Impedance

[Music Man]

When a source(e.g.Mic or guitar)is added to load (Mic Input),how does the
source become affected in transferring its signal?
I believe that circuits work as voltage transfer rather than current
transfer.
What this exactly mean and why do we have things like high-F rolloffs when
cables
are too long in unbalanced systems?
Is Mic output A/C seeing that it's a magnet? How is the voltge "understood"
by the load input?

Thanks

 

[John Fields]

On Mon, 14 Mar 2005 18:54:25 -0000, "Music Man" <j.michael@cwcom.net>
wrote:

When a source(e.g.Mic or guitar)is added to load (Mic Input),how does the
source become affected in transferring its signal?

---
For a resistive source and load, model your source as a voltage source
in series with a resistance and your load as a resistance to ground.
Your circuit then becomes a simple voltage divider:

Vs
|
[Rs]
|
+--->Vl
|
[Rl]
|
GND

Where: Vs is the voltage source
Rs is the source resistance
Vl is the voltage appearing across the load
Rl is the load resistance

And can be described by:

Vs * Rl
Vl = --------- (1)
Rs + Rl

I believe that circuits work as voltage transfer rather than current
transfer.

---
Charge gets transferred, But it's voltage that pushes it around.
---

What this exactly mean

---
What what exactly mean?
---

and why do we have things like high-F rolloffs when
cables are too long in unbalanced systems?

---
The longer the cable, the higher the capacitance, so the lower the
capacitive reactance and the greater the loading on the source
(generator) as frequency increases:


Vs>---[Rs]--->>--+--[Rc]-->>--+---Vl
| |
[Cc] [Rl]
| |
GND>--------->>--+-------->>--+---0V

Rc is the resistance of the cable connecting the source and the
load,and Cc is the distributed cable capacitance shown as lumped, so
Rc and Cc form a low pass filter with the reactance of Cc shunting
high frequencies to ground.
---

Is Mic output A/C seeing that it's a magnet?

---
If it's a passive dynamic microphone, yes. If it's active, It may be
AC riding on DC.
---

How is the voltge "understood" by the load input?

---
The microphone is simply a generator sending a signal to the load and
instantaneous values of the signal can be described by equation (1) if
the system is resistive.

--
John Fields

 

[John Popelish]

Music Man wrote:

When a source(e.g.Mic or guitar)is added to load (Mic Input),how does the
source become affected in transferring its signal?
I believe that circuits work as voltage transfer rather than current
transfer.
What this exactly mean and why do we have things like high-F rolloffs when
cables
are too long in unbalanced systems?
Is Mic output A/C seeing that it's a magnet? How is the voltge "understood"
by the load input?

You may be only interested in the voltage from a source, but before
that voltage can swing across cable capacitance, that capacitance must
be charged twice a cycle by current from the source. The impedance of
the source tells you how much current it can deliver before its output
voltage gets loaded down by a given amount. For instance, if it is
connected to a load impedance equal to its own, its output will fall
to half of the unloaded value. High impedance sources work fine when
connected with short wires (low capacitance) to high impedance
amplifier inputs. But if you want to drive high cable capacitance
without having that capacitance act as a low pass filter, you need to
get the source impedance lower than the impedance of that
capacitance. One way to do this is to put a step down transformer at
the source to lower its voltage while increasing its current
capability (lower its impedance). Amplifiers with so called "low
impedance" inputs are just ones with higher voltage gain to compensate
for the voltage step down.

--
John Popelish

 

[Larry Brasfield]

"Music Man" <j.michael@cwcom.net> wrote in message
news:4235dde1$0$26739$cc9e4d1f@news.dial.pipex.com...

When a source(e.g.Mic or guitar)is added to load (Mic Input),how does the
source become affected in transferring its signal?

The current delivered to the load has to flow
thru the devices constituting the source. This
normally produces a different output voltage
than if the load was not present. In technical
terms, the output impedance of the source and
the load impedance together determine the
signal delivered to the load.

I believe that circuits work as voltage transfer rather than current
transfer.

That can be a useful way to think about very
low impedance sources. But with non-negligible
output impedance and cable characteristics,
that way of thinking can obscure important effects.

What this exactly mean and why do we have things like high-F rolloffs when
cables are too long in unbalanced systems?

The cable capacitance can load some sources
excessively at higher frequency.

Is Mic output A/C seeing that it's a magnet?

Yes. (if it is a typical dynamic microphone)

How is the voltge "understood"
by the load input?

Sounds like a philosophical question. I would
say the power delivered to the load causes
things to happen that one hopes were desired
effects. Then, anthropomorphically speaking,
the input has been understood.

Thanks

You're welcome.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.