cmos IC question

[default]

I'm using some cmos IC gates in a project. All seem to have an
internal protection network to insure that the inputs can't exceed
supply or go more negative than ground with internal diodes.

My question has to do with those protection diode ratings. How much
current can they clamp to prevent exceeding the input voltage rating?

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input. (that's a no-no according to absolute maximum ratings)
HOWEVER, the 12V side is an open collector output with a 10K pull-up
resistor. It works... so apparently the internals diodes don't have a
problem clamping ~400 microamps, but I'd feel better knowing what the
spec. is and haven't found one.

The datasheet shows two sets of protection diodes for every input with
a current limiting resistor to further protect the inputs.

 

[George Herold]

On Tuesday, April 9, 2019 at 7:32:10 AM UTC-4, default wrote:

I'm using some cmos IC gates in a project. All seem to have an
internal protection network to insure that the inputs can't exceed
supply or go more negative than ground with internal diodes.

My question has to do with those protection diode ratings. How much
current can they clamp to prevent exceeding the input voltage rating?

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input. (that's a no-no according to absolute maximum ratings)
HOWEVER, the 12V side is an open collector output with a 10K pull-up
resistor. It works... so apparently the internals diodes don't have a
problem clamping ~400 microamps, but I'd feel better knowing what the
spec. is and haven't found one.

The datasheet shows two sets of protection diodes for every input with
a current limiting resistor to further protect the inputs.

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.

 

[o pere o]

On 9/4/19 13:32, default wrote:

I'm using some cmos IC gates in a project. All seem to have an
internal protection network to insure that the inputs can't exceed
supply or go more negative than ground with internal diodes.

My question has to do with those protection diode ratings. How much
current can they clamp to prevent exceeding the input voltage rating?

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input. (that's a no-no according to absolute maximum ratings)
HOWEVER, the 12V side is an open collector output with a 10K pull-up
resistor. It works... so apparently the internals diodes don't have a
problem clamping ~400 microamps, but I'd feel better knowing what the
spec. is and haven't found one.

The datasheet shows two sets of protection diodes for every input with
a current limiting resistor to further protect the inputs.

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

Pere

 

[George Herold]

On Tuesday, April 9, 2019 at 9:33:32 AM UTC-4, o pere o wrote:

On 9/4/19 13:32, default wrote:
I'm using some cmos IC gates in a project. All seem to have an
internal protection network to insure that the inputs can't exceed
supply or go more negative than ground with internal diodes.

My question has to do with those protection diode ratings. How much
current can they clamp to prevent exceeding the input voltage rating?

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input. (that's a no-no according to absolute maximum ratings)
HOWEVER, the 12V side is an open collector output with a 10K pull-up
resistor. It works... so apparently the internals diodes don't have a
problem clamping ~400 microamps, but I'd feel better knowing what the
spec. is and haven't found one.

The datasheet shows two sets of protection diodes for every input with
a current limiting resistor to further protect the inputs.


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

 

[default]

On Tue, 9 Apr 2019 05:57:21 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Tuesday, April 9, 2019 at 7:32:10 AM UTC-4, default wrote:
I'm using some cmos IC gates in a project. All seem to have an
internal protection network to insure that the inputs can't exceed
supply or go more negative than ground with internal diodes.

My question has to do with those protection diode ratings. How much
current can they clamp to prevent exceeding the input voltage rating?

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input. (that's a no-no according to absolute maximum ratings)
HOWEVER, the 12V side is an open collector output with a 10K pull-up
resistor. It works... so apparently the internals diodes don't have a
problem clamping ~400 microamps, but I'd feel better knowing what the
spec. is and haven't found one.

The datasheet shows two sets of protection diodes for every input with
a current limiting resistor to further protect the inputs.

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.

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

http://www.ti.com/lit/ds/symlink/cd4520b.pdf

Page 4 shows the protection network

 

[George Herold]

On Tuesday, April 9, 2019 at 11:50:13 AM UTC-4, default wrote:

On Tue, 9 Apr 2019 05:57:21 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, April 9, 2019 at 7:32:10 AM UTC-4, default wrote:
I'm using some cmos IC gates in a project. All seem to have an
internal protection network to insure that the inputs can't exceed
supply or go more negative than ground with internal diodes.

My question has to do with those protection diode ratings. How much
current can they clamp to prevent exceeding the input voltage rating?

