Simple push- pull amplifier question

[samy555]

Hi
The following scheme is for real simple push- pull amplifier used with desk PC, I intend to understand.

(1) Why R2 & R3 are so small?? What if I remove them?
(2) The circuit I have been without a power supply and I guessed that the value of it = 12 V. Is my guess was right?
(3) With R4 = 100K the voltage at the collector of Q3 = 5.3V, but when I increased it to say, 330k, the voltage increased to 8.1V. why?

 

[betwixt]

(1) R2 and R3 serve two purposes, they limit the current slightly but more importantly they hep to stabilize the operating point of the output transistors.
(2) 12V is about right, it isn't really critical in that application.
(3) Are you sure R4 is connected like that. It is more normal to bias the input stage from the output voltage so it can stabilize the output voltage.

You should be aware that this is not a good circuit, it is built for lowest cost, not for quality. It may have quite high distortion and certainly will have poor frequency response. It is only capable of producing maybe 300mW from a 12V supply. Ignore anything the manaufacturer writes on it about power ratings, they measure power in special Chinese Watts which are always much bigger than everybody elses Watts. :smile:

Brian.

 

[chuckey]

1. R2 and R3 are small because they are a compromise between power output and thermal stability. The reason they are there is because as the transistor takes more current, it warms up and its leakage current increases, this causes it to get warmer, the leakage current, increases and the transistor could "burn out". The effect of the increased voltage drop across R2 and R3, reduces the bias between the base and the emitter, so cuts the collector current back to a safe value. The higher their value the more thermally stable the amplifier is.

2. Yes

3. R4 supplies a bias current to the base of Q3, if this is reduced, the collector current is reduced, so the voltage drop across R1 + D1 + D2 is reduced, so the collector voltage ( V1 - Volt drop across R1..D2) rises. If the current is zero then the collector of Q3 will be V1.

In general for maximum power output and minimum distortion, the voltage at the junction of R2 and R3 should be V1/2, so 12 V is about right.
Frank

 

[samy555]

Thank you Brian for fast reply

(3) Are you sure R4 is connected like that. It is more normal to bias the input stage from the output voltage so it can stabilize the output voltage.

Yes I'm sure
To Frank and Brian
In fact, I search the internet for a way to design like this circuit but I did not find
Can you help me with some points that I can start with them in order to design such a circuit
Or even give me some links that will help me in that
Thank you alot

 

[betwixt]

This is a good starting point: https://www.circuitstoday.com/250mw-audio-amplifier

The circuit is very similar to yours but note that the first transistor gets it's bias from the output stage. For 12V I would re-introduce the emitter resistors (R2/R3) and increase the output capacitor to 220uF or even 470uF to get better frequency response. The biasing arangement ensures the voltage between R2 and R3 is stabilized, it it tries to rise, the extra base current in the first transistor pulls it down again. If it drops, the bias reduces and it rises again.

If you want to try simple improvements, these are the first steps:
1. It's a class AB amplifier which means the output transistors pass a small residual current even when no signal is present. At the moment there is nothing to stabilize the current and it will vary according to the transistor and diode characteristics. I would suggest you add a third diode in series and wire a 1K variable resistor across all three. Also bond the diodes to the body of the output transistors to give it heat stabilization. Start with the variable resistor at minimum resistance and adjust it with no signal present so that the amplifier draws 10mA. It isn't an ideal solution but it will give lower distortion and also make it more tolerant of temperature variations.
2. At the moment the output signal is present on the voltage being used to bias the input stage and for the reasons I already mentioned, it works against changes in voltage. This is fine for DC, it is negative feedback and just what's needed to keep the DC voltages correct. However, it also feeds back some of the audio and hence reduces the gain. Try splitting the bias resistor in two and connecting a capacitor at the new junction down to ground. The capacitor will make the feedback less responsive to the signal and the gain should increase.
3. The design is capable of far more output power but is limited by the low current transistors (IC = 200mA max), you could try small power transistors instead, even with no other changes you should be able to safely push the output up to 2 - 3 Watts. You will need some lind of heat sink to keep the transistors cool and remember to bond the diodes to the same heat sink to keep the temperature stabilization working.

Good luck!

Brian.

 

[FvM]

The circuit is very similar to yours but note that the first transistor gets it's bias from the output stage.

