Miniature audio amplifier

[jayson07191990]

hey,, I have a miniature audio amplifier the problem is that
I don't know how to read the circuit flow. Can u help me,,,
PLEASE............

 

[cowboybob]

What about the circuit flow do you want to know?

I don't mean to be short, but a signal comes in at the left, traverses through the 3 transistors, getting amplified on its way to the speaker.

A detailed explanation of each component's function, in the course of being an element of an amplifier, and why would result in a very lengthy post.

Please try to pose your question with a little more specificity, such as maybe "What is C1 doing?".

I understand your question, don't get me wrong. I'm not sure you understand the complexity of your question.

 

[jayson07191990]

what I'm trying to say is that how does it work
from the very beginning when the current
is already flowing,, and what is the purpose of the three
transistors ,

does it have a stages in every transistor before it will on?

 

[Brevor]

It's very basic, T1 amplifies the voltage (signal) coming in and then drives T2 and T3 which amplify the current to drive the speaker.

does it have a stages in every transistor before it will on?

I don't understand what that means.

 

[cowboybob]

On closer inspection, this circuit has a fixed bias on the base of T1 of about +600 mV. This effectively puts T1 into full conduction. What that means is that any signal coming into P1 will be swamped out by that bias voltage.

What that means in the end is that the circuit, best I can tell, while looking like an amplifier, is wired wrong for that purpose.

It effectively will do nothing but draw power.

Where did you get the schematic?

 

[Nigel Goodwin]

On closer inspection, this circuit has a fixed bias on the base of T1 of about +600 mV. This effectively puts T1 into full conduction.

No it doesn't, it's NOT got fixed bias, it's biased via negative feedback from the output.

However, it is a really crappy circuit, just like the last one he posted asking the same!.

 

[ericgibbs]

hi jay,
This is your circuit in simulation, it works ok for what it is.

 

[audioguru]

Almost the same amplifier was posted a few days ago on a different website.
The kids in the class should get together and post it only one time.

 

[cowboybob]

My bad.

I didn't do a ERC wiring check. C1 from the original schematic wasn't connected to R1. No input = no output.

My apologies.

On that note, where's a SIM that'll do the wiring for me??!!??

 

[colin55]

Here’s how the circuit works:

When the supply is turned on, current flows though the 8R speaker and through R4 to the base of T2. This pulls the base of T2 towards the 9v rail and the transistor rises to nearly the 9v rail. The voltage on the emitter of T2 is 0.6v lower than the base and this pulls the emitter of T3 towards the 9v rail. The base of T3 is 0.6v lower than the emitter.
This is as far as we can go with the current-path at the moment and we now have to go to T1.
The join of the two emitters has a voltage near the 9v rail and this voltage is passed to the base of T1 via the 82k resistor.
The 82k resistor forms a voltage divider with 12k and the resulting voltage at their join is sufficient to put 0.6v on the base of T1. This turns ON T1 and the voltage between collector and emitter drops to a low value. The exact value will be shown in a moment.
We can now go back to the base of T3 and continue the current-path (also called the voltage path) from the 9v rail to the 0v rail.
T1 pulls the base of T3 towards the 0v rail.
We now have three transistor that all turn on. They are not fully turned on but partially turn on.
The exact amount of “turn-on” for each of the transistors is due to the 83k and 12k biasing components and diodes D1 and D2.
Here’s how the DC coupled amplifier self-adjusts to a state called the QUIESCENT STATE. This is the state where some of the components adjust the “turn-on” of other components and the circuit reaches a point where the voltages settle down and reach a stable value and the current is a constant minimum value.
The voltage at the midpoint of the two output transistors is fairly high and this creates a slightly higher voltage on the base of T1. This turns on T1 slightly more and the voltage on the collector drops. This lowers the voltage on the base of T3 and the emitter voltage drops. This lower voltage is passed to the base of T1 and the transistor turns OFF slightly.
This is how the three transistors adjust themselves to a final value.
The exact final voltage is called a DESIGN VOLTAGE and designer of the circuit want the voltage on the join of the two emitters to be half-rail-voltage.
This allows the circuit to rise and fall and reproduce a waveform without clipping or cutting off the top or bottom of the wave.
To get the circuit to sit with the output (the join of the two emitters) at 4.5v, the values of R2 and R3 have been selected.

We now have the circuit sitting, ready to amplify a signal.
The output stage is called PUSH PULL because one transistor pushes current through the winding of the speaker via the 100u electrolytic and the other transistor pulls current through the speaker via the electrolytic.
You could connect the speaker directly to the output of the stage and remove the electrolytic. The circuit would work just the same.
However if the speaker is connected directly, a voltage of 4.5v will be paced across the speaker and this voltage will cause a current to flow in the winding of the peaker (the voice coil) and the cone will be pulled in. If we try to reproduce a waveform, the cone is already partially pulled-in and it will not reproduce half of the waveform.
In addition, this constant current will heat up the voice coil.
By adding the 100u, we remove the Dc component of the output and only the AC (waveform) will be passed to the speaker.
Now we have to understand how an electrolytic passes energy (current) to the speaker.
If you connect an electrolytic and speaker directly to a supply, you will hear a “plop” This is the electrolytic charging and the charging current flows through the speaker and produces the noise.
But after a very short time the electrolytic is charged and no ore current flows.
Even if you remove the supply and connect it again, no sound will be reproduced because the electrolytic is already charged.
The only way to hear another plop, is to remove the components and short between the power leads.
When the supply is re-applied, you will hear another plop.
To get sound from the circuit, this is what it has to do.
Firstly it has to charge the electrolytic. Then it has to discharge the electrolytic.
As you can see from the circuit, the lower transistor charges the electrolytic and the top output transistor discharges the electrolytic.
Now we have to drive the two transistors so that they charge and discharge the electrolytic.
To charge the electrolytic, T1 turns ON and pulls T3 towards the 0v rail.
This is the easy part.
How do you pull T2 UP so that it discharges the electrolytic?
This is how it is done. It is very clever.
Connected between T2 and T3 are two diodes. Each if these diodes has a voltage drop of 0.6v.
This voltage drop is exactly the same voltage as between the base and emitter of the two transistors in the output.
This means we can directly pull on the base of the top transistor, just like we are directly pulling on the base of the lower output transistor.
Now we have a situation where we can pull down on both transistors and this will turn ON the lower transistor and turn OFF the upper transistor.
This is done when T1 turns ON.
When T1 turns OFF, the top transistor is pulled HIGH via the 1k8.
That’s how it works.

 

[bountyhunter]

Looks like an output stage from an old transistor radio.

 

[Nigel Goodwin]

Looks like an output stage from an old transistor radio.

Old radios were never that bad

 

[Jugurtha]

Hello,

You can search the following terms: "Push-Pull","class AB amplifier", "crossover distortion", "negative feedback", "coupling capacitor", "bypass capacitor", "biasing", "Q point"..