Measuring VSWR in a Simulation?

[Stutzer]

Hi,
I would like to measure the VSWR (or the small signal output impedance)
of a circuit I simulate (using PSpice A/D by Cadence).
What would a circuit look like, that indicates the VSWR of the circuit
loaded with a given impedance, or how would I have to calculate it using
measurable values?

I don't konw if the concept of standing waves applies to infinitely
small lenghts of the "transmission-lines" resp. wires (what is probabely
assumed for simulation).
My actual goal is to get a good matching of the circuits output and the
load.

So far thanks for reading, I hope you can help me.
Regards
D. Stutzer

 

[Kevin Aylward]

Stutzer wrote:

Hi,
I would like to measure the VSWR

Why?

Off hand, I don't see why you would want to do this for most
simulations.

(or the small signal output
impedance) of a circuit I simulate (using PSpice A/D by Cadence).

This is trivial to do. Simply feed in a signal and use the V/I feature
that is present in most spices. In fact, in SS, you just tell a source
its for measuring impedances, and its automatic.

What would a circuit look like, that indicates the VSWR of the circuit
loaded with a given impedance, or how would I have to calculate it
using measurable values?

There is a standard formular, which I would have to go and look up,
but why?

The background to using VSWR is geared around experimental
practicalities. This is irrelevant for spice simulation tools. The only
reason for even knowing the impedances, is to calculate the voltages.
Again, this is not that relevant as spice simply produces the data all
in the wash directly.

I don't konw if the concept of standing waves applies to infinitely
small lenghts of the "transmission-lines" resp. wires

Yep. How the voltage varies with x distance is not a factor in lumped
models.

(what is

probabely assumed for simulation).
My actual goal is to get a good matching of the circuits output and
the load.

If your circuit are LCR type matching circuits, I don't see that VSWR is
any use at all. VSWR is an intermediate calculation that is simply
redundant with simulation tools.

You need to ask the queston, what am I trying to do?

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Stutzer]

Hello Kevin,
First: Congratulations and Thanks for SuperSpice.

I already thought my question would be a little unprecise, but I thought
my problem would become clear. Sorry.

Kevin Aylward wrote:

Stutzer wrote:

Hi,
I would like to measure the VSWR

Why?

Off hand, I don't see why you would want to do this for most
simulations.


(or the small signal output
impedance) of a circuit I simulate (using PSpice A/D by Cadence).


This is trivial to do. Simply feed in a signal and use the V/I feature
that is present in most spices. In fact, in SS, you just tell a source
its for measuring impedances, and its automatic.

Well, I don't know. The circuit actually is a FM-oscillator. It has an
audio (ac voltage) input, a dc voltage source to feed the oscillator and
amp and one output port supposed to drive an antenna.
The goal is to get a good match of the antenna to the output of the
"oscillator" near it's resonance frequency.

As far as I remember, the output impedance of a 2-port is calculated as:
Zout=Vout/Iout , given Iin=0
, where Vout (resp. Iout) are applied at the output port with an
ac-voltage source (reps. current).

How does this concept apply to the oscillator circuit? Does it at all?
Or how can measure Zout?
What sources must be "replaced with a wire" (the AC (audio) source, the
DC (battery) source or both?)?


One I knew the output impedance (at certain freqs) of the "antenna-port"
I would know how to design an LC-matching-circuit to match the antenna.
- But I don't want to build the circuit just to measure the
output-impedance.

So please tell me: What am I trying to do exactly? And how do I do it?

Thank you for your help and best regards,
D. Stutzer

What would a circuit look like, that indicates the VSWR of the circuit
loaded with a given impedance, or how would I have to calculate it
using measurable values?


There is a standard formular, which I would have to go and look up,
but why?

The background to using VSWR is geared around experimental
practicalities. This is irrelevant for spice simulation tools. The only
reason for even knowing the impedances, is to calculate the voltages.
Again, this is not that relevant as spice simply produces the data all
in the wash directly.


I don't konw if the concept of standing waves applies to infinitely
small lenghts of the "transmission-lines" resp. wires


Yep. How the voltage varies with x distance is not a factor in lumped
models.

