Design an audio power amplifier

[PhysicoRaj]

Hello,
I was dissecting an old TV when I came across a set of speakers. They're 8 ohms, though I'm not sure if they're 8w or 6w..

I decided to make my own portable speakers (mono), which could be connected by means of an aux cable to my smartphone.

I don't know what's the output voltage of my smartphone headphone jack.. So please shed some light on that..

Basically I need to design a 6W power amplifier.. I have 9v batteries (Vcc).

I did some basic calculations and found out the quiescent current can fry my bias network if I used a Class A.
So I thought of a class B, and since I don't want Trafos, I am determined on a complementary pair push pull.

What power transistor would suit my application the best? How do I design the circuit.. Can the push pull provide both voltage and current gain? Do I need any other stage before this push pull?
Is cross-over distortion a major problem?

Thanks for your time and help..
Regards

 

[Svein]

A 6W amplifier can be had as an IC. Some examples: http://www.reconnsworld.com/audio_8wattamp.html, http://www.rason.org/Projects/icamps/icamps.htm and http://www.circuitdiagramworld.com/amplifier_circuit_diagram/Car_Amplifier_Circuit_2x40W_based_TDA8560Q_10013.html .

 

[PhysicoRaj]

Thanks a lot.. They can make my work a lot compact. I may use this..
But if the design using push pull isn't too complicated, I would like to try my luck on making my own, or even, out of curiosity to know how to design such circuits.
So any info on the design is also welcome..
Thanks again.

 

[Svein]

But if the design using push pull isn't too complicated, I would like to try my luck on making my own, or even, out of curiosity to know how to design such circuits.

Applause!

I can direct you to some pages, but please start with something simple and low power. When you have done that, I can give you some links to higher power amps.

• http://www.discovercircuits.com/A/a-audioamp1.htm, some simple circuits and some not so simple.
• http://www.circuitstoday.com/tag/audio-amplifier-circuits, several versions.
• http://www.zen22142.zen.co.uk/Circuits/Audio/audio.html, several circuit.

 

[PhysicoRaj]

Thanks :) I noted down some circuits. As I observed it seems that almost all the circuits had a preamplifier before the complementary pair.. Is that necessary? Because the output of a jack is low, I may be needing some gain before the power amp.. With this cleared I can get started.

 

[PhysicoRaj]

What's the use of Q2 in the attachment?

 

[Svein]

What's the use of Q2 in the attachment?

Q2 is the driver for the output stage.

 

[PhysicoRaj]

Ok Thank you.. I decided to drive a push pull using OpAmp as preamplifier (I think it does the role of Q1 and Q2)..and raise the Q point of the push pull using a preset.

Thanks for the help!

 

[Svein]

I decided to drive a push pull using OpAmp as preamplifier

OK, but be careful - not all OpAmps are capable of driving the output transistors or handling the large amplitudes needed. Some advice and circuits:

• http://sound.westhost.com/project76.htm.
• http://www.allaboutcircuits.com/textbook/experiments/chpt-6/class-b-audio-amplifier/.

 

[PhysicoRaj]

Hmm I'll look into it.. Ty

 

[PhysicoRaj]

Opamp supplies have always confused me. The LM383 needs a +Vcc and ground in place of the -Vee as in the case of LM741.
When I'm dealing with batteries, how do I connect the LM383? Can I give the battery negative to ground?
http://www.reconnsworld.com/audio_8wattamp.html

 

[davenn]

Can I give the battery negative to ground?

yes

many op-amps work using a split rail +V, 0V, -V

in the case of a single rail supply ( + and -) the - equals GND /negative of battery or other PSU

 

[PhysicoRaj]

in the case of a single rail supply ( + and -) the - equals GND /negative of battery or other PSU

OK thanks.. Because in my lab we worked with the op amps in +V 0 - V and the instructor used to warn us not to interchange the 0 and - V.. So I was doubtful about the single rail system.
Is a battery different from a dual power supply? Ex, what would be the -ve of a 12v battery at, 0 or - 12v? Because testing out dual supplies, my multimeter reads 24v between the terminals. That means a battery should be of +12,0 form.. Am I right?

 

[davenn]

OK thanks.. Because in my lab we worked with the op amps in +V 0 - V and the instructor used to warn us not to interchange the 0 and - V.. So I was doubtful about the single rail system.

Many of the dual rail op-amps will happily work using a single rail supply
others are specifically designed to use with a dual rail ( split supply) only ... .the datasheet for the particular op-amp gives this info

Is a battery different from a dual power supply?

yes ... it is a single rail supply

Ex, what would be the -ve of a 12v battery at, 0 or - 12v?

the negative terminal is 0V relative to the positive terminal

Because testing out dual supplies, my multimeter reads 24v between the terminals.

that's correct and you have an associated 0V terminal. so with your multimeter negative ( black) lead on the 0V term. and the red lead on the +12V term. you will see +12V on the meter
and with your multimeter negative ( black) lead on the 0V term. and the red lead on the -12V term. you will see -12V on the meter

you can make a dual rail supply from a single or 2 batteries like this ...

single battery or other single rail supply

2 batteries ...

Note that in the first one, two resistors are being used to produce a voltage divider
this is for small currents only as the resistors produce current limiting
so if you had, say, a 12V battery, you could produce a +6V, 0V, -6V supply

the second dual rail supply uses 2 batteries with the addition of voltage regulator IC's to produce a stable outputcheers
Dave

 

[PhysicoRaj]

Ah.. That's enough explanatory :)
And one last thing.. How do I know the output of my smartphone headphone connector? DC level, peak AC, etc? Because most LM383 circuits set a gain of 100, and I want to be able to vary that gain to drive the 8 ohms speaker at 6-8 watts.

