From: Jonathan Kirwan
Subject: Re: can't understand this concept
X-Newsreader: Forte Agent 1.92/32.572
NNTP-Posting-Date: Tue, 29 Oct 2002 22:15:52 GMT
Organization: AT&T Broadband
Date: Tue, 29 Oct 2002 22:15:52 GMT
On Tue, 29 Oct 2002 10:55:36 -0500, "David Jones"
>I have tried to learn electronics several times and given up. Once
>again I'm trying and I figured out what was frustrating me so much.
Sometimes, it's just a matter of getting past a mental block.
Looking at lots of schematics can help.
>In most circuits there's a connection like in the picture of the
>schematic I attached to this post(its in between R1 and the switch).
For starters, binaries (images, for example) are posted to
alt.binaries.schematics.electronic. If you use ASCII art, this
is how I'd redraw it:
LED \ / D1
\ / SCR1 \
\ / D2 \ R1
--- / 820
Often, it's better to draw this even more compactly (which means
more horizontally.) But I chose this vertical for a reason.
For understanding a circuit, it's helpful if you arrange things
so that electron flow is from bottom to top and signal flow is
from left to right. (No signal flow to speak of, here, so it's
mostly vertical looking.)
I also chose to break the physical/visual connection between the
cathode of D2 and one end of the R1. That wire is important for
the circuit to work, but it also can distract you from
understanding it. You get caught up in "all the connections"
and fail to realize that they don't help you get the point any
easier. The above is the same as before, but the removal of
that one 'wire' might help just a little bit.
>In that schematic, when the switch is off, the gate of the SCR
>is negative because only the negative wire is connected to it.
Yes, via R1. And since there is no real current flow going on,
there is no real voltage drop across R1, so the voltage is about
the same on both sides of R1. Or as you say, the gate is
essentially connected to the circuit's (-).
>Then when the switch is closed the gate on the SCR is connected
>to both a negative and a positive source. So which is it?
>negative or positive? Well from the way the circuit works I know
>its positive but can't figure out why. Why?
Once the switch is connected, you have a straight connection
from (+) to (-) via D1, R2, and R1. D1 will drop a little
voltage itself, but the remainder will be divided up across R2
Phone systems can be a little complex, so the exact voltage may
vary some when you close the switch. In the US, I believe a
"loaded" (off-hook) phone line still presents more than 8V. In
this case, there should be more than 3V remaining across R1.
And probably more than that.
But none of this really gets at your question. You seem
confused about what the voltage should be, at a point between
two resistors connected across a voltage source. I think you
imagine that there is some kind of *fighting* going on and that
one must somehow win the war and one must lose it. Instead,
what happens is that the voltage at that midpoint becomes a
value somewhere in between.
Let's assume for a moment that D1 isn't there and this is only
about R1 and R2. Let's say that both R1 and R2 are 1000 ohms.
Let's also say that the (+) to (-) voltage span is 9V. Wouldn't
you expect to see something about halfway inbetween and the
point between the two resistors?
Normally, you just look at the voltage against some other
reference. It's common to consider the (-) that reference. So
anything exactly at the same potential as the (-) point will be
assigned 0 as the voltage. Using the above 9V for the
difference between (+) and (-) means that the (+) point will be
called +9V. So at the point between two 1000 ohm resistors
connected across the (+) and (-) will be +4.5V, by that
Now, it could be that the two equal resistors are both 10,000
ohms, instead. The voltage would still be same, between them.
Or both could be 100 ohms. The voltage between would still be
the same. (Assuming that the voltage supply could handle the
So, with 1200 ohms and 820 ohms, you'd expect the point between
to be a little 'closer' to the (-) than to the (+), which is the
general truth of the matter. It *is* a little closer. The
calculation is: 9V*(820/(1200+820)) or about +3.65V (assuming
9V, of course.) That's simply the proportion that 820 ohms is
to the total resistance.
So when the switch is closed, the voltage moves quickly towards
> also why does it need a D2? doesn't it already have a diode?
Imagine that the diode isn't there and that the SCR cathode is
connected directly to (-). Now suppose that there is a little
bit of 'leakage' current making it through R1 via the SCR gate.
(Nothing is perfect.) This current could develop a slight
voltage across R1, causing the gate to be slightly biased. It's
a risk factor.
To reduce the risks of accidental triggering, 'stacking' the SCR
on top of a diode pretty much ensures that R1 pulls the gate
*below* the cathode voltage. The diode requires something like
0.55V or more across it, so the SCR cathode can't directly
observe (-), but only more than 1/2 a volt above that -- if at
all. That should allow R1 to hold the SCR *hard* to the off
condition. It's just an extra step, cheaply applied.
That's one way to imagine why, anyway.