(what is

probabely assumed for simulation).
My actual goal is to get a good matching of the circuits output and
the load.



If your circuit are LCR type matching circuits, I don't see that VSWR is
any use at all. VSWR is an intermediate calculation that is simply
redundant with simulation tools.

You need to ask the queston, what am I trying to do?

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Tom Bruhns]

Stutzer <spam1front1@gmx.ch> wrote in message news:<40aa3534$1_1@news.tiscalinet.ch>...
....

As far as I remember, the output impedance of a 2-port is calculated as:
Zout=Vout/Iout , given Iin=0
, where Vout (resp. Iout) are applied at the output port with an
ac-voltage source (reps. current).

How does this concept apply to the oscillator circuit? Does it at all?
Or how can measure Zout?

A common way to evaluate the output impedance of a source, when
turning the source off would alter the output impedance or when a load
very far from the load for which the circuit was designed would cause
trouble, is to load the source with two different resistances (both
near the optimal load) and observe the change in voltage. If you wish
to find the complex output impedance, things get more complicated; in
doing the test for a real source, you can add a tuneable series
reactance and adjust for peak output into a resistive load, and the
reactance you have added will be equal in magnitude but opposite in
sign from the source's reactance; you can then use the two resistors
to do the "load pull" measurement of the resistive part. Then if e1
and e2 are the output voltages associated with loads R1 and R2
respectively,

Rsource = (e2-e1)*R1*R2/(e1*R2-e2*R1)

But your problem is that you'd have to do a sweep of the reactive
element to use this method for a simulation, and that's a bit tedious.
One should be able to do the same thing, assuming a linear,
time-invariant system, by using at least three different known loads
which vary both the resistance and the reactance. It may be
computationally easier if you do it with + and - delta R from nominal,
and + and - jX from nominal. You're looking for three unknowns
(Vsource, Rsource and Xsource), so you need three equations; thus the
minimum three loads.

Cheers,
Tom

 

[Jim Thompson]

On Tue, 18 May 2004 10:42:33 +0200, Stutzer <spam1front1@gmx.ch>
wrote:

Hi,
I would like to measure the VSWR (or the small signal output impedance)
of a circuit I simulate (using PSpice A/D by Cadence).
What would a circuit look like, that indicates the VSWR of the circuit
loaded with a given impedance, or how would I have to calculate it using
measurable values?

I don't konw if the concept of standing waves applies to infinitely
small lenghts of the "transmission-lines" resp. wires (what is probabely
assumed for simulation).
My actual goal is to get a good matching of the circuits output and the
load.

So far thanks for reading, I hope you can help me.
Regards
D. Stutzer

See the Tools (and Macros) Page of my website, "Return Loss and
VSWR....."

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice480)460-2350 | |
| E-mail Address at Website Fax480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[Stutzer]

Hi Tom,
As I see the problem seems to have become clear now.
I think your approach is surely correct but as you mentioned yourself it
could take a lot of time as I don't have a good "starting point" for
the "recursion".

If I just had to match this single circuit this would not be too much of
a problem. But I think I will come across this problem (of unknown
output impedance) many moore times in the future.

Thus I think there has to be some kind of "macro/script/test-load" that
automizes this procedure or something else that pops out the output
impedance (at best even for sevral freqs.)

Anyway, so far, thanks very much for directing the discussion into this
direction and for your ideas.
Best regards,
D. Stutzer


Tom Bruhns wrote:

Stutzer <spam1front1@gmx.ch> wrote in message news:<40aa3534$1_1@news.tiscalinet.ch>...
...

As far as I remember, the output impedance of a 2-port is calculated as:
Zout=Vout/Iout , given Iin=0
, where Vout (resp. Iout) are applied at the output port with an
ac-voltage source (reps. current).

How does this concept apply to the oscillator circuit? Does it at all?
Or how can measure Zout?


