Amp output Z

Harry Dellamano · Apr 14, 2005

"qrk" <SpamTrap@reson.com> wrote in message
news:8ahr51t14cq0hsunqdrlqkjtr2ftuml62i@4ax.com...

I'm in process of building up a dual AMD Opteron computer. I'll pass
on details if you want. It will be a week or so until I get my
benchmarks tested out. I'm interested to see what happens with the
better memory access speed.

Mark

Hey Mark, now that is exciting. Keep us updated with hardware and
performance.
JT has mentioned that PC Club has some good hardware. Who is your source?

Cheers,
Harry

The Phantom · Apr 14, 2005

On Thu, 14 Apr 2005 00:06:16 +0100, John Woodgate <jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that The Phantom <phantom@aol.com
wrote (in <0q4r51pg9l8v7sin5c9jenvd4v6opa9k90@4ax.com&gt about 'Amp
output Z', on Wed, 13 Apr 2005:

In the real amp it must be something other a pure resistance of 10 ohms
to account for the voltages he measures.

I see you have assumed that they are accurate. But do you not think that
your inference from them is, on reflection, improbable? How could such
an inductance arise?

As I said in another post, this problem is not well described.

In his original post, Harry said:

"The output voltage of the amp is
10.0Vrms at 400 Hz with no loads connected."

Then in another post, he said:

"Hi John,
I am looking for Zo not Ro. I cannot measure the open circuit output
voltage because the amp is unstable with no load. This is a customers design
that I have to redesign. He believes that the output Z is about 10 ohms
which is easy to measure by Ro=dVo/dRL. I believe it to be 70 ohms but
cannot prove it. The spec requirement is less than 25 ohms."

So, which is it? Is the output voltage of the amp with no loads connected equal to 10.0
Vrms, or is the amp unstable with no loads connected? My best guess is that in the OP he
was talking about the simulated amp, and in the "Hi, John" post he was talking about the
*real* amp. It would help avoid confusion if he gave more detail and tell us whether he's
talking about the simulation or the real amp.

If that 10 ohm resistor is in place, then you can't have the voltages he gives us for E2A
and E2B without some inductance in the circuit.

But if you allow the 10 ohms to become .54874 ohms, and the open circuit voltage to be
14.927 volts, then you can get the 14.81 volts with a 600 ohm load, and 14.5 volts with a
300 ohm load. And in this case, it's actually 14.81 V @ 6.747 degrees, and 14.5 V @
13.301 degrees. But, he's not seeing the phase angle. Too bad; if he were, that would
help figure out what's going on.

Pooh Bear · Apr 14, 2005

qrk wrote:

On Tue, 12 Apr 2005 02:30:32 GMT, "Harry Dellamano"
harryd@tdsystems.org> wrote:

"qrk" <SpamTrap@reson.com> wrote in message
news:fgam51dbabvmr86kfrnfcpie1rev9bsrqm@4ax.com...
On Mon, 11 Apr 2005 22:00:13 GMT, "Harry Dellamano"
harryd@tdsystems.org> wrote:

Ok, so I got this amplifier with an internal output resistance of 10
ohms
and we are outside of all feedback loops. It is coupled to the 600 ohm
resistive load thru a 5.6uF capacitor. The output voltage of the amp is
10.0Vrms at 400 Hz with no loads connected. If I measure the Zo with SPICE
using a swept 1 amp current source at the output, I get about 71 ohms
which
is the Xc of the coupling cap. If I use a DVM on the real circuit and
measure the output voltage at two different loads, 600 ohms and 300 ohms,
I
get about 13.0 ohms using Ro=((E1-E2)/(E1/600)-(E2/300)). I need to
measure
the Zo using a DVM and not Ro. By observation it must be 71 ohms. What
must
I do??

Harry,
Few things are suspect.
Your equation has some sign and parentheses problems.
It might be Ro=(E1-E2)/((E2/300)-(E1/600))

Your equation is for pure resistance. This is a complex number
problem. Use Spice to measure the output amplitude like you would with
a DMM. I'll bet you end up with the same wrong answer.

