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From: "petrus bitbyter"
References: <firstname.lastname@example.org> <email@example.com>
Subject: Re: A Simple Circuit Design Challange
X-Newsreader: Microsoft Outlook Express 5.50.4807.1700
Date: Fri, 25 Oct 2002 20:55:50 GMT
NNTP-Posting-Date: Fri, 25 Oct 2002 22:55:50 MET DST
Organization: chello broadband
If you really want to find *all* possible failures, we may start some
thinking. You may name it "analizing" if you like.
Consider in1. You may have an open with out1 or shorts to all of the
remaining ins and outs. Total five possible failures.
Now consider in2. There is the same amount of failures but we accounted for
a short to in1 already. So we have to add four possible failures.
In the same way you can add three failures for in3.
Now consider out1. We accounted for the open to in1 and the shorts to the
other ins but there may be shorts to out2 and out3. Add two failures.
Finally add one for a short between out2 and out3.
If I counted well, we have fifteen possible failures. With a led for every
one of them you may well illuminate a Christmas tree.
To find or exclude all fifteen off them in one step will be a real
challange. But even if it is possible, I doubt if you can find someone to do
it. It's neither simple nor practical.
Most developers today will call for a micro. With the skills and tools
readily available, it will be the best choice. Nevertheless a solution using
off the shelf non-programmable electronic components is also possible.
The idea of keeping one connection high and check the others is a good one
to start with. But is has a severe drawback. When you have a short you may
connect two outputs with different levels together. You will get unwanted
results and may even blow your measuring device. To prevent this, we will
use open collector (or open drain) outputs of which one is pulled low and
the others are kept high.
We will complete the measurement in five steps so we need a device that
produces five succeding 0's within all 1's. A five bits shiftregister (or
the first four stages of an eight bits) will do it. To initialize this
shiftregister you need a four inputs NAND gate to have a 0 ready when the
first four bits of the register are all 1's. Furthermore you will need the
open collector/drain buffers mentioned already and a clock. Lets say a 555
running at 1kHz will do. So far we have a chipcount of four. We connect the
open open collector/drain outputs (with their pull-up resistors) to five of
the ins/outs. The sixth one needs only a pull-up resistor.
To find the failures we will start looking at the opens. The corresponding
ins and outs should have the same level otherwise there is an open. We can
check this out with a XOR-gate. It produces a high output when the inputs
are unequel. You will have to look whether or not the XOR-gates used, can
provide enough current to blink a LED. We may have to add inverting buffers
as many logical devices can sink more current then they can source.
A short between to points will be recognised when both are low. So we have
to add twelve NOR-ports to check for all possible shorts. An OR-port may be
better as we can use the current sinking rather then the source.
Total chip count eight or nine depending on the need of inverters.
I guess this will do.
"Dan" schreef in bericht
> Nice thoughts on the subject. The chopper idea is definately a unique
> one. I have been toying with comparators to make this work based upon
> DC references set up on the three lines and then checking them with
> respect to one another. I can make this work with two inputs and two
> outputs. But the funkiness comes into play when adding the third line
> and then keeping any of the others from false triggering and giving
> indication incorrectly.
> Thanks NT!!
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