A common way to evaluate the output impedance of a source, when
turning the source off would alter the output impedance or when a load
very far from the load for which the circuit was designed would cause
trouble, is to load the source with two different resistances (both
near the optimal load) and observe the change in voltage. If you wish
to find the complex output impedance, things get more complicated; in
doing the test for a real source, you can add a tuneable series
reactance and adjust for peak output into a resistive load, and the
reactance you have added will be equal in magnitude but opposite in
sign from the source's reactance; you can then use the two resistors
to do the "load pull" measurement of the resistive part. Then if e1
and e2 are the output voltages associated with loads R1 and R2
respectively,

Rsource = (e2-e1)*R1*R2/(e1*R2-e2*R1)

But your problem is that you'd have to do a sweep of the reactive
element to use this method for a simulation, and that's a bit tedious.
One should be able to do the same thing, assuming a linear,
time-invariant system, by using at least three different known loads
which vary both the resistance and the reactance. It may be
computationally easier if you do it with + and - delta R from nominal,
and + and - jX from nominal. You're looking for three unknowns
(Vsource, Rsource and Xsource), so you need three equations; thus the
minimum three loads.

Cheers,
Tom

 

[Kevin Aylward]

"Stutzer" <spam1front1@gmx.ch> wrote in message
news:40aa3534$1_1@news.tiscalinet.ch...

Hello Kevin,
First: Congratulations and Thanks for SuperSpice.

I already thought my question would be a little unprecise, but I
thought
my problem would become clear. Sorry.

Kevin Aylward wrote:
Stutzer wrote:

Hi,
I would like to measure the VSWR

Why?

Off hand, I don't see why you would want to do this for most
simulations.


(or the small signal output
impedance) of a circuit I simulate (using PSpice A/D by Cadence).


This is trivial to do. Simply feed in a signal and use the V/I
feature
that is present in most spices. In fact, in SS, you just tell a
source
its for measuring impedances, and its automatic.

Well, I don't know. The circuit actually is a FM-oscillator. It has an
audio (ac voltage) input, a dc voltage source to feed the oscillator
and
amp and one output port supposed to drive an antenna.
The goal is to get a good match of the antenna to the output of the
"oscillator" near it's resonance frequency.

As far as I remember, the output impedance of a 2-port is calculated
as:
Zout=Vout/Iout , given Iin=0
, where Vout (resp. Iout) are applied at the output port with an
ac-voltage source (reps. current).

How does this concept apply to the oscillator circuit? Does it at all?
Or how can measure Zout?

By measuring an input impedance at the output node.

Replace the load by an AC voltage source with a dc value exactly equal
to the DC voltage at the circuit output node when the load is connected.
This will be zero if ac coupled, if so, you will need a high value bleed
resister to avoid a floating node at the source. Make sure all other ac
sources are zero and do an ac run. Then plot V(source)/I(source).


Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Stutzer]

Hello Jim,
Thanks for the hint. Actually I already came accross your web-page while
I was searching on the internet. Unfortunately I couldn't find the
msim.prb file then..now I found a pspice.prb which probabely has the
same function in pspice.
I will try out your macro this afternoon.

Thank you very much
Regards
D.Stutzer

Jim Thompson wrote:

On Tue, 18 May 2004 10:42:33 +0200, Stutzer <spam1front1@gmx.ch
wrote:


Hi,
I would like to measure the VSWR (or the small signal output impedance)
of a circuit I simulate (using PSpice A/D by Cadence).
What would a circuit look like, that indicates the VSWR of the circuit
loaded with a given impedance, or how would I have to calculate it using
measurable values?

I don't konw if the concept of standing waves applies to infinitely
small lenghts of the "transmission-lines" resp. wires (what is probabely
assumed for simulation).
My actual goal is to get a good matching of the circuits output and the
load.

So far thanks for reading, I hope you can help me.
Regards
D. Stutzer


See the Tools (and Macros) Page of my website, "Return Loss and
VSWR....."

...Jim Thompson

 

[Stutzer]

Kevin Aylward wrote:

"Stutzer" <spam1front1@gmx.ch> wrote in message
news:40aa3534$1_1@news.tiscalinet.ch...

Hello Kevin,
First: Congratulations and Thanks for SuperSpice.

I already thought my question would be a little unprecise, but I

thought

my problem would become clear. Sorry.

Kevin Aylward wrote:

Stutzer wrote:


Hi,
I would like to measure the VSWR

Why?

