Coupling class A amplifier with class AB amplifier

Thread Starter

juanpa

Joined Apr 20, 2019
7
Hi!! First post and I usually don´t write in English, so I hope to do a good job explaining my problem.

I´m trying, for educational purposes, to mount a little audio amplifier, joining a class A amplifier to amplify the voltage of a signal (from a smartphone) and using a class ab amplifier to gain power and move a speaker.

I create a simulation using Falstad.com/circuit simulator (you can see it here)

Questions:
  1. How can I elevate the input impedance of the class ab amplifier?? I think that this is the problem that makes the system doesn´t work at all... As you see in the simulation when I join the two amplifiers the output voltage of the first one falls absolutely...
  2. It would be better include some emitter resistors in the class ab amplifier? I think that it could help me to mantain the stability
  3. How can I calculate the output capacitor? I know that it has to charge during the first part of the signal, but i didn´t find information about how to calculate it.

The initial specifications are:
  • Speaker power: 2Wrms
  • Source voltage: 12V
  • Smartphone output signal: 4Vp (we will use a potentiometer before the first amp to regulate the amount of signal).
  • Smartphone Output impedande: about 400 Ohmns.
Thanks!
 

Attachments

KL7AJ

Joined Nov 4, 2008
2,208
Hi!! First post and I usually don´t write in English, so I hope to do a good job explaining my problem.

I´m trying, for educational purposes, to mount a little audio amplifier, joining a class A amplifier to amplify the voltage of a signal (from a smartphone) and using a class ab amplifier to gain power and move a speaker.

I create a simulation using Falstad.com/circuit simulator (you can see it here)

Questions:
  1. How can I elevate the input impedance of the class ab amplifier?? I think that this is the problem that makes the system doesn´t work at all... As you see in the simulation when I join the two amplifiers the output voltage of the first one falls absolutely...
  2. It would be better include some emitter resistors in the class ab amplifier? I think that it could help me to mantain the stability
  3. How can I calculate the output capacitor? I know that it has to charge during the first part of the signal, but i didn´t find information about how to calculate it.

The initial specifications are:
  • Speaker power: 2Wrms
  • Source voltage: 12V
  • Smartphone output signal: 4Vp (we will use a potentiometer before the first amp to regulate the amount of signal).
  • Smartphone Output impedande: about 400 Ohmns.
Thanks!
Apply negative feedback to the AB stage.
 

MrChips

Joined Oct 2, 2009
19,269
Questions:
(1) How can I elevate the input impedance of the class ab amplifier?? I think that this is the problem that makes the system doesn´t work at all... As you see in the simulation when I join the two amplifiers the output voltage of the first one falls absolutely...
88Ω for bias resistors is way too low for this application. Increase these. Remove the two 10μF coupling capacitors and replace them with a single transistor driver stage. You are missing the driver stage. What you need is a preamp stage, driver stage, and output stage.
(2) It would be better include some emitter resistors in the class ab amplifier? I think that it could help me to maintain the stability.
Yes, emitter resistors will provide negative feedback and thermal stability.
(3)How can I calculate the output capacitor? I know that it has to charge during the first part of the signal, but i didn´t find information about how to calculate it.
The value of the output capacitor will determine the output's low frequency response. Treat this as a first order high pass filter.
 

MisterBill2

Joined Jan 23, 2018
3,845
In addition to the suggestions in post #3, I suggest adding a capacitor across the emitter resistor of the first stage. That will allow it to deliver a bit more power. AND, of course the output of the class "A" stage will drop when it is required to deliver some power. Even with the best impedance matching the output voltage will drop because power is being transferred.
 

Thread Starter

juanpa

Joined Apr 20, 2019
7
Thanks to all. I will try all your sugestions... and I will publish the results
 

crutschow

Joined Mar 14, 2008
23,291
I suggest adding a capacitor across the emitter resistor of the first stage.
That will, of course, significantly increase the first stage distortion since it eliminates the local negative feedback, so there's a tradeoff between gain and distortion.
To get a gain increase and still have some negative feedback, you can use two emitter resistors in series and only bypass one of them.
 

Thread Starter

juanpa

Joined Apr 20, 2019
7
Hi again!

I applied some of the advices, but I can´t obtain a working circuit yet...

As you can see in the schematic I added a drive transistor, I eliminated the two capacitors and I added the two little emitter resistors to the output looking for some temperature stabilization. (THE IMAGE DOESN´T HAVE THE WIRE BETWEN THE COLLECTOR OF Q1 AND THE BASE OF Q2, I have a question about that).

