output to input.
As you connect voltage to the output, the diode conducts it onto the input.  Then, if the regulator is enabled and the voltage on the input actually becomes high enough to operate the regulator, the error amp sees the voltage is too low and fully turns on the internal FET.  This allows any voltage present on the output to feed back through to the input, as long as the voltage is lower than the normal regulated voltage.  In other words, the circuitry inside the regulator is all operational the same way it would be if the (too low) voltage was applied to the input.  The only difference is that the input voltage is actually coming from the output, through the FET.  But the regulator doesn't know that.  If you were to raise the output voltage higher than the regulation point, the FET would turn off and you'd see the diode drop, and be unable to drive very much load off the input without burning out the internal diode.

This is all conjecture.  Everyone please let us know if you have a different guess or knowledge of what actually happens when voltage is fed to the output of this regulator.

[dannyf]

What is the expected voltage of the regulator input when no battery is connected and regulator output is at 3.3V from an separate supply?

It is unknowable for any unknown regulator.

The safest bet would be to use a diode across the regulator; or to change your design;