# How to design the A-B amplifier? I mean I have understood the working but how do i choose the values?

We have a college project to make an A-B amplifier. I have a basic layout of the transistor amplifier. Plus i have chosen the transistors I want. There are : 2N4401 (NPN) and 2N4403 (PNP).

1)My source is a song being played from my mobile. Here I have taken an AC Voltage of 16pk. 2) How do I modify my circuit for it.I realised Higher the Vcc value I take, the output increases more. Why? And what would be the ideal value? 3)Have I selected the wrong diodes? If I have so which one is suitable for the transistors I have chosen? 4)Why is my positive output waveform increasing with each increasing second?Also why is my Ie of npn decreasing?

EDIT 1: I Have changed the circuit after comparing it more carefully with the suggestions other users have given me. The Schematic is attached below with the output. I noticed other users complaining that my questions have been made blindly. Hence, I have been more precise now. My Input here is a song being played from my mobile/smartphone. I will connect to the amplifier using it's audio jack. Thus, the song will be played on the speaker (here RL). Firstly, when I make it on a breadboard the input here will be the song right? So, will the song even play (w/t distortion). Cause, Here I have taken a sine wave input and according to the simulation it is just increasing the amplitude. Secondly, As pointed out by @jock my knowledge of the amplifier design is limited. I have seen probably all the videos relating to this amplifier on youtube . So, I know it's working but I am really confused about how to design one. Is there anywhere I can read up on this?

• There are many sets of values that will work here. To narrow it down, what is the value of RL and what power are you hoping to get there? So the first step to choosing the values it to know the output specification of the amp. – owg60 Nov 4 '16 at 14:04
• @owg60 hi, I have kept RL as a speaker with 8 ohm resistance and 10W is the output power i want. – shodan Nov 4 '16 at 14:09
• If you mean 10Wrms, (into 8 ohms) you're stuck already, you don't have enough supply voltage. – user_1818839 Nov 4 '16 at 14:21
• @BrianDrummond This is a random pic on the net. However this is my layout. Like i said in my question I am unable to calculate any of the values of RL. Isn't there any kind of procedure I have to follow for finding the values of this Amp. – shodan Nov 4 '16 at 14:31
• @BrianDrummond I have changed the question. – shodan Nov 4 '16 at 18:45

You've said the following that I can understand, regarding the design:

• Accept mobile output. (Which I can only assume means "headphone" jack.)
• Use 2N4401 and 2N4403 for output BJTs. (That's specific enough.)
• Class AB amplifier (using a complementary emitter follower topology, taken as a specification gathered from your schematic.)

You've said the following that I cannot understand, regarding the design:

• "mobile" and "AC Voltage of 16pk" -- "Mobile" usually means "headphone" output, to me. But headphones are typically $32\:\Omega$, mostly resistive, and generally you don't see more than perhaps $1.5\:\textrm{V}_{RMS}$ there. Certainly, I can't recall ever seeing anything like $16\:\textrm{V}_{PEAK}$. Your AC simulation voltage source is set up that way. But I can't for the life of me guess why. Are you expecting some prior stage to provide that kind of signal? Where did this come from, exactly?
1. My source is a song being played from my mobile. Here I have taken an AC Voltage of 16pk.

I don't understand much of this. See above.

1. How do I modify my circuit for it?

By designing it, first.

1. I realised Higher the Vcc value I take, the output increases more. Why?

Outputs often increase when they have access to increased voltage rail magnitudes.

1. And what would be the ideal value?

In practice, it's a matter of compromise.

1. Have I selected the wrong diodes? If I have so which one is suitable for the transistors I have chosen?

Lots of options. One is to use diode connected BJTs. Another is diodes. In either case, you may also want to add a resistor. You could also use a VBE multiplier, with or without Early Effect compensation. In most cases where more power is involved, you may want to thermally couple them. In this case, it perhaps matters not so much.

1. Why is my positive output waveform increasing with each increasing second?

Not sure what you are noticing, even with the new diagram.

1. Also why is my Ie of npn decreasing?

Not sure that curve is present, either.

From the datasheets on the BJTs you've selected, these are TO-92 packaging (which is anywhere from $200\:\tfrac{^\circ \textrm{C}}{\textrm{W}}$ to $325\:\tfrac{^\circ \textrm{C}}{\textrm{W}}$ -- and I'm not talking about SOT-23 here.) This basically means they can't dissipate much. Even with just $100\:\textrm{mW}$, you may see internal temperatures at anywhere from $20\:^\circ \textrm{C}$ to $30\:^\circ \textrm{C}$ rise over ambient. And that's only $100\:\textrm{mW}$.

You will need to reserve about $1.2\:\textrm{V}$ to keep both the NPN and the PNP in their active and not saturated regions. Assuming you were willing to run your BJTs at $50\:^\circ \textrm{C}$ over ambient (about a $\tfrac{1}{5}\:\textrm{W}$ each) you still probably wouldn't get any better than $150\:\textrm{mW}$ into an $8\:\Omega$ speaker when using the $5\:\textrm{V}$ rail that such an output power requires. That's how inefficient this design is likely to be.