Off hand, I don't see why you would want to do this for most
simulations.



(or the small signal output
impedance) of a circuit I simulate (using PSpice A/D by Cadence).


This is trivial to do. Simply feed in a signal and use the V/I

feature

that is present in most spices. In fact, in SS, you just tell a

source

its for measuring impedances, and its automatic.

Well, I don't know. The circuit actually is a FM-oscillator. It has an
audio (ac voltage) input, a dc voltage source to feed the oscillator

and

amp and one output port supposed to drive an antenna.
The goal is to get a good match of the antenna to the output of the
"oscillator" near it's resonance frequency.

As far as I remember, the output impedance of a 2-port is calculated

as:

Zout=Vout/Iout , given Iin=0
, where Vout (resp. Iout) are applied at the output port with an
ac-voltage source (reps. current).

How does this concept apply to the oscillator circuit? Does it at all?
Or how can measure Zout?


By measuring an input impedance at the output node.

Replace the load by an AC voltage source with a dc value exactly equal
to the DC voltage at the circuit output node when the load is connected.
This will be zero if ac coupled, if so, you will need a high value bleed
resister to avoid a floating node at the source. Make sure all other ac
sources are zero and do an ac run. Then plot V(source)/I(source).

Allright! Now I see what you mean. Is it generally true, that the (small
sig.) output-impedance of a port is equal to the input-imp. measured at
that very same port if it is driven in the way you explained?
Anyway thanks very much for the idea. I'm going to try it out this evening.

Regards,
D. Stutzer

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Ric]

Kevin Aylward <kevin.aylwardEXTRACT@anasoft.co.uk> wrote:

"Stutzer" <spam1front1@gmx.ch> wrote in message
news:40aa3534$1_1@news.tiscalinet.ch...
Hello Kevin,
First: Congratulations and Thanks for SuperSpice.

I already thought my question would be a little unprecise, but I
thought
my problem would become clear. Sorry.

Kevin Aylward wrote:
Stutzer wrote:

Hi,
I would like to measure the VSWR

Why?

Off hand, I don't see why you would want to do this for most
simulations.


(or the small signal output
impedance) of a circuit I simulate (using PSpice A/D by Cadence).


This is trivial to do. Simply feed in a signal and use the V/I
feature
that is present in most spices. In fact, in SS, you just tell a
source
its for measuring impedances, and its automatic.

Well, I don't know. The circuit actually is a FM-oscillator. It has an
audio (ac voltage) input, a dc voltage source to feed the oscillator
and
amp and one output port supposed to drive an antenna.
The goal is to get a good match of the antenna to the output of the
"oscillator" near it's resonance frequency.

As far as I remember, the output impedance of a 2-port is calculated
as:
Zout=Vout/Iout , given Iin=0
, where Vout (resp. Iout) are applied at the output port with an
ac-voltage source (reps. current).

How does this concept apply to the oscillator circuit? Does it at all?
Or how can measure Zout?

By measuring an input impedance at the output node.

Replace the load by an AC voltage source with a dc value exactly equal
to the DC voltage at the circuit output node when the load is connected.
This will be zero if ac coupled, if so, you will need a high value bleed

If the o/p is ac coupled, it makes no difference what DC voltage you
apply to your source - it doesn't HAVE to be zero.

If the o/p is dc-coupled, it's probable that removing the original load
will affect the quiescent voltage, so you'd want to determine what that
voltage is WITHOUT the load, and then duplicate that.

resister to avoid a floating node at the source. Make sure all other ac
sources are zero and do an ac run. Then plot V(source)/I(source).

Is there a reason why you'd want to do this in preference of the usual
"pump an AC Amp" into the output?

--
Rick

 

[Kevin Aylward]

"Rick" <rik_nntp@dsl.pipex.com> wrote in message
news:FGGqc.16756$k4.338340@news1.nokia.com...

Kevin Aylward <kevin.aylwardEXTRACT@anasoft.co.uk> wrote:



"Stutzer" <spam1front1@gmx.ch> wrote in message
news:40aa3534$1_1@news.tiscalinet.ch...
Hello Kevin,
First: Congratulations and Thanks for SuperSpice.

