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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

DannyS
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  • 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. – Andy aka Jun 01 '18 at 17:40
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  • 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).
    • – G36 Jun 01 '18 at 19:29
  • https://electronics.stackexchange.com/questions/301617/bjt-input-ac-signal-amplitude/301642#301642 this one https://electronics.stackexchange.com/questions/310471/help-understanging-ac-signal-connected-to-a-charged-capacitor-coupling-capacito and this https://electronics.stackexchange.com/questions/368614/why-cant-class-a-amp-drive-8-ohm-speaker-with-just-one-bjt/368660#368660 – G36 Jun 01 '18 at 19:34
  • @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. – DannyS Jun 01 '18 at 19:37
  • 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 – G36 Jun 01 '18 at 19:44
  • https://electronics.stackexchange.com/questions/376592/how-to-select-resistor-for-emitter-follower-when-driving-speaker/376630#376630 – G36 Jun 01 '18 at 19:46