From: firstname.lastname@example.org (Allan Herriman)
Subject: Re: can't understand this concept
Date: Wed, 30 Oct 2002 01:50:12 GMT
Organization: Agilent Technologies
NNTP-Posting-Date: Wed, 30 Oct 2002 01:50:11 +0000 (UTC)
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On Tue, 29 Oct 2002 22:15:52 GMT, Jonathan Kirwan
>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
> \ R2
> / 1200
> | |
> | o
> --- \
> \ / SCR1 \
> --- o
> |\ |
> | '-------+
> | |
> --- |
> \ / 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.)
You've drawn it upside down then. Electrons carry a negative charge,
so they flow from - to +.
>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.