Mixing TTl and CMOS
Roger Hanscom
Anybody out there care to explain what is happening to my 5-volt
(7805) regulated power supply on a TTL circuit, when I add one
CMOS chip?? The TTL is basically a counter/display circuit, and
I'm trying to feed it (cleaned up) 60 Hz (from AC) to count. A 4093
does a pretty good job of cleaning up the rectified 60 Hz, but for
convenience (lazy??) I supply the 4093 with the TTL +5 and ground.
Any of the 60 Hz signal over 5+ (it's a little less than 10 volts!)
gets into the +5 supply. That's a pretty mean ripple, and it's no
surprise that the TTL ceases to function under these conditions.
I know, I know....I should be using a 74LS14 for this, and I did
switch over to one, but now I'm curious. Also the 4093 seems far
more robust. I could feed the 4093 with about 10v, but then the
resulting waveform varies from 0 to about 10v. There must be an
easy way to handle this!!
roger r...@lll-lcc.llnl.gov
Roy Smith
> Any of the 60 Hz signal over 5+ (it's a little less than 10 volts!)
> gets into the +5 supply.
Why don't you clamp the 60 Hz signal with a 5V ziener diode before
feeding it to the CMOS/TTL logic section? Something like:
|\ R1 R2
---) (------------| >----/\/\--+---------------/\/\-----> To Logic
110v ) ( 10v 60 Hz |/ D1 _|_|
---) (--+ |/ \ 5V ziener
| ---
V |
V
D1 is a half-wave rectifier. R1 is some appropriate small value
for current limiting, as determined by the current ratings of D1 and the
ziener. R2 is another optional current limiter; it's probably not
necessary if the first gate is CMOS. You probably want the first gate to
be a schmidt trigger, to avoid stuttering with the slow risetime pseudo
square wave you're generating. To make sure the 60 Hz signal really gets
clipped at 0V on the low end, you might want to add a load resistor and/or
a clamping diode to ground in parallel with the ziener.
--
Roy Smith, Public Health Research Institute
455 First Avenue, New York, NY 10016
r...@alanine.phri.nyu.edu -OR- {att,cmcl2,rutgers,hombre}!phri!roy
"Arcane? Did you say arcane? It wouldn't be Unix if it wasn't arcane!"
Henry Spencer
>does a pretty good job of cleaning up the rectified 60 Hz, but for
>convenience (lazy??) I supply the 4093 with the TTL +5 and ground.
>Any of the 60 Hz signal over 5+ (it's a little less than 10 volts!)
Feeding almost any chip a signal at a voltage beyond the power rails
is a big no-no. In this case, what is happening is that the protective
diodes on the chip are shunting the excess voltage away from the chip,
in the process dumping it onto the supply rail. If the input signal is
hefty enough that it makes a significant difference to your +5, you should
not be relying on those poor little protective diodes to carry it. The
prognosis is for very short chip life. You need to either run the 4093
at a higher voltage or clip the incoming signal to +5. Try a series
resistor, maybe 10k, followed by a diode, say a 1N914, to the +5 rail
(cathode on the +5). The diode will take most of the stress off the
on-chip protective diodes, and the resistor will limit current to keep
the diode alive and keep your +5 from bouncing around.
There are probably better ways, but what do you expect from a software
guy dabbling in hardware? :-)
--
If OSI is the answer, what is | Henry Spencer at U of Toronto Zoology
the question?? -Rolf Nordhagen| uunet!attcan!utzoo!henry he...@zoo.toronto.edu
Dana H. Myers
>
>I'm trying to feed it (cleaned up) 60 Hz (from AC) to count. A 4093
>does a pretty good job of cleaning up the rectified 60 Hz, but for
>convenience (lazy??) I supply the 4093 with the TTL +5 and ground.
>Any of the 60 Hz signal over 5+ (it's a little less than 10 volts!)
>gets into the +5 supply. That's a pretty mean ripple, and it's no
>surprise that the TTL ceases to function under these conditions.
I'm a little surprised the 4093 doesn't cease to function permanently
under these conditions... What appears to be happening is you are exceeding
the Absolute Maximum Rating for input voltage, and it clamps into the
power supply rail. The 'Absolute Maximum' ratings aren't jokes. Limit the
60 Hz waveform to LESS THAN THE SUPPLY VOLTAGE. It also sounds like your
10 VAC 60Hz is coming from a low impedance source (like a maybe directly
from a transformer winding?) and is modulating the power rail, using the
substrate of the 4093 as a diode.
>I know, I know....I should be using a 74LS14 for this, and I did
>switch over to one, but now I'm curious. Also the 4093 seems far
>more robust. I could feed the 4093 with about 10v, but then the
>resulting waveform varies from 0 to about 10v. There must be an
>easy way to handle this!!