You need to deal with complex numbers since your series capacitor
introduces a significant phase angle at 400 Hz. At 400 Hz your output
impedance is 10-j71 Ohms.

Mark
Hi Mark,
If I can only measure and vary the 600 ohm resistive load then all voltages
and currents that I can measure are in phase and complex math is useless.
This yields Zo=13 ohms but it sounds wrong. I like your 70 ohms above but
don't know how to go about measuring it.
OT Got to get a new SPICE puter. You were a big help the last time so I
may bug you again.
Thanks,
Harry

Harry,
If you can change the output capacitor to a large value, like
10,000uF, then you can treat this like a pure resistance voltage
divider. Or, connect up a network analyzer and measure directly.

Even making the cap 100uF would reduce the term to practical insignificance ! I's
certainly be tempted to do that as a quick and easy solution to any measurement
issues.

Graham

Pooh Bear · Apr 14, 2005

The Phantom wrote:

On Thu, 14 Apr 2005 00:06:16 +0100, John Woodgate <jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that The Phantom <phantom@aol.com
wrote (in <0q4r51pg9l8v7sin5c9jenvd4v6opa9k90@4ax.com&gt about 'Amp
output Z', on Wed, 13 Apr 2005:

In the real amp it must be something other a pure resistance of 10 ohms
to account for the voltages he measures.

I see you have assumed that they are accurate. But do you not think that
your inference from them is, on reflection, improbable? How could such
an inductance arise?

As I said in another post, this problem is not well described.

In his original post, Harry said:

"The output voltage of the amp is
10.0Vrms at 400 Hz with no loads connected."

Then in another post, he said:

"Hi John,
I am looking for Zo not Ro. I cannot measure the open circuit output
voltage because the amp is unstable with no load. This is a customers design
that I have to redesign. He believes that the output Z is about 10 ohms
which is easy to measure by Ro=dVo/dRL. I believe it to be 70 ohms but
cannot prove it. The spec requirement is less than 25 ohms."

So, which is it? Is the output voltage of the amp with no loads connected equal to 10.0
Vrms, or is the amp unstable with no loads connected? My best guess is that in the OP he
was talking about the simulated amp, and in the "Hi, John" post he was talking about the
*real* amp. It would help avoid confusion if he gave more detail and tell us whether he's
talking about the simulation or the real amp.

Amen to that. I think the 10 ohms output resistance is a guesstimate too.

If that 10 ohm resistor is in place, then you can't have the voltages he gives us for E2A
and E2B without some inductance in the circuit.

But if you allow the 10 ohms to become .54874 ohms, and the open circuit voltage to be
14.927 volts, then you can get the 14.81 volts with a 600 ohm load, and 14.5 volts with a
300 ohm load. And in this case, it's actually 14.81 V @ 6.747 degrees, and 14.5 V @
13.301 degrees. But, he's not seeing the phase angle. Too bad; if he were, that would
help figure out what's going on.

Multimeters don't see phase angle - lol !

Graham

Tony Williams · Apr 14, 2005

In article <NrC6e.8223$jd6.3292@trnddc07>,
Harry Dellamano <harryd@tdsystems.org> wrote:

Ok, so I got this amplifier with an internal output resistance
of 10 ohms and we are outside of all feedback loops. It is
coupled to the 600 ohm resistive load thru a 5.6uF capacitor.
The output voltage of the amp is 10.0Vrms at 400 Hz with no
loads connected. If I measure the Zo with SPICE using a swept 1
amp current source at the output, I get about 71 ohms which is
the Xc of the coupling cap. If I use a DVM on the real circuit
and measure the output voltage at two different loads, 600 ohms
and 300 ohms, I get about 13.0 ohms using
Ro=((E1-E2)/(E1/600)-(E2/300)). I need to measure the Zo using a
DVM and not Ro. By observation it must be 71 ohms. What must I
do??

Reverse the procedure Harry. Set the device at
Ov out, inject a known current into the output
terminal, and measure the resultant voltage.

Use an ammeter if possible, otherwise use a large
value resistor.

