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pocket sized guitar amplifier

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Bit overly complicated for the OP though :D no need for the current sources - how about replacing them with one resistor, and two resistors with a bootstrap gap for a simpler design?, and run the sim again?.

Following on with the 'simpler' theme, drop the long tailed pair as well, and use a simple PNP at the front end (NFB to emitter).
i thought about that, but that would compromise the open loop gain, and he'd really need to add the bias diodes then, as well as increasing the idle current (the thing is battery powered) eliminating the diff amp? AG's circuit seems to do that pretty well. very cool AG, using the speaker impedance itself as part of the feedback network...
 
Bootstrapping increases the output voltage swing and as I show here it triples the open-loop voltage gain.
When the open-loop voltage gain is increased then negative feedback is more effective to reduce distortion and reduce the output impedance for better damping of speaker resonances.
 

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gentlemen you have all been a helpful bunch, i admit that i lack most of the knowledge in this subject, but believe me i am very eager to learn so please support me, i am very thankful to all of you. now i really wanted to finish this now. could you guide me step by step pls. lets say i will go with the class ab amplifier designed by audioguru but i will start with solid blank, all i have is a 9v battery, 3w 8 ohm speaker and i want a voltage gain of 20. what is the first step in designning? what transistor is best for this?
 
i thought about that, but that would compromise the open loop gain, and he'd really need to add the bias diodes then, as well as increasing the idle current (the thing is battery powered) eliminating the diff amp?

Try it and see what the sim says then?.

Personally I'd prefer to see a Vbe multiplier for the bias so you can adjust it - I've never liked the crude two diodes thing :D

AG's circuit seems to do that pretty well. very cool AG, using the speaker impedance itself as part of the feedback network...

It's not really part of the feedback mechanism - it's for bootstrapping - it's a VERY, VERY common (old circuit) - it's done for cost and size, as it saves one resistor and one electrolytic.

It was probably more popular in the 60's than since, back when components were fairly expensive.

In this case it reduces the component count considerably - which is what that rather crude example is all about.
 
As I said earlier in my post #9, 3W into 8 ohms needs a voltage swing of 13.85V. The amplifier output is not perfect so it should have a supply that is at least 16V. That is not possible from a 9V battery.
BUT two 0.8W amplifiers powered from 9V can be bridged to provide 2.8W to 3W into 8 ohms. But then the bootstrapping cannot be done with the output coupling capacitor and the speaker, instead it must be done with an additional resistor and capacitor. When two amplifiers are bridged then the output DC voltage from each amp are the same so the output coupling capacitor is not needed. A bridged amplifier uses the first amp to drive one wire of the speaker and the second amp drives the second wire of the speaker. Then the effective voltage swing is almost doubled and the currrent is also almost doubled. Double-double= 4 times the power.

Bridged amplifiers need one of them to have an inverted input. This simple amplifier inverts so the first amplifier can simply feed the input of the second amplifier.
Set the gain of the first amplifier to 10 and set the gain of the second amplifier to 1 then the total gain is 20.

3W into 8 ohms produces a peak current of 866mA which is not possible with the little transistors I showed. Use TIP31 and TIP32 power transistors for the output transistors. Their current gain is low so the current for the driver transistors must be increased.
An ordinary little 9V alkaline battery might not be able to produce peak currents of 866mA. Use six AA alkaline battery cells in series instead.

Wait a minute! YOU are supposed to design this amplifier, NOT ME so I wish you good luck and I hope I provided enough info.
 
Here is a hint about using a resistor and capacitor for bootstrapping and how the first amplifier and second amplifier connect together and connect to the speaker.
 

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Personally I'd prefer to see a Vbe multiplier for the bias so you can adjust it - I've never liked the crude two diodes thing

i prefer Vbe multipliers as well, but for simplicity, the diodes are ok. also, when you have the bias at the output transistors, instead of earlier at the driver or predriver stage (which none of the amps we have discussed have) you have to really play with the resistor values to get the Vbe multiplier to work right as you're only dropping 1.0-1.4V across it. i didn't want to have the OP wrestling with that if it was avoidable.

It's not really part of the feedback mechanism - it's for bootstrapping - it's a VERY, VERY common (old circuit) - it's done for cost and size, as it saves one resistor and one electrolytic.
i know what bootstrapping is... a) i'm old enough to remember what they are, and b) some audio manufacturers seem to still use them in current production home audio equipment (Denon still uses them). i have a tendency to avoid frequency dependent networks in the signal path wherever possible which was the reason for the current source in the spot where a bootstrap could have been used. the reason for the dotted boxes in my example schematic are because other things can go there, a simple resistor in place of Q3, and a bootstrap in place of Q4. or on the other extreme, better current sources could also go there...
 
Wait a minute! YOU are supposed to design this amplifier, NOT ME so I wish you good luck and I hope I provided enough info.
;););););););););) he snookered both of us AG.....

scripted13, you now must give US at least honorable mentions on your project......
 