8 Ohm speaker running at 6W means approx 10V peak output voltage. So I need to know the input so that I can play with the gain.

 

[Svein]

I want to be able to vary that gain to drive the 8 ohms speaker

Put a potentiometer in front of the input.

 

[davenn]

So I need to know the input so that I can play with the gain.

Put a potentiometer in front of the input.

yes

just like this, using the circuit you linked to...

Dave

 

[PhysicoRaj]

But wouldn't that be just like attenuating the input and having same gain.. What if I needed more than the full input voltage, I need more gain right? So I thought knowing the output of the jack would be better.

I searched everywhere but in vain.. I can't find the 3.5mm jack specs.

Thanks..

 

[davenn]

But wouldn't that be just like attenuating the input and having same gain..

Yes.
That is the way it is done on pretty much every amplifier ever built
so if the amp has a gain of 100, you can work out the output power relative to the input power

you ONLY get max output from the LM383 ( or whatever other amp) WHEN its input is driven fully

What if I needed more than the full input voltage,

you cannot overdrive the input to the amp it will just produce an unlistenable distorted output and possibly damage the amp chip
the full output of the headphone jack of your phone/media player could easily have enough output to overdrive the amp chip
so that is why the volume control ( attenuator) goes on the input to the amp

 

[davenn]

I searched everywhere but in vain.. I can't find the 3.5mm jack specs.

what jack specs ?

the max output from your smartphone ?
that will be in the phone spec sheet

 

[meBigGuy]

http://www.kenrockwell.com/apple/iphone-5/audio-quality.htm is an analysis of the 3.5mm audio jack output. It is probably similar to any smartphone.

The smartphone output generally has a gain control, so you can send it to a fixed gain amplifier. For example iphone docks change volume (generally) by sending commands to the iphone.

 

[PhysicoRaj]

Yes.
That is the way it is done on pretty much every amplifier ever built
so if the amp has a gain of 100, you can work out the output power relative to the input power

you ONLY get max output from the LM383 ( or whatever other amp) WHEN its input is driven fully
you cannot overdrive the input to the amp it will just produce an unlistenable distorted output and possibly damage the amp chip
the full output of the headphone jack of your phone/media player could easily have enough output to overdrive the amp chip
so that is why the volume control ( attenuator) goes on the input to the amp

OK I get it..

 

[PhysicoRaj]

http://www.kenrockwell.com/apple/iphone-5/audio-quality.htm is an analysis of the 3.5mm audio jack output. It is probably similar to any smartphone.

The smartphone output generally has a gain control, so you can send it to a fixed gain amplifier. For example iphone docks change volume (generally) by sending commands to the iphone.

Exactly what I wanted! Thanks a lot!

 

[PhysicoRaj]

In case of a complementary push pull, the pnp transistor requires -Vee. And the load needs a ground.
How do I power this using a single battery? Can -Vee and ground of the load be given to - ve of battery? (see image)

 

[PhysicoRaj]

Image is here..

 

[meBigGuy]

You have to create a virtual ground at half the battery voltage and AC couple the input and output signals.

 

[PhysicoRaj]

You mean like this:

 

[davenn]

You mean like this:

yes but remember that that circuit, when I posted it, I said that using a resistor voltage divider limits the available current
you would be better off using the other circuit I showed you where you use 2 batteriesDave

 

[PhysicoRaj]

The other circuit features two regulators, what if I omit them.. If I have the battery of the voltage required?

 

[davenn]

The other circuit features two regulators, what if I omit them.. If I have the battery of the voltage required?

yes that is OK, the reg's in that circuit were just a progression for if a well regulated supply was needed ... feeding IC's for example

 

[PhysicoRaj]

yes that is OK, the reg's in that circuit were just a progression for if a well regulated supply was needed ... feeding IC's for example

Thanks

 

[PhysicoRaj]

This amp IC (in the attachment) , is expected to deliver around 8W to an 8ohm load. That means the rms output would be 8v.

1) looking at the circuit, the gain must be decided by the 150k & 4.7k divider, which gives a gain of ~30. (Am I right?)
I learned the headphone jack has a max output of 1v rms. This means the output of the amp would be 15v at half input. This exceeds the figure given by the amp manufacturer. Am I wrong with the gain?

2) What is the use of those diodes going from the output to the supply? Are they a kind of protection? Do I need to incorporate them compulsarily?

Thanks.

 

[PhysicoRaj]

The amplifier :

 

[davenn]

1) looking at the circuit, the gain must be decided by the 150k & 4.7k divider, which gives a gain of ~30. (Am I right?)
I learned the headphone jack has a max output of 1v rms. This means the output of the amp would be 15v at half input. This exceeds the figure given by the amp manufacturer. Am I wrong with the gain?

so you can adjust those resistor values to give a useable gain
Also I think I mentioned in an earlier post, you should have a volume control in the input for gain adjustment so that you don't overdrive the input to the IC and cause distortion

2) What is the use of those diodes going from the output to the supply? Are they a kind of protection? Do I need to incorporate them compulsarily?

these diodes are for the protection of the IC from voltage spikes on the output
Note for a single rail PSU, as you are using, here is a valid connection for them ...

also note, this chip is capable of around 12-14W for a 4 Ohm load and up to 18W for a 8 Ohm loadDave

 

[PhysicoRaj]

Thanks. I am worried about the output because I have to run my 8ohm speaker at 8w, without knowing what the amp is ousting at the moment.. (can't measure the output on my DMM since it's not a sine). I'll have to play with the volume control and gain, and hope my load won't smoke off.
Is there a way to know the output?