A scope would be handy..... the dual beam can show
the phase shift between the injected current and
the voltage across Zo, and a check for a decent
sinewave across Zo verifies that Zo is linear with
current, (or not).

--
Tony Williams.

Tony Williams · Apr 14, 2005

On 14 Apr, tonyw@ledelec.demon.co.uk wrote:

Reverse the procedure Harry. Set the device at
Ov out, inject a known current into the output
terminal, and measure the resultant voltage.

You can also get some idea of the value of Ro.

switch
_________ +----o/o----+
| Device | | |
| +--------+-[10ohms]--+--[Ammeter]--[R]--<Vgen
| | | |
| ===C | |
| | | +-->
| [Ro] | Vout to DVM/scope.
|____|____| +-->
| |
0v+--------------------+-----------------<

Switch the 10 ohms in/out, adjusting Vgen to keep the
current constant. Measure the difference in Vout.
Simple calcs to get a rough idea of Ro (and C).

--
Tony Williams.

Fred Bartoli · Apr 14, 2005

"Tony Williams" <tonyw@ledelec.demon.co.uk> a écrit dans le message de
news:4d5b5d7fc5tonyw@ledelec.demon.co.uk...

In article <NrC6e.8223$jd6.3292@trnddc07>,
Harry Dellamano <harryd@tdsystems.org> wrote:
Ok, so I got this amplifier with an internal output resistance
of 10 ohms and we are outside of all feedback loops. It is
coupled to the 600 ohm resistive load thru a 5.6uF capacitor.
The output voltage of the amp is 10.0Vrms at 400 Hz with no
loads connected. If I measure the Zo with SPICE using a swept 1
amp current source at the output, I get about 71 ohms which is
the Xc of the coupling cap. If I use a DVM on the real circuit
and measure the output voltage at two different loads, 600 ohms
and 300 ohms, I get about 13.0 ohms using
Ro=((E1-E2)/(E1/600)-(E2/300)). I need to measure the Zo using a
DVM and not Ro. By observation it must be 71 ohms. What must I
do??

Reverse the procedure Harry. Set the device at
Ov out, inject a known current into the output
terminal, and measure the resultant voltage.

Use an ammeter if possible, otherwise use a large
value resistor.

A scope would be handy..... the dual beam can show
the phase shift between the injected current and
the voltage across Zo, and a check for a decent
sinewave across Zo verifies that Zo is linear with
current, (or not).

He he, did you or Harry see news:425cf269$0$32031$626a14ce@news.free.fr

Direct measurement is still the better way.

--
Thanks,
Fred.

Harry Dellamano · Apr 14, 2005

"Fred Bartoli"
<fred._canxxxel_this_bartoli@RemoveThatAlso_free.fr_AndThisToo> wrote in
message news:425cf269$0$32031$626a14ce@news.free.fr...

"Harry Dellamano" <harryd@tdsystems.org> a écrit dans le message de
news:NrC6e.8223$jd6.3292@trnddc07...
Ok, so I got this amplifier with an internal output resistance of 10
ohms
and we are outside of all feedback loops. It is coupled to the 600 ohm
resistive load thru a 5.6uF capacitor. The output voltage of the amp is
10.0Vrms at 400 Hz with no loads connected. If I measure the Zo with
SPICE
using a swept 1 amp current source at the output, I get about 71 ohms
which
is the Xc of the coupling cap. If I use a DVM on the real circuit and
measure the output voltage at two different loads, 600 ohms and 300 ohms,
I
get about 13.0 ohms using Ro=((E1-E2)/(E1/600)-(E2/300)). I need to
measure
the Zo using a DVM and not Ro. By observation it must be 71 ohms. What
must
I do??
Harry

Harry,

I've not seen someone suggest the obvious: why don't you directly measure
the output impedance?

Terminate your amp to 600R to make it stable. Inject some current into the
output through a 1K or more resistor or a current source.

Measure the voltage and phase and then correct for the output load (600R
resistor + current source impedance).

Any measurement error will have much less impact than with your method.

--
Thanks,
Fred.

Hey Fred, you have a winner.

