Really, really silly voltage divider question...

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TomWaterman

Well-known member
Joined
Jun 4, 2004
Messages
1,151
Location
The Shire, UK
So I'm going through some stuff in the Handbook for Sound Engineers and see a formula I had not used before for the god damn, most simple thing ever - the voltage divider.

The formula is for calculating actual attenuation when including source (Z1) and load (Z2) impedances.

R1 and R2 are the divider resistors.

No what what I plug-in the formula doesn't provide me with a result I would expect...

Anyone used it before? I'm sure its something glaringly obvious but I can't see it!!

normal_Actual_Attenuation.gif


What do you guys get if?

R1 and R2 = 2k5
Z1 = 50 ohms
Z2 = 20k

Thanks muchly
Tom
 
Hmmm thanks Wayne I must be doing something awfully wrong with the equation as it is way way off....

OK - what is the (4) in the formula? That was throwing me way off.

I'm a dummy.
-T
 
It doesn't state.

It is in reference to a volume control and the book provides another equation that does not involve the source impedance Z1.

That equation provides an answer of 6.55dB loss.

The book then goes on to state that if Z2 is sufficiently high enough the equation can be simplified to:

Vout = Vin * R2/(R1+R2)...which is the normal and often inaccurate divider equation.

Then it states that attenuation in dB equals the formula I posted.

It totally threw me.
Seeing as its log multiplier is 10 you would assume power

Either way I'm not sure I follow its use.
-T
 
[quote author="mediatechnology"]But what's that 4 outside the brackets/inside the log doing? I've never seen that equation before.[/quote]

I have no idea - like you, if I skip the (4) I get 6.93dB...but when I first did the equation as I see it, I freaked as I got something approaching 30dB!!

There's gotta be something up with that (4). Is that some form of function I don't know about??

Maybe we shouldn't try...at least I feel better now, for a minute I thought I was the only one!!! Cheers Wayne.

CJ - Is Tremaine still around on the WWW?
I got it last time but then lost the PDF.

-T
 
[quote author="TomWaterman"][quote author="mediatechnology"]Well, the power loss in the example I drew would be ~3dB.[/quote]
Banished... To the Ethernet!

:green:
-T[/quote]...should we re-state that halving of power = 3dB drop in voltage, whereas halving in voltage = 6dB drop in voltage... since drop in power into a resistive load is always the amount of drop in linear voltage squared...?

Keith
 
What page in which edition of the HB for Sound Engineers does this appear? The formula as presented is bizarre.

But working from simple parallel and series calculations shown by mediatech I get the same answer, 6.638663dB of attentuation based on the open-circuit unloaded voltage behind the 50 ohm generator Z.

But if what we are really asking is what is the attenuation between the 20k load hooked up directly, and the result of inserting the divider, then it is a different answer---we have a delta of atten between the two situations. And the attenuation with a direct connection is 20 log (20k/(20k +50) = 0.0216876...dB.

So the net effect of the attentuator is 6.616975 dB of attentuation.

If one calculates based on powers you should and do get the same answer.
 
[quote author="SSLtech"]should we re-state that halving of power = 3dB drop in voltage, whereas halving in voltage = 6dB drop in voltage... since drop in power into a resistive load is always the amount of drop in linear voltage squared...?[/quote]

Absolutely Keith! I can't believe I wrote that. Was a quick type after Waynes comment without thinking...

I edited my post - you can leave my stupidity in your quote or choose to banish it into the ether! :green:
 
[quote author="TomWaterman"]Brad its on p642 chapter 22...

Thanks for the explanation.[/quote]

In the Third Edition btw.

EDIT: The formula in the HB is WRONG. It's hard to see where the error is, but it's probably a typo.

The correct formula with the definitions of the Z's and R's is

20log [(R2*Z2)/(R2*Z2 + Z1*R2 + Z1*Z2 + R1*R2 + R1*Z2)],

again for the attentuation between an open-circuit generator and the overall circuit with attenuator and load Z.


