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3 dB bandwidth

P

Pooh Bear

Jan 1, 1970
0
keith said:
You're as pig-ignorant as Guy! Decibels are a log ratio of *anything*.
Crap, what are they teaching engineers these days?

I'm not sure but it isn't electronics as you and I know it !

Some yrs ago I was the liaison guy between my client ( a UK audio manufacturer
) and Salford University who run a specialist electronic course in the
electro-acoustic area. We ended up offering them a one-year placement for one
of their students since they like to keep industry ties active. Indeed we were
going to repeat the exercise to but then the company was placed in a situation
where it had to cease trading.

Whilst I was at Salford, I saw some of their labs and work assignments. One
was a mic pre-amplifier. I had to point out to the lecturer who was their
liaison guy with me that it was actually an instrumentation amplifer they
were learning about !

I had other occasions to criticise the student's level of relevant tuition (
but the student himself was pretty ok I should add ). This lecturer sighed and
admitted their electronics lecturer was more into heavy electrical so what did
you expect ?

I doubt that the skill of discrete design is taught at all these days.


Graham
 
P

Pooh Bear

Jan 1, 1970
0
Tony said:
The Decibel is a Unit of Attenuation,

It's just *attenuation* now is it ? Snip rest of trolling garbage.

Holy Shit !

Graham
 
A

Adrian Tuddenham

Jan 1, 1970
0
keith said:
You're as pig-ignorant as Guy! Decibels are a log ratio of *anything*.

Have a look at "Radio Designer's Handbook" by F. Langford-Smith (first
published 1934). In my 4th Edition copy (1953) the whole of Chapter 9
is devoted to the subject of decibels.

It is quite clear that only the power ratio is directly described by
decibels, if you wish to express the ratio of other electrical
quantities in decibels you "must involve the resistance". (Page 807)

Crap, what are they teaching engineers these days?

I have no idea, I did my training in the 1960s.
 
W

Winfield Hill

Jan 1, 1970
0
Pooh Bear wrote...
I doubt that the skill of discrete design is taught at all these days.

Harvard teaches it in their very popular Physics 123 course, using
the book Paul and I wrote. They also teach it in the summer, when
most of the students are from elsewhere and come to Cambridge just
to take the course.

And many other schools use our book to teach how an engineer thinks
about electronics circuit design.
 
M

Mike Monett

Jan 1, 1970
0
Tony Williams wrote:

[...]
The Decibel is a Unit of Attenuation, defined in
terms of the ratio of power levels only.

Attenuation = 10 * log10 (Pin/Pout) in Decibels.

If (and only if) the input/output impedances are identical,
then the Attenuation (in Decibels) can be calculated as the
log10 ratio of the *square* of the currents or voltages.

2 2
Attenuation = 10 * log10 (Vin/Vout) in Decibels.

Or, Attenuation = 20 * log10 (Vin/Vout) in Decibels.
/|\
|____ Notice where the squares went to?

That calculation using (Vin/Vout) produces units of dB
only when the input/output impedances are identical.

For example. An audio power amplifier requires an input
of 1mW into 600 ohms, for an output of 1W into 15 ohms.
The voltage/current ratios are 0.77/3.87 and 1.29/258.

10 * log10 (1mW/1W) is -30, Decibels of Attenuation.

20 * log10 (0.77/3.87) is -14, Units of Nothing.

20 * log10 (1.29/258) is -46, Units of Nothing.

Only the power ratio produces legitimate units of dB.

Tony Williams.

Thanks, Tony.

As you point out, the calculation using voltage is identical to the
calculation using power:
Or, Attenuation = 20 * log10 (Vin/Vout) in Decibels.
/|\
|____ Notice where the squares went to?

We can examine the -3dB point in a RC low pass filter. This gives the
half-power point as well as a phase angle of 45 degrees, so it is a
useful and meaningful parameter.

Although the source impedance changes as the frequency is varied, we
assume the scope or voltmeter has negligible loading on the signal.

The voltage ratio is then 1/sqrt(2), which gives a power ratio of

(1/sqrt(2))^2 = 0.5

so

10 * log10((1/sqrt(2))^2) = -3.0103 dB

and

20 * log10(1/(sqrt(2))) = -3.0103 dB

As you point out
Or, Attenuation = 20 * log10 (Vin/Vout) in Decibels.

Since power is voltage squared, the two equations are identical and
give identical results.

So the legitimate unit for voltage calculations is the dB.

Mike Monett
 
M

Mike Monett

Jan 1, 1970
0
Pooh said:
It's just *attenuation* now is it ? Snip rest of trolling garbage.

Holy Shit !

Graham

Graham,

Please be a little more tolerant. As you know, gain can also be
understood as a form of attenuation. In other words, gain = - attenuation

Tony Williams is one of the oldest members of sed, and he rarely gets
involved in stupid arguments such as this. If you look through the
archives, you will find his contributions are often quite brilliant and
certainly far above the general level we now experience in this
newsgroup.

Tony is not a troll, and it is ungenerous to refer to his post in such a
manner.

Mike Monett
 
J

Jim Thompson

Jan 1, 1970
0
Pooh Bear wrote...

Harvard teaches it in their very popular Physics 123 course, using
the book Paul and I wrote. They also teach it in the summer, when
most of the students are from elsewhere and come to Cambridge just
to take the course.

And many other schools use our book to teach how an engineer thinks
about electronics circuit design.

But MIT doesn't. MIT teaches basic skills, not compendia of circuits
to store away for later use.

...Jim Thompson
 
G

Guy Macon

Jan 1, 1970
0
Tony said:
The Decibel is a Unit of Attenuation, defined in
terms of the ratio of power levels only.

Attenuation = 10 * log10 (Pin/Pout) in Decibels.

If (and only if) the input/output impedances are identical,
then the Attenuation (in Decibels) can be calculated as the
log10 ratio of the *square* of the currents or voltages.

2 2
Attenuation = 10 * log10 (Vin/Vout) in Decibels.

Or, Attenuation = 20 * log10 (Vin/Vout) in Decibels.
/|\
|____ Notice where the squares went to?

That calculation using (Vin/Vout) produces units of dB
only when the input/output impedances are identical.

For example. An audio power amplifier requires an input
of 1mW into 600 ohms, for an output of 1W into 15 ohms.
The voltage/current ratios are 0.77/3.87 and 1.29/258.

10 * log10 (1mW/1W) is -30, Decibels of Attenuation.

20 * log10 (0.77/3.87) is -14, Units of Nothing.

20 * log10 (1.29/258) is -46, Units of Nothing.

Only the power ratio produces legitimate units of dB.

<applause>

--------------------------------------------------------

Adrian said:
Have a look at "Radio Designer's Handbook" by F. Langford-Smith (first
published 1934). In my 4th Edition copy (1953) the whole of Chapter 9
is devoted to the subject of decibels.

It is quite clear that only the power ratio is directly described by
decibels, if you wish to express the ratio of other electrical
quantities in decibels you "must involve the resistance". (Page 807)

<more applause>
 
P

Pooh Bear

Jan 1, 1970
0
Jim said:
But MIT doesn't. MIT teaches basic skills, not compendia of circuits
to store away for later use.

...Jim Thompson

Don't you consider it somewhat foolish to consider one to the exclusion of
the other though ? *Both* are required IME to do a decent job. No point
re-inventing the wheel constantly. Indeed many or even most practical
circuits have been around for decades.

The really nice bit is tweaking a popular configuration to enhance
performance.

Graham
 
J

Jim Thompson

Jan 1, 1970
0
Don't you consider it somewhat foolish to consider one to the exclusion of
the other though ? *Both* are required IME to do a decent job. No point
re-inventing the wheel constantly. Indeed many or even most practical
circuits have been around for decades.

The really nice bit is tweaking a popular configuration to enhance
performance.

Graham

My point is, if you don't know fundamentals, what good is it showing
circuits?

To wit, look at all the folderol that continues here on these NG's
about blinking LED's. Not once have I observed a calculation, just
things like, "I took off two turns (on the inductor) and it started
working". No one has a clue WHY. That's NOT design... that's
hacking.

...Jim Thompson
 
M

Mike Monett

Jan 1, 1970
0
Mike said:
Since power is voltage squared, the two equations are identical and
give identical results.

So the legitimate unit for voltage calculations is the dB.

Mike Monett

I guess I'd have to turn the argument around and say the dB is
dimensionless. If you want the result in power, use 10*log. If you want
the result in volts or amps, use 20*log.

Assuming the same impedance, the results are identical. They refer to the
same thing. Here's a small table:

dB Volts Watts
-3dB 1/sqrt(2) 0.5
-6dB 1/2 1/4
-10dB 1/sqrt(10) 1/10
-20dB 1/10 1/100
-40dB 1/100 1/10000
-60dB 1/1000 1/1000000

You can also tie the dB to a specific unit. For example, dBm refers to
milliwatts, and dBV to Volts. In these examples, the dimension is
defined by the unit.

So where's the problem?

Mike Monett
 
J

Jim Thompson

Jan 1, 1970
0
I guess I'd have to turn the argument around and say the dB is
dimensionless. If you want the result in power, use 10*log. If you want
the result in volts or amps, use 20*log.

Assuming the same impedance, the results are identical. They refer to the
same thing. Here's a small table:

dB Volts Watts
-3dB 1/sqrt(2) 0.5
-6dB 1/2 1/4
-10dB 1/sqrt(10) 1/10
-20dB 1/10 1/100
-40dB 1/100 1/10000
-60dB 1/1000 1/1000000

You can also tie the dB to a specific unit. For example, dBm refers to
milliwatts, and dBV to Volts. In these examples, the dimension is
defined by the unit.

So where's the problem?

Mike Monett

None whatsoever, except for those who argue to hear their head
resonate ;-)

...Jim Thompson
 
M

Mike Monett

Jan 1, 1970
0
Jim said:
[...]
So where's the problem?

Mike Monett

None whatsoever, except for those who argue to hear their head
resonate ;-)

...Jim Thompson

Thanks, Jim. It's amazing how these threads can go on forever and not
solve anything. But when you entered the fray, hopefully that will put a
stop to the nonsense. Only an idiot would be foolhardy enough to argue
with you.

Let's get back to design problems. I have a problem. I want to make a
very high impedance buffer to drive a so-called 24-bit adc (actually only
19 effective bits). The desired voltage range is whatever will be
compatible with the adc. Perhaps 2.5V.

The problem is I have not been able to find a suitable op amp with bias
currents in the fA region, and common mode rejection good enough to use
with a 19 bit adc.

For example, the LMC6482 is spec'd at around 20fA, but has only 80dB or
so common mode rejection. I believe this is only good enough for 12 bits.

Do you have any suggestions on the perfect op amp to use?

Mike Monett
 
F

Fred Bloggs

Jan 1, 1970
0
I guess I'd have to turn the argument around and say the dB is
dimensionless. If you want the result in power, use 10*log. If you
want the result in volts or amps, use 20*log.

Assuming the same impedance, the results are identical. They refer to
the same thing. Here's a small table:

dB Volts Watts -3dB 1/sqrt(2) 0.5 -6dB 1/2 1/4
-10dB 1/sqrt(10) 1/10 -20dB 1/10 1/100 -40dB 1/100
1/10000 -60dB 1/1000 1/1000000

You can also tie the dB to a specific unit. For example, dBm refers
to milliwatts, and dBV to Volts. In these examples, the dimension is
defined by the unit.

Ahhh- good example of a very common misperception and confusion.

In the case of using dB to characterize *amplifier* performance, there
is an implied unit of normalization, and that is the amplifier
performance at "midband". You are taking Log ratios of gains, and not
voltages or watts. The dB is always computed as
20*Log(Gain/Gain,midband) regardless of whether the gain is a voltage or
power gain. To see this, note that a power gain, Gp, is
Gp=Pout/Pin=(Vout^2/Ro)/(Vin^2/Rin), and therefore Pg/Pg,mb=
(Vout/Vout,mb)^2=((Vout/Vin)/(Vout,mb/Vin))^2=(Gv/Gv,mb)^2, so that
20*Log(Gp/Gp,mb)=2*20*Log(Gv/Gv,mb) or Gp,dBmb=2*Gv,dbmb. In words, the
power gain loss relative to power gain at midband is 2x the voltage gain
loss relative to voltage gain at midband in dB. Then a plot of amplifier
voltage gain normalized to midband has its vertical dB scale multiplied
by 2 to show power gain rolloff with frequency. When the Gv is down 3dB,
the Gp is down 6dB, when Gv is down 6dB, the Gp is down 12dB, etc. This
should show you that just because you are talking about power does not
mean you mindlessly apply 10*Log(.).
 
J

Jim Thompson

Jan 1, 1970
0
Jim said:
[...]
So where's the problem?

Mike Monett

None whatsoever, except for those who argue to hear their head
resonate ;-)

...Jim Thompson

Thanks, Jim. It's amazing how these threads can go on forever and not
solve anything. But when you entered the fray, hopefully that will put a
stop to the nonsense. Only an idiot would be foolhardy enough to argue
with you.

