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class of amplifier


prateeksikka
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hi all!
Does anyone have the idea about class of an amplifier.i know that there are class A,class B,class C amplifiers depending on the region of their operation.But i wish to know that is it upto the user to operate the transistor in any region or do we purchase it by the name of say "class A amplifier purchase"
prateek

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Opamps usually operate their output transistors in class-AB for low distortion. If you look at the spec's of a low-power opamp like the LM324 or LM358, you will see that they operate their output transistors with a very low idle current in class-B, which causes crossover distortion. Their datasheet shows how to add a resistor to the output to bias one of the output transistors in class-A for low distortion.

The design of the circuit, not the market and not the operator determines the class of operation of the transistors.

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prateeksikka,
To answer your question;
Class A the transistor conducts for the full cycle drawing a lot of current the efficiency <30%, but it's low noise and a good quality amplifier.

Class B the transistors are only on when a signal is present and draw no power when there's no signal, this is high noise because the transistor needs 0.7 to turn on crossover distortion occurs when the waveform goes near 0.7V but it's more efficient and it runs cool.

Class AB is a happy medium,  the transistors are biased on with a very small current to keep their bases above 0.7V and get rid of any crossover distortion and keep the efficiency reasonable by only using a small current.

Class C is used only in RF amplifiers that don't rely on the amplitude of the signal to transmit information for example CW FM, FSK and PFM, its also used in some inverters. The transistor is only on for half the cycle and it's  either switched on or off, capacitors and inductors in the output filter the harmonics and produce a sine wave, it's fairly noisy but can be 90% efficient.

Class D is fairly new and its used in high efficiency audio amplifiers. Like class C the transistor is only switched on and off but this is done at a high frequency and if PWM with the input signal. For example an audio amplifier with a bandwidth of 20kHz will require a switching frequency of >100kHz this is then connected to the load via a low pass filter made from inductors and capacitors. The result is a high efficiency audio amplifier but the main disadvantages are noise (some of it RF) and a complex circuit is required. This is improving all the time and you can now buy specialist IC for this purpose.

My question:
How do I calculate the value of Rbias in this class AB amplifier?

I thought it would be self biasing but the DC output level is far too low and this severely reduces the output drive capability.

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Hi Alun,
Ibias needs to have a low voltage drop for it to turn-on Tr3 fully when the output swings high. Frequently a bootstrap arrangement drives Ibias higher than the supply voltage (with a capacitor from the output) for a good output swing.

Maybe Ibias has too much voltage drop or maybe you can't adjust Rbias for a half-supply voltage at the output because Tr2 leaks and needs a base-emitter resistor.  ???

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Thanks audioguru,

I've tried a BJT with a zener for Ibias but is wasn't very good so I used a JUGFET with the gate joined to the source as a current source and it worked. I've managed now to get it to work quite well I got 6Vp-p from a 9V supply.

The problem is I had to mess around adjusting the value of Rbais to get it to work and when I change the gain I need to mess around with adjusting Rbais. I want to know if there is an easy way to calculate its value using the other resistor values and transistor betas?

Oh by the way here's a copy of the schematic with the values I used.

The transistors are:
NPN: BC546
PNP: BC560

post-0-14279142202908_thumb.png

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Does anyone have the idea about class of an amplifier.i know that there are class A,class B,class C amplifiers depending on the region of their operation.

Here you go:

  • Asymmetrical
       

    •          
    • Class A1: Duty Cycle (DC)= 360 deg         
    • Class A2: 360 > DC >= 180 deg   

         
Class AB1: DC= 360 deg (+/-)         
Class AB2:  360 > DC > 180 deg (+/-)         
Class B: DC= 180 deg (+/-)   


[*]Balanced    [*]Class C: DC < 180 deg (Note: No distinction between balanced and asymmetrical) [*]Class D: DC= 180 deg (+/-) Active devices driven between saturation and cutoff. Raw output is symmetrical square wave. [*]Class E: DC= 180 deg Single ended. Active device driven between saturation and cutoff.  Includes parallel dump capacitor to drive load. Raw output is square wave. [*]Class F: DC= 180 deg Uses quarter wavelength T-Line (at fundamental) with parallel LC tuned circuit to synthesize a square wave output.

