Switchable Hi-Z input impedance, how ? - Page 56 - diyAudio
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Old 30th April 2013, 01:50 PM   #551
tinitus is offline tinitus  Europe
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so, because the jfets have output resistors after output cap, it means my buffer also need input cap ?

well, I dont have cap on hi-Z input jfet
are you saying its not 1M ohm at all, since there is no cap ?
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Old 30th April 2013, 01:56 PM   #552
AndrewT is offline AndrewT  Scotland
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if you remove the resistors completely then the cap on the output of the preceding stage acts as your DC block.
You can keep the series resistor as a base stopper but a base stopper should be attached to the base, not the output of the FET.

If the two stages were connected with a removable interconnect then adding a grounding resistor to take any leakage past the DC blocker is good practice. This grounding resistor (at both ends of the interconnected equipment) can usually be >1M
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Old 30th April 2013, 01:58 PM   #553
AndrewT is offline AndrewT  Scotland
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When you do this you will discover that the bias resistor values are drawing VERY high current though the BJT.
Post540 told you this, but you chose to ignore his comment.
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Old 30th April 2013, 02:09 PM   #554
tinitus is offline tinitus  Europe
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Quote:
Originally Posted by AndrewT View Post
...but you chose to ignore his comment.
no, I made many changes after that .... and many you havent seen or heard about

man, you make feel like Im a dog
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Old 30th April 2013, 03:56 PM   #555
JMFahey is offline JMFahey  Argentina
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Dear Tinitus, you are very confused, becuse you are mixing 2 very different amplifier stages and considering them the same (which they are not):

1) *that* FET stage is self biased (or "cathode biased" to be more precise) and it's gate needs to be at 0V or Ground potential, so no input cap needed.

2) bipolar transistors need forward bias at the base, in that case you are "artificially" feeding it around, say, 8V DC, so you **need** to DC isolate from the previous stage or your calculations will be way off.

3) you do use a 1uF output cap after the FET stage (fine) ... but you mess with that by adding the unnecessary resistors after it.
Just pull them and you'll be fine.
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Old 30th April 2013, 05:21 PM   #556
tinitus is offline tinitus  Europe
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Quote:
Originally Posted by JMFahey View Post
Just pull them and you'll be fine.
will try that...thank you, sir

ehh...any chance the jfet stage is a current drive curcuit
I think the big man once said that 'might' happen, given certain circumstances

btw, input and output impedance isnt just a simple matter of looking at a few resistors in paralel, is it ?
I think I saw a bit more complicated calcuation the other day

anyway, I got the new att pots today...and that funny chickenhead knob
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Old 30th April 2013, 08:07 PM   #557
tinitus is offline tinitus  Europe
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found another effort explaning input output impedance
hope its ok to post here

http://www.pdx.edu/nanogroup/sites/w...mpedance_9.pdf

http://www.physics.ohio-state.edu/~g...mer04/Lec6.pdf

http://people.seas.harvard.edu/~jone.../bjt_amps.html
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Old 30th April 2013, 08:27 PM   #558
tinitus is offline tinitus  Europe
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and this 'statement'

There are two "tricks" to answering those questions with transistor amplifiers.

The first trick is to understand that the current-transfer ratio (hFE) of a transistor effectivly multiplies the resistance at its emitter. So, to find the impedance looking into the base, you calculate the effective resistance at its emitter, multiply this by hFE, and then add the internal base resistance.

The second trick is to realize that the result of the calculation above is typically orders of magnitude larger than the other resistances connected to the base of the transistor and can therefore be ignored.In other words, the input impedance of a transistor amplifier is usually very close to the impedance of its bias network alone. Indeed, bias networks are very often designed so that this is the case.

The output impedance is a question of how much the output voltage changes with output current: ΔV/ΔI. The transistor itself is essentially a current source, and whatever current it is passing is shared between the various resistances connected to the emitter. Therefore, the output impedance is equal to the net emitter resistance, not including the resistance of the external load.
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Old 30th April 2013, 08:38 PM   #559
JMFahey is offline JMFahey  Argentina
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Quote:
In other words, the input impedance of a transistor amplifier is usually very close to the impedance of its bias network alone.
Big blanket statement which *sometimes* is true, many times is not, by a *long* margin.

I suggest you start flying low and go step by step; otherwise you'll very often find blanket statements such as the one you quote, which should contain the word "if" many times, then confusing you because some other "authority" says what *looks* like the opposite ..... and you can't tell them apart because you lack the fine detail of what's happening.
Maybe they don't bother to explicitly mention them because they think you "already know" them and don't want to repeat themselves .... but you probably didn't read 3 pages (or 2 chapters or another book) earlier where they explained *why*.
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Old 30th April 2013, 09:21 PM   #560
tinitus is offline tinitus  Europe
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yes, I know its never that 'simple'...but thanks anyway

btw, aren't the 470/2k2 resistors on jfet output there to 'set the gain'.....?

maybe I should replace with a series resistor
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