"state" of a pic pin when pic is off - Page 1
Forum section: electronics / general

author: Simon
author_rank: Unknown
timestamp: October 12, 2011, 09:44:05 am
content: I know this will be a tough one, but does u-chip guarantee that pins will have a certain state when the pic is off ? It would be helpful if my PWM output pin when high impedance when the pic is not powered but i think maybe I'm hoping for a bit much ?

author: PeterG
author_rank: Frequent Contributor
timestamp: October 12, 2011, 10:15:34 am
content: That is a very interesting question. I am thinking a dso and dmm could answer the question, just program the output pins and see what happens when you power it off.regards

author: Simon
author_rank: Unknown
timestamp: October 12, 2011, 10:16:51 am
content: yea, i suppose so, I just wondered if micro chip gave any specific data on the situation to avoid fluke ocurances

author: jahonen
author_rank: Super Contributor
timestamp: October 12, 2011, 10:40:06 am
content: Remember that ESD protection diodes are still there, so if you apply voltage to an IO-pin when the device is powered off, you can inadvertently power your MCU via an IO-pin.Regards,Janne

author: Simon
author_rank: Unknown
timestamp: October 12, 2011, 11:45:49 am
content: well any voltage applied will be through a high value resistor, if it were to start the pic that would be great

author: PeterG
author_rank: Frequent Contributor
timestamp: October 12, 2011, 12:08:08 pm
content: If you apply voltage to a pin, you may power up the pic however you can easily blow the pin io by sinking to much current (done it myself). I generally use a weak pulldown resistor on the output so the output is known when it is powered down, if required.Regards

author: ejeffrey
author_rank: Super Contributor
timestamp: October 12, 2011, 02:45:22 pm
content: DC or AC impedance? At low voltage, the impedance is high but not super high, given by the diode characteristic. As the voltage applied to the output pin increases, the diode starts to conduct, and the DC impedance drops. When the current is high enough to turn on the chip, the DC resistance starts going up again, but AC signals get to see the decoupling capacitor.If you really need high impedance when the circuit is off, you may need to use an external circuit like a relay or an open-collector driver.

author: Simon
author_rank: Unknown
timestamp: October 12, 2011, 02:48:56 pm
content: well I'm interested in the DC impedance 0-3V range so from what you say it will be high, that's good, yes i can use a buffer of some sort

author: Unknown
author_rank: Unknown
timestamp: October 12, 2011, 02:58:50 pm
content: Most vendors specify that no voltage should be applied to any of the other pins if the chip is not powered, so I would assume this unless the datasheet explicitly states otherwise.

author: Bambur
author_rank: Regular Contributor
timestamp: October 13, 2011, 09:14:16 am
content: Quote from: alm on October 12, 2011, 02:58:50 pmTrue. Applying any voltage to I/O pin when VDD is off may result in SCR-latchup and chip damage. Be aware

author: Psi
author_rank: Super Contributor
timestamp: October 13, 2011, 09:40:57 am
content: I dunno if this helps, but most (if not all) micros go high-z on i/o pins when you hold them in a reset state.(The only exception might be the programming output pins on some micros)

author: Simon
author_rank: Unknown
timestamp: October 13, 2011, 10:17:44 am
content: Quote from: Psi on October 13, 2011, 09:40:57 amso your saying keep the micro powered and use the MCLR pin to turn it on and keep the pins on high Z, that is something I had consicered, infact the other way around this is to just use a mosfet to pull the the MCLR low when it sees the same voltage i was going to put onto the output to kick start the circuit (boot it)

author: deephaven
author_rank: Frequent Contributor
timestamp: October 13, 2011, 01:41:39 pm
content: If you don't mind it drawing a tiny bit of power all the time, why not put it in sleep mode and have it wake up by some external signal?

