What is the function that transistor in the circuit?

[samy555]

Hi

I want to know what is the role of the transistor Q3 in the circuit, and why 1N4148 Diode?
I read the following but didnot uvderstand so much:
http://talkingelectronics.com/projects/HearingAid-2/HearingAid-2.html

The 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.

thanks

 

[KrisBlueNZ]

This is a crude AGC (automatic gain control) circuit.

The gain of the first stage is affected by the base bias, which is controlled by the voltage on the 10 uF capacitor, through the 1M base resistor. Reducing this voltage will reduce the gain of the input stage.

The 1 uF capacitor, the diode, and the base-emitter junction of the third transistor form a crude signal detector. The voltage on the base is 0V on average, because the DC condition is set by the 22K volume potentiometer. So the signal coupled from the 1 uF capacitor extends equally above and below 0V.

When its amplitude exceeds about 1.4V peak-to-peak, the diode and the transistor base-emitter junction start to conduct on the peaks. The diode conducts on the negative peaks; the transistor conducts on the positive peaks. This keeps the voltage centred around 0V.

Every time the transistor's base-emitter junction conducts, the base current also causes the transistor to conduct from collector to emitter, which reduces the voltage on the 10 uF electrolytic due to current flow through the 1k5 resistor. This in turn reduces the gain of the input stage, and therefore, the signal level at the 1 uF capacitor.

Once the loud signal disappears, the 10 uF capacitor charges back up through the 100k resistor and the gain increases back to what it used to be.

So the circuit acts to keep the signal at the 1 uF capacitor below 1.4V peak-to-peak, by backing off the input stage gain when it reaches or tries to exceed that level.

 

[samy555]

This is a crude AGC (automatic gain control) circuit.

The gain of the first stage is affected by the base bias, which is controlled by the voltage on the 10 uF capacitor, through the 1M base resistor. Reducing this voltage will reduce the gain of the input stage.

The 1 uF capacitor, the diode, and the base-emitter junction of the third transistor form a crude signal detector. The voltage on the base is 0V on average, because the DC condition is set by the 22K volume potentiometer. So the signal coupled from the 1 uF capacitor extends equally above and below 0V.

When its amplitude exceeds about 1.4V peak-to-peak, the diode and the transistor base-emitter junction start to conduct on the peaks. The diode conducts on the negative peaks; the transistor conducts on the positive peaks. This keeps the voltage centred around 0V.

Every time the transistor's base-emitter junction conducts, the base current also causes the transistor to conduct from collector to emitter, which reduces the voltage on the 10 uF electrolytic due to current flow through the 1k5 resistor. This in turn reduces the gain of the input stage, and therefore, the signal level at the 1 uF capacitor.

Once the loud signal disappears, the 10 uF capacitor charges back up through the 100k resistor and the gain increases back to what it used to be.

So the circuit acts to keep the signal at the 1 uF capacitor below 1.4V peak-to-peak, by backing off the input stage gain when it reaches or tries to exceed that level.

Excellent answer thank you
But what if the diode is not used
Transistor Q3 will not conduct in the negative part of the voice signal?

 

[KrisBlueNZ]

Excellent answer thank you
But what if the diode is not used
Transistor Q3 will not conduct in the negative part of the voice signal?

Q3 doesn't conduct on the negative part of the signal at all. It conducts when its base goes positive by 0.7V with respect to its emitter (which is grounded).

The diode conducts on the negative part of the signal, and forces the negative peaks of the signal voltage to be no further negative than -0.7V. This defines the signal voltage where the transistor starts to conduct as being 1.4V peak to peak, i.e. the negative peaks are at -0.7V (clipped by the diode) and the positive peaks are at +0.7V (clipped by the transistor, which discharges the 10 uF capacitor in the process, because of the current flow into the base). I hope that's a bit clearer

 

[samy555]

The diode conducts on the negative part of the signal, and forces the negative peaks of the signal voltage to be no further negative than -0.7V.

What if ther is no diode and the negative peaks of the signal voltagego further negative than -0.7V.? will that damage the transistor?
In short what is the harm of the lack of that diode?

thank you very much

 

[KrisBlueNZ]

The transistor won't be damaged as long as the base doesn't go below about -7V, which it cannot because the power supply voltage is only 3V so the maximum signal at Q2 collector is only 3V peak to peak.

