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#1 |
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diyAudio Member
Join Date: Aug 2006
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Hello fellow DIYers,
I have made every effort to understand Steve Bench's regulated power supply (found here), but I cannot figure out everything that is going on in this circuit and would appreciate an assist from those smarter than me. What determines the current through the shunt regulators? Steve lists the plate current as 65ma and the grid current as 11ma. Curves for the SV571-160 at 900v with a 30v grid voltage show a plate current of ~135ma (the high line condition described in the article). What prevents the shunt regulator and VR tubes from drawing all of the current? Trying to figure out how to calculate the current through the shunt/VR tubes and current that goes into the next stage through the choke. Thanks!
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#2 |
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diyAudio Moderator
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The actual shunt elements are the triodes. The current through the VR tubes is limited to the grid current of the 572/811. If the conduction of the shunt got too high, the output voltage would drop and the VR string extinguish.
It's a very clever circuit, but I expect that from Stef.
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If there's a sucker born every minute, where do the rest of them come from? |
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#3 | |
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diyAudio Member
Join Date: Dec 2003
Location: San Diego
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The OD tubes provide a fixed voltage drop from the rail to the SV572 grid. As the rail rises so does the grid voltage, therefore the 572 conducts more current. As the 572 conducts more current, the voltage drop across the power transformer, diode, and inductor increases. This ultimately limits the current, and serves the same purpose as the series resistor (or CCS) used in a typical shunt reg.
Quote:
Sheldon Last edited by Sheldon; 13th February 2011 at 03:54 PM. |
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#4 |
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expert in tautology
diyAudio Member
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The tubes selected as the pass elements are so-called "zero bias class B triodes" which means that they are essentially "off" with no bias voltage present on the grids. They will draw current whenever driven positive, so the grid will act as a sink for the bottom end of the VR tube string. (he calls it A2 in his article, since it is set to be "on", not biased off as in a P-P class B output stage) Ordinarily there would be a resistor to ground at the bottom of the VR tube string, and a proportional representation of the output voltage would appear on the shunt reg tube.
I'd say that the actual voltage appearing on the plates of these tubes will have to be determined more or less empirically, since the operating points of each tube will be a bit different... the current will be proportional to the vdrop with and without the vreg tube operating, simple ohms law... Seems to me that the difference in the two tubes chosen was because of the different mu, and the 572 will handle more current - so it seems like they are pulling quite a bit of current... The VR tube string is a fixed voltage, which in turn is applied to the grid, which turns the tube on to some degree... IF the output voltage goes up, this pulls the voltage on the bottom of the VR tube string & grid UP, which turns ON the tube, pulling the voltage down. Conversely if the output voltage drops, the voltage on the grid drops reducing the current drawn by the 572 or 811 which makes the voltage on the junction of the plate and choke float back up. It's just that simple. In the Steve Bench article on the design he describes in some detail the design and principles of operation. He also more or less says that you have to work within the limitations of the current available from this supply and that you more or less have to select parts to get the voltage out that you want... Hope this helps? _-_-bear PS. in theory one could use any fixed voltage reference circuit in place of the VR tubes - one could use a string of zeners for example, or a composite of zener + Mosfet for example...
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_-_-bear http://www.bearlabs.com [...2SJ74 Toshiba bogus asian parts - beware! ] -- Btw, I don't actually know anything, FYI --
Last edited by bear; 13th February 2011 at 03:59 PM. Reason: They posted before I did! ha ha... |
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#5 | |
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diyAudio Member
Join Date: Aug 2006
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Quote:
Very much appreciate everyone who took the time to help me understand this circuit. |
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