I took a 10Vrms voltage source at 400Hz and connected it thru a 1K resistor
to the 600 ohm load. Call the voltage on the voltage source Va and the other
side of the 1k resistor Vb then Zo=Vb/I1k = Vb*1K/(Va-Vb). Using a DVM I got
73 ohms. If I had a higher 400Hz voltage source I could have used higher
than 1K and got closer to 71 ohms. A constant current source would yield
great results but this is close enough.
Thanks for your help.
Harry

Harry Dellamano · Apr 14, 2005

"Fred Bartoli"
<fred._canxxxel_this_bartoli@RemoveThatAlso_free.fr_AndThisToo> wrote in
message news:425e31df$0$22099$626a14ce@news.free.fr...

"Tony Williams" <tonyw@ledelec.demon.co.uk> a écrit dans le message de
news:4d5b5d7fc5tonyw@ledelec.demon.co.uk...
In article <NrC6e.8223$jd6.3292@trnddc07>,
Harry Dellamano <harryd@tdsystems.org> wrote:
Ok, so I got this amplifier with an internal output resistance
of 10 ohms and we are outside of all feedback loops. It is
coupled to the 600 ohm resistive load thru a 5.6uF capacitor.
The output voltage of the amp is 10.0Vrms at 400 Hz with no
loads connected. If I measure the Zo with SPICE using a swept 1
amp current source at the output, I get about 71 ohms which is
the Xc of the coupling cap. If I use a DVM on the real circuit
and measure the output voltage at two different loads, 600 ohms
and 300 ohms, I get about 13.0 ohms using
Ro=((E1-E2)/(E1/600)-(E2/300)). I need to measure the Zo using a
DVM and not Ro. By observation it must be 71 ohms. What must I
do??

Reverse the procedure Harry. Set the device at
Ov out, inject a known current into the output
terminal, and measure the resultant voltage.

Use an ammeter if possible, otherwise use a large
value resistor.

A scope would be handy..... the dual beam can show
the phase shift between the injected current and
the voltage across Zo, and a check for a decent
sinewave across Zo verifies that Zo is linear with
current, (or not).

He he, did you or Harry see news:425cf269$0$32031$626a14ce@news.free.fr

Direct measurement is still the better way.

--
Thanks,
Fred.

Hey Fred, you have a winner.

I took a 10Vrms voltage source at 400Hz and connected it thru a 1K resistor
to the 600 ohm load. Call the voltage on the voltage source Va and the other
side of the 1k resistor Vb then Zo=Vb/I1k = Vb*1K/(Va-Vb). Using a DVM I got
73 ohms. If I had a higher 400Hz voltage source I could have used higher
than 1K and got closer to 71 ohms. A constant current source would yield
great results but this is close enough.
Thanks for your help.
Harry

The Phantom · Apr 14, 2005

On Thu, 14 Apr 2005 14:43:31 GMT, "Harry Dellamano" <harryd@tdsystems.org> wrote:

Reverse the procedure Harry. Set the device at
Ov out, inject a known current into the output
terminal, and measure the resultant voltage.

Use an ammeter if possible, otherwise use a large
value resistor.

A scope would be handy..... the dual beam can show
the phase shift between the injected current and
the voltage across Zo, and a check for a decent
sinewave across Zo verifies that Zo is linear with
current, (or not).

He he, did you or Harry see news:425cf269$0$32031$626a14ce@news.free.fr

Direct measurement is still the better way.

--
Thanks,
Fred.

Hey Fred, you have a winner.
I took a 10Vrms voltage source at 400Hz and connected it thru a 1K resistor
to the 600 ohm load. Call the voltage on the voltage source Va and the other
side of the 1k resistor Vb then Zo=Vb/I1k = Vb*1K/(Va-Vb). Using a DVM I got
73 ohms. If I had a higher 400Hz voltage source I could have used higher
than 1K and got closer to 71 ohms. A constant current source would yield
great results but this is close enough.
Thanks for your help.
Harry

Assuming this is the correct schematic:

+---[10R]---------+------TPA
| |
| [5.6ľF]
| |
[GEN] +------TPB
| |
| [600R]
| |
+-----------------+
| |
GND GND

Are you saying you injected the signal at TPB?