If you remove the diode, the AGC will not work as cleanly as it should, because when the signal exceeds 1.4V peak to peak, it will not be clipped evenly, and its average voltage will go below zero. The circuit will allow signals that are constantly greater than 1.4V peak to peak to get through.

This AGC circuit is very crude and will not perform in a very well-defined way anyway, but removing the diode will make it worse.

 

[samy555]

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

 

[KrisBlueNZ]

Q3 will only conduct on the peaks of the signal, so each time it conducts, it needs to discharge the 10 uF capacitor by a significant amount. That's why the resistor from Q3 collector is 1.5k. When the overload has gone, the gain needs to return to normal, but not too quickly. That's why the 100k resistor is such a high value. There is a recovery delay after the loud signal, before the gain opens up again.

 

[samy555]

Q3 will only conduct on the peaks of the signal, so each time it conducts, it needs to discharge the 10 uF capacitor by a significant amount. That's why the resistor from Q3 collector is 1.5k. When the overload has gone, the gain needs to return to normal, but not too quickly. That's why the 100k resistor is such a high value. There is a recovery delay after the loud signal, before the gain opens up again.

Its ok, I understood by 50%
But why would we want to discharge quickly and charge slowly, note that the charging process does not stop during the discharging process?

I'm sorry for that many questions
thanks

 

[KrisBlueNZ]

As soon as the signal becomes too loud (i.e. amplitude exceeds 1.4V peak to peak), we want the gain to be reduced. One peak in the waveform will cause the transistor to conduct for, say, 1 ms, which will reduce the capacitor voltage slightly, and reduce the gain slightly. When the next cycle of the audio waveform comes along, if the gain is still too high, the transistor will conduct briefly again, and discharge the capacitor a bit more. After a few cycles, the gain should be reduced enough that the signal has been limited to 1.4V peak to peak.

Once the gain has been reduced to the point where the signal has been limited to 1.4V peak to peak, the circuit needs to gradually increase the gain again, in case the input signal has gone quiet again and the gain needs to be increased to make quiet sounds audible. This happens fairly gradually.

While the capacitor is charging up, and the gain is slowly increasing, if a sound comes through that is louder than 1.4V peak to peak, the transistor will conduct again and reduce the gain. Only if the input remains quiet will the capacitor charge up fully, and give maximum gain.

So the circuit is like a tug-of-war, between the 100k resistor which is slowly trying to increase the gain, and the transistor which is fairly quickly reducing the gain every time a signal comes through that is louder than the threshold.

It doesn't matter that the capacitor is charging slowly all the time. The transistor's discharge circuit has only a 1k5 resistor, and it can easily overpower the tiny charging current coming into the capacitor.

 

[samy555]

As soon as the signal becomes too loud (i.e. amplitude exceeds 1.4V peak to peak), we want the gain to be reduced. One peak in the waveform will cause the transistor to conduct for, say, 1 ms, which will reduce the capacitor voltage slightly, and reduce the gain slightly. When the next cycle of the audio waveform comes along, if the gain is still too high, the transistor will conduct briefly again, and discharge the capacitor a bit more. After a few cycles, the gain should be reduced enough that the signal has been limited to 1.4V peak to peak.

Once the gain has been reduced to the point where the signal has been limited to 1.4V peak to peak, the circuit needs to gradually increase the gain again, in case the input signal has gone quiet again and the gain needs to be increased to make quiet sounds audible. This happens fairly gradually.

While the capacitor is charging up, and the gain is slowly increasing, if a sound comes through that is louder than 1.4V peak to peak, the transistor will conduct again and reduce the gain. Only if the input remains quiet will the capacitor charge up fully, and give maximum gain.

So the circuit is like a tug-of-war, between the 100k resistor which is slowly trying to increase the gain, and the transistor which is fairly quickly reducing the gain every time a signal comes through that is louder than the threshold.

It doesn't matter that the capacitor is charging slowly all the time. The transistor's discharge circuit has only a 1k5 resistor, and it can easily overpower the tiny charging current coming into the capacitor.

Thank you from the bottom of my heart
I understand very well how the circuit works
You are a wonderful man
Thank you

 

[KrisBlueNZ]

I understand very well how the circuit works

Great!

You are a wonderful man

That's sweet. Your cheque is in the mail BTW