Peak to Peak

Ppppfffftttttt..... I was *always* right on this point of course !

Graham

Reply to
Eeyore
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possible

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Since when is "a few tens of microseconds or so" instantaneous?
Reply to
John Fields

Eeyore a écrit :

Obviously (hem) not.

V = a sin(wt) V'= a w cos(wt) V''=-a w^2 sin(wt) w=sqrt(-V''/V)

a= V''/sqrt(-V''/V)

-- Thanks, Fred.

Reply to
Fred Bartoli

message

need

i.e.

possible

output

The measurement is instantaneous but it takes a few us to calculate the result.

Graham

Reply to
Eeyore

You've come up with a smarter method still ?

I'm intruiged. I need to think about that.

Graham

Reply to
Eeyore

Emitter resistor provides dynamic biasing (holds the current down as the transistor heats and tries to pull more) . The emitter resistor also would hold the signal down (degenerative feedback) without an AC bypass for signal.

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Reply to
Fred Bartoli

I don't check my work very carefully for newsgroup postings, since the penalty for being wrong is zero. I'm a lot more careful when it matters, which is how I get to sell rev A of most of my boards.

John

Reply to
John Larkin

As regards being wrong, I and one of my design engineers are working on the architecture of a new product. We visited the customer last week and got an idea of his problems, and we promised him a proposal by this coming Friday. Accordingly, we have allocated Monday, Tuesday, and Wednesday specifically to being confused and wrong as much as we possibly can, and we're doing mighty fine so far. I've been learning stuff about NCOs that I never suspected, and discovered that you can buy a might fine 16-bit DAC for $3 nowadays and that Xilinx has some ultracool logic blocks for free.

Sometimes it pays to be wrong.

John

Reply to
John Larkin

--
Yup, if you learn from your misteaks.

Yes, I know...
Reply to
John Fields

Damnation, you're right. I was assuming the frequency was known.

Bob M.

Reply to
Bob Myers

No, he's not right. Well, if we consider that's a sine wave.

Here's what I replied to Graham:

But: a^2 = V^2 + (V'/w)^2

Then: a=sqrt(V^2 - V'^2 * V/V'')

--
Thanks,
Fred.
Reply to
Fred Bartoli

John Larkin wrote in news: snipped-for-privacy@4ax.com:

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Nanner nanner boo bo.

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Reply to
Chris Foster

"WAZ" wrote in news:1156269157.387710.12830 @p79g2000cwp.googlegroups.com:

No, that is RMS or Root Mean Square.

To calculate peak, mult RMS by 2 x (Square root of 2) (approx 1.414)

Then to get paek-to-peak, multiply peak by 2

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Reply to
Chris Foster

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No.  

    Peak = RMS * sqrt(2)
Reply to
John Fields

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