amplifier question

Discussion in 'Homework Help' started by Adsa, Apr 3, 2012.

Mar 10, 2012
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i am still very new to amplifier the design. I am assigned to construct a 3 stage amplifier to drive a speaker with a power output of 2w. My question is should the 2w consist of high voltage , high current, or a balance of bothin order to drive the speaker better? Is there any rule of thumb to get ideal approximate values?

2. MrChips Moderator

Oct 2, 2009
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3,451
Use Ohm's Law and the formula for power.

$P = I^2 R$

If R = 8 ohms

I = 0.5A
V = 4V

Mar 10, 2012
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thanks , didnt realise it was that simple.

4. MrChips Moderator

Oct 2, 2009
12,624
3,451
All of electronics is simple, once you know the basic rules, like Ohm's Law.

5. bountyhunter Well-Known Member

Sep 7, 2009
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Since most speakers are 8 Ohms, start with the assumption that is the load. Remember RMS power to the load is .707 x peak sine wave value, so design for a peak to peak output swing about 3X times the voltage that RMS calculation gives you.

6. Audioguru New Member

Dec 20, 2007
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3 stages or 3 transistors? 3 stages with 4 transistors?
Google is full of hundreds of these amplifiers.

A little common-emitter transistor at the input is a voltage amplifier. It drives a complementary (NPN and PNP) output transistor pair that are emitter-followers to provide current gain. Then only 3 transistors are needed but it has only 2 stages.
Another transistor can be the 3rd stage as a preamp for a low level microphone.

Mar 10, 2012
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3 stages and class b. Im planning on having the last power stage using darlington pairs. What i cant quite understand is one speaks of voltage and current gain, but isnt current gain my only concern? i mean if i can drive the required current through my resistor (0.5 amps 8 ohm resistor) wont voltage become of the required value(v=ir) in order to achieve the required 2w?

Dec 26, 2010
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Your problem lies partly in semantics (the meaning of wording, more or less), but there are some real issues here. Some amplifier structures, particularly the very widely used emitter follower can provide significant current gain, but have a voltage gain somewhat less than unity.

Many amplifiers nevertheless use emitter follower output stages preceded by voltage amplifier stages working in common emitter, because amongst other things it is convenient to obtain a low output impedance from the emitter follower, and class B (better, AB) amplifiers can readily be made using two complementary NPN and PNP transistors - other ways being possible.

Last edited: Apr 5, 2012
9. Audioguru New Member

Dec 20, 2007
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A class-B amplifier produces horrible crossover distortion. Audio amplifiers use class-AB not class-B.

The amplifier has an input transistor and a driver transistor that provide voltage gain. If the input signal is 150mV RMS line-level which is 0.21V peak and you need an output of 2W into 8 ohms then the peak output current is 0.7A and the peak output voltage is 5.7V. The voltage gain needed is 5.7V/0.21V= 27 times.

2W into 8 ohms can easily be done with ordinary little power transistors like TIP31 and TIP32. Their minimum current gain is 25 at 1A and is a little higher at the peak output current of 0.7. Then their driver transistor needs a current of only 25mA. The input transistor can have much lower current.

The supply voltage for the amplifier will be about 14.5VDC.

Mar 10, 2012
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i have another small question. For the final stage, the power stage, how will i ensure equal gain from both transistors ? Is there like a chip with pnp and npn mounted together to have equal beta?

11. Audioguru New Member

Dec 20, 2007
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The output transistors each have an emitter resistor so that both have almost the same base-emitter voltage and current.
It doesn't matter if their beta is different because they are emitter-followers where the output voltage does not change if the beta is more than the minimum amount needed.

Here is a very simple 2W power amplifier. It has a driver and an output so it has only 2 stages.
Add an ordinary single transistor at the input for more voltage gain. It will be the 3rd stage.

Caculate the resistor values and input capacitor value.

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12. WBahn Moderator

Mar 31, 2012
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You should design your circuits, when possible, so that they only have to have some minimum beta in order to allow for manufacturing tolerances, if nothing else.

Don't forget to consider power dissipation in your output stage (each component individually) and provide proper heat sinking, if needed.

Mar 10, 2012
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thanks for all your help i managed to make the power stage work correctly. Now im stuck and have no idea what the problem is. I designed a ce stage, but the gain is way lower then calculated and i cant figure why.

IE from simulation is 6.8ma so i figured gain should be:

Av=(1000//160) /r'e

and r'e is 25mv / 6.8ma = 3.7

therefore a gain of almost 38 times

But the gain is only of 13.

Am i missing something obvious or am I doing everything totally wrong?

14. Audioguru New Member

Dec 20, 2007
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The emitter voltage is about +4.03V which is much too high. The collector voltage is about +8.2V. The saturation voltage of the transistor is about 0.1V. Then the collector can swing only 2V peak.
At such a high gain the transistor has extremely high distortion. I reduced the input a lot and it is still very distorted and the output level is impossible to measure.

1) Reduce the value of the emitter resistor and remove its bypass capacitor.
Then the transistor will have more output voltage swing if the base is re-biased. it will also have some negative feedback which reduces distortion.
2) Increase the value of the load, maybe to 10k ohms so it doesn't short the output of the transistor.
3) Learn a simulation program. The LTspiceIV program I use is free from Linear Technology.

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Mar 10, 2012
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I changed the impedance of the last stage changed the design of the ce and now is working fine . I simulated it with what you recommended and its great thanks! I also constructed it and it worked fine with a sine wave.

Now i have one final problem (hopefully). I need to test the amp with real sound through a speaker. so i tried through my laptop jack. I was told it should be around 0.3v, but I cant get it to peak more then 30mV. Any idea to why this is? Im using a wire from an old set of headphones to transfer the signal. Also the amp has a high input impedance so that should not be a problem.