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 Robert Baer 06-15-2013 06:19 AM

Printer port question

According to the original IBM documentation viz IBM Technical
Reference First Edition August 1981 page D-34 Parallel Printer Adapter
schematic, the following pins of the DB-25 printer connector are
exclusively inputs:
pin 13 is +SLCT (select, high for "1")
pin 11 is +BUSY (busy, high for "1")
pin 12 is +PE (paper out, high for "1")
pin 10 is -ACK (acknowledge, low for "1"; eg: inverted logic)

So, how come with computer NOT CONNECTED to anything (NO power at
all), one sees that pins 10, 11 and 12 are SHORTED TO GROUND?

Please explain how a printer can possibly work with that condition.

 Jeff Strickland 06-15-2013 03:24 PM

Re: Printer port question

"Robert Baer" <robertbaer@localnet.com> wrote in message
> According to the original IBM documentation viz IBM Technical Reference
> First Edition August 1981 page D-34 Parallel Printer Adapter schematic,
> the following pins of the DB-25 printer connector are exclusively inputs:
> pin 13 is +SLCT (select, high for "1")
> pin 11 is +BUSY (busy, high for "1")
> pin 12 is +PE (paper out, high for "1")
> pin 10 is -ACK (acknowledge, low for "1"; eg: inverted logic)
>
> So, how come with computer NOT CONNECTED to anything (NO power at all),
> one sees that pins 10, 11 and 12 are SHORTED TO GROUND?
>
> Please explain how a printer can possibly work with that condition.

Because they are held high by the printer that's not there? The printer
pulls these lines high.

 Robert Baer 06-15-2013 05:05 PM

Re: Printer port question

Jeff Strickland wrote:
>
> "Robert Baer" <robertbaer@localnet.com> wrote in message
>> According to the original IBM documentation viz IBM Technical
>> Reference First Edition August 1981 page D-34 Parallel Printer Adapter
>> schematic, the following pins of the DB-25 printer connector are
>> exclusively inputs:
>> pin 13 is +SLCT (select, high for "1")
>> pin 11 is +BUSY (busy, high for "1")
>> pin 12 is +PE (paper out, high for "1")
>> pin 10 is -ACK (acknowledge, low for "1"; eg: inverted logic)
>>
>> So, how come with computer NOT CONNECTED to anything (NO power at
>> all), one sees that pins 10, 11 and 12 are SHORTED TO GROUND?
>>
>> Please explain how a printer can possibly work with that condition.

>
>
>
> Because they are held high by the printer that's not there? The printer
> pulls these lines high.
>
>
>
>
>
>
>

On the surface,that _might_ sound reasonable..
Except.
The schematic shows INPUTS to a TTL gate for each one.
On TTL logic, an input must be pulled down (below 1.4V) for a low to
be asserted; a floating input on a gate acts like a high.

 Paul Drahn 06-15-2013 08:41 PM

Re: Printer port question

On 6/14/2013 11:19 PM, Robert Baer wrote:
> According to the original IBM documentation viz IBM Technical Reference
> First Edition August 1981 page D-34 Parallel Printer Adapter schematic,
> the following pins of the DB-25 printer connector are exclusively inputs:
> pin 13 is +SLCT (select, high for "1")
> pin 11 is +BUSY (busy, high for "1")
> pin 12 is +PE (paper out, high for "1")
> pin 10 is -ACK (acknowledge, low for "1"; eg: inverted logic)
>
> So, how come with computer NOT CONNECTED to anything (NO power at all),
> one sees that pins 10, 11 and 12 are SHORTED TO GROUND?
>
> Please explain how a printer can possibly work with that condition.

Perhaps you have a computer with a version 2 parallel printer port. Not
quite the same as the original IBM port.

Paul

 Paul 06-15-2013 08:46 PM

Re: Printer port question

Robert Baer wrote:
> Jeff Strickland wrote:
>>
>> "Robert Baer" <robertbaer@localnet.com> wrote in message
>>> According to the original IBM documentation viz IBM Technical
>>> Reference First Edition August 1981 page D-34 Parallel Printer Adapter
>>> schematic, the following pins of the DB-25 printer connector are
>>> exclusively inputs:
>>> pin 13 is +SLCT (select, high for "1")
>>> pin 11 is +BUSY (busy, high for "1")
>>> pin 12 is +PE (paper out, high for "1")
>>> pin 10 is -ACK (acknowledge, low for "1"; eg: inverted logic)
>>>
>>> So, how come with computer NOT CONNECTED to anything (NO power at
>>> all), one sees that pins 10, 11 and 12 are SHORTED TO GROUND?
>>>
>>> Please explain how a printer can possibly work with that condition.