I already thought my question would be a little unprecise, but I
thought
my problem would become clear. Sorry.

Kevin Aylward wrote:
Stutzer wrote:

Hi,
I would like to measure the VSWR

Why?

Off hand, I don't see why you would want to do this for most
simulations.


(or the small signal output
impedance) of a circuit I simulate (using PSpice A/D by Cadence).


This is trivial to do. Simply feed in a signal and use the V/I
feature
that is present in most spices. In fact, in SS, you just tell a
source
its for measuring impedances, and its automatic.

Well, I don't know. The circuit actually is a FM-oscillator. It has
an
audio (ac voltage) input, a dc voltage source to feed the
oscillator
and
amp and one output port supposed to drive an antenna.
The goal is to get a good match of the antenna to the output of the
"oscillator" near it's resonance frequency.

As far as I remember, the output impedance of a 2-port is
calculated
as:
Zout=Vout/Iout , given Iin=0
, where Vout (resp. Iout) are applied at the output port with an
ac-voltage source (reps. current).

How does this concept apply to the oscillator circuit? Does it at
all?
Or how can measure Zout?

By measuring an input impedance at the output node.

Replace the load by an AC voltage source with a dc value exactly
equal
to the DC voltage at the circuit output node when the load is
connected.
This will be zero if ac coupled, if so, you will need a high value
bleed

If the o/p is ac coupled, it makes no difference what DC voltage you
apply to your source - it doesn't HAVE to be zero.

Ho hummm...

If the o/p is dc-coupled, it's probable that removing the original
load
will affect the quiescent voltage,

Yep.

so you'd want to determine what that
voltage is WITHOUT the load, and then duplicate that.

Nope. This will upset the bias condition as it is when it is in actual
use. Suppose removing the load makes the output transistor clip of
bottom out, what happens then? You want to measure the impedance at the
same bias conditions as it is when operating normally. Anything else is
meaningless.

It might even be preferable to keep the load in circuit and have the
source in series with it.

resister to avoid a floating node at the source. Make sure all other
ac
sources are zero and do an ac run. Then plot V(source)/I(source).

Is there a reason why you'd want to do this in preference of the usual
"pump an AC Amp" into the output?

Yes. It highlights the fact that you don't have to do this in SS,
although its always a bit disconcerting to see mega volts on the
schematic nodes.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Joe McElvenney]

Hi,

The measured value of the output impedance is not normally
used to design a network to follow such a circuit - the 'source-Z
MUST equal load-Z' theorem is not always applicable. Common
practice is to offer a load impedance that will provide the
required output power at the given collector voltage swing.
Remembering that a class-C stage is very much like an on-off
switch, the usual cookbook formula is -

RL = (Vcc - Vb)^2 / 2.Po ; Vcc - Vb is the collector swing,

where Vcc is the supply and Vb the DC base bias voltage.

So if you set your sights on 100mW with a 12V supply and 3V of
base bias, you are looking at 400 ohms or so.

For true RF power devices, the manufacturer will specify an
output impedance which is usually a few tens of ohms in series
with a capacitive reactance of the same order. This 'Z' is then
used to design the matching network. All of this is reminiscent
of the 'optimum value' used in old tube power amps which was a
compromise between power output and distortion and not the actual
output impedance.


Cheers - Joe

 

[Stutzer]

Hello Joe,
Thanks a lot for your posting!
It is really pointed towards the core of my problem. Probably your
formula will satisfy my needs completely. It also helps a lot
considering the question "what am I trying to do?".
I will try it out later.

Best regards
D. Stutzer

Joe McElvenney wrote:

Hi,

The measured value of the output impedance is not normally
used to design a network to follow such a circuit - the 'source-Z
MUST equal load-Z' theorem is not always applicable. Common
practice is to offer a load impedance that will provide the
required output power at the given collector voltage swing.
Remembering that a class-C stage is very much like an on-off
switch, the usual cookbook formula is -

RL = (Vcc - Vb)^2 / 2.Po ; Vcc - Vb is the collector swing,

where Vcc is the supply and Vb the DC base bias voltage.

So if you set your sights on 100mW with a 12V supply and 3V of
base bias, you are looking at 400 ohms or so.

