# Flashing led circuit explanation

Discussion in 'Electronic Basics' started by Ian H., May 4, 2004.

1. ### Ian H.Guest

Could someone please provide an alternate explanation of how this
flashing LED circuit works? The description on the site is doubtless
correct but I can't get a grip on it. Thanks!

Ian

2. ### John PopelishGuest

There are 2 keys to understanding how this works. One is that the
base to emitter acts as a diode (conducts in one direction, only).
The other is that when voltage changes at one end of a capacitor very
fast, the voltage at the other end must also change by the same
amount, regardless of what voltage drop there is across the cap when
this sudden swing started.

Let's also assume that the power has been applied for some time and
cycles have been going on long enough to stabilize, with TR1 on and
TR2 just about to turn on. If the explanation can get us back to this
point, we will have followed a full cycle.

R2 and R3 are each capable of holding one of the transistors on
indefinitely, so the question becomes, what turns them off. The only
other thing connected ot the bases are the cross coupling capacitors
from the opposite transistor. With TR1 on, the voltage on its
collector is near zero (if we assume the bottom rail is called zero
volts or ground). So R2 is charging the right end of C1 positive
while the left end is grounded. Once the right end of C1 rises to
about +.6 volts, the base emitter junction of TR2 becomes forward
biased and detours all current from R2 away from C1 so that this
current turns on TR2.

As TR2 switches on, the right end of C2 swings from almost 9 volts to
nearly zero. This sudden negative swing must also appear on the left
end of C2, but starting from wherever the voltage was before the
swing. Since we have been assuming that TR1 started out by being on,
its base voltage must have been about +.6 volts, so the negative swing
passing through C2 drives this to almost -8.4 volts. This has two
immediate effects. The current through R3 that had been keeping TR1
on is now detoured to C2, instead and TR1 turns off, allowing its
collector voltage to be pulled up by current through L1 and R1. But
the collector voltage cannot instantaneously rise all the way to +9
because the right end of C1 cannot rise above a diode drop from zero,
because of the conducting junction in TR2. So the current through R1
simply adds to the base bias current passing through TR2, turning it
on very completely.

After several R1*C1 time constants have passed, the current through R1
has charged the left end of C1 up to a high enough voltage that L1 is
almost completely starved of current, and the additional base current
to TR2 is an insignificant addition to that supplied through R2, so
TR2 is just on with its steady state bias.

As this settling of voltage across C1 has been taking place, a much
slower charging process has been taking place across C2. Current
through R3 (boosted by the large negative voltage on the left end of
C2 has been starting a ramp up in the voltage on the base of TR1.
But, since the time constant is 10 times longer longer for this
charging process than the one that involved R1
(4.7k*47uF=.22 seconds versus 470*47uf=.022 seconds)
The base voltage on TR1 will remain below ground for some time.

Eventually, the current through R3 will bring the voltage on the base
of TR1 0.6 volts higher than ground and that current will suddenly
detour to the base. At that moment, TR1 will turn on and the second
half cycle begins with C1 driving the base of TR2 way below ground
which turns off TR2, allowing the current from R4 to overdrive TR1 on
via C2, etc. and C1 begins the long climb back to +.6 volts.

3. ### Ian H.Guest

Thanks for the reply John! A few questions if you don't mind :

You said : "Once the right end of C1 rises to about +.6 volts, the
base emitter junction of TR2 becomes forward biased and detours all
current from R2 away from C1 so that this current turns on TR2."

How or why is the current "detoured" away from C1?

Also : "As TR2 switches on, the right end of C2 swings from almost 9
volts to nearly zero. This sudden negative swing must also appear on
the left end of C2, but starting from wherever the voltage was before
the swing."

Why does this negative swing have to appear on the left end of C2? My
understanding of capacitors is that they store and release charge.

Lastly, could you clarify : "But the collector voltage cannot
instantaneously rise all the way to +9 because the right end of C1
cannot rise above a diode drop from zero, because of the conducting
junction in TR2."

Thanks very much!

Ian H.

4. ### benchGuest

what would happen if R1 and R4 where increased to 470K,
I know the leds will not light, but assuming we took the leds
out and just look at the waveform, would it continue to
oscillate ?