Bridged amplifiers need one of them to have an inverted input. This simple amplifier inverts so the first amplifier can simply feed the input of the second amplifier.
Set the gain of the first amplifier to 10 and set the gain of the second amplifier to 1 then the total gain is 20.

isnt it supposed to be, 10 to the first and 2 to the second for it to have 20 total gain?
 
im constantly amazed by how easy for you to design something like that, but how did you arrived with the values of resistors? cause that is my main problem with my design, i have to defend it and explain why i used that or those particular resistors. please help. thanks

many of us here have been fixing or designing amplifiers for a long time. if you fix a lot of them, you get an understanding of what works and what doesn't. when you design amplifiers, you understand what makes them work, and in my case, i begin with a "ballpark" design, which i then take and test (usually with the same simulator (LTSpice) that you have seen AG and i use. i use the simulator to make quick changes to part values, or add more features (like the current sources or bias network) and get an estimate (the simulator isn't perfect) of how well it will work (if it works at all). then when everything seems to be right, i build it. that's the real test of whether it will work, and whether it will work to my expectations. for instance, if you look at the circuit i posted second (one of the transistors was backwards in the first copy of it), the circuit didn't work at all with +/-4.5V as a power supply. i had to lower the resistance values in the current sources to get them working, because the "ballpark" design was actually drawn with +/-30 or 40 Volt rails in mind. then i had to change one of the resistors in the diff amp to 1 ohm (actually a piece of wire would suffice there) to get the diff amp working properly with the low supply voltage. i had to lower the current setting resistors (which increases the current) in order to get enough current flowing through the diff amp and Q7 for the amplifier to work properly.

also, since i've repaired a lot of amplifiers, i know that resistors in a particular part of the circuit are usually within a certain range, because these circuits are common to most amplifiers, and i see basically the same amplifier circuits with a few minor variations all the time. in home audio these days, this type of amplifier is called "Lin" after the name of the researcher at RCA that came up with this type of amplifier in 1956. transistors were new technology back then, and most designers kept on using basically the same amplifier circuits they had used for tube amps, just with transistors. but transistors have two polarities, and Lin came up with an amplifier that takes advantage of that. so you will find circuits like the one i posted called "Lin topology amplifiers". you can read a part of an interview with H.C Lin about his amplifier design here: http://www.semiconductormuseum.com/Transistors/RCA/OralHistories/Lin/Lin_Page3.htm
 
if the voltage gain of my circuit is 200, and my ac input is 100mv, will my ac outpput be 20v? can the speaker handle this much voltage? will the speaker be damaged?
 
isnt it supposed to be, 10 to the first and 2 to the second for it to have 20 total gain?
No, because bridging doubles the gain.
You want the second amplifier to simply invert the output of the first amplifier with no gain (its gain must be 1).
 
the gain of the first part is -10, the gain of the second is -1. multiply those you get +10, so the output (across the speaker) is a difference of 20 because the outputs are 180 degrees out of phase. the total AC voltage between the outputs is twice what the voltage between either output and ground would be. each amplifier output stage only "sees" half of the speaker impedance (i.e. with an 8 ohm load, each output stage is effectively driving a 4 ohm load, and hypothetically the middle of the voice coil winding is a virtual ground)..... hmm maybe that's too much for you to absorb.... but i'll leave it for you to ponder anyway...
 
No, the output can't exceed the supply (and in practice will be at least a couple of volts short of it), if it tries to exceed that the amp will simply clip it.

good sir, what do you mean by clipping? can you show me the formula for the voltage out if it has a large voltage gain. thanks
 
good sir, what do you mean by clipping? can you show me the formula for the voltage out if it has a large voltage gain. thanks

I've no idea what formula might apply (if any?) - but with a 9V supply it's pretty obvious that the output can't go higher than 9V, or lower than 0V - and in practice it's more like 8V and 1V (at best).

So when the positive peak reaches 8V (or so) it can't increase any more so will continue to output 8V until the level drops lower again. The same applies at the negative side of the signal as well - this effectively 'clips' the signal, turning a sine wave in to a square wave (and is how you generally do such a conversion as well).
 
I've no idea what formula might apply (if any?) - but with a 9V supply it's pretty obvious that the output can't go higher than 9V, or lower than 0V - and in practice it's more like 8V and 1V (at best).

So when the positive peak reaches 8V (or so) it can't increase any more so will continue to output 8V until the level drops lower again. The same applies at the negative side of the signal as well - this effectively 'clips' the signal, turning a sine wave in to a square wave (and is how you generally do such a conversion as well).

thanks anyway sir nigel, you have been very helpful.
 
Hee, hee. Sir Nigel is a Knight!

An amplifier output clips and produces severe distortion when the input level is too high which causes the output to try to go higher than it can.

Here is an example of no clipping, a little bit of clipping, more clipping and pretty bad clipping:
 

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Hee, hee. Sir Nigel is a Knight!

We can be over here :D

Although I don't think my chances are terribly good!.

An amplifier output clips and produces severe distortion when the input level is too high which causes the output to try to go higher than it can.

Here is an example of no clipping, a little bit of clipping, more clipping and pretty bad clipping:

Nice symmetrical clipping AG!.
 
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