Yeah, you could use an LM311 comparator as a zero crossing detector,
run it off of the +5 supply, use a resistive voltage divider on the 10VAC
to get the amplitude down to about 1 VAC, couple this via a .01 - .1 uF
cap to an input of the comparator, bias both of the comparator pins to the
same voltage (1/2 of +5 is fine) through a 10k resistor. Use some positive
feedback on the comparator to avoid oscillations.
*****************************************************************
* Dana H. Myers KK6JQ | Views expressed here are *
* (213) 337-5136 (ex WA6ZGB) | mine and do not necessarily *
* da...@locus.com | reflect those of my employer *
*****************************************************************
Dana H. Myers
>In article <28...@lll-lcc.UUCP> r...@lll-lcc.UUCP (Roger Hanscom) writes:
>>I'm trying to feed it (cleaned up) 60 Hz (from AC) to count. A 4093
>>does a pretty good job of cleaning up the rectified 60 Hz, but for
>>convenience (lazy??) I supply the 4093 with the TTL +5 and ground.
>>Any of the 60 Hz signal over 5+ (it's a little less than 10 volts!)
>>gets into the +5 supply...
>
[ .... perfectly correct observation about clamping diodes deleted ...]
>at a higher voltage or clip the incoming signal to +5. Try a series
>resistor, maybe 10k, followed by a diode, say a 1N914, to the +5 rail
>(cathode on the +5). The diode will take most of the stress off the
>on-chip protective diodes, and the resistor will limit current to keep
>the diode alive and keep your +5 from bouncing around.
The 1N914 conducts at about .6 to .7 V of forward voltage, thus clamping
the input voltage to 5.6 to 5.7 volts, which is in excess of the 'Absolute
Maximum' rating of CMOS (and most other chips) which is Vdd + .5V (5.5V
for most folks). The substrate diodes will still be taking the hit. I
suggested a zero-crossing detector, but using a 5V Zener to clamp it
may be easiest.
>There are probably better ways, but what do you expect from a software
>guy dabbling in hardware? :-)
Well, Henry, with the disclaimer, I won't let the above oversight impinge
upon your near legendary status as the Omniscient One :-)
W Swan
>Anybody out there care to explain what is happening to my 5-volt
>(7805) regulated power supply on a TTL circuit, when I add one
>I supply the 4093 with the TTL +5 and ground.
>Any of the 60 Hz signal over 5+ (it's a little less than 10 volts!)
> roger r...@lll-lcc.llnl.gov
What is probably happening is that the CMOS device has a diode clamp on its
input, provided for some static protection roughly as follows (I believe the
RCA databooks show the actual input circuit):
+5
|
__|__
^
/ \
`-+-' |
| |--
in: 0----*------------| (input FET)
|--
|
When you drive the input higher than the +5 rail, the diode becomes
forward biased and starts to pull the +5 line up. I believe the 7805 has
only a pass transistor - it regulates the voltage by reducing the current
it sources, not by sinking "excess" current, so if you are able to put
enough current through the clamp diode (likely in a system small enough
to be powered by a 7805), you could pull the power supply rail voltage up.
One bet for a simple fix is a resistor divider on the input from the 10v
AC signal. If I understand right and you're getting 0 to 10 volts (that
is, 10v p-p on 5V DC) now try, say, a 4.7k/5.3k resistor divider to get 0
to 4.7 volts.
Henry Spencer
> The 1N914 conducts at about .6 to .7 V of forward voltage, thus clamping
>the input voltage to 5.6 to 5.7 volts, which is in excess of the 'Absolute
Hmm, right you are. I'd been working with something slightly more tolerant
and had forgotten how tight the CMOS limits are.
In practice, limiting the current with a series resistor is probably enough
to avoid serious trouble -- the absolute-maximum rating is for a low-impedance
source -- but it's still dubious practice. Tighter clamping or a different
approach would be better.
Jim Cathey
>In article <28...@lll-lcc.UUCP> r...@lll-lcc.UUCP (Roger Hanscom) writes:
>>The TTL is basically a counter/display circuit, and
>>I'm trying to feed it (cleaned up) 60 Hz (from AC) to count. A 4093
> Yeah, you could use an LM311 comparator as a zero crossing detector,
>run it off of the +5 supply, use a resistive voltage divider on the 10VAC...
What will probably work even better for the application is to wire up a
simple RC oscillator that runs near 60Hz using a CMOS schmitt-trigger
inverter (three components) and hook a resistor from the timing cap to
the 10VAC from the line (to serve as an external synchronizer). This
way most hash on the power line will be filtered out so that you don't
end up counting the spikes every time the refrigerator kicks in. I
saw this trick in an EDN magazine a few years ago.