For true RF power devices, the manufacturer will specify an
output impedance which is usually a few tens of ohms in series
with a capacitive reactance of the same order. This 'Z' is then
used to design the matching network. All of this is reminiscent
of the 'optimum value' used in old tube power amps which was a
compromise between power output and distortion and not the actual
output impedance.


Cheers - Joe

 

[Ric]

Kevin Aylward <kevin.aylwardEXTRACT@anasoft.co.uk> wrote:

"Rick" <rik_nntp@dsl.pipex.com> wrote in message
news:FGGqc.16756$k4.338340@news1.nokia.com...

If the o/p is dc-coupled, it's probable that removing the original
load
will affect the quiescent voltage,

Yep.

so you'd want to determine what that
voltage is WITHOUT the load, and then duplicate that.

Nope. This will upset the bias condition as it is when it is in actual
use. Suppose removing the load makes the output transistor clip of
bottom out, what happens then? You want to measure the impedance at the
same bias conditions as it is when operating normally. Anything else is
meaningless.

Yep. I agree - I came to the same conclusion shortly after I'd posted
my reply. My method produces the impedance of the unloaded circuit.

It might even be preferable to keep the load in circuit and have the
source in series with it.

I think the best bet is to leave the load in, but not fiddle around
with this voltage source stuff - just pump 1 Amp AC in, an accept
that the resulting impedance is the parallel combination of the
circuit's output impedance and the load impedance. Working back
to the circuit's o/p impedance is then trivial if the load is passive.

Is there a reason why you'd want to do this in preference of the usual
"pump an AC Amp" into the output?

Yes. It highlights the fact that you don't have to do this in SS,

I'd swap a bucket of "convenience" features if your software offered
either of:
1. Ability to temporarily disable any given components in the schematic
without having to delete them.
2. Override component model parameters from within the schematic without
having to edit the part library first. I hate having components called
"hi beta", "hi beta zero tempco" - I just want to change the damned
parameter!

Does SS do these?

--
Rick

 

[Jim Thompson]

On Wed, 19 May 2004 12:01:30 +0200, Stutzer <spam1front1@gmx.ch>
wrote:

Hello Jim,
Thanks for the hint. Actually I already came accross your web-page while
I was searching on the internet. Unfortunately I couldn't find the
msim.prb file then..now I found a pspice.prb which probabely has the
same function in pspice.
I will try out your macro this afternoon.

Thank you very much
Regards
D.Stutzer

[snip]

Correct. I need to update my page. Since I did that page OrCAD
bought MicroSim and then Cadence bought OrCAD. Fortunately only the
file names have changed.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice480)460-2350 | |
| E-mail Address at Website Fax480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[Tom Bruhns]

Oh dear. That was an invitation for me to get on my soapbox.

You should view Spice as an _ANALYSIS_ tool, NOT a "design" tool. I
think Spice is dangerous in the hands of folk who don't have a basic
understanding of the circuits they are trying to design. How does
that apply in this situation? Well, I would hope that you have a
pretty good idea about the source impedance of the circuit you are
dealing with, and the optimal load as well. Is it an output from a
feedback-controlled amplifier, or is it from something which is
open-loop? Is it from the emitter circuit or the collector circuit?
If you're trying to put a load on the oscillator itself, what will a
heavy load do? (Hint: it's generally NOT good for frequency
stability, because small percentage variations in the load can still
be significant.) Fairly often, the optimal load for a circuit is an
impedance which is far different from the circuit's output impedance:
"optimal" is NOT always maximum power transfer.

It's the philosophy of why you should do various things in the circuit
that Spice generally won't help with, unless you are exceptionally
observant about the simulation results, and try a wide range of
things.

In addition, having a good basic "feel" for the circuit will help you
immeasurably in knowing when your simulation results are incorrect, as
they often can be. That can happen because of Spice limitations, and
because your model does not match reality sufficiently well.

I trust this won't open too big a can of worms...I hope instead it's
food for thought about the design process.

Cheers,
Tom



Stutzer <spam1front1@gmx.ch> wrote in message news:<40ab2b9f_1@news.tiscalinet.ch>...

