thanksThe Hearing aid circuit above is a 3-stage arrangement, using transistors 1, 2 and 4. The third transistor discharges a 10u electrolytic when audio passes through the circuit and is not part of amplifying the signal.
We will concentrate on the operation of the third transistor.
The 10u is initially charged via the 100k resistor and the voltage on the 10u is passed to the base of the first transistor to provide maximum gain.
When audio is passed through the circuit, any waveforms above 0.6v are detected by the third transistor to turn it on briefly. This action partially discharges the 10u via the 1k5 resistor. The lower voltage on the 10u is passed to the first transistor to reduce its gain.
In this way, any loud signals are not amplified as much as weak signal and the circuit will pick up very faint sounds while it will not be overloaded by loud signals.
The output is connected to high-impedance earphones.
they do two things:
1. The two PN junctions (diode and B-E junction) clip the sound level to prevent it becoming too loud.
2. Q3 is unbiased, it gets turned on by the signal being rectified. when it turns on, it's collector voltage drops and the bias to Q1 is reduced, forming a crude automatic gain control.
Brian.
hi
sorry, diode 4148 is used highspeed switching diode.
sorry, I meant removig the 1N4148 diodeyou can try it by removing q3.
This is a kind of Automatic Gain Control / Automatic Volume Control.
If the Audio signal is strong enough, Q3 will start conducting. This in turn will reduce voltage at its collector and, hence at the base of Q1. So Q1 will have a lower gain. This ensures that the audio level will not get too strong for comfortable listening.
The BE junction of Q3 will limit the audio signal about 0.7Vpeak for positive amplitude. The diode will do the same for negative amplitudes. So, audio level after Q2 is limited to 0.7Vpeak (or 1.4 V peak-to-peak, if you prefer).
Thank you i understand it
I have another question:
Why charging the 10u cap through a big 100K resistance (t= 1.1 R C = 1.1 sec) and discharge it through a relatively small 1.5 k resistance (t = 17msec), Why choose those values of resistors?
thanks alot
My question is what will happen if left transistor 3 and removed the diode?
Diode is used in a negative clamp circuit with Vbe of Q3 as a positive rectifier
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