# question about d.b. meter circuit diagram

Discussion in 'Electronic Basics' started by Dion, Jan 12, 2004.

1. ### DionGuest

Hello all,
I am a bit puzzled as to how the sound measuring circuit works
http://ourworld.compuserve.com/homepages/Bill_Bowden/page8.htm
Specifically my queries are as follows. Why are the values of the
chain of resistors which determine references to the comparators
where chosen to be 2.2k, 0.56k, 0.75k, and 3k. Why is 0.56k
considered 3d.b above 2.2k? Also, is it the best choice to use
a speaker as an input device, wouldn't an electret microphone be
better, and also is it possible to further increase the measurement range,
so that we get not just 70d.b but 70 to 100 d.b or 70 to 120 d.b ?

2. ### Ian BellGuest

The 3k, 0.75k, 0.56k and 2.2k resistors form a potential divider fed from
the 12 v supply. The voltage across the 2.2k is therefore:

(12 * 2.2)/(2.2 + .56 +.75 + 3) = 4.7 volts

The voltage from the .56k to ground is similarly:
(12 * (2.2 + .56) )/(2.2 + .56 +.75 + 3) = 5.08V

The ratio of these voltages is not 3dB so I agree with you.

I uspect the designer just used what was to hand rather than what was
'best'. it is a simple but not very good design IMHO.

and also is it possible to further increase the measurement range,
Yes, but it would probably easier to do a redesign using a bargraph LED chip
which has the comparators built in.

Ian

3. ### Bill BowdenGuest

Actually, it's about 2dB per step. The voltage at the 2.2k
is the reference for the lowest sound which is about 4.05 volts.
the voltage at the top of the 560 resistor will be about 1
volt higher so that the ratio is 4.05/5 = 0.8 or about 2dB.
The voltage at the top of the 750 resistor will be about
1.4 higher or 6.4 volts so the ratio is 5/6.4 = .78 which
is a little more than 2dB. I just used parts I had on hand so
the values are not exact. But the difference from 2 to 3
on a log scale is not much.

You can set the 500k gain pot to decrease the sensitivity
and read higher volume levels but this thing was designed to
monitor room noise within a narrow range of about 60 to 70dB.

An electret mic would probably be better than the speaker
and would not need the transistor. Most electrets have a
built-in amplifier so the output is about a half volt.
You might want to reduce the gain pot to 10k or so since
the mic will supply a much larger signal and the op-amp
will not have to boost the voltage much.

-Bill

4. ### DionGuest

o.k.
I understand you are using 20log V2/V1 , this is fine, thanks.
When you say
If I simply reduce the gain of the opamp feedback by 10
(by changing the 500k resistor to 50k), will that automatically
change our lights by 10db. I.e. if one light on indicates 70db,
can we say that by reducing the op-amp gain by 10, the first
light will now be 80db ?

5. ### DboweyGuest

Dion posted this and other examples:
Lesson #2: That should be 10dB, as in deci-Bells.

Don

6. ### Bill BowdenGuest

10dB is a voltage ratio of 3.162, so the pot could be
reduced to 500/3.162 = 158K or maybe a standard 150 or 200K.
But you still need to calibrate the lower light with
a known sound level. I have no idea what could be
used to produce a known 80dB sound level.

But the range will be the same so you would only read

You could add another LM324 so you have 7 lights spaced
3db and cover 80 to 98 dB.

The resistor values would be:

3.3K
2.2K
1.5K
1.1K
780
551
390
1K

-Bill

7. ### dionGuest

10dB is a voltage ratio of 3.162, so the pot could be
ah ha, you did not fall into my little trap. Reducing the gain by 10
will be a 20 db change. Well done!
I assume you are always aiming for the total of 10k for
good balance of op amp inputs.  