>>
>>
>>
>> Because they are held high by the printer that's not there? The printer
>> pulls these lines high.
>>
>>
>>
>>
>>
>>
>>

> On the surface,that _might_ sound reasonable..
> Except.
> The schematic shows INPUTS to a TTL gate for each one.
> On TTL logic, an input must be pulled down (below 1.4V) for a low to
> be asserted; a floating input on a gate acts like a high.
>

I would question how you determined "SHORTED TO GROUND".

A short to ground, will read zero on any range on your ohmmeter.
Including the diode range.

If you use the diode range on the multimeter, and you're seeing
a diode drop (0.5 to 0.7V), then, it's not a "SHORT TO GROUND",
is it ?

Remember that, when a hardware circuit is not powered, the
VCC and GND appear shorted together. The protection network
on the input of the gate, then becomes involved while you're
diodes to VCC and GND (to prevent overshoot, to prevent
undershoot). And a few gates are protected by what we used
to call "up up up down", which is three diodes to one rail,
and one diode to the other rail. Which allows larger
excursions past the rail in one case than the other.

So when you characterize a circuit, you have to know something
about the internal protection structure, whether it's one
stage or two, has a series resistor between stages, and so on.
You need to know the protection structure, to interpret the
results. In some cases (inputs "open to the world"), you'll
actually see external protection diode networks, intended
to provide a measure of ESD protection perhaps. The multimeter
can end up probing the protection structure, instead of telling
you something about actual circuit operation.

You can use a variable power supply, and a resistor, to do
your own crude characterization of the circuit while it
is running. On a TTL gate, perhaps the knee is at 1.3V
on the input. Applying a voltage below 1.3V, through a
resistor, should sink a small current (400uA on a TTL
gate of modern vintage, 10uA on CMOS maybe). Applying a
voltage above 1.3V might source a small current, and the
current will likely be a smaller flow than in the logic 0
test case. It's a bit like running a curve tracer,

So there are a few things you can try, with a bit of
crude equipment. And your friends Thevenin and Ohm's Law
("It's The Law").

*******

If you see an actual shorting strap on the PCB, then
that's another matter. I don't know anything about
printer ports, so can't really comment on that.
Status inputs usually need to remain functional,
especially if they mediate the transfer rate in
some way. Shorting all the status signals would
not make sense. But if this is a question of
characterization of an unpowered logic gate,
go back and look at it again.

Paul

 Jeff Strickland 06-15-2013 09:42 PM

Re: Printer port question

"Robert Baer" <robertbaer@localnet.com> wrote in message
> According to the original IBM documentation viz IBM Technical Reference
> First Edition August 1981 page D-34 Parallel Printer Adapter schematic,
> the following pins of the DB-25 printer connector are exclusively inputs:
> pin 13 is +SLCT (select, high for "1")
> pin 11 is +BUSY (busy, high for "1")
> pin 12 is +PE (paper out, high for "1")
> pin 10 is -ACK (acknowledge, low for "1"; eg: inverted logic)
>
> So, how come with computer NOT CONNECTED to anything (NO power at all),
> one sees that pins 10, 11 and 12 are SHORTED TO GROUND?
>
> Please explain how a printer can possibly work with that condition.

Why do you care what the parallel port is doing? They have not sold a
parallel printer in almost a decade. Surely if you have a parallel printer,
you deserve a new one for Father's Day.

 Jeff Strickland 06-15-2013 09:48 PM

Re: Printer port question

"Robert Baer" <robertbaer@localnet.com> wrote in message
> Jeff Strickland wrote:
>>
>> "Robert Baer" <robertbaer@localnet.com> wrote in message
>>> According to the original IBM documentation viz IBM Technical
>>> Reference First Edition August 1981 page D-34 Parallel Printer Adapter
>>> schematic, the following pins of the DB-25 printer connector are
>>> exclusively inputs:
>>> pin 13 is +SLCT (select, high for "1")
>>> pin 11 is +BUSY (busy, high for "1")
>>> pin 12 is +PE (paper out, high for "1")
>>> pin 10 is -ACK (acknowledge, low for "1"; eg: inverted logic)
>>>
>>> So, how come with computer NOT CONNECTED to anything (NO power at
>>> all), one sees that pins 10, 11 and 12 are SHORTED TO GROUND?
>>>
>>> Please explain how a printer can possibly work with that condition.

