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Diode-Resistor AND gate question

  • Thread starter Christopher Collins
  • Start date
C

Christopher Collins

Jan 1, 1970
0
An AND gate can be made of two diodes and a resistor:

+5V
|
R
|
A -|<|--+---- OUT
|
B -|<|--+

Here, R is a resistor and the |<| thingies are small diodes.

As long as either of the inputs A and B is connected to ground
(logical 0),
the output will be a 0 as well (not exactly zero volts, since there's
a
voltage drop over the diode). If both inputs are open, or connected to
logical 1, the output will be a 1 as well. Hence, the output is the
logical
and of the inputs.

My question: What's the point of the resistor here? It seems to me
this could work without it.

Christopher
 
B

Brian

Jan 1, 1970
0
Without the resistor, when both diodes go to a logical "1", there would be
nothing to pull the output to a logical "1" (or +5 volts in this case).
 
H

Hal Murray

Jan 1, 1970
0
An AND gate can be made of two diodes and a resistor:

+5V
|
R
|
A -|<|--+---- OUT
|
B -|<|--+

Here, R is a resistor and the |<| thingies are small diodes.
[snip]

My question: What's the point of the resistor here? It seems to me
this could work without it.

Without the resistor, how much current would you get through
one of the diodes when the inputis at 0V?
 
R

Rich Grise

Jan 1, 1970
0
An AND gate can be made of two diodes and a resistor:

+5V
|
R
|
A -|<|--+---- OUT
|
B -|<|--+

Here, R is a resistor and the |<| thingies are small diodes.
[snip]

My question: What's the point of the resistor here? It seems to me
this could work without it.

Without the resistor, how much current would you get through
one of the diodes when the inputis at 0V?

That's simple. All of it. ;-)
 
J

John Popelish

Jan 1, 1970
0
Christopher said:
An AND gate can be made of two diodes and a resistor:

+5V
|
R
|
A -|<|--+---- OUT
|
B -|<|--+

Here, R is a resistor and the |<| thingies are small diodes.

As long as either of the inputs A and B is connected to ground
(logical 0),
the output will be a 0 as well (not exactly zero volts, since there's
a
voltage drop over the diode). If both inputs are open, or connected to
logical 1, the output will be a 1 as well. Hence, the output is the
logical
and of the inputs.

My question: What's the point of the resistor here? It seems to me
this could work without it.

Christopher

The best way for you to understand the function of the resistor is to
add a model of the load of the logical inputs that this and gate
drives.

If they are cmos logic, they look something like capacitance ot the
two supply rails. In that case, what is the source of charging
current to pump these capacitors positive when the diodes are reverse
biased?

If the load is TTL, each load gate contains some internal pull up
resistance since the inputs are current operated (the input is a logic
zero when you pull current out pf the input till the input voltage
falls below about 1.2 volts). In that case, the diode pull up
resistance may be redundant.

So the answer is, it depends...
 
G

Glenn Gundlach

Jan 1, 1970
0
Brian said:
Without the resistor, when both diodes go to a logical "1", there would be
nothing to pull the output to a logical "1" (or +5 volts in this case).

We used to call this MML or Mickey Mouse Logic. Handy for CMOS (no
loading) and slow signals like door open switches.
gg
 
W

Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
Rich Grise said:
An AND gate can be made of two diodes and a resistor:

+5V
|
R
|
A -|<|--+---- OUT
|
B -|<|--+

Here, R is a resistor and the |<| thingies are small diodes.
[snip]

My question: What's the point of the resistor here? It seems to me
this could work without it.

Without the resistor, how much current would you get through
one of the diodes when the inputis at 0V?

That's simple. All of it. ;-)

Yeah. Besides, how didja come up with this "AND Gate" stuff? ;-)
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/diodgate.html
 
W

Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
Glenn Gundlach said:
We used to call this MML or Mickey Mouse Logic. Handy for CMOS (no
loading) and slow signals like door open switches.
gg

Didja hear that Michael Eisner is going to be leaving Disney?

Because it's a Mickey Mouse Operation. ;-)
 
R

Roger Johansson

Jan 1, 1970
0
Yeah. Besides, how didja come up with this "AND Gate" stuff? ;-)
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/diodgate.html

Why did you give us this url ?

To show a totally faulty way of doing it?

Look at the first circuit on that web page.
If both inputs are high there is no problem, the output is high, it is
directly connected to the plus 6 Volt power supply.

If input A goes low, what happens?

Unless it can take out the power supply and kill it, nothing happens to
the output. It is still directly connected to the plus 6 Volt power
supply.

If the input to A is made up of a switch to ground it will burn the
diode, or if a very sturdy diode is used, the power supply will burn.
 
