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question about noise detector

F

fred

Jan 1, 1970
0
The noise detector can be found here

http://www.redcircuits.com/Page16.htm

I was wondering what benefits do we get by using
two op-amps in this circuit, because gain can also
be increased simply by increasing the feedback
resistor, i.e. R9.

Thanks!!
 
M

Mikal Hadvik

Jan 1, 1970
0
Fred,

Without actually doing the math, I'm reasonably sure that the designer
faced this problem: Op amps like the LM358 don't have high enough
gain/bandwidth product to take this much gain in one gulp and maintain
adequate bandwidth for the job at hand.

Mikal Hadvik
Decade Engineering
www.decadenet.com
 
The noise detector can be found here

http://www.redcircuits.com/Page16.htm

I was wondering what benefits do we get by using
two op-amps in this circuit, because gain can also
be increased simply by increasing the feedback
resistor, i.e. R9.

Thanks!!

As the second post said, you get less bandwidth.

This is because the ratio of the op-amp's ("open loop") gain to the
actual ("closed loop") gain is a measure of the quality i.e. lack of
distortion i.e. bandwidth.

If you increase R9 then this ratio is reduced, the output is "softer"
at the extremes because the feedback to the op-amp's negative-pin is
less (via the bigger resistor) and the corrective-effect (that forces
the output to follow-truly the input) is reduced.

Taken to the limit, the op-amp with an infinite (i.e. no) feedback
resistor will have a very distorted output because the gain might be
10E6 in the middle but only 10E5 and reducing toward either limit. If
the feedback is hard i.e. a short circuit then the output will be true
right up to the limit i.e. the gain "loss" from 10E6 down to unity is
replaced with by a fidelity increase of "true tracking" to one part in
10E-6 i.e. one part in a million accuracy for 0.999999 of the output
swing.

(Of course it is also a shame to waste the othe half of the LM385)

Robin
 
F

fred

Jan 1, 1970
0
I have seen circuits before when the feedback ratio i.e.
R9/R5 (or R6,R7) is 10M/10K, which gives a gain
of 1000. This is quite resonable as 1000 is very far
away from the op-amp limits.

Here is another question:

When we change the switch position from R7 to R6 to R5
we are changing the gain by a factor of 10 every time because
R7 = 560
R6 = 5.6K and
R5 = 56K
therefore the db scale should be changing by equal amounts
as well, i.e. 50, 70, 90, so why instead of 90 we have 85?
 
fred said:
I have seen circuits before when the feedback ratio i.e.
R9/R5 (or R6,R7) is 10M/10K, which gives a gain
of 1000. This is quite resonable as 1000 is very far
away from the op-amp limits.

Here is another question:

When we change the switch position from R7 to R6 to R5
we are changing the gain by a factor of 10 every time because
R7 = 560
R6 = 5.6K and
R5 = 56K
therefore the db scale should be changing by equal amounts
as well, i.e. 50, 70, 90, so why instead of 90 we have 85?

A 10M feedback is not a good idea because the

10M
---\/\/\/\----
|
|
/\/----|---
|
----- stray C
-----
|
-----
---
-

extra CR delay might be enough to make it "ring" or even oscillate.
Another "feature" would be that the input Z at the op-amp input is
proportional to the feedback resistor (and inversly proportion to the
open loop gain) i.e. a large FB resistor makes that node more likely
to "pickup" or "cross-talk" Again; self oscillation-prone from
more-distantly-related-outputs.

Even if the circuit "works" on the bench, it may not in the field i.e.
if the board gets grubby and/or condensation settles on it.

**************************************

My guess is that the reduction in bandwidth caused by the 56,000 / 560
combination caused the 5db loss.

Robin
 
F

fred

Jan 1, 1970
0
A 10M feedback is not a good idea because the

10M
---\/\/\/\----
|
|
/\/----|---
|
----- stray C
-----
|
-----
---
-

extra CR delay might be enough to make it "ring" or even oscillate.
Another "feature" would be that the input Z at the op-amp input is
proportional to the feedback resistor (and inversly proportion to the
open loop gain) i.e. a large FB resistor makes that node more likely
to "pickup" or "cross-talk" Again; self oscillation-prone from
more-distantly-related-outputs.

Even if the circuit "works" on the bench, it may not in the field i.e.
if the board gets grubby and/or condensation settles on it.

**************************************

My guess is that the reduction in bandwidth caused by the 56,000 / 560
combination caused the 5db loss.

Any ideas why we need R4, loading the microphone will surely
reduce the input voltage, and this means we will need more
amplification, so this is strange ?
 
M

Mikal Hadvik

Jan 1, 1970
0
R4 presents an insignificant load to the microphone. It's there
primarily to provide a DC bias current path for the op-amp's input
pin. In combination with the input coupling cap, it incidentally
limits low-frequency response to about 16Hz in the first gain stage.