Hi Tom,
As I see the problem seems to have become clear now.
I think your approach is surely correct but as you mentioned yourself it
could take a lot of time as I don't have a good "starting point" for
the "recursion".

If I just had to match this single circuit this would not be too much of
a problem. But I think I will come across this problem (of unknown
output impedance) many moore times in the future.

Thus I think there has to be some kind of "macro/script/test-load" that
automizes this procedure or something else that pops out the output
impedance (at best even for sevral freqs.)

Anyway, so far, thanks very much for directing the discussion into this
direction and for your ideas.
Best regards,
D. Stutzer

....

 

[Kevin Aylward]

"Rick" <rik_nntp@dsl.pipex.com> wrote in message
news:SPHqc.16782$k4.338601@news1.nokia.com...

Kevin Aylward <kevin.aylwardEXTRACT@anasoft.co.uk> wrote:

"Rick" <rik_nntp@dsl.pipex.com> wrote in message
news:FGGqc.16756$k4.338340@news1.nokia.com...

If the o/p is dc-coupled, it's probable that removing the original
load
will affect the quiescent voltage,

Yep.

so you'd want to determine what that
voltage is WITHOUT the load, and then duplicate that.

Nope. This will upset the bias condition as it is when it is in
actual
use. Suppose removing the load makes the output transistor clip of
bottom out, what happens then? You want to measure the impedance at
the
same bias conditions as it is when operating normally. Anything else
is
meaningless.


Yep. I agree - I came to the same conclusion shortly after I'd posted
my reply. My method produces the impedance of the unloaded circuit.

It might even be preferable to keep the load in circuit and have the
source in series with it.

I think the best bet is to leave the load in, but not fiddle around
with this voltage source stuff - just pump 1 Amp AC in, an accept
that the resulting impedance is the parallel combination of the
circuit's output impedance and the load impedance. Working back
to the circuit's o/p impedance is then trivial if the load is passive.


Is there a reason why you'd want to do this in preference of the
usual
"pump an AC Amp" into the output?

Yes. It highlights the fact that you don't have to do this in SS,

I'd swap a bucket of "convenience" features if your software offered
either of:
1. Ability to temporarily disable any given components in the
schematic
without having to delete them.

Well, you can just move them to a blank piece of the page. (use the
rubberband toolbar button to temporally disable the rube). However...

You can do exactly what you wish. Double-click on the symbol to popup
its set-up dialog. Use the PCB data tab, and lo and behold, check boxes
to include in the spice and/or pcb netlist.

2. Override component model parameters from within the schematic
without
having to edit the part library first. I hate having components
called
"hi beta", "hi beta zero tempco" - I just want to change the damned
parameter!

Does SS do these?

Sort off. There is the docked preview window (top left main window)
where you can edit the model directly (but it does via the lib), or you
can edit the model from the symbol, set-up dialog. However, this effects
all symbols with that model name. However...

You can have 3 sets of models, easily *switchable*. The model tab of the
set-up has checks for weak, nominal and strong. If you switch this from
its default of normal, it will try and find that model variation, or
copy the nominal model. You can then change those weak or strong model
parameters and those parameters will only effect other models that are
also connected to that specific model type. This allows quick check out
of different model variations simply by clicking those buttons.
However...Changing these weak and strong parameters will effect the
automatic WC runs for that model type as it uses this data. However...
you shouldn't really have to do what you suggest above, as that is what
the auto WC (worst case) is *designed* to do for you already. It already
has sensible WC values set-up, e.g. max/min hfe, etc. I have tried to
anticipate what is required so that the user don't have to worry about
these things. The WC reruns should give a pretty good (overestimate) of
what variations will occur without the user having to manually do this.


Oh.. I reduced the price a bit today...

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Kevin Aylward]

"Kevin Aylward" <kevin.aylwardEXTRACT@anasoft.co.uk> wrote in message
news:VVNqc.7050$YS4.1466@doctor.cableinet.net...

Sort off. There is the docked preview window (top left main window)
where you can edit the model directly (but it does via the lib), or
you
can edit the model from the symbol, set-up dialog. However, this
effects
all symbols with that model name. However...