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input. (that's a no-no according to absolute maximum ratings)
HOWEVER, the 12V side is an open collector output with a 10K pull-up
resistor. It works... so apparently the internals diodes don't have a
problem clamping ~400 microamps, but I'd feel better knowing what the
spec. is and haven't found one.

The datasheet shows two sets of protection diodes for every input with
a current limiting resistor to further protect the inputs.

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.

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

http://www.ti.com/lit/ds/symlink/cd4520b.pdf

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.

 

[default]

On Tue, 9 Apr 2019 11:43:21 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Tuesday, April 9, 2019 at 11:50:13 AM UTC-4, default wrote:
On Tue, 9 Apr 2019 05:57:21 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, April 9, 2019 at 7:32:10 AM UTC-4, default wrote:
I'm using some cmos IC gates in a project. All seem to have an
internal protection network to insure that the inputs can't exceed
supply or go more negative than ground with internal diodes.

My question has to do with those protection diode ratings. How much
current can they clamp to prevent exceeding the input voltage rating?

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input. (that's a no-no according to absolute maximum ratings)
HOWEVER, the 12V side is an open collector output with a 10K pull-up
resistor. It works... so apparently the internals diodes don't have a
problem clamping ~400 microamps, but I'd feel better knowing what the
spec. is and haven't found one.

The datasheet shows two sets of protection diodes for every input with
a current limiting resistor to further protect the inputs.

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.

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

http://www.ti.com/lit/ds/symlink/cd4520b.pdf

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.

 

[jfeng@my-deja.com]

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.

 

[default]

On Tue, 9 Apr 2019 11:43:21 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Tuesday, April 9, 2019 at 11:50:13 AM UTC-4, default wrote:
On Tue, 9 Apr 2019 05:57:21 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, April 9, 2019 at 7:32:10 AM UTC-4, default wrote:
I'm using some cmos IC gates in a project. All seem to have an
internal protection network to insure that the inputs can't exceed
supply or go more negative than ground with internal diodes.

My question has to do with those protection diode ratings. How much
current can they clamp to prevent exceeding the input voltage rating?

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input. (that's a no-no according to absolute maximum ratings)
HOWEVER, the 12V side is an open collector output with a 10K pull-up
resistor. It works... so apparently the internals diodes don't have a
problem clamping ~400 microamps, but I'd feel better knowing what the
spec. is and haven't found one.

The datasheet shows two sets of protection diodes for every input with
a current limiting resistor to further protect the inputs.

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.

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

http://www.ti.com/lit/ds/symlink/cd4520b.pdf

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.
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.

 

[default]

On Tue, 9 Apr 2019 14:25:38 -0700 (PDT), "jfeng@my-deja.com"
<jfeng@my-deja.com> wrote:

>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.

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.

 

[John Larkin]

On Tue, 09 Apr 2019 07:32:12 -0400, default <default@defaulter.net>
wrote:

I'm using some cmos IC gates in a project. All seem to have an
internal protection network to insure that the inputs can't exceed
supply or go more negative than ground with internal diodes.

My question has to do with those protection diode ratings. How much
current can they clamp to prevent exceeding the input voltage rating?

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input. (that's a no-no according to absolute maximum ratings)
HOWEVER, the 12V side is an open collector output with a 10K pull-up
resistor. It works... so apparently the internals diodes don't have a
problem clamping ~400 microamps, but I'd feel better knowing what the
spec. is and haven't found one.

The datasheet shows two sets of protection diodes for every input with
a current limiting resistor to further protect the inputs.

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

 

[Jasen Betts]

On 2019-04-09, default <default@defaulter.net> wrote:

I'm using some cmos IC gates in a project. All seem to have an
internal protection network to insure that the inputs can't exceed
supply or go more negative than ground with internal diodes.

My question has to do with those protection diode ratings. How much
current can they clamp to prevent exceeding the input voltage rating?

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input. (that's a no-no according to absolute maximum ratings)
HOWEVER, the 12V side is an open collector output with a 10K pull-up
resistor. It works... so apparently the internals diodes don't have a
problem clamping ~400 microamps, but I'd feel better knowing what the
spec. is and haven't found one.

The datasheet shows two sets of protection diodes for every input with
a current limiting resistor to further protect the inputs.

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.

