Voltage Shifting Question

I've been looking at a circuit which produces sawtooth oscillations. It
does this by sending a constant current to a capacitor which is part of a
circuit containing a 555 timer. The capcitor generates a ramp and is
periodically reset by the 555 timer. The way the circuit is wired, the
ramp ranges from -12v to -4v. I understand why this works the way it
does. However, the final output of the circuit produces a ramp which
ranges from -4v to +4v.

The circuit is wired as follows:

1. The ramp capacitor (.01uf) has the negative side connected to a -12v
source and it's positive side to a constant current.
2. The ramp capacitor is also connected to a non-inverting pin of an opamp
(LM324)
3. The output of the opamp feeds back to the inverting pin.
4. The output of the opamp is also connected in series to a 1 uf
capacitor, a 10k resistor and ground.

There are misc. other connections which connect this portion of the
circuit in with a 555 but I'm hoping this is enough information to answer
my question. I can also try to provide a drawing, but again I am hopeful
that won't be necessary.

The output of the circuit is taken between the 1 uf capacitor and the 10k
resistor. It's at this point that the ramp has been changed from a
-12v/-4v ramp to a -4v/+4v ramp. My question is simply this, why is this
happening? I could understand how the second capacitor might introduce a
phase shift, but I don't understand what's causing the voltage shift.

Any help will be appreciated.

Thanks.

 

The 1uF capacitor provides what is known as "AC coupling" at the output.
The AC component of the signal is faithfully coupled but the DC level is
shifted. This is standard practice in (for example) multi-stage amplifiers
to isolate the biasing networks between stages. The resistor to ground
establishes the average output voltage at 0V (ground). A DC voltage of 8V
develops across the capacitor providing the level shift from -12/-4 up
to -4/+4. The size of the capacitor must be large enough that it's
reactance is small enough at the operating frequency for the phase shift to
cause minimal degradation to the wave shape.

 

Subject: Re: Voltage Shifting Question

A basic high pass RC network always behaves this way. The AC wavform stays the
same but is level shifted such that the average voltage is equal to the voltage
that the resistor is conected to, in this case 0v. If you were to return the
resistor to -8v the wave would go back to -12v to -4v.

 

What? The other 12 answers weren't good enough?

You need to go to the public library.

Good Luck!
Rich

 

Considering the other answers were unrelated to this particular question,
I would say additional responses were warranted.

Not a bad idea, although your response would have been more helpful if you
had mentioned specific books to look for. I own a couple different
electronics books and I would be happy to look at others. It's entirely
possible that I simply overlooked something in one of my books which would
have answered my question, but I didn't see the answer so it seemed
entirely appropriate to pose the question to a 'basics' newsgroup.
Thankfully there were a couple other posters who actually bothered to
answer the question.

 

Capacitors are, basically, two plates, that are put close enough so that
any charge imbalance can affect the other plate, causing a similar but
opposite charge imbalance.

By changing the amount of charge, you can change the voltage across the
plates. Capacitance is defined by this:

C = Q/V

That means that if there is Q more colombs of charge on one side of a
capacitor than the other, then with a capacitor of size C farads, you
get a voltage difference of V volts.

However, there is a wrinkle. In order to change the voltage across the
capacitor, you need a source of charge. With your setup, when you change
the voltage quickly, the resistor to ground on the other side restricts
the flow of charge onto the capacitor. Thus, the other side of the
capacitor can't keep adding and subtracting charge as quickly as the
opamp side. That means that the voltage across the capacitor cannot
change quickly, which means that the voltage on the other side tracks
the voltage on the opamp side.

The side opposite of the opamp will eventually equalize with ground if
the opamp side is left unchanged, but no, you have to keep changing it
don't you?

What this all means is that a capacitor can be used to track changes in
voltage, even though there is no DC connection between the two sides.

So, the question is, what is the absolute voltage on the other side of
the capacitor from the opamp? Well, the answer is that it is anyplace
you want it to be. Only the changes will be reflected.

Thus, with your connection through the resistor to ground, the voltage
will start off at ground, and move with the capacitor. So, if the opamp
starts at 0V, and is discharged to -12V, the other side of the capacitor
will start at 0V, and go down somewhere near -12V. Then, when the opamp
goes up to -4V from there, the other side will track it, but the whole
time, it's slowly equalizing voltage with ground, moving charge across
the resistor. After a few cycles, the average value will be at ground,
and it will be oscillating between 4 and -4 volts.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.