>>
>>
>>
>> Because they are held high by the printer that's not there? The printer
>> pulls these lines high.
>>
>>
>>
>>
>>
>>
>>

> On the surface,that _might_ sound reasonable..
> Except.
> The schematic shows INPUTS to a TTL gate for each one.
> On TTL logic, an input must be pulled down (below 1.4V) for a low to be
> asserted; a floating input on a gate acts like a high.
>

If there is a gate array, then the gate is low until a printer comes along
to pull the inputs high when the condition becomes true. If the printer is
available, the select line goes high, if the printer is busy, the busy line
goes high, if whatever is happening, the line goes high, or it goes low in
the case of the not-logic.

I don't understand the question. I really don't understand why you think
something is broken on a port that has not been used in any current
technology in the past 15 or 20 years. I guess that's not true, I'm using a
Brother MFC that is connected to the parallel port. It could be connected to
a USB port, but I'm running out and I had a parallel cable laying around.

 Paul 06-15-2013 10:39 PM

Re: Printer port question

Jeff Strickland wrote:
>
> "Robert Baer" <robertbaer@localnet.com> wrote in message
>> Jeff Strickland wrote:
>>>
>>> "Robert Baer" <robertbaer@localnet.com> wrote in message
>>>> According to the original IBM documentation viz IBM Technical
>>>> Reference First Edition August 1981 page D-34 Parallel Printer Adapter
>>>> schematic, the following pins of the DB-25 printer connector are
>>>> exclusively inputs:
>>>> pin 13 is +SLCT (select, high for "1")
>>>> pin 11 is +BUSY (busy, high for "1")
>>>> pin 12 is +PE (paper out, high for "1")
>>>> pin 10 is -ACK (acknowledge, low for "1"; eg: inverted logic)
>>>>
>>>> So, how come with computer NOT CONNECTED to anything (NO power at
>>>> all), one sees that pins 10, 11 and 12 are SHORTED TO GROUND?
>>>>
>>>> Please explain how a printer can possibly work with that condition.
>>>
>>>
>>>
>>> Because they are held high by the printer that's not there? The printer
>>> pulls these lines high.
>>>
>>>
>>>
>>>
>>>
>>>
>>>

>> On the surface,that _might_ sound reasonable..
>> Except.
>> The schematic shows INPUTS to a TTL gate for each one.
>> On TTL logic, an input must be pulled down (below 1.4V) for a low to
>> be asserted; a floating input on a gate acts like a high.
>>

>
>
> If there is a gate array, then the gate is low until a printer comes
> along to pull the inputs high when the condition becomes true. If the
> printer is available, the select line goes high, if the printer is busy,
> the busy line goes high, if whatever is happening, the line goes high,
> or it goes low in the case of the not-logic.
>
> I don't understand the question. I really don't understand why you think
> something is broken on a port that has not been used in any current
> technology in the past 15 or 20 years. I guess that's not true, I'm
> using a Brother MFC that is connected to the parallel port. It could be
> connected to a USB port, but I'm running out and I had a parallel cable
> laying around.

People use printer ports for GPIO functions. For example,
I have a JTAG programmer cable, that plugs into a parallel
appropriate pins on the interface. And I don't think my
adapter, works on the end of a USB to printer dongle either - it
has to be a real parallel port. I use a PCI Express card
with parallel port connector, to move the JTAG cable to
my current machine. That's because parallel ports are
out of style now, so you pay extra to get them. This is
the card I'm using.

That card uses OXPCIe952. Which apparently has more functions,
than there are connectors on my card.

http://www.plxtech.com/products/uart...ldocumentation

Paul

 Robert Baer 06-16-2013 05:56 AM

Re: Printer port question

Paul Drahn wrote:
> On 6/14/2013 11:19 PM, Robert Baer wrote:
>> According to the original IBM documentation viz IBM Technical Reference
>> First Edition August 1981 page D-34 Parallel Printer Adapter schematic,
>> the following pins of the DB-25 printer connector are exclusively inputs:
>> pin 13 is +SLCT (select, high for "1")
>> pin 11 is +BUSY (busy, high for "1")
>> pin 12 is +PE (paper out, high for "1")
>> pin 10 is -ACK (acknowledge, low for "1"; eg: inverted logic)
>>
>> So, how come with computer NOT CONNECTED to anything (NO power at all),
>> one sees that pins 10, 11 and 12 are SHORTED TO GROUND?
>>
>> Please explain how a printer can possibly work with that condition.