After playing during long time with a simulator to understand the influence of the components I tried to do some calculations:

  • Without input signal I have to have 6V at the output of the AB amplifier (before the capacitor) so I need 6.7V at the base of the NPN transistor Q3 (it has to be in active region, with minimun Ic), so i need 5.3V falling in the first resistor of the diodes branch.
  • So now i need to calculate how much current have to pass by this branch. Looking for tension estability I choose a current 10 times bigger than the maximal current that have to go to the npn Q3 transistor base... How much is it? If you look my specifications (first post) I need a peak current of 0.6A (2Wrms; 8Ohmns), so my max Ib is 0.6/200= 3mA.
  • Then, I need a 30mA current to flow by the diodes branch.
  • So 5.3V falling in a resistor with 30mA of current flowing... I need a 176Ohmns resistor to polarize with the correct voltage the Q3 base.
  • The diodes have enough current to work in good conditions, so we can say that there are 1.4V between them.
  • So we have 12V-5.3-1.4= 5.3V at the base of the PNP Q4 transistor. And the same voltage at the collector of the driver transistor.
  • The driver transistor has to be in active mode all the time, so I have to calculate its quiescent point to be in the middle of the charge line, so i need 5.3V/2 between the collector and the emitter terminals.
  • If driver transistor´s Vce is 2.65 volts then the rest of the voltage has to fall in the emitter resistor. As we know we have a Icq of 30mA, that is, a Ieq of 30mA... so we can calculate the Re; Re = 2.65/30mA = 86 Ohmns.
  • If I have 2.65volts at the emitter and the driver transistor is in active region then the Vb has to be at least at 2.65+0.7= 3.35volts.
  • The DC voltage of the driver transistor´s base is given by the collector resistor of the preamplifier stage. But that resistor have another priorities in the design of he class A previous amplifier, so I would put a resistor betwen them to reduce the voltage... but it doesn´t work well. I tried with values about 1K to 10K. I tried to add a negative feedback as the suggestion, but I don´t know how to use it.
And by now i have a circuit that doesn´t work. I think that I have something that my teacher usually define as a "GCE" or Giant Concept Error. But I don´t know where is it. I never achieve a good behaviour or enough power in the speaker (8 Ohmns).

I have been looking for references, but I didn´t find a complete example of a design like this, with the maths about the election of the components. I am sure that there is in some place... Any suggestion to look for?

I can´t understand in a intuitive manner how can i adjust the correct quiescent current in the base of the power transistors Q3 and Q4.

I also want to ask about the value of the output coupling capacitor. If I treat it as a component of the high pass filter that forms with the speaker I can use a 220uF capacitor, that give me about 100Hz of cut frecuency. It´s good for me. But, what about of the energy that it has to give to the Q4 transistor when it starts to work? Is there any influence or only in the Voltage limit of the capacitor?

Again, thanks in advance for your suggestions. I hope that i write all the necesary information.

I attach a image with my simulation (I disconnect the input generator and the speaker to see how the polarization works)

upload_2019-4-25_15-9-38.png
 

MisterBill2

Joined Jan 23, 2018
3,845
You have done a good analysis of the circuit that you have, and indeed it shows that the simplest circuits do not always work. That is why actual high signal quality, low distortion amplifiers have quite a few more components and several more active devices. It is also why some simple circuits only "sort of" work.
 

Jony130

Joined Feb 17, 2009
4,967
Your emitter resistors (8Ω) are way too big.

Also try read this
http://www.kennethkuhn.com/students/ee351/power_amplifier1.pdf
And try to read this thread also
https://forum.allaboutcircuits.com/threads/audio-amplifier-design-basic-questions.104719/page-6#post-804580
https://forum.allaboutcircuits.com/threads/class-ab-amplifier-determining-the-real-maximum-output-voltage.156364/#post-1351368
https://forum.allaboutcircuits.com/threads/compensating-diodes-used-on-push-pull-amplifier.101749/#post-766570
https://forum.allaboutcircuits.com/threads/class-ab-push-pull-amplifier-total-current-drain-on-supply.152597/page-2#post-1310639

I also want to ask about the value of the output coupling capacitor. If I treat it as a component of the high pass filter that forms with the speaker I can use a 220uF capacitor, that give me about 100Hz of cut frecuency. It´s good for me. But, what about of the energy that it has to give to the Q4 transistor when it starts to work? Is there any influence or only in the Voltage limit of the capacitor?
No, you do not have to worry about it because the output capacitor is in series with RL. And 100Hz is quiet hight for a audio circuit.
 

