# Flasher circuit

Discussion in 'Electronic Basics' started by [email protected], Sep 13, 2003.

1. ### Guest

Hi, I'm trying to understand the following simple flasher circuit:

http://students.washington.edu/ksf/circuit.JPG

I don't really know what I'm doing here, but I think I have a vague
idea of how it works... I assume you have to take the forward
resistance of the LED into account, because otherwise I don't think
the circuit would do anything. (Let's say a voltage drop of 2V and
resistance 5 ohms.) When the circuit is first turned on, Q1 and Q2
turn on, because there isn't enough voltage across the 2M resistor to
keep Q1 off. C1 begins charging up (the wrong way!) through the 1K
resistor, approaching a steady-state voltage. (Assuming the power
supply is 4.5V, hFE for both transistors is 100, and the above LED
values, I calculate this to be about -1.4V.) However, any capacitor
voltage above -1.3V is enough to keep Q1 off once it is turned off,
because that would produce a voltage over 3.8 (4.5-0.7) across the 2M
resistor. Thus, I'm guessing that the capacitor gets charged to near
-1.4V, Q1 turns off, the capacitor discharges to -1.3, and the cycle
repeats. This would account for the "flasher" behavior, since during
the on cycle, C1 is charging through the smaller 1K resistor and
during the off cycle, it's discharging through the 2M resistor. If
this is correct, then I have some questions:

Why is the capacitor oriented the wrong way in the circuit?

When the capacitor is charged between -1.3 and -1.4V, it seems
entirely arbitrary whether the transistors are off or on and the
capacitor is charging or discharging, so what decides when the circuit
changes states?

TIA, Karl

2. ### John PopelishGuest

You are right that you can not make the simplifying assumption that
the LED has a constant voltage drop, independent of current. The
changing voltage drop across the LED is the feedback voltage to the
input of the two stage amplifier.
Oops. This is exactly what you cannot assume. This circuit varies
the LED current from almost zero (where the voltage drop will be much
less than 2 volts) to severe overload current (limited only by the
current carrying capability of the second transistor), where the wire
bond resistance and the die resistance will come into play. I can
only guess whether 5 ohms is a good approximation of those, unless I
made a measurement of voltage and current, simultaneously during the
peak, or extrapolated from the LED data sheet.
A clearer way to think of this state is that the capacitor can be
assumed to start with no voltage drop across it. If we guess that Q2
will be slammed on (we'll check this later), then it is safe to also
assume that with the battery voltage used up across the series
combination of the LED and Q2, then the voltage across the LED must be
quite a bit more than the .6 volts or so it takes to turn on Q1 via
the 1K resistor. The small current through R2 is not of much
consequence in the presence of this >2 volt LED drop feeding the base
of Q1 through 1K. With Q1 driven on by this current, the capacitor
starts to build up a voltage drop that gradually decreases this base
current. But while the base current flows, Q1 is on, and dumps base
current to Q2, so our original guess that Q2 starts by being blasted
on looks good.
So far, so good.
You generally have it right. As soon as the base drive for Q1 gets
low enough to let Q2 come out of saturation, the resistive and diode
drop of the LED suddenly decreases, and the voltage at both ends of
the capacitor head positive (assuming this happens fast enough that
the voltage across the capacitor does not have time to change much).
This produces a positive feedback situation that snaps Q1 completely
off, and continues to a little reverse bias.
It has so little voltage across it that it will probably work in
either orientation. Try it the other way and see if the timing cycle
changes. If it gets faster, it means that leakage current is
interfering less with the slow charge time. I agree that it is
backwards.
The capacitor does its fast charge through Q1's base. It does its
slow discharge through R2. Q1's base has to be reverse biased during
this process to isolate the transistor. When the base of Q1 starts to
have enough forward bias to start turning on Q2, the current from R2
has to be gig enough to get the 'switch on' positive feedback going
before all its current has been detoured, or the transistors will just
stall at some slight conduction and the capacitor will discharge no
further (because all of R2's current is going to the base, instead of
the capacitor).

I would add an emitter resistor to Q2 to reduce the peak base current
from Q1 and limit the peak LED current. 10 to 47 ohms ought to work.

3. ### Watson A.Name - Watt SunGuest

This circuit should have a resistor betwen the C of Q1 and the B of Q2
to limit the current. Often these flasher circuits seem to have the
caps in backwards, and if you check the voltage across them, the
voltage reverses.

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