Common collector as output stage - problems

[ZaeemAhmad]

I want to design an multistage audio amplifier with discrete transistors and have already design the first two common emitter stages, the last one is to be a common collector stage. But when I connect an 8 ohm speaker the voltage gain drops considerably. Is it a problem? Also tell me how to design the common collector stage for this 8 ohm speaker (2W). What peak to peak value of voltage do I need?
Thanks!

 

[flatulent]

This stage should be a push-pull type. Also, the bias current should be large.

 

[ZaeemAhmad]

So you mean that it is not possible for a single common collector stage to provide gain nearly 1, when a load of 8 ohms is attached to its output (i.e. it will always loose gain and peak-to-peak swing)? Also, tell me how much gain will I need from the first two CE stages if I connect a normal dynamic mic to its input to drive an 2W speaker?

 

[FvM]

A picture is worth a thousand words

 

[ZaeemAhmad]

I have attached the circuit diagram. I am actually making an audio amplifier using discrete configuration stages (Mic used is simple dynamic type). The first two are CE, with a gain of approx. 170 (not in dBs) each. Now, I don't know what configuration to connect between the last CE and the 8 ohm speaker. I tried common collector, but it didn't work (a large decrease in output voltage swing!). The DC level output at each stage is 4.5V. and collector current is 0.136 mA. Beta assumed is 100. Please tell me what to connect in the "question mark" region (on the left of 8 ohm, 2W speaker), so that I get maximum swing at the output. (Do not consider the values of capacitors now). If you draw the circuit diagram, then I will be much pleased. Thanks!
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[Audioguru]

This is a very simple power amplifier circuit:

 

[ZaeemAhmad]

Thanks AudioGuru!
I've tested your circuit in Proteus and it gives only (about) 1V peak to peak at the output with (about) 460mV at the input. Will this voltage gain of about 2 be enough to amplify the signal from the mic to a 2W speaker? Also tell me why have you connected the output of the amplifier to the top of 270k resistor? Is it for feedback? And lastly, which software have you used to simulate your circuit (because I am interested in it)?
Do you have any suggestions regarding my circuit?
Thanks!

 

[Audioguru]

Your proteus simulation software is wrong!
I use the free sim software called LTspice IV from Linerar Technology.

The output is 4.6V p-p at clipping with an input of 0.6V peak. Then the voltage gain is 4.6/1.2= 3.8.

The mic needs a voltage gain of 4.6V/0.014V= 329 so my power amplifier with a gain of only 3.8 is far from enough gain.
Your preamp with a gain of 170 will provide plenty of gain to drive my power amplifier.

Of course the 270k resistor provides AC and DC negative feedback. Without negative feedback the distortion is horrible.

 

[ZaeemAhmad]

Thanks a lot for your quick reply!
I have downloaded LTSpice IV and will start using it in some time.
So, you mean that I use my 170 gain preamplifier and use your power amplifier to get the final output, hmmm...
I would like to know how can I adjust the output resistance of the push-pull stage to about 4 ohms. Also tell me how to adjust its input impedance. And if you don't mind, I would like to know that how can I design my own power amplifier (just like yours, but with some different values, like those of gain and current, because I am somewhat confused about the 2N3904 transistors biasing, without collector and emitter resistance). Please give me some hints regarding the design...
Thanks!

 

[Audioguru]

You do not adjust the output resistance of a modern amplifier to match the resistance odf a speaker. If you do then you throw away half the output power of the amplifier. Old vacuum tube amplifiers matched their output resistance to the resaistance of the speaker.

Modern amplifiers have an output impedance of 0.04 ohms or less to provide excellent damping of the resonances of the spreaker.

If you want to use a 4 ohm speaker with an amplifier designed to drive an 8 ohm speaker then it must be re-designed to provide double the output current, or the power supply voltage can be reduced so that the max current into a 4 ohm speaker is the same as the max current into an 8 ohm speaker.

The DC collector load of the 2N3904 transistor in my amplifier is 330 ohms + 330 ohms + the resistance of two diodes + 33 ohms.

The AC impedance load at the collector of the 2N3904 transistor is about 33k since the 220uf capacitor provides "bootstrapping" (positive feedback) to the two 330 ohm resistors.

The gain of my power amplifier is changed by adjusting the value of the 39k input resistor and the source resistance.

 

[FvM]

I am somewhat confused about the 2N3904 transistors biasing, without collector and emitter resistance

The amplifier has a simple voltage controlled current feedback by the 270k resistor. It's bias point thus depends on the 2N3904 current gain, which varies of course between transistor exemplars and slightly with temperature. But the circuit has the advantage of providing a maximum output swing. A emitter resistor, as present in your preamp would give a more stable bias point but also reduce the output swing.