You can have 3 sets of models, easily *switchable*. The model tab of
the
set-up has checks for weak, nominal and strong. If you switch this
from
its default of normal, it will try and find that model variation, or
copy the nominal model. You can then change those weak or strong model
parameters and those parameters will only effect other models that are
also connected to that specific model type. This allows quick check
out
of different model variations simply by clicking those buttons.
However...Changing these weak and strong parameters will effect the
automatic WC runs for that model type as it uses this data.
However...
you shouldn't really have to do what you suggest above, as that is
what
the auto WC (worst case) is *designed* to do for you already. It
already
has sensible WC values set-up, e.g. max/min hfe, etc. I have tried to
anticipate what is required so that the user don't have to worry about
these things. The WC reruns should give a pretty good (overestimate)
of
what variations will occur without the user having to manually do
this.

I posted an err.. "update" to this feature on this today. If the
explicit WC model already existed, switching types would use and load
that model. However, if they didn't exist, the buttons wouldn't use/load
in the auto generated WC model version values that the WC reruns use for
this condition. This is now fixed.

 

[Peter O. Brackett]

Stutzer:

Measuring VSWR with Spicees...

The most direct way of doing this with any Spice is simply to build
[simulate] a VSWR [Return Loss]
bridge circuit as part of your simulation and include it in your simulation
runs.

The output of the VSWR bridge is what you are looking for.

In it's simplest form a VSWR bridge is simply a purely passive resistance
Wheatstone bridge but you may
get fancy with your simulations and create active bridges in Spice with Op
Amps, etc... if you wish.

--
Peter


"Stutzer" <spam1front1@gmx.ch> wrote in message
news:40a9cc6d$1_2@news.tiscalinet.ch...

Hi,
I would like to measure the VSWR (or the small signal output impedance)
of a circuit I simulate (using PSpice A/D by Cadence).
What would a circuit look like, that indicates the VSWR of the circuit
loaded with a given impedance, or how would I have to calculate it using
measurable values?

I don't konw if the concept of standing waves applies to infinitely
small lenghts of the "transmission-lines" resp. wires (what is probabely
assumed for simulation).
My actual goal is to get a good matching of the circuits output and the
load.

So far thanks for reading, I hope you can help me.
Regards
D. Stutzer

 

[Stutzer]

Hello Peter,
First of all, thanks for your interest.
I really don't want to offend you, but are you sure about this?

I ask this, because this "natural approach" was my first as well. But it
did not seem to work with the (very few and simple) circuits I tried.

From a theoretical point of view: How can one distinguish between the
part of the voltage (or power) that is due to the "backwards" travelling
waves and the "foreward"-part when you have just one (infinitely small)
point (resp. part of the transmission line/wire) where you can measure
the (total) voltage? To me this seems impossible, in general.
But I can imagine to implement such a circuit in a way appropriate for
simulation, by use of "Transmission-Lines" to insert a delay.

Please correct me where I'm wrong
Regards
D.Stutzer


Peter O. Brackett wrote:

Stutzer:

Measuring VSWR with Spicees...

The most direct way of doing this with any Spice is simply to build
[simulate] a VSWR [Return Loss]
bridge circuit as part of your simulation and include it in your simulation
runs.

The output of the VSWR bridge is what you are looking for.

In it's simplest form a VSWR bridge is simply a purely passive resistance
Wheatstone bridge but you may
get fancy with your simulations and create active bridges in Spice with Op
Amps, etc... if you wish.

--
Peter


"Stutzer" <spam1front1@gmx.ch> wrote in message
news:40a9cc6d$1_2@news.tiscalinet.ch...

Hi,
I would like to measure the VSWR (or the small signal output impedance)
of a circuit I simulate (using PSpice A/D by Cadence).
What would a circuit look like, that indicates the VSWR of the circuit
loaded with a given impedance, or how would I have to calculate it using
measurable values?

I don't konw if the concept of standing waves applies to infinitely
small lenghts of the "transmission-lines" resp. wires (what is probabely
assumed for simulation).
My actual goal is to get a good matching of the circuits output and the
load.

So far thanks for reading, I hope you can help me.
Regards
D. Stutzer