Maker Pro
Maker Pro

Need help to understand sound level meter circuit


Ted Wilson

Jan 1, 1970
Yuk - what a horrible circuit!

The answer to your question is that C1 is there to bypass R1 and couple the
output of the microphone onto the base of T1.

Basically, circuit operation is as follows:

T1 and its associated components provide voltage gain to the signal from the
microphone and T2, together with C4 effectively provide half-wave
rectification of the amplified signal on T1 collector.

The larger the output from the microphone, the more positive the voltage on
T2 emitter will be and hence the more LEDs will be illuminated, starting
with LD1, turned on by T3, and progressively lighting each LED to the right
until all are lit.

Each of transistors T4 thru T7 has a threshold voltage one diode drop above
the one to its left.

I assume from your question that you are a relative newcomer to electronics
and, if so, I would not recommend this circuit as an indication of how to do

It is very poorly defined in terms of its DC bias conditions and both its
zero level and sensitivity will be prone to temperature drift. Furthermore,
RV1 which is there presumably to set the zero threshold level changes the
sensitivity at the same time - awful!

Hope this helps

Ted Wilson

Ted Wilson

Jan 1, 1970
A good candidate for one of the 'bad circuits' sections in the next edition
of A of A I'd say.



Jan 1, 1970
mk said:

I started to try and explain the purpose of C1 and then I realised that the
circuit is too awful for me to actually offer a sensible explanation.

I'll try.

C1 is intended to stop DC current (pedants hold your tongues) from flowing
into the base of T1 but allows AC to flow. R2 provides a path for DC biasing
of T1.

The circuit in the microphone will cause a DC voltage of maybe 2V to appear
on the +pin. (You would need the mic spec to estimate it better).

Current flowing down R2 will cause current to flow from base to emitter of
T1 and this will cause a larger current to flow from collector to emitter.
T1 will settle at a point where the voltage on R4 (caused by the emitter
current) + the base emitter voltage of T1 is equal to the Mic+ voltage - R2
* base current.

The problem is that the base emitter voltage depends on the current and the
temperature and will vary from one transistor to another. The gain of the
transistor will vary with temperature and current. If you want stable
biasing you need to make the voltage drop across R2 small compared with the
BE voltage changes and the drop across R4 large compared with same. (0.1V
and 1V are reasonable).

In your circuit I guess that with no signal from the microphone and RV1 at
125k a voltage of about 0.9V is intended on T2 base. If the supply is 3V
this means 10.8uA collector current in T1.
The BC547 is specced to have a gain of between 110 and 800 at Ic = 2mA so
lets say its 100 at 10uA.

We can work out (roughly) how T1 will set up:
assume Vbe = 0.55
Vb = 2 - 4.7E6 * Ib (mic volts - drop on R2)
Vb = 10E4 * Ib * 100 + 0.55 (volts across R4 + Vbe)

2 - 4.7E6 * Ib = 10E4 * Ib * 100 + 0.55

so Ib = 0.254uA and the drop across R2 is 1.19V
The voltage across R4 is 0.254 V.
The pot will need to be set to 14.6k to get 0.9V on T2 base.

Now back to that capacitor - if its a reasonable small electrolytic it will
be specced for perhaps 3uA leakage current - which is 10x what we think the
base current will be.

So the answer to your initial question is that C1 probably provides a path
for the base bias current for T1 but it will depend on temperature, how good
the capacitor is, the gain of the transistor .....

I have been thinking about all of this too, and there is one thing I was thinking
we should have considered: the impedance of C1 can be found. If we assume
the frequency to be 1000Hz, the 1/jwc is the impedance, and substituting in for
values we get (1/(2*pi*1000*1e-6)) = 159ohms. Shouldn't we be substituting
this into our ac equivalent circuit?