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The simplistic Salas low voltage shunt regulator
The simplistic Salas low voltage shunt regulator
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Old 6th August 2011, 04:31 PM   #4111
Joshua_G is offline Joshua_G  Israel
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Quote:
Originally Posted by AndrewT View Post
sensing at the load or locating the regulator at the load reduces the impedance of the connection between the regulator and the load.
Reduced impedance gives the effect you require, Dumping a load back emf into the regulator.
Thanks, Andrew.
Only I'm not clear about the way sensing wires reduce the impedance of the connection between the regulator and the load.
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Old 6th August 2011, 04:49 PM   #4112
keantoken is offline keantoken  United States
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They don't reduce the impedance of the connection. They "effectively" eliminate the wiring resistance in series with the regulator by moving the feedback point directly to the load, so the regulator compensates for the extra resistance.

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Old 6th August 2011, 04:54 PM   #4113
merlin el mago is offline merlin el mago
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Quote:
Originally Posted by keantoken View Post
They don't reduce the impedance of the connection. They "effectively" eliminate the wiring resistance in series with the regulator by moving the feedback point directly to the load, so the regulator compensates for the extra resistance.

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Old 6th August 2011, 10:56 PM   #4114
Joshua_G is offline Joshua_G  Israel
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Quote:
Originally Posted by keantoken View Post
They don't reduce the impedance of the connection. They "effectively" eliminate the wiring resistance in series with the regulator by moving the feedback point directly to the load, so the regulator compensates for the extra resistance.

- keantoken
Thanks.
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Old 8th August 2011, 08:46 AM   #4115
hesener is offline hesener  Germany
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The simplistic Salas low voltage shunt regulator
Default off by a factor of 20...

Quote:
Originally Posted by ikoflexer View Post
Can you please post details about your setup for measuring the output impedance?
Hi, for this test the current source was set at 500mA and the output voltage at 12V. I then used a electronic load (selfmade) which was drawing 300mA, and overlaying an AC load current of 20mA. This load has a 1ohm resistor to measure the current so I could easily measure the AC current amplitude. I then used a soundcard with interface to change the frequency of that signal from 20Hz to 22kHz. At the same time, I measured the AC voltage at the output, and it was -50dBu or thereabouts, flat across the frequency span (with no signal, hovering around -90dBu).

And when you do the math, just like I redid just now, you'll find out that the output impedance of this regulator is -50dBu (=2.45mV)/20mA = 0.12ohm, quite lousy. Honestly, I do not know where I did that mistake in the first place - obviously the 4mOhm claimed earlier is false. Shame on me.....
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Old 8th August 2011, 10:12 AM   #4116
AndrewT is offline AndrewT  Scotland
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Your calculator has a software error. Reprogamme it.

4mOhm to 0r120 is 30 times, not factor of twenty.
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Old 8th August 2011, 10:22 AM   #4117
Salas is offline Salas  Greece
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The simplistic Salas low voltage shunt regulator
30 times out of Spice prediction for the V1 topology? R.A.P was right to look for the garbage can when arguing simulators it seems. Anyway, I will be off to an island for a week or so, see ya guys, and if you confirm what is the best method and results it will be nice to see then. Maybe the cabling needs be sensed also.
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Old 8th August 2011, 10:53 AM   #4118
iko is offline iko  Canada
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The simplistic Salas low voltage shunt regulator
Quote:
Originally Posted by hesener View Post
Hi, for this test the current source was set at 500mA and the output voltage at 12V. I then used a electronic load (selfmade) which was drawing 300mA, and overlaying an AC load current of 20mA. This load has a 1ohm resistor to measure the current so I could easily measure the AC current amplitude. I then used a soundcard with interface to change the frequency of that signal from 20Hz to 22kHz. At the same time, I measured the AC voltage at the output, and it was -50dBu or thereabouts, flat across the frequency span (with no signal, hovering around -90dBu).

And when you do the math, just like I redid just now, you'll find out that the output impedance of this regulator is -50dBu (=2.45mV)/20mA = 0.12ohm, quite lousy. Honestly, I do not know where I did that mistake in the first place - obviously the 4mOhm claimed earlier is false. Shame on me.....
I see. The reason I asked is because I've tried such measurements some time ago. Can you post the schematic of your electronic load?
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Old 8th August 2011, 11:00 AM   #4119
jackinnj is offline jackinnj  United States
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The simplistic Salas low voltage shunt regulator
Quote:
Originally Posted by hesener View Post
Hi, for this test the current source was set at 500mA and the output voltage at 12V. I then used a electronic load (selfmade) which was drawing 300mA, and overlaying an AC load current of 20mA.
The methodology which Walt Jung used for supply impedance measurement can be found on his website: http://waltjung.org/PDFs/Regs_for_High_Perf_Audio_1.pdf

The AP analyzers allow you to measure the impressed current by comparing the voltage drop across the internal protection resistor. It's a good idea to measure the current explicitly since the signal generator output impedance isn't constant over frequency.

To make everything consistent I like to use the current and voltage he used in his articles, i.e. 15VDC with 150mA load and 50mA impressed current.
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Old 8th August 2011, 03:01 PM   #4120
iko is offline iko  Canada
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The simplistic Salas low voltage shunt regulator
That would be Fig. 2c. It is quite surprising that an AP can drive the regulator straight. I had built a mosfet load whose gate was driven by one of my HP sine wave generators.

One question, he says there's a 2.5V RMS over the 50R load, which works out nicely to the reported 50mA RMS current. But you're talking about 15V DC and 150mA DC load, which would suggest a 100R load. Am I getting something wrong here?
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