If the 5.6 uF capacitor *isn't* part of the amp, why don't you inject it at TPA?

I would have thought you would want the output impedance of the amp, not the output
impedance of the amp+cap in series.

Harry Dellamano · Apr 14, 2005

"The Phantom" <phantom@aol.com> wrote in message
news:te3t51plr0uf3uqdk0vb0dn1gpqvcdbtl8@4ax.com...

On Thu, 14 Apr 2005 14:43:31 GMT, "Harry Dellamano" <harryd@tdsystems.org
wrote:

Reverse the procedure Harry. Set the device at
Ov out, inject a known current into the output
terminal, and measure the resultant voltage.

Use an ammeter if possible, otherwise use a large
value resistor.

A scope would be handy..... the dual beam can show
the phase shift between the injected current and
the voltage across Zo, and a check for a decent
sinewave across Zo verifies that Zo is linear with
current, (or not).

He he, did you or Harry see news:425cf269$0$32031$626a14ce@news.free.fr

Direct measurement is still the better way.

--
Thanks,
Fred.

Hey Fred, you have a winner.
I took a 10Vrms voltage source at 400Hz and connected it thru a 1K
resistor
to the 600 ohm load. Call the voltage on the voltage source Va and the
other
side of the 1k resistor Vb then Zo=Vb/I1k = Vb*1K/(Va-Vb). Using a DVM I
got
73 ohms. If I had a higher 400Hz voltage source I could have used higher
than 1K and got closer to 71 ohms. A constant current source would yield
great results but this is close enough.
Thanks for your help.
Harry

Assuming this is the correct schematic:

+---[10R]---------+------TPA
| |
| [5.6ľF]
| |
[GEN] +------TPB
| |
| [600R]
| |
+-----------------+
| |
GND GND

Are you saying you injected the signal at TPB?

If the 5.6 uF capacitor *isn't* part of the amp, why don't you inject it
at TPA?

I would have thought you would want the output impedance of the amp, not
the output
impedance of the amp+cap in series.

The coupling cap is part of the total amp, outside of the feedback loop,

so we only have access to where the amp ties to the load, which is TPB. Not
a typical design.
Harry

The Phantom · Apr 14, 2005

On Thu, 14 Apr 2005 18:59:38 GMT, "Harry Dellamano" <harryd@tdsystems.org> wrote:

"The Phantom" <phantom@aol.com> wrote in message
news:te3t51plr0uf3uqdk0vb0dn1gpqvcdbtl8@4ax.com...
On Thu, 14 Apr 2005 14:43:31 GMT, "Harry Dellamano" <harryd@tdsystems.org
wrote:

Reverse the procedure Harry. Set the device at
Ov out, inject a known current into the output
terminal, and measure the resultant voltage.

Use an ammeter if possible, otherwise use a large
value resistor.

A scope would be handy..... the dual beam can show
the phase shift between the injected current and
the voltage across Zo, and a check for a decent
sinewave across Zo verifies that Zo is linear with
current, (or not).

He he, did you or Harry see news:425cf269$0$32031$626a14ce@news.free.fr

Direct measurement is still the better way.

--
Thanks,
Fred.

Hey Fred, you have a winner.
I took a 10Vrms voltage source at 400Hz and connected it thru a 1K
resistor
to the 600 ohm load. Call the voltage on the voltage source Va and the
other
side of the 1k resistor Vb then Zo=Vb/I1k = Vb*1K/(Va-Vb). Using a DVM I
got
73 ohms. If I had a higher 400Hz voltage source I could have used higher
than 1K and got closer to 71 ohms. A constant current source would yield
great results but this is close enough.
Thanks for your help.
Harry

Assuming this is the correct schematic:

+---[10R]---------+------TPA
| |
| [5.6ľF]
| |
[GEN] +------TPB
| |
| [600R]
| |
+-----------------+
| |
GND GND

Are you saying you injected the signal at TPB?

If the 5.6 uF capacitor *isn't* part of the amp, why don't you inject it
at TPA?