Thread Starter

juanpa

Joined Apr 20, 2019
7
For now I have only do a general and quickly lecture, but tomorrow I read all the links. Specially the first one...

Thanks a lot to all, I promise to publish here my results.
 

Thread Starter

juanpa

Joined Apr 20, 2019
7
Hi to all again.

I have a new circuit. Using the documents that you showed I can do a circuit that works, but it needs a better adjustments...

With the circuit attached I have a good output signal, about 1W. It has distorsion, but it doesn´t mind me for now. The circuit is in "DC mode", with the input disconnected.
class a class b amplifier too much current.JPG

The problem is that I have too much current in the class A amplifier, with 108mA in the collector branch... and it would be a current amplifier, not a voltage amplifier as I need.
After trying a lot of things I putted those resistors in the base of the driver to put a DC point that make the driver works good.

I calculated the input impedance seen in the base of the driver:
  • Dinamyc resistor = 25mV/75.6mA = 0.343 Ohmns
  • The impedance of the emitter branch of the driver is 67.09 ohms
  • With hfe=40 the input impedance would be (40+1)*(0.33+67.09)=2764.7 Ohms
  • With the 1K and 510 resistors I obtain a Zin of 300 Ohms aprox
I think that I have to create a previous stage with ten times this input impedance. In a class A amplifier this would match with the Rc resistor, so I use a 3300 resistor... but the driver stage needs much more current. So I was trying with values and calculus and the 47Ohmns resistor was the unique value that worked good.

I think I need another stage, so I tried to calculate it, but I don´t find the way. I tried to put a NPN transistor with hfe of 200 to control the current of the driver stage, but I didn´t find the correct polarization network. Any suggestion??

I don´t put all the maths of my design for simplicity, but I followed the document of Kenneth Kuhn referenced above. It´s a closed loop gain amplifier, but the AB calculus give my the neccesary information.

Thanks in advance!!
 

Jony130

Joined Feb 17, 2009
4,967
If you are using Qdriver as a PNP transistor. Use it as a CE amplifier. So try to switch 82R resistor with the Qdriver.
And recalculate the circuit.
And why did you decide to use 108mA in the first stage?
 

Thread Starter

juanpa

Joined Apr 20, 2019
7
Only with that current I could do the circuit worked... I didn´t decide it at all ;D

Now I´m trying to put a NPN transistor in the driver stage, but I have some problems too. I find without problems the way to polarice it with a resistor from Vcc to the base, but when I try to put a signal it starts to saturate.

For my previous calculus the current by the diodes branch have to be 75.6mA. If I use a hfe 40 transistor then I would have a 75.6/40=1.89mA base current in my NPN driver transistor.

Now I use the 0.65V that falls between the base and the emitter, so I have a (12Vcc-0.65V)/1.89mA=6005.2Ohmns resistance --> 5K6... As I put in the schematic attached.

But If there´s only 0.65V at the base of the driver, how much voltage could vary without put the driver in cutoff mode? I am trying to calculate a class A amplifier to adapt to this situation, but for now I fall again...

Thanks!
 

Attachments

MisterBill2

Joined Jan 23, 2018
3,845
For the class "A" amplifier the resistors for emitter and collector are far to low. try 470 ohms and 100 ohms, those numbers are more reasonable.
 

Jony130

Joined Feb 17, 2009
4,967
But If there´s only 0.65V at the base of the driver, how much voltage could vary without put the driver in cutoff mode? I am trying to calculate a class A amplifier to adapt to this situation, but for now I fall again..
This is why you should use a global negative and bootstrap the 82R resistor. But it seems that you are afraid of using the global negative feedback.
Also, the negative feedback will fix your problem with the driver transistor biasing.

http://www.learnabout-electronics.org/Amplifiers/amplifiers55.php (Bootstrapping.)

Because as you probably notice the CE amplifier do not like to drive high current demanding loads and at the same time provide a large voltage swing.
For a 1W of power at the 8Ω load you need a peak voltage equal to :

And what voltage swing did you need?
Well, let us see Vpeak =√(2*1W*8Ω) = 4V which gives us 8V peak to peak. So you only left with 4V headroom from the 12V supply, not a lot.
This is why it is not very wise to use a 1-ohm emitter resistor at the output stage. This resistor is in series with RL resistor so you are loosing there around of 0.7V.
 
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