LED flasher 150209

[pittsallen@usa.net]

Hello Electronics Circuits,

Have been working with a circuit that alternates a signal
between two sides of a circuit. The schematics is at
http://reprise.com/host/circuits/images/flasher2.gif
I got this circuit from Bill Den Beste out of Beaverton OR.
(also attached)

I think I understand how this works but would offer
an explanation of the circuit with the 2N3904 to
make sure I've got it.

I always think of current as traveling from the VCC+
to ground so that where I start. The current could
go on the left or the right side depending on the
variations in the components but lets say the current
is drawn to the left side first. The current thru
the 39k resistor hits the base in the left side transistor
and allows current to flow from the collector
to the emitter thru the left side transistor.
This does two things: 1. It draws current thru
the left side LED which lights up and 2.
starts the 10uF capacitor charging.
When the capacitor reaches its maximum capacitance
its discharges. This sends a charge to the base
of the right side transistor base which makes
current flow thru the right side of the circuit
lighting the right side LED and charging the
right side 10uF capacitor. When the right side capacitor
maxes out it sends a charge to the left side
transistor base and the flow moves back to
the left side. So a loop is begun and the
current goes back and forth.
At first the function of the 39k resistor
was not understood because it seems that
the capacitors would charge and which
ever one maxes out first would the
activate the opposite transistor base without the 39k resistor.
Perhaps the function of the 39k resistor
is to get things going by feeding a current
to the base before the one of the capacitor fully charges.

Without the 39k resistors perhaps
(maybe an opportunity for an experiment)
both capacitors would charge and which
ever one reached max capacitance first
would light the opposite LED and
start the charge of the opposite
capacitor.

Is there another function to the 39k
resistor?

Thanks.

Allen Pitts, Dallas Texas

[pebe]

Here is a good explanation of what goes on.

http://www.hobbyprojects.com/multivibra ... table.html

Note that time-contant is not described in the text. In that article it is the value of C1 x R2, or C2 x R 3 (in megohms and microfarads). In your case it is 10µF x 0.039Mohms = 390milliseconds.

As one transistor switches on its collector falls from the supply voltage (Vs) to 0V. That takes the base of the other transistor form +0.7V down to minus Vs-0.7V. The maximun negative voltage permitted on the base is approximately -6V otherwise the BE junction breaks down, so you should limit Vs to +6V max. If you want to use a higher Vs then put a diode in series with each base and the capacitor and resistor it is connected to.

Hope that helps.

[pittsallen@usa.net]

Hello Pebe,

Thanks for your excellent reply. The explanation at 'Multivibrators Tutorials'
is detailed and helpful. One last question and I will go back to my soldering
gun.

In your explanation there is
'As one transistor switches on its collector falls from the supply voltage (Vs) to 0V.'
I think the 'Multivibrators Tutorials' is saying the same thing with
'Due to the rise of current through R1, the voltage across it will increase, causing the collector voltage of Tr1 to fall.'
If the voltage is increasing at R1 would not the voltage at the TR1 collector increase since R1 is connected to the TR1 collector?

Thanks,

Allen in Dallas

[pebe]

Hello Pebe,
..........'Due to the rise of current through R1, the voltage across it will increase, causing the collector voltage of Tr1 to fall.'
If the voltage is increasing at R1 would not the voltage at the TR1 collector increase since R1 is connected to the TR1 collector?

No. When Tr1 is cut off, no current will flow through R1. But as TR1 switches on it saturates, its collector voltage falls to about 100mV and the current through, and the voltage across, R1 is then at its maximum.

I hope that explains it.