The noted 5dB gain discrepancy results, no doubt, from a lack of
sufficient amplifier gain to achieve the ideal calculated closed-loop
gain. Accurate 1000X boost in a single stage is simply asking too much
from a low cost op-amp chip, even if you're willing to give up most of
the bandwidth. The LM358 can barely achieve 60dB (1000X) *open-loop*
gain at 1KHz!

Mikal Hadvik
Decade Engineering
www.decadenet.com
 
fred said:
Any ideas why we need R4, loading the microphone will surely
reduce the input voltage, and this means we will need more
amplification, so this is strange ?

What would happen if R4 was missing? Assume a perfect op-amp (no
current flow into/out of the input). All is switched off: everything
is at ground. Switch-on. The left plate of C1 will jump up to some
voltage set by the resistances. The right plate will jump up too to
the same voltage. This voltage may not be the optimal mid-range bias
for the op-amp.

In reality this voltage will "drift with the wind" dependent on
"leakage" currents/temperature/cosmic ray inpacts etc. You *must* fix
it deliberately at "mid-range" DC wise. R4 does this.

R4 must be as large as necessary to avoid loading the mike. The
mid-range voltage is set by R2/R3 and "passed without loss" through R4
because, the current through R4 will be small (where can it go? The
op-amp input demands very little) so the voltage drop across it will
be very small so the voltage at either end will be just about the same
i.e. mid-range.

Robin
 
F

fred

Jan 1, 1970
0
What would happen if R4 was missing? Assume a perfect op-amp (no
current flow into/out of the input). All is switched off: everything
is at ground. Switch-on. The left plate of C1 will jump up to some
voltage set by the resistances. The right plate will jump up too to
the same voltage. This voltage may not be the optimal mid-range bias
for the op-amp.

In reality this voltage will "drift with the wind" dependent on
"leakage" currents/temperature/cosmic ray inpacts etc. You *must* fix
it deliberately at "mid-range" DC wise. R4 does this.

R4 must be as large as necessary to avoid loading the mike. The
mid-range voltage is set by R2/R3 and "passed without loss" through R4
because, the current through R4 will be small (where can it go? The
op-amp input demands very little) so the voltage drop across it will
be very small so the voltage at either end will be just about the same
i.e. mid-range.

If you think a gain of 1000 is too much for an op-amp, have a look here:
http://hop.concord.org/s1/ext/s1em.html

One more question:
I am have some problems to "accept" the value of R9=56K . This surely
will effect the voltage divided of R2, R3. Don't you think 220K would
have been better (adjusting of course the other resistors)?
 
M

Mikal Hadvik

Jan 1, 1970
0
I dunno how far you want to go with this, but the voltage drop across
R4 *is* significant. The LM358 spec allows 250nA worst-case bias
current, which works out to 25mV DC across R4. With a gain of 100X,
that's 2.5V of DC offset (relative to half of the B1 voltage) at the
output! And we forgot the worst-case input voltage (7mV) and current
(50nA) offset specs! The 2nd stage gain of 25X (ignoring its offset,
too) flings you out to 60V, and you're lost in the realm of Mordor.
This circuit clearly suffers from poor design, even if the prototype
was lucky enough to function as intended.

Of course, there's an input resistor on the other side that develops a
compensating input offset voltage, but that resistance is never equal
in value to R4. At max gain, it's almost a short compared to R4. Good
design would make it always equal to R4, with circuit gain controlled
some other way. Other possible solutions: AC couple the 2nd stage, or
substitute an FET-input op amp.

Mikal Hadvik
Decade Engineering
www.decadenet.com
 
fred said:
If you think a gain of 1000 is too much for an op-amp, have a look here:
http://hop.concord.org/s1/ext/s1em.html

One more question:
I am have some problems to "accept" the value of R9=56K . This surely
will effect the voltage divided of R2, R3. Don't you think 220K would
have been better (adjusting of course the other resistors)?

What I think is just an opinion. It means nothing. You have to do the
calculations to *know* the answer (+ - limits) e.g. as Mikal has
kindly done above.

I found "Design with Operational Amplifiers and Analog integrated
circuits" by Sergio Franco quite clear. I assume you already own the
AofE, if not why not!

It is healthy to assume that everything you see print-wise is wrong,
more or less.

Robin
 
F

fred

Jan 1, 1970
0
Mikal Hadvik said:
I dunno how far you want to go with this, but the voltage drop across
R4 *is* significant. The LM358 spec allows 250nA worst-case bias
current, which works out to 25mV DC across R4. With a gain of 100X,
that's 2.5V of DC offset (relative to half of the B1 voltage) at the
output! And we forgot the worst-case input voltage (7mV) and current
(50nA) offset specs! The 2nd stage gain of 25X (ignoring its offset,

how did you find these values, i.e. 7mV , 50nA ?
too) flings you out to 60V, and you're lost in the realm of Mordor.
This circuit clearly suffers from poor design, even if the prototype
was lucky enough to function as intended.

