A few questions on a transistor as amp

[d4n13l]

Hi,
I've been reading on transistors and I'm focusing on the common emitter configuration now. However I have some questions, newbie level, that I'm not sure about.

Let's suppose I want to drive a 1W speaker with 8 ohms resistance. To get the 1W you need 0.35A and 2.8V on the speaker. Now, if I have a Vcc of 12V, the questions are:

1) To confirm, I'm supposed to set Vc to 1/2Vcc which is 6V. I must set the circuit so that it doesn't exceeds the 8.8V right? given the 2.8V required by the speaker.

2) Does the variations in current caused by the input are going to be absorbed by the speaker and so Vc will stay always the same?

3) I'm not sure how an AC input (through a capacitor) would affect the DC in the base of the transistor. If let's say the input AC voltage is 2 Vpp, and the DC in the transistor's base is set by a voltage divider to let's say 3V. Is the DC going to start oscillating over 4V and 2V?

[innkeeper]

Your thinking DC.. you're not applying DC to an 8-ohm load (speaker).

you are applying AC to a reactive load. and the average reactance is 8 ohms.

to keep this simple, you can work with RMS voltage.  (root mean squared) ac values which can be used to calculate equivalent power to DC.

keeping with this simple concept yes, 2.8Vac rms into an 8 ohm resistive load will be 1W

now... that's AC output not DC output, .. the output capacitor blocks any DC components.
what your caring about is the voltage change that makes up the AC component of the signal, not its static voltage for determining the ac output, though it does matter to prevent clipping.  the voltage could be 100Vdc but the ac component could be 2.8Vac rms and the output would be 1W assuming the dc is blocked by a capacitor)

I am oversimplifying this just to get the point across for ac vs dc.

some video links to amplifier basics

[Audioguru]

A single common emitter transistor driving a speaker is a heater, not an audio amplifier. Almost all audio amplifiers use class-AB (look it up) for the two common collector complementary output transistors.
DC is NEVER applied to a speaker.

Please learn about how to bias a common emitter transistor.

[d4n13l]

Probably I wasn't very clear with my questions before. I know a class A amp is not good for audio but since I don't have a wave generator to see the effect the amp has on the signal I'm using audio purely for educational purposes. Also, I understand that the capacitors block DC and so the speaker only gets AC. I attached a circuit I designed.

I changed the speaker power rating so that I could actually set it up in the protoboard. Let's say the speaker is rated at 0.3W and has 8 ohms resistance. So the speaker requires 0.2A and 1.5V to get the 0.3W.

1) since the Vout can exceed the speaker's 1.5V (for 0.3W), will it get fried or something if it happens? or will it just absorbe what it needs?
2) would it be a better practice to not allow Vout to exceed these 1.5V?
3) conceptually I'm having trouble understanding how is the current distribution. As I understand there is always 0.2A through the transistor (collector to emitter). So when the input causes the current to increase, the speaker takes the excess and so the current through the transistor stays the same?

[Zero999]

R_C has a resistance of  13.25Ω, so the amplifier's output impedance will be equal to that figure, i.e. the output terminal of the amplifier will look like an ideal AC voltage source, in series with a 13.25Ω resistor. When the speaker is connected, the voltage across it will drop, because it will form the lower side of a potential divider.
https://en.wikipedia.org/wiki/Voltage_divider

To make a good amplifier, the output impedance needs to be much less than the speaker, which is one of the reasons why this configuration is not a good one. The best way to do this is to use a common emitter amplifier to provide the voltage gain and an emitter follower for the current gain. Put voltage gain and current gain together and you have a lot of power gain.

[innkeeper]

You would want the RMS voltage to be above that....
but you can only limit the peak voltage in a simplistic amplifier so..... you could limit it to 2.12V Peak to Peak output. (1.414 x 1.5v rms)

one way to do that is to design your circuit so the transistor will clip at the 2.12v point.

as pointed out by someone else, you can pad the output to bring it down to that level so clipping would occur when there's only 2.12v p-p out of the pad/voltage divider.

[gcewing]

If you build that circuit, be aware that things are going to get hot. With a 5V supply, the emitter and collector resistors and the transistor will dissipate 1 or 2 watts between them, so make sure you use adequately rated components and sufficient heatsinking on the transistor. With a 12V supply it will get even hotter.

If you insist on using a class A amplifier to drive a speaker, it's much better to do it through a transformer, like this:


Taken from https://www.electronics-tutorials.ws/amplifier/amp_5.html

The transformer converts the low impedance of the speaker to something much higher (typically a few k ohms) which is a better match for the relatively high output impedance of the common emitter amplifier. Another way to say it is that the transistor can operate with a much lower quiescent current, so it wastes far less power.

Circuits like that were used extensively in the past, but nowadays it's much more usual to find a push-pull output stage like this, which is effectively two emitter followers connected together:


Taken from https://www.electronics-tutorials.ws/amplifier/amp_6.html

This kind of circuit is very efficient, because the transistors can be biased to pass very little current when there is no input signal.

[d4n13l]

Thanks! I didn't know you could use the transformer like that.

I'm also satisfied with answers I found in electronic principles by A. Malvino. I liked the way it's explained much more than other books I tried before.