# Another Basic Question

Discussion in 'Electronic Basics' started by Jonathan Mohn, Feb 5, 2004.

1. ### Jonathan MohnGuest

I've been trying to understand the LED flasher circuit at the bottom of
http://members.shaw.ca/roma/twenty-three.html. I'm not quite there yet.

On the right side, midway down the schematic, I find a symbol that looks
like this: /|/|/ . What does that mean? Also, where the base current
would come into the transistor, I find the letters NC. Does that mean "no
current"??

Can anyone help me understand how this circuit works?

Thanks!

-Jonathan

2. ### John PopelishGuest

I think that is supposed to be the voltage waveform you would see if
you looked at that node with an oscilloscope. A transistor hooked up
this way is not operating within its maximum specs. It is operating
in reverse base emitter junction breakdown. I think that the
operation is also very variable unit to unit. An interesting
experiment but not a very useful design.

3. ### Dan AkersGuest

Jon wrote;
<snip>
"On the right side, midway down the schematic, I find a symbol that
looks like this:   /|/|/ . What does that mean? Also, where the base
current would come into the transistor, I find the letters NC. Does that
mean "no current"??
Can anyone help me understand how this circuit works?"
____________________________________
Re;
The "NC" means "not connected"; in your case the base of the transistor
is not connected. It can also mean, in reference to a push-button
switch or the like, "normally closed". The sawtooth symbol is an
illustration of the waveform that you should see at the emitter of the
transistor relative to the circuit ground. I'm not sure exactly how
this oscillator works, although I suspect that it takes advantage of the
minute interelectrode capacitance of the transistor. Good luck...

-Dan Akers

4. ### Sir Charles W. Shults IIIGuest

Well, first off- that little squiggle is the waveform you would see on
an oscilloscope. It shows that the voltage is rising until the transistor
breaks into conduction. Second, the NC means "no connection." Third, this
is a type of relaxation oscillator. What happens is simple- the voltage on
the capacitor slowly rises and during that time, the LED is off.
Once the voltage rises over a specific point (which will depend on that
transistor), the transistor breaks into reverse conduction- it "shorts out"
sort of. This is not the normal mode of operation for a bipolar transistor!
Once the transistor goes into conduction, the capacitor discharges
rapidly through the transistor and the LED flashes. The transistor,
meanwhile, sees the voltage go below its conduction level ans switches back
off. Now the capacitor can charge once more.
Think of a toilet float that, when the tank is full, automatically
flushes again. That is what is happening here. When the tank (capacitor)
is full, the flapper valve (transistor) opens and dumps the water (current).
Then it fills and the cycle starts all over again.

Cheers!

Chip Shults

5. ### Jonathan MohnGuest

Thanks very much!

Wouldn't it also work if I set the circuit up as follows:

1) turn the transistor around so that the emitter is connected to ground,
and the collector is connected to the positive end.

2) put the resistor and the LED on the emitter side of the transistor,
connecting the LED's negative end to ground.

3) pump up the voltage to just over the collector/emitter break-down point.

I believe the voltage would have to be much higher (the spec sheet shows
Vceo to be 40V), so the resistor attached to the LED would have to be bumped
up accordingly. Other than that, wouldn't it work the same? Would I
destroy the transistor?

That also raises another question -- the spec sheet does not show a value
for Veco. Is the emitter/collector break-down voltage generally far less
than the collector/emitter break-down voltage? Is there a general rule for
how much less?

-Jonathan

6. ### Sir Charles W. Shults IIIGuest

point.

I tend to think that this will destroy the transistor unless you are
very careful. Still, this is a really cheap part and it's best to try it
empirically. Make one and see. But remember that even if it does work, you
may be doing some damage that will limit its useful life quite a bit. I
haven't tried it myself.
You definitely would have to increase the resistance. However, if the
internal resistance of the transistor was higher in this mode of conduction,
it could seriously limit the total brightness you will get out of the LED in
the first place, and it might not light well at all.
I will have to dig out a book and see- but that is a good question. Win
Hill probably could tell you right off the cuff.
My stuff is in boxes at the moment- going through a nasty divorce and I
had to get everything packed and out in a hurry. I'll try to get to the
proper books and find the answer.
Cheers!

Chip Shults

7. ### Jonathan MohnGuest

Thanks for your responses!

Please don't trouble yourself on that last question, though. I'm going to
sit down with a couple of different types of transistors and simply test
them. I'll hook up my DMM to read current, and then simply dial the voltage
up until it starts rushing through the transistor.

Pretty interesting stuff!

-Jonathan

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