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#1 |
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diyAudio Member
Join Date: Dec 2001
Location: Melbourne, Australia
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Parts and eagerness abound,
But knowledge is lacking. Sadly, I wait. Gents, I'm trying to teach an old dog (me) new tricks. But I'm trying to learn a bit of knowledge first before I embark on a path than may lead to danger. To this end, I've been trying to grasp some of the Zen variations by reading various tutorials available on the net, and I thinks I've got a bit of a handle on Son of Zen (SOZ) and the variation 6 (ZV6). But I've been looking at variation 7 (ZV7-R), where an active current source is used and I'm back in the dark ages again. So, is there a good place to read about active current sources and why they work? I've looked for this on the site, and it generally assumes knowledge and is asking for improvements (or showing them). Also, is it possible to inductively load ZV6 without resorting to an active current source? Would this also provide efficiency benefits along the lines of variation 7 (ZV7-T), without the added complications (for me)?
__________________
Jont. "It is impossible to build a fool proof system; because fools are so ingenious." |
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#2 | |
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diyAudio Member
Join Date: Jan 2003
Location: ancient Batsch , behind Iron Curtain
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Quote:
after that ,shoot with questions
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my Papa is smarter than your Nelson ! tnx to thread ; Cook Book ; PSM LS Cook Book ; Baby Diyaudio FORUM ; BAF Forum & Gallery;I'm dumb
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#3 | |
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The one and only
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Quote:
sources written by Walt Jung. Lesser examples can be found in pieces at www.passdiy.com
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#4 |
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diyAudio Member
Join Date: Feb 2001
Location: Columbia, SC
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You're familiar with the idea of a voltage regulator, right? An ideal voltage regulator would deliver any arbitrary amount of current from 0 amps to infinite amps while maintaining an absolutely steady voltage. That's why it's a voltage regulator--the voltage remains constant, no matter what, but the current varies.
Now...call the current source by its alias: current regulator. What's a current regulator do? It locks the current and lets the voltage vary. Suppose you want a steady 1 amp. If you give a 1A current source a 1 ohm load, it will develop exactly 1V across the load. Simple application of Ohm's Law: I*R=E...1A*1 Ohm=1V. But what happens if you give it a 2 Ohm load? It will do whatever it has to do to force 1A through the load. In this case, it will develop 2V of output. 1A*2 Ohms=2V. An ideal current source could develop 1kV across a 1k resistor, simply because you told it to deliver 1A, no matter what. Naturally, real current regulators have limitations, just as voltage regulators do. They can only deliver current up to a certain voltage, after which they run out of steam. It's all a matter of perspective. Think of a current regulator (aka current source) as an "upside down" voltage regulator and all will be clear. Grey |
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#5 | ||||
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diyAudio Member
Join Date: Dec 2001
Location: Melbourne, Australia
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Quote:
Quote:
The problem is that I'm still in kindergarden, and you guys (even dumbing it down for me) are still talking at high-school level. Quote:
Quote:
I'm pretty sure I could get some parts together and copy what someone else has already done. But I'm not really into "monkey see, monkey do" - so even if I do copy someone's work, I'd like to know why I'm doing it. I'll come back to this when I know a bit more about what questions to ask. When? Well, hopefully before "..., the sun will cool to a white dwarf, and after that the universe will experience heat death."
__________________
Jont. "It is impossible to build a fool proof system; because fools are so ingenious." |
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#6 | |
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diyAudio Member
Join Date: Jan 2003
Location: ancient Batsch , behind Iron Curtain
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Quote:
are you on broadband? if yes-mail me and I'll shoot you with few forbidden books
__________________
my Papa is smarter than your Nelson ! tnx to thread ; Cook Book ; PSM LS Cook Book ; Baby Diyaudio FORUM ; BAF Forum & Gallery;I'm dumb
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#7 |
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diyAudio Member
Join Date: Dec 2001
Location: Melbourne, Australia
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Zen mod - 2 mails sent. The second contains my (unlimited) email address. I think hotmail is limited to 2Mb.
__________________
Jont. "It is impossible to build a fool proof system; because fools are so ingenious." |
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#8 |
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diyAudio Member
Join Date: Feb 2001
Location: Columbia, SC
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The thing to note about current sources is that the control element (grid/base/Gate) is held constant. As such, the pass device isn't amplifying a signal, it's passing a steady current.
It's just a question of sitting down and staring at it for a while. Grey |
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#9 | |
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diyAudio Member
Join Date: Nov 2006
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Quote:
Voltage is pressure, amperage is the water flow rate. The mathematical equations for the two systems are largely identical. I found the DIY op amps article to be quite helpful in describing exactly how that current source works. For bipolar transistors, the .65 volts is a constant, it is the voltage gate to source required to turn the device on. As long as you are willing to do a lot of reading, and ask questions, you will get this stuff no problem. But don't hesitate to build some of this stuff even though you don't understand everything that is going on -- you will probably learn a lot more if you build some simple circuits and look at what is going on with them--via a DC meter and an oscilloscope if you can get your hands on one. I can only learn so much by just reading. JJ |
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#10 | |
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diyAudio Member
Join Date: Jul 2006
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Quote:
Let me try to answer your question: As you may already know that a diode will start conduct when a forward voltage of 0.66 voltage is applied to it. The B-E junction of Q3 acts like a diode. When the voltage across B-E of Q3 gets to about 0.66V, then current start to flow through the B-E junction and the current is amplified and causes a current between C-E. Since the current at C of Q3 is supplied from two fixed resistors R11 and R12, if current passes through Q3, the voltage will drop. When the voltage drops, the gate voltage of Q2 will also drop, current passing through Q2 will drop too, so does the current passing through R1. This will cause Q3 to conduct less current in turn. So Q3 is keeping the DC current of Q2 to a constant level. The value of R1 will determine the current of Q2. The voltage across R1 is always 0.66V. The current of Q2 is equal to 0.66V / R1. (The part no. are based on Zen-V4.pdf) I would say that this forum is better than any book can supply. Feel free to list your questions here. (That will also benefit other people who are too shy to ask.) |
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