 

[default]

On Tue, 09 Apr 2019 17:32:05 -0700, John Larkin
<jjlarkin@highland_snip_technology.com> wrote:

On Tue, 09 Apr 2019 07:32:12 -0400, default <default@defaulter.net
wrote:

I'm using some cmos IC gates in a project. All seem to have an
internal protection network to insure that the inputs can't exceed
supply or go more negative than ground with internal diodes.

My question has to do with those protection diode ratings. How much
current can they clamp to prevent exceeding the input voltage rating?

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input. (that's a no-no according to absolute maximum ratings)
HOWEVER, the 12V side is an open collector output with a 10K pull-up
resistor. It works... so apparently the internals diodes don't have a
problem clamping ~400 microamps, but I'd feel better knowing what the
spec. is and haven't found one.

The datasheet shows two sets of protection diodes for every input with
a current limiting resistor to further protect the inputs.

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.

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.

http://www.ti.com/lit/ds/symlink/cd4520b.pdf

http://www.ti.com/lit/ds/symlink/cd4520b.pdf

 

[George Herold]

On Tuesday, April 9, 2019 at 6:01:02 PM UTC-4, default wrote:

On Tue, 9 Apr 2019 11:43:21 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, April 9, 2019 at 11:50:13 AM UTC-4, default wrote:
On Tue, 9 Apr 2019 05:57:21 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, April 9, 2019 at 7:32:10 AM UTC-4, default wrote:
I'm using some cmos IC gates in a project. All seem to have an
internal protection network to insure that the inputs can't exceed
supply or go more negative than ground with internal diodes.

My question has to do with those protection diode ratings. How much
current can they clamp to prevent exceeding the input voltage rating?

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input. (that's a no-no according to absolute maximum ratings)
HOWEVER, the 12V side is an open collector output with a 10K pull-up
resistor. It works... so apparently the internals diodes don't have a
problem clamping ~400 microamps, but I'd feel better knowing what the
spec. is and haven't found one.

The datasheet shows two sets of protection diodes for every input with
a current limiting resistor to further protect the inputs.

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.

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

http://www.ti.com/lit/ds/symlink/cd4520b.pdf

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.
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.

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

George H.

 

[John Larkin]

On Wed, 10 Apr 2019 05:53:32 -0400, default <default@defaulter.net>
wrote:

On Tue, 09 Apr 2019 17:32:05 -0700, John Larkin
jjlarkin@highland_snip_technology.com> wrote:

On Tue, 09 Apr 2019 07:32:12 -0400, default <default@defaulter.net
wrote:

I'm using some cmos IC gates in a project. All seem to have an
internal protection network to insure that the inputs can't exceed
supply or go more negative than ground with internal diodes.

My question has to do with those protection diode ratings. How much
current can they clamp to prevent exceeding the input voltage rating?

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input. (that's a no-no according to absolute maximum ratings)
HOWEVER, the 12V side is an open collector output with a 10K pull-up
resistor. It works... so apparently the internals diodes don't have a
problem clamping ~400 microamps, but I'd feel better knowing what the
spec. is and haven't found one.

The datasheet shows two sets of protection diodes for every input with
a current limiting resistor to further protect the inputs.

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.

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.

http://www.ti.com/lit/ds/symlink/cd4520b.pdf

http://www.ti.com/lit/ds/symlink/cd4520b.pdf

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

 

[whit3rd]

On Tuesday, April 9, 2019 at 8:50:13 AM UTC-7, default wrote:

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input.

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.

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

Texas Instruments CD4520B
Dual up counter

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.

 

[whit3rd]

On Tuesday, April 9, 2019 at 3:01:02 PM UTC-7, default wrote:

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.

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.

 

[George Herold]

On Thursday, April 11, 2019 at 5:16:49 PM UTC-4, whit3rd wrote:

On Tuesday, April 9, 2019 at 3:01:02 PM UTC-7, default wrote:

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.

No, not if you want to have a 12V signal amplitude. That was about burning

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.

the chip up, not about signal integrity.
It's the value in 'absolute maximum' for current, and relates to bond wire
current carrying capacity.

 

[default]

On Thu, 11 Apr 2019 14:13:01 -0700 (PDT), whit3rd <whit3rd@gmail.com>
wrote:

On Tuesday, April 9, 2019 at 8:50:13 AM UTC-7, default wrote:

For instance, I'm running one part of the circuit at 8 volts and
another at 12. It is necessary to input a 12volt output to a 8 volt
input.

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.

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

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)

Texas Instruments CD4520B
Dual up counter

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.

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.

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.

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.