+----------------+
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! II CC ! TAF-C8; Spokane, WA 99220
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! II CCCCCC ! (509) 927-5757
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ke...@m.cs.uiuc.edu
You *can* exceed the `absolute maximum ratings' of CMOS devices if you
take care to limit current; the protection diodes will generally handle
about 10 mA without any problems.
For conditioning the power line to a logic signal, I'd recommend
something like
+5
|
100K
| Schmitt
12 Vrms ---- 0.1 uF --- 1 meg ---+-----+----- trigger ----- 60 Hz squarewave
from power | |
xformer .01 uF 100K
| |
GND GND
The Schmitt trigger may be a 4093 or 4584, or may be built from a 4050
or 4001. The 4001 circuit looks like:
.------ 100K -----------------------.
| ___ |
Input ---- 22K ----+---+--| \ ___ |
| | >o---------+--| \ |
+--|___/ 1/4 4001 | | >o---+--- Output
+--|___/ 1/4 4001
(Any non-inverting gate configuration will work).
Kevin, KE9TV
ke...@cs.uiuc.edu
ftp...@acad3.fai.alaska.edu
I used a similar circuit to generate the sync pulse for a lighting control
system using TRIAC's as the main switching elements. The transformers I used
had severe phase shift. I now use AC input optocouplers (H11AA4 or equivalent)
with series input resistors in both legs (hot and neutral). Using TRIAC driver
optocouplers (MOC3010 or equivalent) on the outputs means the logic is isolated
at both ends from the mains to 2500V or so.
Philip Munts N7AHL
NRA Extremist, etc.
University of Alaska, Fairbanks
John Whitmore
>>In article <28...@lll-lcc.UUCP> r...@lll-lcc.UUCP (Roger Hanscom) writes:
>>>I'm trying to feed it (cleaned up) 60 Hz (from AC) to count.
>>>gets into the +5 supply...
>>
>>resistor, maybe 10k, followed by a diode, say a 1N914, to the +5 rail
>
>the input voltage to 5.6 to 5.7 volts, which is in excess of the 'Absolute
The diodes aren't exactly 'substrate' diodes, but are protection
diodes (i.e. tthey're not just there by accident!). These diodes are
perfectly OK with currents of .1mA and below (see RCA CMODatabook)
so the proposed 10k resistor can be changed to a 100k resistor and
make just about everyone happy. The absolute max voltage spec is for
a low impedance voltage source; at high temperatures, that much
overvoltage could burn out the protection diodes by going 'way over
their current rating. If you limit the current, this
danger is averted.
I am known for my brilliance, John Whitmore
by those who do not know me well.
ke...@m.cs.uiuc.edu
Philip Munts, N7AHL:
>I used a similar circuit to generate the sync pulse for a lighting
>control system using TRIAC's as the main switching elements. The
>transformers I used had severe phase shift. I now use AC input
>optocouplers (H11AA4 or equivalent) with series input resistors in
>both legs (hot and neutral). Using TRIAC driver optocouplers (MOC3010
>or equivalent) on the outputs means the logic is isolated at both ends
>from the mains to 2500V or so.
If you need to have the clock in phase with the power line, that's the
only way to fly. Nevertheless, H11AA4's are perhaps a trifle exotic
for a hobbyist who's just trying to get a 60 Hz clock for free out of
the power supply, and who's working mostly with junk parts; the
clock/counter application that the original poster described sounded
like that, and also fairly obviously doesn't care about being
precisely in phase with the power line. The circuit I described has
total cost approximately zero to a hobbyist who knows what he's doing.
BTW, DigiKey does stock H11AA4's; they're about five bucks in unit
quantities (I have mislaid my latest catalog, so I can't quote the
current price).
Kevin, KE9TV
ke...@cs.uiuc.edu
Bruce Walker
> (7805) regulated power supply on a TTL circuit, when I add one
> CMOS chip?? The TTL is basically a counter/display circuit, and
> I'm trying to feed it (cleaned up) 60 Hz (from AC) to count. A 4093
> does a pretty good job of cleaning up the rectified 60 Hz, but for
> convenience (lazy??) I supply the 4093 with the TTL +5 and ground.
> Any of the 60 Hz signal over 5+ (it's a little less than 10 volts!)
> gets into the +5 supply. That's a pretty mean ripple, and it's no
> surprise that the TTL ceases to function under these conditions.
Put a 1 Meg resistor in series with the input to the 4093. You
probably were allowing too much current to flow from the AC input
through the protection diodes in the 4093 to the +5 supply.