Are you prepared for the paltry output that using these BJTs would normally imply?

(Also, as a side question that will rise up later on, how do you expect to handle the stereo output from your mobile? Will you select just one side or the other or do you require summing of the two channels into one?)

• @DeathstrokeTwelved Despite that, perhaps I may yet look forward to seeing an improved question and the chance to fully revise my answer as appropriate. – jonk Nov 5 '16 at 5:16
• I have edited the question. – shodan Nov 5 '16 at 5:20
• @DeathstrokeTwelved Just noticed. I changed my reply and added some questions at the end. – jonk Nov 5 '16 at 6:05
• I will make two circuits one for the left audio signal and one for the right audio signal. Though, could it possible if we could get a mono output from a stereo input? Also, can't I run the transistor at 25°C (room temperature) Won't the power dissipated be more at 25°? – shodan Nov 5 '16 at 8:39
• @DeathstrokeTwelved You can make two separate circuits or you can combine the inputs into one mono output. Different things to do. But either approach can be achieved with effort and thought. You misunderstand me about the temperature. The BJTs will, by themselves, heat up. They will get hot. Sure, run them at room temperature. But their temperature will rise. How much depends on just how hard you work them and how you arrange air circulation. Their internal temp will be a lot higher than their surface temp. Just happens. If you want them cooler, you get bigger BJTs and/or add heat sinks. – jonk Nov 5 '16 at 8:44

How do I go about the calculations?

The procedure is this: -

Decide what you want the output load to be (speaker load impedance value) and what (reasonably) undistorted sinewave power you want delivered to it. From those numbers the power rails can be decided upon and then you have a choice - split rails and no bulky output capacitor or a single rail (and 0V) with a bulky output capacitor.

When you have got this far you have to possibly recognize the limitations in your original choice of output transistors. This may cause you to go back to square 1 and re-state what you want.

Once all this is cleared up, an EE would try to work out how high the push-pull emitter resistor values can be to get the specified full power output. A higher value emitter resistor ultimately reduces output distortion and reduces the possibility of thermal runaway.

Then it's just a case of seeing what the current gains of the output transistor will be (under full load) and choosing R1 and R2 to provide enough bias current through the diodes to get low cross-over distortion. Adding another diode in series with the existing two is not unheard of.

At some point in the process, it is possible, that to achieve your design aims, it will be better to use a darlington configuration for your output transistors. Nobody can advice this yet.

It might also be necessary to use a technique known as boot-strapping in order to maximize signal output level. Nobody can advise this yet.

To start with I propose to feed the transistors not from in between the two biasing diodes thereby preventing signal distortion due to the diode properties. This is only a little change and does not influence on the calculations. The -VCC is not marked in the picture.

First a simple verification of the requirements and the limits of the transistors.

You have a supply voltage of 12 Volt. Consider the input signal as a simple sinewave. The peak value of the sine wave can not be more then 12V. Forget about the little details for the moment.

So the RMS value is more or less 8,5V Apart from the transistor limitations and other details your power limits already at 9W.

Now we look at the transistors.

Look in the datasheet for the max ratings of the transistors. From there you can calculate the max. allowable power and transistor dissipation. Keep a close eye on the limits. Max current is mostly to much your max power is also to be taken into consideration.

From here try first and come back with questions showing what you have done.

Taking a look at the new schematic.

First of all you are making it more difficult for yourself than you should. Keep the output transistors NPN / PNP whith the emitors directing towards the load. Also for now do away with the output capacitor and use a symetric input voltage. Keep the system symetrical. And then look at the simulated output again and find the limitations.

Taking a look at the comments.

Read the comments and respond to them by updating your question. Do not remove history. Show the progress.

• Hi, Thanks for the answer. I have decided to reduce my speaker to a 0.5W and 8ohms. The maximum Power rating of the 2N4401 is 1.5W (datasheet). I have considered Vcc to be around 8V. Are the values I have taken are fine? – shodan Nov 4 '16 at 15:06
• Show what you have done with how and why. Adjust your question and come back. Just asking through comment is not going to work. – Decapod Nov 4 '16 at 15:06
• Decapod, Here is The Link: electronics.stackexchange.com/questions/267591/… – shodan Nov 4 '16 at 18:10
• I changed the question itself.:) – shodan Nov 4 '16 at 18:41
• See my updated answer – Decapod Nov 4 '16 at 18:52

Your scope traces look like you have some issues to deal with. This video on Vbe multipliers will help you Vbe multiplier and class B output. The problem with using diodes to offset the Vbe of the output transistors is that they will never have the transistors and diodes will never have the same current going through them. Although the diode curve is very steep it will has a slight voltage difference with current that you need to take care of.

• Just noticed Q2 has emitter and collector reversed. – owg60 Nov 4 '16 at 23:16