That pretty much covers it. The main difference is that classes A -- C use the active device as a voltage controlled resistor. In these cases, the active device dissipates quite a bit of heat (as does any resistor). Classes D -- F use the active devices as switches, so that conditions of max current coincide with minimum voltage across the device, and max voltage occurs when the device is drawing leakage current only. This keeps dissipation low. These classes all produce square waves as raw output, not sine waves or parts of sine waves.
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Hi Alun,
You are severely straining your poor little output transistors that have an absolute max current rating of only 100mA.
To get 6Vp-p into your 8 ohm load, each output transistor will have a peak collector (emitter) current of 375mA. You are lucky they survived and their hFE drops like a rock above about 50mA.

Our 2W Mini-box amp project is a good example of bootstrapping:
http://www.electronics-lab.com/projects/audio/003/index.html
It uses the speaker as a resistor in a voltage divider with R8. C7 supplies the swinging output voltage to the upper end of R8 and raises the voltage to well above the supply voltage on positive swings so that Q3 has lots of continuous base current. Since both ends of R8 have about the same signal level, the current is constant (very high impedance) and therefore Q2 has lots of gain. Q2 can saturate to a much lower voltage than your darlington. A resistor can replace the speaker in this design making R8 half its existing value and the new resistor half as well. Then the speaker can be cap-coupled to the output as normal without the current of Q2 in it.

Even the input of the amplifier is bootstrapped with C4 for a very high input impedance.
The project will probably oscillate without C5, your amp doesn't because the darlington has its transistors' capacitance multiplied.

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Hi Miles,
Here's another class to add to your collection.  ;D Class-H.
It switches the power supply voltage of an amp between a low voltage for good efficiency at low power output, and a higher supply voltage during loud peaks in the output.
This Philips car amp IC even has a built-in voltage doubler:
http://www.semiconductors.philips.com/pip/TDA1562.html

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Hi Alun,
You are severely straining your poor little output transistors that have an absolute max current rating of only 100mA.
To get 6Vp-p into your 8 ohm load, each output transistor will have a peak collector (emitter) current of 375mA. You are lucky they survived and their hFE drops like a rock above about 50mA.


Yes the transistors did get quite warm, I'll replace them with more chuncky ones.
The open loop gain probably stayed quite high at the high output current because the darlington gives the circuit it's very high gain (>20,000) not the output transistors that carry the load current.

I think I've solved my problem of claculating Rbais.
If the we want the DC output voltage to be half the supply and the voltage across Rbias will always be the Rb (voltage of the darlington) then we have a potential divider consisting of Rf and Rbias.

The value given from the formula is not too different from the value I used, the differences are probebly due to the voltage drop of Ibias and the current taken by the darlington. I might try one of those proper constant current diodes and see if it's any better.

What do you think about removing Re and adding a 22ohm resistor between D1 and D2 to keep the baising the same and reduce the output impedance furter?

post-0-14279142204137_thumb.gif

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Our 2W Mini-box amp project is a good example of bootstrapping:
http://www.electronics-lab.com/projects/audio/003/index.html
It uses the speaker as a resistor in a voltage divider with R8. C7 supplies the swinging output voltage to the upper end of R8 and raises the voltage to well above the supply voltage on positive swings so that Q3 has lots of continuous base current. Since both ends of R8 have about the same signal level, the current is constant (very high impedance) and therefore Q2 has lots of gain. Q2 can saturate to a much lower voltage than your darlington. A resistor can replace the speaker in this design making R8 half its existing value and the new resistor half as well. Then the speaker can be cap-coupled to the output as normal without the current of Q2 in it.

That's an interesting circuit, I've noticed it doesn't have any emmiter resistors, is this to reduce the output impedance?

What do you think the open loop gain is?
My circuit is more of an experiment than anything I was seeing how negative feedback increases the bandwidth and reduces nois and distortion. My objective was to build an amplifier with as higher gain as possible and then reduce it to a fairly low value and compare test results with open and closed loop.


Even the input of the amplifier is bootstrapped with C4 for a very high input impedance.
The project will probably oscillate without C5, your amp doesn't because the darlington has its transistors' capacitance multiplied.


I think my amplifier will still have quite a high bandwidth, and I am concerned about oscilation if I use better transistors, a small capacitor in the feedback loop should do the job.


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Hi Alun,


Yes the transistors did get quite warm, I'll replace them with more chuncky ones.

Good, but chunky transistors have less gain so Ibias will need to be higher.

I think I've solved my problem of claculating Rbais.

