Re: Cat Alarm, pages 23 and 24, August 2013:
Thanks for the previous help! I've got a few more
questions if you don't mind answering them:
1) I think I understand everything in the schematic
except for the RF transmitter pins. Why is the out of the
LMC555 connected to the ground of the transmitter?
Also, what is the input pin used for? I tried looking at the
datasheet but it didn’t explain too much.
2) I have started to try and tackle the receiver
schematic, as well (Figure 4A in the August column). I
looked up the 4013N chip and discovered that it is a
flip-flop, and read about what each of the pins do. I still
do not understand, however, why the two Q\ output pins
are connected in Figure 4A.
3) Also, why are the clock and data inputs connected
to ground? I know the clock input is rising edge triggered,
but is this merely an alternative form of triggering the trip?
For the data input, I am entirely confused what this does.
4) Why are there two resistors (R1 and R2) between
the RF receivers out and the 4013N's pins? Couldn't this be
accomplished with just one?
Thanks again and if you know of any good resources
on these topics for beginners, I would really appreciate it
Bill Blackburn
1) The output of the 555 goes high and low because
it is connected as an oscillator. When the output is low, it
is like a switch to ground which applies power to the
transmitter. The transmitter input is connected to VCC, so
it will produce maximum output when it is active.
2) The Q and Q/ outputs are connected in order to
provide more drive to the transistor Q1. It is possible to do
this when the two devices are on the same silicon chip, so
they are identical. If you were to try this on separately
packaged devices, the switching times would be different
and the power dissipation would be increased.
3) The clock and data inputs are not used, so they are
connected to ground. It is not wise to leave any input
floating because stray electric fields can change the logic
level and give unexpected results. I am using the 4013 as
a set-reset flip-flop. The output of the receiver sets the Q
output high and the reset switch sets the Q/ high.
Whatever logic level is on the data input when the clock
goes high is transferred to the Q output. The only time I
use the data input is when I want a toggle. If you connect
the Q/ to D, then every time it is clocked it changes state.
4) You are right; R2 should have connected to the
reset line because it is otherwise floating. Mea culpa.
I have not looked into resources for beginners
but I am sure readers will have some suggestions;
I will let you know.
Re: ESD Instrument Circuits, page 16, September 2013:
Mr. Kincaid asked about measuring static electricity:
"Two foils will separate when charged due to electric
repulsion, but I don't know how to implement that. Does
anyone have any ideas?"
I do! Ahem ... I might! What about reflecting a beam
of light off one of the leaves which are shiny metal to start
with; if I recall properly, displacement of the leaf would
create a proportional displacement of the beam. Said
beam is hitting — depending on the position of the leaf —
somewhere between the edge and the center of a loupe
lens. The lens is located over a photoresistor. The light
beam ending up striking the photoresistor varies in its area
depending on its position, varying the actual resistance of
the photoresistor. Appropriate amplification shall be
implemented downstream, of course.
Would that work? Run with it if you want; I am just
glad I could finally think of something — anything — for a
Nuts & Volts question. LOL!
I’m always glad to read your columns. I signed up this
year, thanks to having acquired an iPad which lets me read
N&V at my leisure.
Stephen Young
Yes, that will work. It will be nonlinear and subject
to vibration, but lacking any other suggestion it is a
good one.
MAILBAG
November 2013 23