Of course, there's an input resistor on the other side that develops a
compensating input offset voltage, but that resistance is never equal
in value to R4. At max gain, it's almost a short compared to R4. Good
design would make it always equal to R4, with circuit gain controlled
some other way. Other possible solutions: AC couple the 2nd stage, or
substitute an FET-input op amp.

What about R9=56K, this will have an effect on the divider of R2/R3.
I assume that because it's only a dc change, it will not matter much, is
this the correct analysis? I assume R12 is similar to R4 i.e. it allows
a dc offset to develope on C4. Next point, can we use an npn instead,
simply moving R13 and D1 to the top and R12 to the bottom, will that
work the same?

I think this circuits offers quite in interesting feature or "trick" if you
like,
which is the voltage divider, to avoid the +-Vcc requirement of op-amps,
but I the LM358 can work down to 0V so maybee it would have worked
without this ?
 
M

Mikal Hadvik

Jan 1, 1970
0
If you think a gain of 1000 is too much for an op-amp, have a look here:

I didn't mean to say that it's impossible, only that you can expect
little bandwidth or gain accuracy. A 741 type op amp configured for
1000X gain has a worst-case bandwidth of 437Hz, based on National's
437KHz minimum unity-gain bandwidth spec for their LM741A. If you
accept their "typical" BW spec, you still get only 1,500Hz BW at
1000X! Sometimes that's okay, but it's not, in my judgement, okay for
a sound level meter. Unless you attend rock concerts, of course, which
can easily demolish your auditory bandwidth...
One more question:
I am have some problems to "accept" the value of R9=56K . This surely
will effect the voltage divided of R2, R3. Don't you think 220K would
have been better (adjusting of course the other resistors)?

Lower values in the feedback divider are generally better, to minimize
effects of stray C and flatten frequency response, but the amp has
trouble driving its own feedback network if you go too low. As long as
the loop is closed, amplifier pin 2 tracks pin 3, so bias divider
loading isn't a big factor. You might be able to decrese the value of
C2 if you raise feedback divider impedance, but it's not worth the
effort.

Mikal Hadvik
Decade Engineering
www.decadenet.com
 
M

Mikal Hadvik

Jan 1, 1970
0
how did you find these values, i.e. 7mV , 50nA ?

http://www.national.com/ds/LM/LM158.pdf
I think this circuits offers quite in interesting feature or "trick" if you
like,
which is the voltage divider, to avoid the +-Vcc requirement of op-amps,
but I the LM358 can work down to 0V so maybee it would have worked
without this ?

It's true that LM358 input common mode range extends to zero volts,
and you can squeak by with inputs that swing slightly below ground
(i.e. microphones), but the output side must be biased well above
ground to accommodate large output signal swings.

BTW, another possible fix for the first-stage DC offset problem is to
insert a capacitor in series to the wiper of SW1A. That brings your DC
gain down to unity, while preserving AC gain. It doesn't help with
gain accuracy or bandwidth issues.

Mikal Hadvik
Decade Engineering
www.decadenet.com
 
F

fred

Jan 1, 1970
0
how did you find these values, i.e. 7mV , 50nA ?
http://www.national.com/ds/LM/LM158.pdf


It's true that LM358 input common mode range extends to zero volts,
and you can squeak by with inputs that swing slightly below ground
(i.e. microphones), but the output side must be biased well above
ground to accommodate large output signal swings.

could we assume that the negative
cycle is not important and use only the positive cycle, the positive cycle
will not be distorted and we will have a rectifing effect,
would this change the reading as there is no load requirement ?
 
M

Mikal Hadvik

Jan 1, 1970
0
could we assume that the negative
cycle is not important and use only the positive cycle, the positive cycle
will not be distorted and we will have a rectifing effect,
would this change the reading as there is no load requirement ?

Since LM358 outputs can swing very nearly to ground, it's not
unrealistic to attempt a half-cycle amplifier scheme. I haven't done
this, but I would be on the lookout for saturation effects- where the
output "sticks" to the ground rail long after it should be moving
positive on each cycle, or even shoots positive during part of the
(clipped) negative excursion. You'll need a scope for this test. If
the LM358 fails, you should be able to find modern rail-to-rail I/O
amps that behave.

I'm not sure I understand the question about load requirement. The
data sheet shows a very kinky Vo vs. Io plot for the output current
sinking case, so you have to consider loading very carefully if you
want the output to swing all the way to ground. It's probably best to
make Q1 an NPN and rearrange the LED drive circuit as necessary, or
eliminate Q1 and drive the LED through R13 directly from the final
amplifier output pin, because LM358s can source a good 10mA or so from
the positive rail. If you do this, calibration to known SPL is
certainly necessary. Calibration seems unavoidable in any case,
because the sensitivity spec for the microphone is not given by the
designer.

Mikal Hadvik
Decade Engineering
www.decadenet.com
 
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