cmos IC question

How about a link to data sheet, or some part numbers. Did you measure the 400 uA? As a WAG you can probably dissipate ~100mW in the diode/series R. Try blowing some up maybe? (How about a resistor divider to drop the 12 volts to 8?)

George H.

If it is an open collector output, just put the pull-up to +8V instead of +12V.

Pere

Right... I thought of adding that, but it seemed too obvious. GH

Can you answer the question? I already know how to return the pull-up to the 8 volt side - but want to keep the 12V swing for other circuits using that same signal line. I'd rather not add anything since the build is such that I'd prefer to stay out of it... Likewise, I could add a zener clamp as the easiest bulletproof solution, but if the internal diodes are happy with it why bother?

Texas Instruments CD4520B Dual up counter

Datasheet acquired from Harris Semiconductor SCHS067D - Revised Mrch 2004

Page 4 shows the protection network

Huh, what does the dashed line between the diodes going to the positive supply mean? Do you have some series resistance between the 12V output and 8 V input? Otherwise you've either fried the first diode, or maybe it's current limited by the driver. What's the voltage at the input to the 8 V part? Typical currents are listed as ~6mA... I'm guessing that's not enough to blow the first diode, but I really don't know.

George H.

The 12 volt output is at the collector of a transistor and there is

10K pull-up resistor (to 12V) limiting current to the transistor when it is on, and input to the 8 volt cmos input when the transistor is off. (ground is common in both circuits)

The 12V output is NOT a high-side switch and can only supply what the

10K allows. A whopping 1.2 milliamps to the open collector transistor, and less than 400 microamps to the protection diode of the cmos input.

As I understand it, those diodes are part of the electrostatic protection scheme, but there's lots of cases where a cmos circuit can be shut off (Vdd in this case is zero volts) and the signals still be present on lines feeding inputs (so there are voltages greater than the absolute max) and no damage occurs.

So I figure it is safe enough, but I don't remember ever reading a specification on that input protection diode, and I'm curious.

If it is a onesy project, why not just put a 1N4148 between the input and Vcc? Without even looking at the spec sheet, I am pretty sure it will tolerate more than 1 ma. The last time I looked, they were less than five cents at your hobbyist parts store.

I found what I needed...

I forgot. In ancient times they had these things made from trees called books.

According to the CMOS Cookbook - Don Lancaster: "Any "force fed" currents above or below the supply rails must be limited to 10 milliamperes or less."

I'm good to go.

Well I wouldn't do that because I don't know what else is using that signal. A high value series resistor and a zener diode would be my choice or a high value resistor and diode to Vdd.

But as it turns out any current less than 10 milliamps is fine so I don't need to do anything.

I think those signal diodes are good for 100 milliamps at 50 or 100 volts. I remember testing power supplies many years ago and there was a 1N914 between the remote sensing connection to keep the power supply output down in the event the sense wire wasn't connected. Anyhow, a failure occurred, and I'd left the supply under test running into a dummy load when I went to lunch... That sense diode was actually shorted and passing ~5 amps when I got back. The leads were hot enough to melt solder, but there was no other damage.

Most modern cmos logic chips can handle 50 mA into either of the input ESD diodes, with power on or off. Some (like the Tiny gates) don't have a diode to Vcc, but the equivalent of a 7v zener to ground. That allows, say, a 3.3v chip to accept a 5v logic inputs. That zener couldn't handle 50 mA.

The real hazard isn't usually diode current rating, it's SCR latchup: bias the ESD diodes on a few mA and the whole chip collapses and shorts the power supply. The ancient 4000-series was notorious for latchup. Your 8 and 12 volt supplies hint at 4000s.

Many mixed-signal parts, like DACs, have latchup problems too.

What parts are you using? Post links to the data sheets.

Semi data sheets tend to lie and deliberately hide defects, so thousands of users can independently enjoy discover the bugs. Sometimes one of the application examples near the end of the data sheet, or a 2-point-font footnote, might give hints.

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com

I've heard stories of SCR-like effects caused by exercising the ESD diodes, check the data sheet for the part you have.

--
  When I tried casting out nines I made a hash of it.

Thanks.

I'm tapping into the speedometer circuit on my truck (12V) and have added a GPS hockey puck to supply a 16K pulse/mile to it. It needs 8K so the 4520 divides by 2 and 16 (and a few 4017's for an odometer, and to flash some leds for a light show) I used 8 volts because I have a mess of 7808 chips sitting around and the whole project is built from parts on hand.

The truck is 1991 and has a problem with the transmission speedo drive gear and this was lots cheaper than a transmission fix. The 8K/mile signal is also used by the fuel management computer and necessary for overdrive shifting etc.. There's a 555 astable too to check the operation/calibration and a back-up in case the hockey puck is out to lunch.

Excellent! Don knows lot's more about Cmos than me. :^)

George H.

The 4000B series was pretty good. Just make sure you buy a real, modern B part and not some old ebay 4000A chips.

And don't run any external wiring directly into any CMOS chip. Induced spikes could be a lot bigger than 10 mA. RC filter and schmitt external inputs.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

Any protection diodes will NOT necessarily respect the '12V' signal, they might clamp it

The data sheet says this WILL diode-clamp input to +8V. Either your +12V signal or your +8V supply will suffer, if the chip doesn't burn up.

The (relatively) clean solution is two resistors, making a voltage divider on the (0V, 12V) signal to make (0V, 8V) output. Something like a 10K + 20K ohm will do, depending on what the resistor value of the open collector pullup is. Accurate values are not needed, for normal logic operation, any output above 6V is 'high' enough for that CMOS with 8V applied.

No, not if you want to have a 12V signal amplitude. That was about burning the chip up, not about signal integrity. It's the value in 'absolute maximum' for current, and relates to bond wire current carrying capacity.

I think it's OK. The ~8.5V clamped signal is enough to turn on the 12V logic, and the IC sources less than 10 mA.

George H.

You're correct the 12V will be clamped, I could add a dedicated transistor to drive the 8V input and separate the signal to the engine. (that wouldn't be hard to do)

Ain't nothin' burning up here... the 12V signal can't supply more than 2 milliamps and the internal chip diodes can handle that; but you are right in saying that the 12V signal will suffer as result of the diode.

Or... the chip input only needs point one microamps to bias it on, so a series resistor of 100K should be plenty without affecting the 12V signal integrity too much.