author: Simon
author_rank: Unknown
timestamp: October 13, 2011, 01:54:56 pm
content: that would be ideal, I'm looking into finding a low quiescent current regulator as this project is powered off a car battery

author: deephaven
author_rank: Frequent Contributor
timestamp: October 13, 2011, 02:00:09 pm
content: Quote from: Simon on October 13, 2011, 01:54:56 pmIf you can make sure it's input voltage doesn't go above 16 V. an LT6003CS5 would do it.

author: Simon
author_rank: Unknown
timestamp: October 13, 2011, 02:05:18 pm
content: hm I suppose i could, it's for an automotive use so I am a bit concerned about any spikes etc but i suppose if I put some filtering in front of it it should be ok,

author: Unknown
author_rank: Unknown
timestamp: October 13, 2011, 04:44:08 pm
content: Wouldn't it make sense to use a more robust regulator designed for the nasty automotive environment? Even the continuous voltage is fairly close to 16V, let alone spikes and all the other garbage encountered in automotive electronics.Is quiescent current such an issue? It's not like it has to run for a year on an CR2032 cell. What's the self discharge of a car battery like?

author: Simon
author_rank: Unknown
timestamp: October 13, 2011, 06:49:49 pm
content: Well it does not have to be so low, I think up to 1mA is fine

author: deephaven
author_rank: Frequent Contributor
timestamp: October 13, 2011, 08:02:42 pm
content: How about one of these:http://www.maxim-ic.com/datasheet/index.mvp/id/6336"The device operates from a +3.5V to +30V input voltage, delivers up to 200mA of load current, and consumes only 20µA of quiescent current at no load. The input is +45V transient tolerant"

author: Simon
author_rank: Unknown
timestamp: October 13, 2011, 08:48:09 pm
content: that looks perfect, price isn't bad either

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: October 17, 2011, 09:44:47 am
content: It will typically be equivalent to a high impedance with a pair of inverse-connected diodes to ground (as the Vdd rail will effectively be at ground). Any current applied to the pin will be fed to Vdd via the ESD diode, so depending on what else is connected to Vdd, it may start to power things up, but due to the limited current available it could misbehave in various ways and so not a good idea to rely on to do anything useful. It is also important to have a good brownout reset, as without it, the marginal supply to the pin may get the MCU in to a wierd state which persists when proper power appears.

author: Simon
author_rank: Unknown
timestamp: October 17, 2011, 11:35:19 am
content: well I am looking at feeding a few volts via a 10-100K resistor, there is the gate of a mosfet connected to the pin so this should be powered at >3V anyhow that will signal the rest of the circuit to start

author: ivan747
author_rank: Super Contributor
timestamp: November 08, 2011, 12:07:10 am
content: Quote from: jahonen on October 12, 2011, 10:40:06 amYep, figured that out the hard way. One day I was wondering what the heck was going on then I plugged my USB to TTL serial converter to my PIC-based project. It supplied around 3.7 volts out of the 5V TX line. It was enough to dimly light the power LED, but it couldn't make the PIC work because I had the brownout reset activated so it would go into reset mode at 4.5V.

author: Zero999
author_rank: Super Contributor
timestamp: November 08, 2011, 05:47:04 pm
content: The LM2936-5.0 is a good regulator: low quiescent current and low drop-out but it's limited to only 50mA.What's the operating frequency? You could just connect a resistor in series with the IO pin.

author: ivan747
author_rank: Super Contributor
timestamp: November 09, 2011, 07:08:49 pm
content: If you really care about power consumption, you can use some spike-proof 9V or 12V linear regulator and connect it's output to a switching regulator to get the voltage down to 5V or 3.3V. Texas Instruments and Maxim have some good switching regulators in non-evil packages like SOT-23 and SOIC.You can also omit the linear regulator and use the classic MC34063 with some heavy filtering and spike protection. Remember it can stand 40V at the input, so spikes won't kill it.

author: mobbarley
author_rank: Regular Contributor
timestamp: November 13, 2011, 11:23:15 am
content: Get a regulator for automotive use - they are really rugged designed to survive load dumps and similar events that are possible in cars.