EDIT 2: missed a close-parenthesis.
 
Yeah aside from the dumbest remark I ever made which Keef sorted me on - cheers K!

I'm don't feel so silly for asking. Something about the damn thing just bothered me enough to post.

Glad you guys were smart enough to solve the crime.

Thanks all.

-T
 
By the way---evidently no one ever complained about the first appearance in the first edition, because it is more-or-less reproduced in the third. Both are incorrect, but in the first at least the dimensions are correct (squared Z in both numerator and denominator, thus overall dinesionless).
 
Looks bogus to me. I'd do it like mediatechnology shows: combine series and parallel pairs until you get a simple ratio. bcarso actually knows math so I assume his derivation may be exactingly correct (even so his finger missed a close-paren the first try).

> What do you guys get if?

On thumbs: Half voltage. Quarter power. Either way: 6dB. A bit less due to loading. The 50R is beneath notice. The 20K hanging on 2550||2500 is less than 10%, so less than 1dB. My sliderule is damp but I accept -6.64dB. If the reference is "insertion loss", not enough less to see on my ACVM.

> how Ballou and his proofreaders got that weird expression!

Another author has told me that this publishing house's "typesetting room" is full of underpaid graphic design students wearing iPods poking at PageMaker. And that what little proofreading happens is as likely to increase typos as to decrease them. Proofreading english prose is challenging, technical prose is worse, effective proofing of formulas is real brain-pain and not that division's specialty.

Never trust ONE book. No technical book is 100% perfect.

If you find the same in two books, it may be right, or one book may have cribbed from the other.

> I flat missed the (4) inside the log.

That looks like it fell from the sky. I bet the kid was boppin and groovin to the beat, and cut/pasted the wrong thing to the wrong place.
 
[quote author="mediatechnology"]Good point. The same kids in the typesetting room unfortunately do some of the writing. We've had history textbooks here in Texas state that the Korean war ended with the dropping of an atomic bomb. I didn't remember that and I'm pretty sure the Koreans don't either.[/quote]

LOL I thought Americans just rewrote history in the movies?!!!! hohohoho
:thumb:
 
[quote author="PRR"]Looks bogus to me. I'd do it like mediatechnology shows: combine series and parallel pairs until you get a simple ratio. bcarso actually knows math so I assume his derivation may be exactingly correct (even so his finger missed a close-paren the first try).
[/quote]

It is always humbling to do real algebra once in a while. There are sooo many ways to f&*k up. It took me about three passes at this problem to get all the terms right. I knew my first two attempts were wrong, having done the quick parallel/series crank on Tom's example with the HP15C to begin with. A lot of real mathematicians use symbolic processor programs now for such "trivial" things, and would also be hard-pressed to get the right answer without a machine at hand.

But as mentioned the Ballou formula was reproduced almost unchanged, including the errant 4, between the pre-iPod-era 1st edition of the HB to the 3rd (I don't have the 2nd, but it's probably in there too). So there is substantial egg on the man's face, although consider that he also edited the whole tome. He was a busy man.

One thing about such expressions as that bogus equation: do reality checks if it looks wrong. One glaring one: note the Z1 term in the denominator: what happens if it goes to zero? Well----the attentuation gets infinite! The only way it couldn't would require that the numerator diminish at the same rate, and there's nothing there to allow that. Taking the log just slows things down a bit. Not exactly what we expect from real circuits, and right off the bat we know there is something very wrong.
 
I was using a good old fashioned calculator and expected a little more than 6dB.

However I got about 29dB or something...didn't trust myself to just move on, didn't want to assume the formula was bad either as I have been known to have the odd lapse! LOL

What got me going even more was Wayne's response that he got 6.93 with the equation I posted!!! I was like, how? At least I didn't miss that (4) out. Haha

So is there some sort of appropriate channel to contact Ballou to correct this mistake for the 4th edition?
Focal seem pretty poor at dealing with emails.

-T
 

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