+5
_
|
in +------+ out
10V AC ----------/\/\/\/--------| 4093 |o-------- to TTL
60 Hz 1M0 +------+
|
V
GND
--
Bruce Walker ...uunet!utai!lsuc!isgtec!bmw b...@isgtec.uucp
"Remember Rule Number 79: When the tough get going, the weak get screwed."
ISG Technologies Inc. 3030 Orlando Dr. Mississauga. Ont. Can. L4V 1S8
Phillip Harbison
> In article <86...@oolong.la.locus.com> Dana H. Myers writes:
> > The 1N914 conducts at about .6 to .7 V of forward voltage, thus
> > clamping the input voltage to 5.6 to 5.7 volts, which is in excess
> > of the 'Absolute Maximum' rating of CMOS ...
You might also check out Schottky diodes. I believe they have a much
lower forward voltage drop (as low as 0.2 volts). You might also run
your AC signal through a voltage divider. We used to build this type
of circuit alot in telecom circuits (usually for ring detection). A
typical approach was as follows:
+----/\/\/\-----+--->|--+
| R1 | D1 |
| | |
Vin +---|<--+
| O1 |
| |
+-----------------------+
O1 is the diode half of an opto-isolator. D1 is in the circuit so that
the opto never has to endure much reverse bias. R1 should be sized such
that the current is enough to turn on the opto even at the minimum input
voltage, but high enough to limit the current and keep either diode from
going into meltdown. You will probably have to integrate the output of
the optoisolator, either with an analog integrator or by using the opto
output to fire a one-shot or trigger a flip-flop.
To be really safe, you could put a varistor across the input to safe-
guard against voltages that are outside the expected reasonable limits,
e.g. you designed for 110V-140V but you took a lightning hit. :-(
--
Live: Phil Harbison, Xavax, P.O. Box 7413, Huntsville, AL 35807
Uucp: alv...@xavax.com
Bell: 205-539-1672, 205-880-8951
Michael Kirschner
>
>You *can* exceed the `absolute maximum ratings' of CMOS devices if you
>take care to limit current; the protection diodes will generally handle
>about 10 mA without any problems.
...with the caveat that the "CMOS device" is on the order of a 4000-series
metal-gate device with 5 or 10 micron channel lengths, deep source-drain
diffusions, thick gate oxide, etc., and you don't really care if your
circuit is still working next year.
The purpose of "Absolute Maximum Ratings" is to ensure that the device
functions reliably for years and years; that the device operates outside
the "safe" limits of 4.5v to 5.5v is to ensure yield and reliability.
It's certainly the case that the device will work up to an astoundingly
high voltage (even complex CMOS microprocessors, too. They get lots
faster with higher voltage). However, you severely compromise the
reliability. A 386 may run at 70MHz at 6 or 7 volts Vcc (guess), but after an
hour or so of that you may find some of your AC or DC parameters have
shifted due to hot electron injection. Run it long enough and the
parameters will shift enough to violate some internal or external timing
and your circuit will crash. Current increases, as well. See the next
paragraph.
The dangers of dumping more current thru an internal metal line than it
was designed to withstand (your 10mA thru the input protection diode,
for example) is the danger of electromigration-related open in months
or years. The flow of current actually moves aluminum atoms to create
voids (and hillocks). Eventually metal lines will open up.
Higher power dissipation also results in higher die temperature. Higher
temperatures induce higher failure rates in components by accelerating
thermally-activated failure mechanisms, like various types of bond wire
failure.
All IC manufacturers (I hope) understand the reliability-related aspects
of their products and incorporate some margin into the design in order
to produce a reliable product. Lots of people mistake this margin for
poor binning and lost yield ("Hey! my 10MHz part goes 12MHz!") and end
up pushing the part beyond its specs. This is dangerous and is usually
disclaimed by the manufacturer.
Mike Kirschner: EMAIL SMAIL
...uunet!ingr!apd!obnoid or ! Intergraph Corp. APD Who is garth?
{pyramid,sri-unix}!garth!obnoid or ! 2400 Geng Rd. What are BUBFETS?
garth!obn...@ingr.com (Internet) ! Palo Alto, CA 94303 FONE:(415)852-2392 (W)
obn...@well.uucp ! (415)BUB-FETS (H)
[Disclaimer? Who cares what my opinions are?]
ke...@m.cs.uiuc.edu
I write:
>You *can* exceed the `absolute maximum ratings' of CMOS devices if you
>take care to limit current; the protection diodes will generally handle
>about 10 mA without any problems.
and obn...@garth.UUCP (Mike Kirschner) gives a whole collection of
good reasons *not* to. Reread his posting; he's right -- for the
domain that he's talking about.
