The forward voltage drop of the diodes must match the base-emitter voltage of the outout transistors.
1N4148 little diodes would match 2N3904/2N3906 little transistors but their output power would be very low.
1N400x power diodes would match TIP31/TIP32 medium power transistors.

I have never heard the term "rubber diode" before. We call it a Vbe multiplier transistor.
If it is mounted on the heatsink for the output transistors then it thermally adjusts the current to be constant when temperature changes would make the current much different if it wasn't sensing the temperature.

 

for the two resistors which connect to the base of each transistor, can i choose a 10K ohm resistors?

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people always said that measured the quiescent current. Is it mean that the current draw by the amplifier? Does it measure at the Emitter of the transistor?

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how to calculate the value of the power for the circuit.so that i know what range of 8 ohm speaker i should use.

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from the datasheet below, it shows the Vce(sat) and Vbe(on) value. can someone explain abit on this section? Vce(sat) means the transistor will in saturation mode when it is in 1.2V??



thank you.

 

for the two resistors which connect to the base of each transistor, can i choose a 10K ohm resistors?

You need only one of the resistors. If an NPN voltage gain transistor drives the output transistors low then a resistor to the positive supply will drive them high. The resistor must be calculated to provide enough base current to the output transistors. 10k ohms is far too high.

For example if the supply is +24V and the load is 8 ohms then the transistor emitter current will be 1A if the signal swings to a peak of 8V higher than the +12V quiescent voltage. A medium power transistor like a TIP31 has a minimum hFE of 25 at 1A so its max base current is 40mA or less. The base resistor will have 3.3V across it and 40mA in it so its value must be 82.5 ohms. The resistor can be bootstrapped for increased output and driver transistors can drive the output transistors so that the base resistor can have a much lower current and a much higher value.

people always said that measured the quiescent current. Is it mean that the current draw by the amplifier? Does it measure at the Emitter of the transistor?

The quiescent current of the output transistors (not of the entire amplifier) determines how much crossover distortion is produced. It should be about 25mA to 100mA for a medium power amplifier and can be calculated by measuring the voltage across one emitter resistor.

 

how to calculate the value of the power for the circuit.so that i know what range of 8 ohm speaker i should use.

The supply voltage of an amplifier determines its output power if it is designed correctly. The drive circuit, emitter resistors and output transistor voltage losses must be calculated and subtracted from the supply voltage.
For example, if the supply voltage is +24V, the drive to the output transistors has a peak voltage loss of 4V, the emitter resistors have a peak voltage loss of 0.3V each and the output transistors have a peak voltage loss of 1.5V each then the peak output swing is 12V - 5.8V= 6.26V peak which is 4.4V RMS. The power into 8 ohms is 4.4V squared/8 ohms= 2.4W.

from the datasheet below, it shows the Vce(sat) and Vbe(on) value. can someone explain abit on this section? Vce(sat) means the transistor will in saturation mode when it is in 1.2V??

The transistor has a max Vce saturation voltage loss of 1.2V when its collector current is 3A and its base curremt is 375mA. Its max Vbe is 1.8V when its collector current is 3a and it has enough Vce so that it is not saturated.

 

The base resistor will have 3.3V across it and 40mA in it so its value must be 82.5 ohms.

how do u know it is 3.3V?

The quiescent current of the output transistors (not of the entire amplifier) determines how much crossover distortion is produced. It should be about 25mA to 100mA for a medium power amplifier and can be calculated by measuring the voltage across one emitter resistor.

issit the measuring method is as the figure below? Is it measure with no input signal?


thanks.

 

how do u know it is 3.3V?

I said that if the supply voltage is +24V and the signal swings the emitter of the NPN output transistor 8V peak to +20V and the base-emitter voltage is 0.7V then the base will be at 20.7V and the base resistor will have 3.3v across it.

is it the measuring method is as the figure below?

Yes. Measure the voltage across one emitter resistor then calculate the current.

Is it measure with no input signal?

Yes. Quiescent is when the circuit is resting and is doing nothing.

 

if the signal swings to a peak of 8v higher than the +12v quiescent voltage.

this part can be see by using an oscilloscope?

For example, if the supply voltage is +24V, the drive to the output transistors has a peak voltage loss of 4V, the emitter resistors have a peak voltage loss of 0.3V each and the output transistors have a peak voltage loss of 1.5V each then the peak output swing is 12V - 5.8V= 6.26V peak which is 4.4V RMS. The power into 8 ohms is 4.4V squared/8 ohms= 2.4W.

how can we know "the drive to the output transistors has a peak voltage loss of 4V"?

while for the "emitter resistors have a peak voltage loss of 0.3V each", is it the quiescent voltage across the emitter resistor?

thanks.

 

this part can be see by using an oscilloscope?

Yes, of course.

how can we know "the drive to the output transistors has a peak voltage loss of 4V"?

the 'scope will show the voltage loss. The voltage loss is less if the pullup resistor is bootstrapped.

while for the "emitter resistors have a peak voltage loss of 0.3V each", is it the quiescent voltage across the emitter resistor?

No.
If the emitter resistor is 0.33 ohms then the max peak output current of 1A into an 8 ohm load creates 0.33V across the emitter resistor. If the quiescent current is only 30mA then the 0.33 ohm emitter resistor will have 0.01V across it.

 

ok this is the circuit i build. when i on the power, i inputted 1Vp-p sine wave from signal generator to the input. After that,i measure on the output part. i get a 0.3Vp-p sine wave without crossover distortion. But the volume is really low until hardly to hear the tone sound.

Is it because of my supply too low?

thank you.

 

Your transistors are common-collector (emitter followers) that have no voltage gain. They are the usual output stage of an amplifier. Another stage or two in an amplifier provides voltage gain.

The value of your emitter resistors is way too high at 47 ohms so they make a voltage divider with the speaker. The emitter resistors should be 0.1 ohm to 0.47 ohm each.

 

Your transistors are common-collector (emitter followers) that have no voltage gain. They are the usual output stage of an amplifier. Another stage or two in an amplifier provides voltage gain.

Do u mean by adding another NPN for the upper and PNP for the lower transistor to have the voltage gain?
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after i change the two emitter resistor to 0.47 ohm resistors, the Vpp of the output become higher.which is around 1.12Vp-p. but the output sound still very low. ok, guru mentioned that because my circuit has no voltage gain.

So if i would like to build the voltage gain for it.what should i do?


thank you.

 

Do u mean by adding another NPN for the upper and PNP for the lower transistor to have the voltage gain?

No. Then the output transistors will be darlington transistors with no voltage gain but with a low input current. Then an amplifying transistor can be added to have a high voltage gain.
A common-emitter amplifying transistor can drive both output transistors.

So if i would like to build the voltage gain for it, what should i do?

Didn't your teacher teach about amplifier circuits?
Didn't you see many amplifier circuits in magazines and in books?
Didn't you see some of the thousands of amplifier circuits on the web?