For learning purposes I mounted basic audio amplifier on breadboard, schematic below:

enter image description here

Basically it works, meaning voltage has been slighly icreased as well as current and as a result I hear the sound in small 8 Ohm, 0.5W speaker. Nevertheless I have a few questions. Thanks in advance for detailed answers:

1.Looking at the schamatic we see point B (red dot). If I remove the C2 capacitor I have no output signal. My explanation to this is, without C2 audio signal is "shoted" to ground immediately before it reaches Q1 base. With C2 however, its right plate is constantly charged and discharged and therefore the signal has the chance to reach Q1 base. Is that good explanation?

2.Even though I use samll values resistor across the circuit I have small voltage and current amplification. Especially I would expect high current gain but it's not so. It's enough to put ~100 Ohm on the emitter terminal and I don't hear anything form the speaker. Below scope capture shows voltage levels having the gain of max 3. It did not help if I powered the circuit from 9V battery instead of 3. (changing base bias of cource). So why I was able to achieve only tiny gain?

  1. I am a bit suprised that it works at all, since audio input signal has positve as well as negative components. While positive components are ok, negative swings should stop transistor operation. My explanation is that the negative swings on left capacitor plate rises right capaciotr plate to positive value with magnitute two times bigger and therefore traniststor is ON. Does it make sense?

enter image description here

Blue signal - Audio_Out

Yellow signal - Audio_In

  • \$\begingroup\$ None of it makes sense, sorry. Go study how a capacitor blocks DC and prevents the BJT bias getting shorted out by a pure AC signal with average value of zero. Average value of zero means bias resistors are bypassed and transistor does not switch on unless the peak AC voltage of the signal rises significantly about +0.5 volts. \$\endgroup\$ – Andy aka Jun 1 '18 at 17:40
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    \$\begingroup\$ 1) If you remove C2 from the circuit AC signal is not short. The signal source is left open. 3) This is why you need a DC bias network at BJT base and C2 capacitor. Every "active device" to work properly as an amplifier supplied from a single source need proper bias circuit. When you use BJT as a CE amplifier, you use a voltage divider to bias the active device somewhere in the "linear region" (set the collector at 0.5Vcc). \$\endgroup\$ – G36 Jun 1 '18 at 19:29
  • \$\begingroup\$ @G36, I maybe wasn't clear enough. I should have said if I replace capacitor with short cirrcuit the output is gone. So it looks like base bias ate that input signal? Second, if I make Vc 0.5Vcc output is also dead. But maybe thats because emiter voltage has raised and now there is not enough base voltage to turn it ON. \$\endgroup\$ – DannyS Jun 1 '18 at 19:37
  • \$\begingroup\$ Also, do not forget the load current (speaker current) path. The transistor can only "sink" the current for a positive portion of an input signal. And the path of this current is: C1--->Colector-emiter--> Speaker--->C1. As you can see in this portion of a time the capacitor is discharging and provide a current into the load. But for a negative portion of an input signal transistor current decreases so now the speaker current is provided from the supply via R3-->C1--->speaker (C1 is charging). So this current cannot be larger than 3.3V/400R = 8mA \$\endgroup\$ – G36 Jun 1 '18 at 19:44

If I remove the C2 capacitor I have no output signal.

C2 is your input coupling capacitor meaning it allows your AC signal to pass thru and not any DC current. If you connect your source directly to the rest of the circuit (direct coupling) without C2, you will effectively be shorting the base to ground thru your source and the transistor will no longer be in forward-active mode, which is why you get no signal.

So why I was able to achieve only tiny gain?

This circuit doesn't stand a chance of driving an 8 ohm speaker. You are going to see extreme loading effects. For starters, output impedance is too high and it can't provide enough current to the load. You will need a power amp to drive a speaker. A push-pull stage might work if you design it properly.

I am a bit surprised that it works at all, since audio input signal has positive as well as negative components.

The purpose of your resistive divider at the base is to bias the base above ground at a certain DC voltage. Coupling thru C2 allows you to superimpose your AC signal on top of that DC bias voltage. So you have a little room to swing negative. Might make it more clear if you look at the AC equivalent circuit.

  • \$\begingroup\$ "...you will effectively be shorting the base to ground thru your source" Do you mean that base bias voltage will generate current flow from base to signal source i.e microphone? \$\endgroup\$ – DannyS Jun 3 '18 at 8:56
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    \$\begingroup\$ I mean you'd basically short out R2 thru the source. If you look at the DC equivalent circuit, you short out all AC sources, think of it that way. The capacitor prevents the AC source from shorting out the DC bias while allowing AC signal to pass thru. \$\endgroup\$ – Avid Pro Tool Jun 3 '18 at 9:04
  • \$\begingroup\$ Okay make sens, thanks for the answer I am going to mark your answer as valid one athough I still tink the performance of this amplifier can be improved significantly without use of any op amp or sth else as you suggested. \$\endgroup\$ – DannyS Jun 3 '18 at 9:37
  • \$\begingroup\$ You will also probably need a larger output coupling capacitor. I'm skipping a lot of details though. \$\endgroup\$ – Avid Pro Tool Jun 3 '18 at 9:46
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    \$\begingroup\$ It might help you to do some more research about class A power amp design and loading effects, etc... \$\endgroup\$ – Avid Pro Tool Jun 3 '18 at 9:50

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