I would have thought you would want the output impedance of the amp, not
the output
impedance of the amp+cap in series.

The coupling cap is part of the total amp, outside of the feedback loop,
so we only have access to where the amp ties to the load, which is TPB. Not
a typical design.
Harry

I see. In another posting, you said:

"Hi John,
I am looking for Zo not Ro. I cannot measure the open circuit output
voltage because the amp is unstable with no load. This is a customers design
that I have to redesign. He believes that the output Z is about 10 ohms
which is easy to measure by Ro=dVo/dRL. I believe it to be 70 ohms but
cannot prove it. The spec requirement is less than 25 ohms.
Harry"

I assume the problem here is that you are trying to tell your customer that the output
impedance *can't* be as low as 10 ohms with that capacitor in series with the output.

Not without feedback around it, or an inductor in series with the amp output.

Does he believe now?

Harry Dellamano · Apr 15, 2005

"The Phantom" <phantom@aol.com> wrote in message
news:gmlt51hljjjad2kefie8auf1bope9rjd14@4ax.com...

On Thu, 14 Apr 2005 18:59:38 GMT, "Harry Dellamano" <harryd@tdsystems.org
wrote:

"The Phantom" <phantom@aol.com> wrote in message
news:te3t51plr0uf3uqdk0vb0dn1gpqvcdbtl8@4ax.com...
On Thu, 14 Apr 2005 14:43:31 GMT, "Harry Dellamano"
harryd@tdsystems.org
wrote:

Reverse the procedure Harry. Set the device at
Ov out, inject a known current into the output
terminal, and measure the resultant voltage.

Use an ammeter if possible, otherwise use a large
value resistor.

A scope would be handy..... the dual beam can show
the phase shift between the injected current and
the voltage across Zo, and a check for a decent
sinewave across Zo verifies that Zo is linear with
current, (or not).

He he, did you or Harry see
news:425cf269$0$32031$626a14ce@news.free.fr

Direct measurement is still the better way.

--
Thanks,
Fred.

Hey Fred, you have a winner.
I took a 10Vrms voltage source at 400Hz and connected it thru a 1K
resistor
to the 600 ohm load. Call the voltage on the voltage source Va and the
other
side of the 1k resistor Vb then Zo=Vb/I1k = Vb*1K/(Va-Vb). Using a DVM I
got
73 ohms. If I had a higher 400Hz voltage source I could have used higher
than 1K and got closer to 71 ohms. A constant current source would yield
great results but this is close enough.
Thanks for your help.
Harry

Assuming this is the correct schematic:

+---[10R]---------+------TPA
| |
| [5.6ľF]
| |
[GEN] +------TPB
| |
| [600R]
| |
+-----------------+
| |
GND GND

Are you saying you injected the signal at TPB?

If the 5.6 uF capacitor *isn't* part of the amp, why don't you inject it
at TPA?

I would have thought you would want the output impedance of the amp, not
the output
impedance of the amp+cap in series.

The coupling cap is part of the total amp, outside of the feedback loop,
so we only have access to where the amp ties to the load, which is TPB.
Not
a typical design.
Harry

I see. In another posting, you said:

"Hi John,
I am looking for Zo not Ro. I cannot measure the open circuit output
voltage because the amp is unstable with no load. This is a customers
design
that I have to redesign. He believes that the output Z is about 10 ohms
which is easy to measure by Ro=dVo/dRL. I believe it to be 70 ohms but
cannot prove it. The spec requirement is less than 25 ohms.
Harry"

I assume the problem here is that you are trying to tell your customer
that the output
impedance *can't* be as low as 10 ohms with that capacitor in series with
the output.

Not without feedback around it, or an inductor in series with the amp
output.

Does he believe now?

Naw, the customer is always right, even in this business.
Thanks for your interest, maybe next time I will better explain myself.
Cheers,
Harry
>

qrk · Apr 15, 2005

On Thu, 14 Apr 2005 02:04:18 GMT, "Harry Dellamano"
<harryd@tdsystems.org> wrote:

"qrk" <SpamTrap@reson.com> wrote in message
news:8ahr51t14cq0hsunqdrlqkjtr2ftuml62i@4ax.com...

