shorting a dependent current source

[John Larkin]

On Sun, 02 Dec 2007 20:24:08 -0800, Peter Bennett
<peterbb@somewhere.invalid> wrote:

On Sun, 02 Dec 2007 17:09:23 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sun, 02 Dec 2007 15:59:52 -0800, Peter Bennett
peterbb@somewhere.invalid> wrote:

On Sun, 02 Dec 2007 17:45:06 -0500, "H.S." <hs.samix@gmail.com> wrote:

Jamie wrote:

Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

.

I think maybe something is getting miss interpreted here.

The triangle symbol could be nothing more than an inline AMP meter
taking measurements. or inline current sense device.

That device is a voltage controlled current source and not a measuring
device. The convention used in the text book in question is clear on that.

regards,
HS

If you put a short across the current source, there will be no voltage
across it - so the question becomes how much current does the current
source source or sink when there is no voltage across it?

My common-sense technician's view is that all the current will flow
through the short, and none through the current source.


How can no current flow through a current source?

In fact, 1 amp flows through the current source, and zero amps flow
through the short.

John

But it is a voltage-controlled current course, not a fixed 1 amp
current source - if there is no voltage across it (due to the short)
what current does it pass?

Its current is controlled by V1, which is 1 volt, not that it matters
to the overall problem. But it is a constant 1 amp, since V1 is a
constant 1 volt.

Shorting a current source doesn't change the current it generates.

John

 

[Richard Seriani]

"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in message
news:8uf6l3tfq8hd3ev8ccq85co2b5ts0j9vu5@4ax.com...

On Sun, 02 Dec 2007 16:46:49 -0500, "H.S." <hs.samix@gmail.com> wrote:



If V1=1 volt, then the triangle-doobie dependant source (as I
interpret it) is 1 amp in the down direction. So 1 amp flows through
the 1 ohm resistor, so the voltage drop across R is one volt. So, with
the short not yet applied, the voltage at X is already zero, with a
thevenin impedance of one ohm. So the short does nothing and conducts
no current.

The reasoning applies for any value of V1; it all washes.

Is that right?

John

For a voltage-controlled current source, the output current is the input
voltage times the transconductance, where the transconductance is 1/R. Does
the text mention transconductance?

For a VCIS, with the given input voltage of 1 V and resistance of 1 Ohm,
transconductance equals 1 S (Siemen) and the output current will be 1A. This
will be true regardless of whether the load (across x and y) is shorted,
open, or something in between.

Thus, for this problem, the current source will supply 1A and the load will
have 1A passing through it.

If the text is really trying to present a VCIS and it it showing the
controlling input voltage and resistor physically connected to a symbol for
a current source, that is very confusing as an introduction to the theory of
dependent sources. It sounds like something of a cross between a model of a
VCIS and an actual VCIS circuit.

Richard

 

[H.S.]

Richard Seriani wrote:

"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in message
news:8uf6l3tfq8hd3ev8ccq85co2b5ts0j9vu5@4ax.com...
On Sun, 02 Dec 2007 16:46:49 -0500, "H.S." <hs.samix@gmail.com> wrote:


If V1=1 volt, then the triangle-doobie dependant source (as I
interpret it) is 1 amp in the down direction. So 1 amp flows through
the 1 ohm resistor, so the voltage drop across R is one volt. So, with
the short not yet applied, the voltage at X is already zero, with a
thevenin impedance of one ohm. So the short does nothing and conducts
no current.

The reasoning applies for any value of V1; it all washes.

Is that right?

John


For a voltage-controlled current source, the output current is the input
voltage times the transconductance, where the transconductance is 1/R. Does
the text mention transconductance?

No.

For a VCIS, with the given input voltage of 1 V and resistance of 1 Ohm,
transconductance equals 1 S (Siemen) and the output current will be 1A. This
will be true regardless of whether the load (across x and y) is shorted,
open, or something in between.

Thus, for this problem, the current source will supply 1A and the load will
have 1A passing through it.

If the text is really trying to present a VCIS and it it showing the
controlling input voltage and resistor physically connected to a symbol for
a current source, that is very confusing as an introduction to the theory of
dependent sources. It sounds like something of a cross between a model of a
VCIS and an actual VCIS circuit.

The problem is a sub-part of a bigger problem -- that of finding two
port parameters of the given circuit. I have given the reference in
another post of mine, should you feel interested in looking it up.

I agree with John's explanation though. It agrees with the application
of basic laws when applied to the circuit.

regards,
->HS

Richard

 

[H.S.]

John Larkin wrote:

On Sun, 02 Dec 2007 20:24:08 -0800, Peter Bennett
peterbb@somewhere.invalid> wrote:

On Sun, 02 Dec 2007 17:09:23 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sun, 02 Dec 2007 15:59:52 -0800, Peter Bennett
peterbb@somewhere.invalid> wrote:

On Sun, 02 Dec 2007 17:45:06 -0500, "H.S." <hs.samix@gmail.com> wrote:

Jamie wrote:

Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

.
I think maybe something is getting miss interpreted here.

The triangle symbol could be nothing more than an inline AMP meter
taking measurements. or inline current sense device.
That device is a voltage controlled current source and not a measuring
device. The convention used in the text book in question is clear on that.

regards,
HS
If you put a short across the current source, there will be no voltage
across it - so the question becomes how much current does the current
source source or sink when there is no voltage across it?

My common-sense technician's view is that all the current will flow
through the short, and none through the current source.

How can no current flow through a current source?

In fact, 1 amp flows through the current source, and zero amps flow
through the short.

John
But it is a voltage-controlled current course, not a fixed 1 amp
current source - if there is no voltage across it (due to the short)
what current does it pass?

Its current is controlled by V1, which is 1 volt, not that it matters
to the overall problem. But it is a constant 1 amp, since V1 is a
constant 1 volt.

Shorting a current source doesn't change the current it generates.

John

Agreed. An ideal current source delivers the current no matter what is
potential difference across its two terminals.

->HS