> Perhaps you have a computer with a version 2 parallel printer port. Not
> quite the same as the original IBM port.
>
> Paul

NOT relevant!

 Robert Baer 06-16-2013 06:02 AM

Re: Printer port question

Paul wrote:
> Robert Baer wrote:
>> Jeff Strickland wrote:
>>>
>>> "Robert Baer" <robertbaer@localnet.com> wrote in message
>>>> According to the original IBM documentation viz IBM Technical
>>>> Reference First Edition August 1981 page D-34 Parallel Printer Adapter
>>>> schematic, the following pins of the DB-25 printer connector are
>>>> exclusively inputs:
>>>> pin 13 is +SLCT (select, high for "1")
>>>> pin 11 is +BUSY (busy, high for "1")
>>>> pin 12 is +PE (paper out, high for "1")
>>>> pin 10 is -ACK (acknowledge, low for "1"; eg: inverted logic)
>>>>
>>>> So, how come with computer NOT CONNECTED to anything (NO power at
>>>> all), one sees that pins 10, 11 and 12 are SHORTED TO GROUND?
>>>>
>>>> Please explain how a printer can possibly work with that condition.
>>>
>>>
>>>
>>> Because they are held high by the printer that's not there? The printer
>>> pulls these lines high.
>>>
>>>
>>>
>>>
>>>
>>>
>>>

>> On the surface,that _might_ sound reasonable..
>> Except.
>> The schematic shows INPUTS to a TTL gate for each one.
>> On TTL logic, an input must be pulled down (below 1.4V) for a low to
>> be asserted; a floating input on a gate acts like a high.
>>

>
> I would question how you determined "SHORTED TO GROUND".
>
> A short to ground, will read zero on any range on your ohmmeter.
> Including the diode range.

* YEP! That is EXACTLY what i have; been in electronics for about 60
years. I meant what i said and i said what i meant.

>
> If you use the diode range on the multimeter, and you're seeing
> a diode drop (0.5 to 0.7V), then, it's not a "SHORT TO GROUND",
> is it ?

* Your statement is correct, but i saw ZERO which is a SHORT; see above.

>
> Remember that, when a hardware circuit is not powered, the
> VCC and GND appear shorted together. The protection network
> on the input of the gate, then becomes involved while you're
> diodes to VCC and GND (to prevent overshoot, to prevent
> undershoot). And a few gates are protected by what we used
> to call "up up up down", which is three diodes to one rail,
> and one diode to the other rail. Which allows larger
> excursions past the rail in one case than the other.
>
> So when you characterize a circuit, you have to know something
> about the internal protection structure, whether it's one
> stage or two, has a series resistor between stages, and so on.
> You need to know the protection structure, to interpret the
> results. In some cases (inputs "open to the world"), you'll
> actually see external protection diode networks, intended
> to provide a measure of ESD protection perhaps. The multimeter
> can end up probing the protection structure, instead of telling
> you something about actual circuit operation.
>
> You can use a variable power supply, and a resistor, to do
> your own crude characterization of the circuit while it
> is running. On a TTL gate, perhaps the knee is at 1.3V
> on the input. Applying a voltage below 1.3V, through a
> resistor, should sink a small current (400uA on a TTL
> gate of modern vintage, 10uA on CMOS maybe). Applying a
> voltage above 1.3V might source a small current, and the
> current will likely be a smaller flow than in the logic 0
> test case. It's a bit like running a curve tracer,
>
> So there are a few things you can try, with a bit of
> crude equipment. And your friends Thevenin and Ohm's Law
> ("It's The Law").

* Again,what you say is correct and a good lesson for those that are
weak in electronic theory.
But (again) not relevant.

>
> *******
>
> If you see an actual shorting strap on the PCB, then
> that's another matter. I don't know anything about
> printer ports, so can't really comment on that.
> Status inputs usually need to remain functional,
> especially if they mediate the transfer rate in
> some way. Shorting all the status signals would
> not make sense. But if this is a question of
> characterization of an unpowered logic gate,
> go back and look at it again.

* Powered or unpowered makes no difference; seeing ZERO means ZERO.

>
> Paul

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