The poor bias circuit in post #1 is strongly affected by transistor current gain variations and is only suitable for an output amplifier, if the resistor value is individually selected for each transistor exemplar. The circuit won't seriously be improved by connecting R4 to the output node.

The circuitstoday amplifier implements a voltage divider bias and is somewhat better. The design focus is still on minimal part count rather than stable behaviour.

 

[Audioguru]

The poor little output transistors will be destroyed in that circuit.
For an output that is 9V p-p the peak current in each output transistor is 4.5V/8 ohms= 563mA. But the maximum allowed current in those little transistors is only 200mA and they work poorly above 100mA.

I changed the output transistors then simulated it. It has no negative feedback so its distortion is extremely high but its voltage gain is also very high.

I added negative feedback, an input resistor and added bootstrapping. Now the distortion is fairly low, the gain is normal and the output level is higher.

 

[samy555]

Thank you Brian

Start with the variable resistor at minimum resistance and adjust it with no signal present so that the amplifier draws 10mA. It isn't an ideal solution but it will give lower distortion and also make it more tolerant of temperature variations.

Why 10mA? what if I need the output power to be 250mW?

Try splitting the bias resistor in two and connecting a capacitor at the new junction down to ground. The capacitor will make the feedback less responsive to the signal and the gain should increase.

Which resistor?

3. The design is capable of far more output power but is limited by the low current transistors (IC = 200mA max), you could try small power transistors instead, even with no other changes you should be able to safely push the output up to 2 - 3 Watts.

Yes, you're right
I am surprised that this amp. produces only about 1.5 milliwatts (mW) in the simulation, while in fact it works well and I can guess that it produces more than 200 milliwatts
My question is:
How much should be the collector current of Q1 to have an output power of 250 mW?
thank you very much

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The poor bias circuit in post #1 is strongly affected by transistor current gain variations and is only suitable for an output amplifier, if the resistor value is individually selected for each transistor exemplar. The circuit won't seriously be improved by connecting R4 to the output node.

The circuitstoday amplifier implements a voltage divider bias and is somewhat better. The design focus is still on minimal part count rather than stable behaviour.

Yes, the Chinese like to reduce the cost, and this leads to poor efficiency, but remain easy designs and accessible to novices (Like me) and that's what I like about them.
Hey people, I want expressions, rules and numbers
I read dozens of pages of structural speech. I know about class A B AB and even C, I also know abowt crossover distortion etc.

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Thank you GURU

The poor little output transistors will be destroyed in that circuit.
For an output that is 9V p-p the peak current in each output transistor is 4.5V/8 ohms= 563mA. But the maximum allowed current in those little transistors is only 200mA and they work poorly above 100mA.

yes you are right. The actual transistors are: Q1=S8050, Q2= S8550 and Q3 = C945, but no one of them is in Multisim.10 library.

I changed the output transistors then simulated it. It has no negative feedback so its distortion is extremely high but its voltage gain is also very high.

I added negative feedback, an input resistor and added bootstrapping. Now the distortion is fairly low, the gain is normal and the output level is higher.

Let me guess:
(1) The negative feedback by connecting R1 to the o/p instead of C of Q1 (in your schematic), this reduces the gain but have many advantages.
(2) Bootstrapping through C2
(3) an input resistor is R5 (=12K)
Although I know you will not answer my questions easily
I will not get the answers at least a month
All of this because you are trying to make me learn, and I thank you for that
But I will try to conclude from your announcements and the required answers quickly (a week!!!)
Input resistor reduces the i/p current. If i/p current is 10mA and the amplifier amplifing it 100 time , we get 1 A. With your i/p resistor, you reduced the current gain, so what is the benefit of using R5?

Bootstrapping: I read somewhere that Bootstrapping increase the Zin of the amplifier.
You split R2 (=680 ohm) into two 330 ohm and connect C2 between the output and the split point. Is there a rule for that connection? Why 100uF? And why at the mid point of the 680 ohm?
I promise you I'm going now to read more about Bootstrapping, but until then, I need some help.
Thank you alot

 

[Audioguru]

The 10mA is DC idle current in the output transistors to avoid crossover distortion. It has nothing to do with output power.

The value of the input capacitor in your simulation is WAY TOO LOW! The simulation assumes a source impedance that is zero ohms. Since the input of your simple amplifier is a low impedance then 100nF cuts all frequencies below about 12kHz.
Using 10uF (100 times your value) then in a simulation response down to about 40Hz is good. Your source might actually be 10k ohms or more which reduces the gain and distortion of your amplifier and it reduces the low frequency cutoff. Try your simulation again with a 10k resistor in series with its input capacitor.
My modification and the other amplifier posted have real series input resistors and much higher value input capacitors.

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I did not lookup your Oriental transistors.

Without the series input resistor then negative feedback of the signal through R1 is shorted by the zero impedance of the simulator's signal source. Adding a series input resistor allows negative feedback of the signal which reduces the voltage gain and reduces the distortion of the amplifier. It also produces a much lower low frequency cutoff since the input impedance with the series resistor is much higher.

I guessed 100uF is good for the bootstrap capacitor. Calculating 10Hz and 330/2 ohms gives a capacitor that is 97uF so I was very close.
The bootstrapping by C2 is not at the input so it does not increase the input impedance.

 

[samy555]

thank you guru

Well,,, let me read this answer and delve where long
Let me feel a headache while trying to understand every word

I have "The Art of Electronics (Paul Horowitz_ Winfield Hill)"
Paul has the same way you have
Speaking in general, and makes you try to infer information on your own

On page 97 he talked about Bootstrapping collector load resistors, And gives a figure similar to a large degree to the one you introduced in participate #7
I need to read it carefully then back to discussion
thanks alot

 

[Audioguru]

The bootstrapping circuit shown on page 97 in The Art of electronics is exactly the same as what I did in my modification of your circuit. The other circuit posted uses the speaker and its coupling capacitor to do the bootstrapping.

The important part of bootstrapping was allowing plenty of base current for the NPN output transistor as the signal voltage rose higher. In your circuit, R1 provided less and less base current as the signal voltage rose higher, which reduced the maximum output level and caused severe distortion.

 

[kam1787]

Why use the LM386N-4 ? https://www.circuitstoday.com/wp-content/uploads/2009/03/lm386.pdf

 

[samy555]

The 10mA is DC idle current in the output transistors to avoid crossover distortion. It has nothing to do with output power.

So what parameter that controls the output power?

The value of the input capacitor in your simulation is WAY TOO LOW! The simulation assumes a source impedance that is zero ohms. Since the input of your simple amplifier is a low impedance then 100nF cuts all frequencies below about 12kHz.
Using 10uF (100 times your value) then in a simulation response down to about 40Hz is good.

Yes, this is a great point
When I changed the input cap to 10uF I got 1.64Vp-p at the 8-Ω speaker, while when it was 100nF, I got 326mVp-p (a big difference).
The input of my simple amplifier is a low impedance, I measured it about 330Ω, so:

Your source might actually be 10k ohms or more which reduces the gain and distortion of your amplifier and it reduces the low frequency cutoff. Try your simulation again with a 10k resistor in series with its input capacitor.
My modification and the other amplifier posted have real series input resistors and much higher value input capacitors.

My signal source is a mic connected to the 12V power supply through 39KΩ resistor. If the internal impedance of my mic is 5KΩ, then Zo of my source (in reality) = 5K//39K = 4.4KΩ.
When I add a 10 kilo resistor to the i/p of my amp., the situation is very bad, I got only 26mVpp at the output (the input sig is 10mVp).

Without the series input resistor then negative feedback of the signal through R1 is shorted by the zero impedance of the simulator's signal source. Adding a series input resistor allows negative feedback of the signal which reduces the voltage gain and reduces the distortion of the amplifier. It also produces a much lower low frequency cutoff since the input impedance with the series resistor is much higher.

Good point. In the real Chinese circuit, they use 2.2KΩ in series with 50k potentiometer (act as a volume), now I can understand why 2.2K.
But I do not agree with that this is negative feedback, I think it is +ve FB, cause the o/p signal is inphase with input.

I guessed 100uF is good for the bootstrap capacitor. Calculating 10Hz and 330/2 ohms gives a capacitor that is 97uF so I was very close.
The bootstrapping by C2 is not at the input so it does not increase the input impedance.

Yes it is 97uF, but do C2 sees the two 330 ohm resistors in parallel?

The other circuit posted uses the speaker and its coupling capacitor to do the bootstrapping.

Thank you, this is another good point. I was not to be aware of it alone.