The problem here is that we're dealing with two people with
fundamentally different objectives -- the professional and the
hobbyist.
A professional who exceeds the manufacturers' ratings for the parts
that she uses should be shot. (With the caveat that some
manufacturers' ratings basically state, `we don't actually claim that
this device will work in any real circuit', so you have to read
between the lines a bit.)
The hobbyist, on the other hand, isn't dealing in general with devices
that are critical to the safety of life or property, or even critical
to the reputation of the manufacturer. There is therefore some point
in discussing the considerations of `this rating is an absolute max --
don't even think about violating it or the device will make smoke,'
versus `this rating is a conservative one, and you can push things a
little.'
In the case that I was discussing, that of clamping the input of a
CMOS gate to the supply rails, the basic circuit for either design
looks like
+V
|
-+-
/_\
|
----/\/\/\/---+---> to gate
|
-+-
/_\
|
GND
A professional gets the data sheet on the gate, sees that the voltage
limit is the supply rails, and specifies a clamp made out of 1N4731A
or 1N5229B Zener diodes. (If you're using other than a 5V supply,
substitute the correct diodes for the 4.3V ones mentioned.) She
calculates currents and power dissipations, and estimates the stray
capacitance of the circuit, and chooses the appropriate resistor value
for the speed required. A lot of work, and some parts that the
hobbyist won't have lying around. But we're certain that the circuit
works Come What May.
A hobbyist knows that he won't degrade gate reliability grossly by
having an overvoltage, as long as the current is low enough, so he
clamps with the 1N914's that he has in his junk box, uses a 47K
resistor (also from junk) on the input, maybe also ties a 1K between
the clamp and the gate, and figures that's good enough. Most of the
time, he'll be right. If he isn't, no harm done; he swears, replaces
the gate, and does the clamp over the way the pro did it the first
time around. (After going out and buying the 1N5229B's).
Some projects need to be designed to MIL-SPEC. Some don't.
Kevin, KE9TV
ke...@cs.uiuc.edu
Dan Dunphy
60Hz oscillator with a 555 timer.
David Chapman
>In article <2100...@m.cs.uiuc.edu> ke...@m.cs.uiuc.edu writes:
>>
>>You *can* exceed the `absolute maximum ratings' of CMOS devices if you
>>take care to limit current; the protection diodes will generally handle
>>about 10 mA without any problems.
>...with the caveat that the "CMOS device" is on the order of a 4000-series
>metal-gate device with 5 or 10 micron channel lengths, deep source-drain
>diffusions, thick gate oxide, etc., and you don't really care if your
>circuit is still working next year.
>The dangers of dumping more current thru an internal metal line than it
>was designed to withstand (your 10mA thru the input protection diode,
>for example) is the danger of electromigration-related open in months
>or years. The flow of current actually moves aluminum atoms to create
>voids (and hillocks). Eventually metal lines will open up.
Metal migration is bad enough, but the problem is that it's driven by
current density. So if you get necking at some point, the current
density at that point is a bit higher than anywhere else. And that
causes the metal to migrate faster, which narrows the wire even more,
which increases the current density still more,... A nice positive
feedback loop (if you sell ICs :-).
An amusing story (I won't name names because it will embarrass a rather
well-known company): on our old HMOS process, we spec'd a current density
of 1.0 mA/micron (of wire width). One of our customers, who had a
reputation for pushing process limits, came to us complaining that our
chips were all dying in the field after 6 months. Obviously we had a
reliability problem in our metalization, because they'd tracked the
failures to metal migration (voids and bubbles in metal make more spots
for metal migration, because the wire enters life too narrow, so if you
have a lot of defects, reliability and lifetime drop as does initial sort
yield).
Anyway, our product engineer asked how much current they were sending
through that wire.
Them: "3 mA/micron. Why?"
Us: "You can't do that. The limit is 1 mA/micron."
Them: "Oh, you underestimate the current carrying capacity of your
metallization."
Oh, no we don't. And they had to redesign their chip.
More embarrassment: as a hack they asked us to set up an HMOS process
with two layers of metal, and added a metal mask with some nice wide
wires to supply the necessary current (they could still use the old masks,
saving $10K or more, and perhaps wafers partway through the line). That
was fine, but they made the wires so wide that the extra capacitance slowed
the chip down! I don't know what happened after that, but it still makes
me chuckle because that company caused me a lot of grief when I worked with
them.
--
David Chapman
{known world}!decwrl!vlsisj!fndry!davidc
vlsisj!fndry!dav...@decwrl.dec.com