Hearing aid circuit

[walid1]

Hi
for the following circuit

if i replace R2 by 10 k and C3 by 10u what would youy expect.
someone told me that the circuit operate fine
it is a hearing aid
thanks

 

[mvs sarma]

Hi Walid,

these are my feelings with my limited knowledge.

the circuit should work from a supply down upto 1.8V (2*0.9V) where after the batteries are to be replaced. perhpas you have to maintain the overall gain of the device in the battery supply range. 10k may work well atlower voltage,but with BC549C having so highgain 10K wold be improper. collector has to be main tained at near half the supply voltage as seen across C4. if 10K is used it will fall down. and the voice may be distorted. battery lifemay be shortened.

10uF  in place of 1uF-- may boost low frequencies, but already limited by earlier coupling capacitors at 0.1uF.

sarma

 

[audioguru2]

Hi Walid,
The mic and its load/powering resistor is about 2k ohms. The 1st transistor has negative feedback through its base bias resistor so its bias voltage will drop a little when the resistor value is reduced and its its voltage gain will drop a lot. Its input becomes attenuated.

The remainder of the circuit is terrible. Q3 has base bias current directly from the positive supply. Transitors should never be biased this way unless their hFE is selected and an emitter resistor or another method of DC negative feedback is used. So it is saturated all the time causing severe distortion and low output level. It causes Q4 to also be saturated.

Nearly every circuit from that site has severe errors.

 

[audioguru2]

I forgot, Q2 is also saturated.
The circuit is not a hearing aid, it is a fuzz circuit for an electric geetar.
Have you ever seen a hearing aid without a volume control? New ones have an automatic volume control.

 

[walid1]

Hi sarma

the circuit should work from a supply down upto 1.8V (2*0.9V) where after the batteries are to be replaced.

What is the 0.9V?

perhpas you have to maintain the overall gain of the device in the battery supply range.

Why?

Hi guru

The mic and its load/powering resistor is about 2k ohms. The 1st transistor has negative feedback through its base bias resistor so its bias voltage will drop a little when the resistor value is reduced and its its voltage gain will drop a lot. Its input becomes attenuated.

I read somewhere that the Q1 stage voltage gain = approx. R2/R1,
with R2 = 680K Av= 680/3.3 = 206, where at R2=10, Av = 3 only??????

Q3 has base bias current directly from the positive supply.

No it is biased through R7 = 100K

So it is saturated all the time causing severe distortion and low output level. It causes Q4 to also be saturated.

I forgot, Q2 is also saturated.

please explain it more.

Nearly every circuit from that site has severe errors.

I enjoy seartching the errors

The circuit is not a hearing aid, it is a fuzz circuit for an electric geetar.
Have you ever seen a hearing aid without a volume control? New ones have an automatic volume control.

I'm sure it is a hearing aid without a volume control, later I'll attatched it.

Finally, you use a very good simulator, can you please tell the its name to try get it.

 

[audioguru2]

I read somewhere that the Q1 stage voltage gain = approx. R2/R1,
with R2 = 680K Av= 680/3.3 = 206, where at R2=10, Av = 3 only??????

It is more complicated than that. The source impedance must also be included since if it is low then the AC negative feedback is reduced which makes the AC gain higher.

you use a very good simulator, can you please tell the its name to try get it.

My sim program is SwCAD III free from Linear Technology.

The transistors have a wide range of hFE from 110 to 900. I calculated DC voltages for typical transistors. They are all saturated. Higher gain transistors would be worse:

 

[mvs sarma]

Hi sarma


What is the 0.9V?
that is the voltage of each cell when almost discahrged.(below that we will not abe able to use it purposefully.)



Why?

the range- i mean 3V around when new cells are put and around 1.8 to 2V when the cells are to be repplaced.

 

[audioguru2]

Never mind the battery voltage. This circuit doesn't work.
It is an example of "how to make a transistor amplifier wrong".

 

[walid1]

Hi guru

Never mind the battery voltage. This circuit doesn't work.
It is an example of "how to make a transistor amplifier wrong".

No, the circuit worked
i built it 3 times for me and for some frinds
it was very very sensitive to voice
it picks a very very low voice
at night i put it and hear persons speak with eachother, Believe me.
the only change i do, it was by a chance, is that it operate very well with one not new 1.5v battery
with this batt u will Surprised great ability to capture votes

 

[audioguru2]

The only way I could make the second part of the circuit work with a 1.0V battery is when I added an emitter resistor to Q3 like this: