The side-effects of using Weak PULL-UPS?

[MrNobody]

Hi, I was just thinking about using weak pull-ups in sleep mode to minimize current consumption in PIC12F615.

Lets say in normal operation, I use GP0/AN0 to read ADC and the ADC voltage is around 3V.. When I want to go into sleep mode, according to the datasheet, I can enable weak pull-up to prevent the pin from floating which will minimize current consumption. When I do that, the voltage at that input pin is 5V and if my understanding is correct, it has just turned an input into an 'output' because current is flowing into the 'thing' that i was trying to measure which is 3V (lower voltage). Of course I can add a diode to only allow the current to flow in one direction but that would increase the cost.

Is my understanding of weak pull-ups correct..? Is there a better way to minimize current consumption during sleep mode without imposing a voltage level onto the device which we are trying to measure and also without using external components..?

I don't know, i was just thinking about weak pull-ups and sleep mode.. Just trying to understand and learn more i guess.. Hmm..

 

[3v0]

Maybe I am missing something.

If you do not mind the pins floating while the processor sleeps make all the pins digital input prior to sleeping. Restore when it wakes from sleep.

3v0

 

[Diver300]

As I understand it, analog inputs do not take additional current when there is a floating voltage.

 

[3v0]

As I understand it, analog inputs do not take additional current when there is a floating voltage.

That makes sense. The internal ADC cap will charge to whatever voltage is on the input pin. So there is no advantage to changing the pin back to digital.

3v0

 

[DirtyLude]

Floating inputs are a no-no for a sleeping microcontroller. All datasheets list switching to output for low power operation. It's a trait of the CMOS logic gates that they will eat up current if left to float.

I'm not up to speed on the exact theory, but googling 'CMOS floating input' gives a bunch of leads.

 

[Sceadwian]

The reason floating inputs for sleep is bad is because even in sleep mode the internal schmit trigger is still active, so if there's enough capacitive coupling to another part of the circuit that isn't sleeping or to an external noise source the pin will oscillate which drains power because cmos drains power only when it switches states, for lowest power usage the weak pullups/downs are probably fine.

 

[Russ Hensel]

Analog inputs should never float.

 

[3v0]

Analog inputs should never float.

So make them all digital inputs prior to sleeping ? Inputs will not draw any significant power.

3v0

 

[dougy83]

There seems to be some confusion here regarding digital inputs above; allowing a digital input to float or setting its voltage towards half-supply may increase the supply consumption.

Analogue inputs are allowed to be at whatever voltage they want within the supply range. When an analogue input is enabled, the digital input circuitry is disabled, as is the pullup for that pin (see datasheet ANSEL notes).

As the OP stated, if the pullup is enabled, current will flow out the pin to the voltage source that was being measured.

So it would seem that leaving the pin configured as an analogue input during sleep is the way to go.

 

[MrNobody]

Analogue inputs are allowed to be at whatever voltage they want within the supply range. When an analogue input is enabled, the digital input circuitry is disabled, as is the pullup for that pin (see datasheet ANSEL notes).

Oh.. yeah.. i missed that part about setting input as analog will disable pull-ups..

It did say on the ANSEL section of the datasheet..

Note 1: Setting a pin to an analog input automatically disables the digital input circuitry, weak pull-ups, and
interrupt-on-change if available.

So, if the input pin is kept as analogue input, is it considered as HIGH IMPEDANCE input..? Is it considered floating..? Because in the SLEEP mode section of the datasheet, it states that

I/O pins that are highimpedance
inputs should be pulled high or low externally
to avoid switching currents caused by floating inputs.

 

[dougy83]

I can only assume that they are referring to digital inputs. The analogue inputs are designed to have an analogue (e.g. floating) voltage present when operating, so I can only assume that it's the same during sleep.

 

[3v0]

My thinking has been all over the map on this. Does this logic hold ?

All but one ADC, the selected channel, is disconnected from the cap. Are the input gates of the unselected inputs freely floating! Even when you have a stable voltage on the pin.

If so I understand the following why

Analog inputs should never float.

It seems to me that the best bet is to make the inputs digital and use the weakest pullups that will keep the lines stable.

3v0

 

[dougy83]

My thinking has been all over the map on this.

Watching your posts in this thread is like watching a [good] tennis match.

All but one ADC, the selected channel, is disconnected from the cap.

No channel is connected to the cap except for during the acquisition period preceding the conversion cycle.

It seems to me that the best bet is to make the inputs digital and use the weakest pullups that will keep the lines stable.

A digital input held at a voltage that is neither high nor low can cause shoot-through current in the digital input transistors. Enabling the weak pullup, as the OP pointed out, provides a leakage path for current from VCC to the ~3V input voltage.

Disabling the digital input circuitry stops the problem with the 'floating' input voltage (because the problem - i.e. the digital input - has been removed).

 

[dougy83]

Straight from the horse's rear: https://www.electro-tech-online.com/custompdfs/2009/12/41200a.pdf

In particular, check out page 15/16.