I'm in process of building up a dual AMD Opteron computer. I'll pass
on details if you want. It will be a week or so until I get my
benchmarks tested out. I'm interested to see what happens with the
better memory access speed.

Mark

Hey Mark, now that is exciting. Keep us updated with hardware and
performance.
JT has mentioned that PC Club has some good hardware. Who is your source?

Cheers,
Harry

Harry,
I buy little bits and pieces and build up the system. I like taking
air tools to the chassis to improve air flow by cutting out the stupid
rear fan grills. JT likes PC club because it's down the street from
his house and he doesn't have to put up with the dweebs at Frys. PC
Club also has some pretty good deals. I typically buy from
http://www.newegg.com
Good prices and service most of the time. Their selection of pre-built
computers isn't great.

We're starting to run some tests on the Opteron. The AMD Opteron has
about the same performance as the AMD Athlon for simple standard
32-bit applications. Apps that try load everything in RAM, like Xilinx
FPGA routing tools, run around two times faster due to memory transfer
speed. The Athlon had a very broken memory controller and could only
trickle data in and out of RAM (apx 5 times slower than the memory
rating). I'm amazed Intel didn't fry AMD for that blunder. The Opteron
SSE2 instruction set seems a bit slow, however, I need to compare this
against a P4.

Mark

Jim Thompson · Apr 15, 2005

On Thu, 14 Apr 2005 18:58:19 -0700, qrk <SpamTrap@reson.com> wrote:

On Thu, 14 Apr 2005 02:04:18 GMT, "Harry Dellamano"
harryd@tdsystems.org> wrote:

"qrk" <SpamTrap@reson.com> wrote in message
news:8ahr51t14cq0hsunqdrlqkjtr2ftuml62i@4ax.com...

I'm in process of building up a dual AMD Opteron computer. I'll pass
on details if you want. It will be a week or so until I get my
benchmarks tested out. I'm interested to see what happens with the
better memory access speed.

Mark

Hey Mark, now that is exciting. Keep us updated with hardware and
performance.
JT has mentioned that PC Club has some good hardware. Who is your source?

Cheers,
Harry

Harry,
I buy little bits and pieces and build up the system. I like taking
air tools to the chassis to improve air flow by cutting out the stupid
rear fan grills. JT likes PC club because it's down the street from
his house and he doesn't have to put up with the dweebs at Frys.

I like that term "dweebs" ;-) Like a phrase my father first said back
in 1956... "I knew God made a lot of stupid people, but I didn't know
he put them all in one place". ;-)

I like my local PC Club because it's populated by some young (younger
than my youngest child) nerds who have no problem building to my
requirements.

PC
Club also has some pretty good deals. I typically buy from
http://www.newegg.com
Good prices and service most of the time. Their selection of pre-built
computers isn't great.

We're starting to run some tests on the Opteron. The AMD Opteron has
about the same performance as the AMD Athlon for simple standard
32-bit applications. Apps that try load everything in RAM, like Xilinx
FPGA routing tools, run around two times faster due to memory transfer
speed. The Athlon had a very broken memory controller and could only
trickle data in and out of RAM (apx 5 times slower than the memory
rating). I'm amazed Intel didn't fry AMD for that blunder. The Opteron
SSE2 instruction set seems a bit slow, however, I need to compare this
against a P4.

Mark

480)460-2350 | |
| E-mail Address at Website Fax

rex · Apr 15, 2005

On Thu, 14 Apr 2005 18:59:38 GMT, "Harry Dellamano"
<harryd@tdsystems.org> wrote:

The coupling cap is part of the total amp, outside of the feedback loop,
so we only have access to where the amp ties to the load, which is TPB. Not
a typical design.
Harry

So measuring the impedance with a 70 ohm component (Xc at 400 Hz) in
series with the rest of the amp proves to give an answer in the realm of
70 ohms or greater?

Thanks for wasting our time with attention to issues that never
mattered.

Amp output Z

Harry Dellamano

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