IC powered via non-vcc pin?! - Page 1
Forum section: electronics / projects

author: Gizmo
author_rank: Regular Contributor
timestamp: October 22, 2014, 11:31:45 pm
content: I'm just putting together what is basically an EPROM emulator for an old console. I have a PIC micro running the show, a 1MB SRAM to store the data, a few 245's to separate the host bus while programming and a 4040 counter to count up the low 11 bits of the address, the high bits being taken directly from the PIC.Now, I thought I was being clever, as I have a switchable (on/off) 3.3v regulator, which I'm using to power the 4040 which has 5v tolerant inputs. So while in "programming mode" I disable the host bus, power up the 3.3v regulator for the SD card and 4040 and clock everything up and program the SRAM. Everyone's a winner.When I power down the regulator it all appears to be good, but then when I enable the host bus, activity on the A0 line, which is connected to the 4040, appears to be powering the chip, as I get 3.3v out of the Vcc pin and the signal appears to be clamped to 3.3v. I assumed if the chip wasn't powered then it would do nothing...In short, my question is, if a chip isn't powered, how can it interfere with the signal like this? How can it be powered by a voltage applied to an output pin?!The solution of course is to add another 245 to isolate the 4040, but I was hoping to reduce the part count by being "clever".

author: c4757p
author_rank: Super Contributor
timestamp: October 22, 2014, 11:46:43 pm
content: Here are the input and output circuits of NXP's 74HC series (from here) - see the diodes? Those will conduct current from the pin to the switched supply rail, if the supply rail is below the voltage applied to the pin.5V tolerant inputs will look different, but the outputs won't.

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: October 23, 2014, 02:04:05 am
content: IIRC there are entire series of logic chips with special kind of i/os that don't pass-through power like that.Can anyone confirm or deny this? I seem to recall it.

author: c4757p
author_rank: Super Contributor
timestamp: October 23, 2014, 02:06:25 am
content: I vaguely recall seeing something like that, though, it'd probably be cheaper to pass the signals through either a tristate buffer, or an analog switch (for bidi)

author: calexanian
author_rank: Super Contributor
timestamp: October 23, 2014, 04:57:33 am
content: Common trap for young players. Is it really shut off? Especially with new micros that can be fully running with under a milliamp at 1.7 volts Vdd.

author: Gizmo
author_rank: Regular Contributor
timestamp: October 23, 2014, 07:43:29 am
content: It's definitely off. 0v registered on the reg output until I enable the address bus, then magically the 5v applied to the output leg seems to power everything up! I am happy with the explanation given above as to what is happening. I am going to move to using 2x 8bit tristate binary counters linked together as I can't find any larger counters with a chip enable / output enable.An interesting one to learn anyway! This stuff is all quite new to me so I've got a lot to learn...

author: jahonen
author_rank: Super Contributor
timestamp: October 23, 2014, 08:32:38 am
content: My colleague had unpowered '245 bus buffer (or something similar) in telecommunications board backplane buffer for quite some time, until in some circumstances, the bus failed to operate. Only then the missing power connection was found out. It worked perfectly fine as long as there were enough 1s on the bus signalsRegards,Janne

author: calexanian
author_rank: Super Contributor
timestamp: October 23, 2014, 05:26:39 pm
content: Most micros have shunt protection diodes right on all I/O pins. Classic path to VDD and GND

author: SArepairman
author_rank: Frequent Contributor
timestamp: October 23, 2014, 05:52:48 pm
content: looking at chip level schematics really helps explain some of these weird things.Combine with a shitty brread board and you wasted 1/2 your day !

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: October 23, 2014, 08:43:26 pm
content: Quote from: c4757p on October 23, 2014, 02:06:25 amSN74LVC8T245There, I knew it! It's a 245 with two VCC pins for both sides, it has" The I off circuitry disables the outputs, preventing damaging current backflow through the device when it is powered down."I think this means it won't power itself from logic signals when powered off. Might help the OP if he can mash those chips in there somehow!

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: October 23, 2014, 08:57:01 pm
content: I've heard stories about products that have actually been shipped, passed design reviews and QC, with unpowered CMOS ICs that worked fine- until at some point they didn't.

author: marshallh
author_rank: Unknown
timestamp: October 24, 2014, 12:34:02 am
content: BTW, you can run some MCUs without a ground... for a while

author: Gizmo
author_rank: Regular Contributor
timestamp: October 25, 2014, 08:23:31 am
content: Thanks for all the comments, all very helpful! I've decided to do it properly (like I should have done in the first place!) and have redesigned with some 74HC590 counters cascaded together as these have output enables. So nice and easy to decouple the counter from the bus.

author: TonyStewart
author_rank: Contributor
timestamp: October 25, 2014, 08:43:33 am
content: Now applying power after high currenton a signal line MAY cause anSCR crowbar effect inside the chipfrom Vcc to Vdd in the substrateon unloaded circuits at low current, the ESD diodes protect the chip while pulling up Vcc.due to inherent PNPN layers in the process.The same is true if the Schottky diode clamps exceed their rated current with excess ESD to cause a 1.2V drop.Like an SCR on DC, it latches until power is cycled.In 74HC class the ESR was not low enough at <5V but at 15V it could fry the chip.Not sure if 74LVC2 would also since they have the lowest ESR even at low voltage.,But like ESD it can wound the chip from excess heat until power cycle then it runs ok but rapidly aged.

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: October 26, 2014, 03:48:32 am
content: Quote from: Gizmo on October 25, 2014, 08:23:31 amAFAIK output enables don't change the basic fact of the diodes to vcc and ground.All the OE does is turn off the drive to both output transistors, you don't get rid of the diodes.http://www.ti.com/general/docs/lit/getliterature.tsp?baseLiteratureNumber=scla007&fileType=pdfpage 15 (9 of the document)Then again, I don't know your schematic. What's wrong with the SN74LVC8T245?