Let's get back to design problems. I have a problem. I want to make a
very high impedance buffer to drive a so-called 24-bit adc (actually only
19 effective bits). The desired voltage range is whatever will be
compatible with the adc. Perhaps 2.5V.

The problem is I have not been able to find a suitable op amp with bias
currents in the fA region, and common mode rejection good enough to use
with a 19 bit adc.

For example, the LMC6482 is spec'd at around 20fA, but has only 80dB or
so common mode rejection. I believe this is only good enough for 12 bits.

Do you have any suggestions on the perfect op amp to use?

Mike Monett

So you have a differential signal? Does it have some bias return
point, or totally floating and you have to establish common-mode?

...Jim Thompson
 
F

Fred Bloggs

Jan 1, 1970
0
Let's get back to design problems. I have a problem. I want to make a
very high impedance buffer to drive a so-called 24-bit adc (actually only
19 effective bits). The desired voltage range is whatever will be
compatible with the adc. Perhaps 2.5V.

The problem is I have not been able to find a suitable op amp with bias
currents in the fA region, and common mode rejection good enough to use
with a 19 bit adc.

For example, the LMC6482 is spec'd at around 20fA, but has only 80dB or
so common mode rejection. I believe this is only good enough for 12 bits.

Do you have any suggestions on the perfect op amp to use?

If the manu's are challenged to produce a "perfect buffer" internally,
what hope do you have to do it externally. LTC has the "East Drive"
technology, LTC2484:
http://www.linear.com/pc/productDetail.do?navId=H0,C1,C1155,C1001,C1152,P11229
 
P

Pooh Bear

Jan 1, 1970
0
Mike said:
I guess I'd have to turn the argument around and say the dB is
dimensionless. If you want the result in power, use 10*log. If you want
the result in volts or amps, use 20*log.

Assuming the same impedance, the results are identical. They refer to the
same thing. Here's a small table:

dB Volts Watts
-3dB 1/sqrt(2) 0.5
-6dB 1/2 1/4
-10dB 1/sqrt(10) 1/10
-20dB 1/10 1/100
-40dB 1/100 1/10000
-60dB 1/1000 1/1000000

You can also tie the dB to a specific unit. For example, dBm refers to
milliwatts, and dBV to Volts. In these examples, the dimension is
defined by the unit.

So where's the problem?

Mike Monett

There is of course no problem.

The dB is simply a *mathematical convenience* that allows more meaningfully
understood interpretation of large ratios in particular.

Graham
 
K

Kevin Aylward

Jan 1, 1970
0
Adrian said:
Have a look at "Radio Designer's Handbook" by F. Langford-Smith (first
published 1934). In my 4th Edition copy (1953) the whole of Chapter 9
is devoted to the subject of decibels.

We have moved on from there. The book is dated, and simply not relevant
to the modern world on this issue.
It is quite clear that only the power ratio is directly described by
decibels, if you wish to express the ratio of other electrical
quantities in decibels you "must involve the resistance". (Page 807)

Again, in short, quite nonsense today. dBs, today, are a general term.
Its irrelevant how they may have been first used.

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
 
K

Kevin Aylward

Jan 1, 1970
0
Tony said:
The Decibel is a Unit of Attenuation, defined in
terms of the ratio of power levels only.> Only the power ratio
produces legitimate units of dB.

You are mistaken. You are confusing the possible *first* use of the dB
with what it actually means after the fact.

The dB is a totally general term. It simply signifies that a log of a
number was taken. Thats it. End of story.

The db is not even a real unit, that is, it is not a dimension. The term
under the log must be a unitless. You cant take a function of something
with dimensions. e.g.

I=Io.exp(qV/KT)

note, qV/KT is dimensionless.

db's are applied in many fields, the power use is no more than a
historical incidental.

You are trying to claim that, for example, if it is stated that an
amplifier has 25dB gain more than another, that such a statment is
wrong. You will be pretty much alone on that idea.

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
 
F

Fred Abse

Jan 1, 1970
0
Wrong. Decibels are an expression of the RATIO of two powers,
OR two voltages, and, in most applications, the method being
used (power or voltage ratios) is usually specified, except
in some audiophool crap as marketing hype. "Peak Envelope
Power", indeed.

Nothing wrong with "Peak envelope power", if we're talking about SSB
transmitters. Rather fundamental, in fact.
 
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