I read in sci.electronics.design that The Phantom wrote (in ) about 'Amp output Z', on Wed, 13 Apr 2005:
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?
Compared to the total *resistance* in the chain of 610 ohms, -71j makes a minor contribution to the total load impedance.
Graham
He could also measure 'the other side' of the 5.6uF and back-calculate Rout to a high degree of accuracy essentially ignoring the reactive component for practical purposes.
Graham
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
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
equal to 10.0
in the OP he
about the
whether he's
Amen to that. I think the 10 ohms output resistance is a guesstimate too.
us for E2A
voltage to be
volts with a
14.5 V @that would
Multimeters don't see phase angle - lol !
Graham
Assuming this is the correct schematic:
+---[10R]---------+------TPA | | | [5.6µF] | | [GEN] +------TPB | | | [600R] | | +-----------------+ | | GND GNDAre 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.
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).
You can also get some idea of the value of Ro.
switch _________ +----o/o----+ | Device | | | | +--------+-[10ohms]--+--[Ammeter]--[R]-- | [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" a écrit dans le message de news: snipped-for-privacy@ledelec.demon.co.uk...
He he, did you or Harry see news:425cf269$0$32031$ snipped-for-privacy@news.free.fr
Direct measurement is still the better way.
I see. "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?
"Fred Bartoli" wrote in message news:425cf269$0$32031$ snipped-for-privacy@news.free.fr...
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"Fred Bartoli" wrote in message news:425e31df$0$22099$ snipped-for-privacy@news.free.fr...
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, 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
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
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 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.
Mark, Keep me posted... particularly in regard to comparisons to our previous benchmarks. Thanks!
...Jim Thompson
Naw, the customer is always right, even in this business. Thanks for your interest, maybe next time I will better explain myself. Cheers, 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.
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