Yes, Rbias is in a voltage divider with Rf.

I might try one of those proper constant current diodes and see if it's any better.

Have you seen the high cost of those things? All they are is a FET with a gate-source short like you have now. Try a couple of bootstrap resistors with a cap.

What do you think about removing Re and adding a 22ohm resistor between D1 and D2 to keep the biasing the same and reduce the output impedance further?

Guaranteed thermal runaway of the output transistors! The negative feedback already makes the amp's output impedance extremely low. If you load the output and its voltage swing tries to drop, the negative feedback tells the open-loop gain, "Oh yeah? C'mon gain, help out!"  ;D
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That's an interesting circuit, I've noticed it doesn't have any emmiter resistors, is this to reduce the output impedance?

I think all circuits like that should have emitter resistors to avoid thermal runaway. This one gets away with it because of its low output power.

What do you think the open loop gain is?
My circuit is more of an experiment than anything. I was seeing how negative feedback increases the bandwidth and reduces noise and distortion. My objective was to build an amplifier with as high a gain as possible and then reduce it to a fairly low value and compare test results with open and closed loop.

Its gain isn't nearly as high as an opamp as seen by its fairly high distortion. An opamp has a fairly high open-loop noise and a distortion of maybe 5%. Its open-loop bandwidth is only 10Hz. With negative feedback reducing the gain from a few million down to 20, the bandwidth extends and the noise and distortion reduce by about the same amount. At a gain of 10, the OPA134 opamp has distortion so low you can't measure it at 0.00008%! It can amplify the RF of an AM radio station! Add a complimentary pair to the output and inside the negative feedback loop of one of those babies and you'll have a very good amp.
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Well if I use BC546Cs for the darlington the gain would be a whopping 176,400 to 640,000, which beats the uA741's 20,000 to 200,000, but it will be never a good as the OPA134's gain of 1000,000. ;D

How does making a darlington pair affect the bandwidth?

The BC546C has an Ft of 300MHz, would this stay the same?

If this is the case then the circuit might have an open-loop gain of say 200,000 at 1Hz and 1 at 300MHz.

About your idea to use an op-amp, are you sure it will be able to provide enough current to drive the complementary pair at large output swings?

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Hi Alun,
An Englishman named John Linsey Hood developed a "Liniac" small amp in the 70's. It used a darlington and current source for extremely high gain and very low distortion with negative feedback. He reported that the darlington multiplied its own small capacitance and it created a low frequency pole when it combined with the high impedance load. Therefore its frequency response was limited and he used it in single-frequency oscillators.
You might be able to find the article in a library archive of Wireless World magazine.

Your darlington has a current gain very high but its voltage gain won't be as much. With its high impedance load its response will rolloff at a fairly low frequency.

The Ft of a transistor is the frequency that its voltage gain is only one, like a piece of wire. So in theory, a 300MHz transistor has a gain of 10,000 at 30kHz. In practice, you'd be lucky to get a gain of 50, a darlington might have a voltage gain of 2500 at 6kHz.

Using an opamp for gain, you would need voltage-wasting darlingtons or a compound PNP-NPN complimentary pair as output current amplifiers for about 50W into 4 ohms or 30W into 8 ohms. For just a few watts and a 12V battery, a single pair of BD135/BD136 would be fine.  ;D 

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Hi Alun,
An Englishman named John Linsey Hood developed a "Liniac" small amp in the 70's. It used a darlington and current source for extremely high gain and very low distortion with negative feedback. He reported that the darlington multiplied its own small capacitance and it created a low frequency pole when it combined with the high impedance load. Therefore its frequency response was limited and he used it in single-frequency oscillators.
You might be able to find the article in a library archive of Wireless World magazine.


Is there any chance it looked similar to my circuit? ;D


Your darlington has a current gain very high but its voltage gain won't be as much. With its high impedance load its response will rolloff at a fairly low frequency.


Is this because the parasitic capacitences are multiplyed too?

But hang on a second, if the both the gains and capacitences are multiplyed then won't this cancel out?

If an amplifier has a 6bB roll of caused by a capacitor, and we multiply its gain by 100 and the capacitor's value by 100 then the unity gain bandwidth should remain the same. This makes sense to me as the MPSA14 has an Ft 0f 125MHz.


The Ft of a transistor is the frequency that its voltage gain is only one, like a piece of wire. So in theory, a 300MHz transistor has a gain of 10,000 at 30kHz. In practice, you'd be lucky to get a gain of 50, a darlington might have a voltage gain of 2500 at 6kHz.