C

Clarence

Jan 1, 1970
0
Roger Johansson said:
Why did you give us this url ?

To show a totally faulty way of doing it?

Look at the first circuit on that web page.
If both inputs are high there is no problem, the output is high, it is
directly connected to the plus 6 Volt power supply.

If input A goes low, what happens?

Unless it can take out the power supply and kill it, nothing happens to
the output. It is still directly connected to the plus 6 Volt power
supply.

If the input to A is made up of a switch to ground it will burn the
diode, or if a very sturdy diode is used, the power supply will burn.


The link shows an "OR" gate. There seems to be a disconnect in the definition.

It is possible to have an "AND" gate or an "OR" gate using the same components.

There is also the possibility that there may be some confusion in the
definition of a 1 and 0. Let us assume a 1 = positive voltage above 1V. 0 =
less than 1V.

An "AND" for positive logic has a pull up to VCC and the cathodes of the
diodes connected so that pulling the anode of either diode to a "LOW" (GND by
definition) will give one diode drop (nominal 0.7V) at the output. GND +
(nominal 0.7V)

AN "OR" for positive logic (as shown in the link,) has the resistor pulling
down, and the diodes connected with the cathodes common to the resistor.
Pulling the Anode of either diode to VCC results in an output of VCC - (Nominal
0.7V)

Is that any clearer?
 
R

Roger Johansson

Jan 1, 1970
0
Clarence said:
The link shows an "OR" gate.

No, it doesn't. If anything it is an always gate, the output is always
high because it is directly connected to the plus 6 Volt supply.

How could it ever be anything but high, unless the power supply is turned
off, or breaks down?
AN "OR" for positive logic (as shown in the link,) has the resistor
pulling down, and the diodes connected with the cathodes common to the
resistor. Pulling the Anode of either diode to VCC results in an output
of VCC - (Nominal 0.7V)

Is that any clearer?

No. And I can't believe that you are trying to defend that circuit as
anything but a big mistake.

Don't you see that the output is directly connected to plus 6 Volt?

Compare how the connection to plus 6 Volt is drawn in the the following
circuits. The connection is drawn exactly alike in the first circuit, so
there can be no doubt that the output is connected to plus 6 Volt.
 
C

Clarence

Jan 1, 1970
0
Roger Johansson said:
No, it doesn't. If anything it is an always gate, the output is always
high because it is directly connected to the plus 6 Volt supply.

How could it ever be anything but high, unless the power supply is turned
off, or breaks down?

Ah-ha! I see what you mean. I thought it referred to the design maximum
voltage. But IF it is actually a power supply (battery?) it is really wired
wrong. But since it was an educational web site, perhaps the questions should
be "WHAT is WRONG with this picture?"
 
R

Roger Johansson

Jan 1, 1970
0
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/diodgate.htm
Ah-ha! I see what you mean. I thought it referred to the design
maximum voltage. But IF it is actually a power supply (battery?) it is
really wired wrong. But since it was an educational web site, perhaps
the questions should be "WHAT is WRONG with this picture?"

This web page is obviously not intended to show examples of faulty
circuits, so it is simply a mistake.
 
R

Roger Johansson

Jan 1, 1970
0
So I was too subtle?


Maybe just a bit too fast, from being wrong to being subtle.

Remember that we are writing in an international newsgroup, where you
cannot be too subtle without a big risk of being misunderstood. So we
need to establish what is right and wrong clearly first, and then you can
be subtle as much as you like.
 
C

Clarence

Jan 1, 1970
0
Roger Johansson said:
Maybe just a bit too fast, from being wrong to being subtle.

Remember that we are writing in an international newsgroup, where you
cannot be too subtle without a big risk of being misunderstood. So we
need to establish what is right and wrong clearly first, and then you can
be subtle as much as you like.

I was not wrong, the web site was!
Remember the diagram was NOT mine.
I did not refer to it other than to say it was connected as an "OR" which is
is.
It has as you pointed out, a fatal flaw.
What I wrote was entirely correct.

Don't blame me if you didn't understand.
English IS the common language here.
 
R

Rich Grise

Jan 1, 1970
0
Didja hear that Michael Eisner is going to be leaving Disney?

Because it's a Mickey Mouse Operation. ;-)

I heard that Mickey wants to divorce Minnie on the grounds of
insanity because she's fucking Goofy.

Cheers!
Rich
 
W

Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
Rich Grise said:
I heard that Mickey wants to divorce Minnie on the grounds of
insanity because she's fucking Goofy.

Cheers!
Rich

Owww, you get a howler for that one!! </Harry Potter>
 
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