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Cry,!

This is the second circuit which I want to analyze to learn how to calculate the RC time constant of each stage, but they seem to end with fail!

About this circuit:
1: I want to know if it is a good idea to design and run the LED's in cutoff/saturation state or no (Suppose I want to start and design it from zero, stage to stage. ie designing and running a led by a transistor then adding more stuff...).
2: I want to know how to calculate the Time constant for the circuit?


Hi again,


You seem disheartened a bit here and you shouldnt allow yourself to
feel that way because as you learn you can pick up better and better
ways to understand circuits and that's what it's all about. I find it
great that you are this interested in these circuits so i took a little time
to write out a quick analysis the way you might go about attacking a
circuit like this...

It appears that you just happened to pick slightly harder circuits
to analyze for your first analysis tries and that's part of the
problem here. You should first review how time constants work in
simpler circuits, then progress to this one. You may want to work
with some transistors first too. This will give you the groundwork
experience to understand how to attack this circuit.


A QUICK ANALYSIS

Let's say that we approximate the Vbe drops of the transistors to 1v
to make it simpler (normally this is roughly 0.7v but 1v is easier to
write and you can always go over it later with 0.7v).
Let's also remove the LED's so that we only have the collector resistors
in the circuit. This also helps simplify the analysis for the first
try. I'll call the right side of a capacitor RS and left size LS
to make the writing a bit faster too. We will also approximate the
saturation voltage of the transistors to be 0v rather than the typical
0.2v or slightly higher.

We start the analysis off in a typical state of operation, where Q1 is
turned off and Q2 is turned on. This means current is flowing through
R1 into the base of Q2 and C1 is charged to about 11v with 12v on LS
and 1v on RS. C2 is charged to about 11v with LS at -11v and RS at
about 0v. This means Q1 is cut off and C2 is discharging through R2
and Q2 (actually trying to charge in the opposite direction).
After C2 charges up enough, R2 begins to turn Q1 on. Q1's collector
voltage starts to drop quickly and that means less current through
C1 which in turn turns off Q2. Q2's collector voltage rises quickly
and C2 conducts more current which keeps Q1 on.
Eventually C1 discharges and Q2 again turns on and Q1 turns off which
completes one cycle.

That rough analysis tells us that the time constants R1C1 and R2C2
are important. It also tells us that R1 must be low enough to
be able to keep Q2 turned on enough to keep the collector low.
We also see that the time constant will be the result of the capacitor
discharging through R1 (or R2) as the voltage goes through roughly
11v of change with a source of 23v (the supply voltage plus the
capacitor voltage). We can approximate this to 11v and 22v.
Now we know that a capacitor charging through a resistor with a supply
voltage of Vs volts is:
Vc=Vs*(1-e^(-t/RC))

and we know that Vc=11v and Vs=22v so we have:
11=22*(1-e^(-t/RC))

Now to quickly simplify a bit we divide both sides by 22 and we get:
0.5=1-e^(-t/RC)

This is an equation with two variables t and RC so lets solve for t...

Subtracting 1 from both sides we get:
-0.5=-e^(-t/RC)

and multiplying both sides by -1 we get:
0.5=e^(-t/RC)

Now take the natural log of both sides we get:
-0.693147=-t/RC

and again multiply both sides by -1 and we get:
0.69=t/RC (approximately)

Now multiply both sides by RC and we get:
0.69*RC=t

so we have solved for t:
t=0.69*RC

Recognizing that as 1/2 cycle, the full cycle takes twice that
long with R1=R2 and C1=C2 so we get:
t=2*0.69*RC

which is approximately:
t=1.4*RC

and the frequency in Hertz is 1/t, so we get:
F=1/(1.4*RC)

and that's an estimate of the running frequency.

There are a few things to note also...

1. The transistors have to be able to be saturated with the choice of
R1 (and R2), but not driven too hard or they wont turn off.
2. A lot of the burden of the timing is put on the choice of capacitor.
3. Since the timing depends on the ratio of roughly twice the supply
voltage to roughly half the supply voltage, the frequency should be
relatively stable with supply voltage, but large changes will still affect
the circuit however because Vbe is constant.
4. One of the resistors R1 or R2 can be made slightly different in value to
ensure the oscillator starts quickly. This will of course affect the timing
unless the respective capacitor is changed slightly too.
 
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