??? you must be mistaken, I've never seen a single transistor with such a high gain even at DC, unless you're talking about a darlington of course.


Using an opamp for gain, you would need voltage-wasting darlingtons or a compound PNP-NPN complimentary pair as output current amplifiers for about 50W into 4 ohms or 30W into 8 ohms. For just a few watts and a 12V battery, a single pair of BD135/BD136 would be fine.  ;D 



What do you think of these circuits?

I don't like the first as I doubt the op-amp will have a high enough gain to get rid of the crossover distortion under load, I prefer the second as there shouldn't be any crossover distortion in the first place.

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Hi Alun,
In 1980, telephone interconnect came to Canada and the sound systems company I worked for decided to sell and install phone systems. Instead of selling the same ones as other companies they decided on a phone system from a little-known Korean company called Goldstar (now called LG Electronics). It had the worst pcbs and soldering that I have ever seen! The phones were all speakerphones and had the same amp circuit as your 1st one, except with much more gain (and therefore much less negative feedback).
They had terrible crossover distortion! I had to go around and add a resistor to the circuit to let the opamp drive the load for low signals but only the ones customers complained about. I would modify a few in one company and they called back the next week requesting the rest also be done. Whew, lotsa work! :o

I have modified your 1st circuit including the resistor at the output transistors. I also changed the circuit around so it has more gain for use with a CD player or tuner. More gain reduces its tendency to oscillate. I didn't use inverting like you did so that its input impedance wouldn't be too low or its neg feedback resistor be too high (phase shift).

Your 2nd circuit has the opamp driving the output transistors and their 220 ohm biasing resistors. An opamp doesn't have enough output current to do both.  ;D

post-1706-14279142204841_thumb.gif

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Nice circuit audioguru.

How did you calculate the values you used?

What do the extra capacitors do?

That 68ohm resistor is connected accros from the input to the output of the complementary pair, as it's non-inverting this will give posative feedback could this lead to oscilation or even latchup?

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Hi Alun,
My circuit has proper negative feedback, the transistors don't invert, they are just followers.

I very carefully matched the 39k feedback resistor to the 33k in series with the two paralleled 10k resistors so any high input current opamp can be used without offset voltage.

I also carefully matched the 20Hz rolloff of your 1000uF output cap into the 8 ohm load with the 1.5k resistor and 4.7uF cap, and the 0.22uF input cap with the 33k reference resistor. There is a trick about matching the 3 capacitors so that the amp produces a perfect square wave without tilt at low frequencies. Now I think the 4.7uF should be about 2.2uF for that to occur. Try it, changing the cap changes the tilt (overall phase shift).

Oh, oh! Both you and I forgot to include a supply bypass cap.  ;D

The 47uF cap should fully-charge (98%) by the two 10k resistors in about 1.2 seconds for good filtering.

The gain is 1 plus (39k divided by 1.5k) so a 100mV input gives a 2.7V output (about 7.6Vp-p).

The opamp drives the load through the 68 ohm resistor until 10.3mA flowsand the resulting 0.7V turns on the transistors.

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So the 68ohm resistor is to get rid of the cross-over distortion my concern was that because the complementry pair has a gain of 1 the resistor would give positive feedback and lead to latch or oscillation why won't this happen?

Yes it does need a capacitor across the supply, 100uf should do.

Thanks for explaining it so well, I'll build this when I get the time.

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Hi Alun,
The complimentary emitter-followers won't oscillate due to positive feedback into them through the 68 ohm resistor:
1) An emitter-follower isn't perfect so will have a max voltage gain of maybe 0.95 with a low resistance load. It is less than 1.0 so it can't oscillate.
2) The open-loop output impedance of the opamp is between 10 ohms and 100 ohms depending on which opamp. Its output impedance forms a voltage divider with the 68 ohm resistor further reducing positive voltage feedback through the 68 ohm resistor.

I haven't tried it for many years but I remember my little amplifier performing very well:
1) The output transistors can't have thermal runaway since the 68 ohm resistor bypasses any base leakage current.
2) Very high output voltage swing since the transistors don't have emitter resistors.
3) Very low crossover distortion since the opamp drives the load until the current allows the output transistors to be fairly linear.
4) Extremely low idle current since the output transistors are turned off.  ;D

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