Apart from this interesting circuit details, Audioguru has anwered your exclamation mark question, I think. You need an output stage with sufficient current gain to drive the low impedance speaker. A single transistor class A stage would cause high power dissipation at the quiscent current required for linear operation. So a class AB push-pull output stage as presented is the solution. There are of course many possible variants of this basic concept.

 

[ZaeemAhmad]

Thanks a lot people (especially AudioGuru)!
Actually now I am interested in designing my own power amplifier and I want to know how you select those resistors values (i.e. why those, why not some other values). I shall be very thankful if you guide me or give me a reference to some book or article regarding the "inside" of the power amplifier design (preferably online) (in simple and understandable words for a beginner like me).
Thanks!

 

[Audioguru]

The peak output current into an 8 ohm speaker is 2.3V peak/8= 288mA. The hFE of a 2N4401/2N4403 is typically 150 at 288mA so a base current of 288/150= 1.92mA is good. The 330 ohm resistors have a total of 1.85V across both so their current is 1.85V/660 ohms= 2.8mA which is a little more than is needed.

The 2N3904 transistor has a typical hFE of 240 at 2.8mA so a base current of 2.8mA/260= 10.8uA is fine.

The 270k resistor has 2.9V across it so its current is 10.7uA.

The 270k resistor provides negative DC feedback so if the gain of the 2N3904 transistor is low then it will not turn on as much which makes the output voltage rise which provides more base current so it turns on more. If the 2N3904 transistor has high gain then the opposite happens. Then the gain of the transistor does not affect the circuit much.

 

[ZaeemAhmad]

Actually, you see, I am new to designing Audio Amplifiers. What I want to know is how things work, so that I can design them the way I want. The working of (my) CE Stage and the complementary push-pull region of the output stage is clear to me. I am confused about the working of (refer to the circuit uploaded by AudioGuru in this post):
1) The first 2N3904, whose collector is directly connected with the lower diode of the push-pull region.
2) The feedback resistor (270k) and how it actually helps in what it does.
3) The two 330 ohms resistors. What are their function and why have you connected a capacitor at the interconnecting node to the right of 270k feedback?
4) Why have you connected 33 ohm resistor between the diodes?
4) What do you mean by bootstrapping (in simple words)?

If you answer my questions, I shall be very thankful to you all!

 

[Audioguru]

Actually, you see, I am new to designing Audio Amplifiers. What I want to know is how things work .....

Sorry. I am not a teacher of basic electronics.
My circuit is just an extremely simple power amplifier circuit.

The boostrap capacitor allows the 330 ohm resistors to apply enough base current to the 2N4401 NPN output transistor for it to go to a high enough voltage.
My simulation shows what happens without it.

The 33 ohm resistor provides additional voltage drop so it and the two diodes turn on both output transistors a small amount to eliminate crossover distortion.
My simulation shows crossover distortion without it.

 

[ZaeemAhmad]

Nonetheless AudioGuru, you really helped me a lot. I appreciate your efforts.
I shall ask further questions, if I feel some problem!
Thanks to all!

 

[stanislavb]

Hi,
I got better result with R3=100 Ohm. I simulate in PSpice 9.2

 

[Audioguru]

Hi,
I got better result with R3=100 Ohm. I simulate in PSpice 9.2

Then the output transistors are both turned on a lot at the same time with a current of 34mA and the resulting high current makes them hot and quickly drains the battery. With 33 ohms the crossover distortion is nearly eliminated and the current in the output transistors is only 4mA.
Maybe R3 should be 39 ohms.

 

[ZaeemAhmad]

I have three questions regarding your simple power amplifier:
1) If I change my VCC to +9V, how would you recalculate the values of R2 and R6 (i.e. 330 ohm resistors)
2) What is the name given to the lower 2N3904 transistor (is it known as the driver?)?
3) How have you calculated the value of bootstrapping capacitor and why connect it in the middle of two equal valued resistors, and why not directly at the base of 2N4401?
Also tell me, what value of the signal source should I set to depict the behavior of a common dynamic mic at normal voice amplitude.
Thanks!

 

[Audioguru]

I have three questions regarding your simple power amplifier:
1) If I change my VCC to +9V, how would you recalculate the values of R2 and R6 (i.e. 330 ohm resistors)

Use Ohm's Law, the calculated peak current in the speaker and the typical hFE of the output transistors.

2) What is the name given to the lower 2N3904 transistor (is it known as the driver?)?

It is the voltage amplifier.

3) How have you calculated the value of bootstrapping capacitor and why connect it in the middle of two equal valued resistors, and why not directly at the base of 2N4401?

I guessed at its value. It must operate properly at the lowest expected frequency.
The transistors operate from current, not voltage so the bootstrap capacitor keeps the current in the 330 ohm resistor constant that feeds the base of the 2N4401 transistor. The voltage at the junction of the two 330 ohm resistors actually swings higher than the supply voltage.

Also tell me, what value of the signal source should I set to depict the behavior of a common dynamic mic at normal voice amplitude.

5mV to 10mV RMS.