I have two important questions remained and I hope to get a straight answer them:
(1) In the figure below, and from the view of DC biasing: Did you say that is a good design if VA = VB = 0.5 VCC?
(2) What factor that mainly controls the output power?

Thank you guru very much, I've enjoyed the discussion with you.

 

[Audioguru]

So what parameter that controls the output power?

Power= RMS voltage swing x current. Increase the power by increasing the power supply voltage or bridging two amplifiers. increase the power by increasing the current by reducing the impedance of the speaker. Most car speakers are 4 ohms so the current is twice as high as an 8 ohm speaker. The speakers in my car are 2 ohms and the amplifiers are bridged so the power is fairly high.

When I changed the input cap to 10uF I got 1.64Vp-p at the 8-Ω speaker, while when it was 100nF, I got 326mVp-p (a big difference).
The input of my simple amplifier is a low impedance, I measured it about 330Ω

There is more than one coupling capacitor and they all cut low frequencies. Make the cutoff frequency of each RC at about 10Hz for good bass down to about 40Hz.

My signal source is a mic connected to the 12V power supply through 39KΩ resistor. If the internal impedance of my mic is 5KΩ, then Zo of my source (in reality) = 5K//39K = 4.4KΩ.

The very low input impedance of your amplifier is loading down the output from the mic as a voltage divider. The input impedance of the amplifier should be 5 times to 10 times the source impedance.

But I do not agree with that this is negative feedback, I think it is +ve FB, cause the o/p signal is inphase with input.

Positive feedback makes an oscillator, not an amplifier. The output emitter followers do not change the phase. The collector of the first transistor has the opposite phase to its base so the feedback is negative.

Does C2 sees the two 330 ohm resistors in parallel?

Not really. The top resistor connecting to the positive supply is 330 ohms and the lower resistor connected to the base of the NPN output transistor is in series with the resistance of the base of that transistor. It is best to have the capacitor value too large.

I have two important questions remained and I hope to get a straight answer them:
(1) In the figure below, and from the view of DC biasing: Did you say that is a good design if VA = VB = 0.5 VCC?

No, because the output swing is not symmetrical. It has bootstrapping going positive but not going negative. My amplifier DC output is 6.46V, not 6.0V then my maximum output swing is symmetrical.

(2) What factor that mainly controls the output power?

Voltage swing x current. A bridged amplifier uses two amplifiers with opposite phase. The voltage swing across the speaker is almost doubled then the current is also almost doubled. Then the power is about 3.5 times more than a single amplifier driving the same load impedance.

 

[zorro]

R2 and R3 should be (usually are) PTC (positive temperature coefficient) resitors in order to effectively achieve thermal stability.
Regards

Z

 

[Audioguru]

R2 and R3 should be (usually are) PTC (positive temperature coefficient) resitors in order to effectively achieve thermal stability.

Never.
The diodes should touch the output transistors for thermal stability. When the output transistors get hot they heat the diodes then the forward voltage of the diodes is reduced which reduces the idle current in the transistors which cools them.

 

[betwixt]

Agreed, PTC is exactly the opposite of what is needed. The temperature coeficient of a silicon diode is almost the same as that of the B-E junction in the output transistors so if you make sure they are at the same temperature they will compensate for each other. This is what I explained in post #5.

Brian.

 

[zorro]

Agreed, PTC is exactly the opposite of what is needed.

Do you mean that NTC would be right?

Never.
The diodes should touch the output transistors for thermal stability.

Of course, this is right. But with TO-92 plastic packages (standard 2N3904/3906 are TO-92, variants exist in SOT-223 that would be much better, but they have different part numbers), hardly the junction of the diodes can approach the junction temperature of the transistors. Temperature compensation is only partial.
Stabilization with PTC is partial as well.
I don't know how much is the "part" of "partial" in this case.

Regards

Z

 

[Audioguru]

I do not know if PTC resistors are made with such a low value to be the emitter resistors. In the olden days incandescent light bulbs would have been used. Some amplifiers had an incandescent light bulb in series with the speaker. If the speaker was shorted then the light bulb would light which makes its filament very hot which increases its resistance so the amplifier didn't burn up.

This amplifier needs power transistors, not those tiny ones mounted on a heatsink and the diodes or a biasing transistor also mounted on the heatsink.