Designing a transistor amplifier. Part B.

Power stage.

After competing the preamplifier it’s time to design the power section of our amplifier. I want this section to be implemented as a Class B amplifier. We are going to select the Push-Pull configuration for the power transistors.

pp-1

The following is intended as an educational aid and not a professional effort.

To the circuit above we can see the working principle of a Push-Pull amplifier. The amplifier is consisted of two complementary transistors. An NPN and a PNP. Each transistor amplifies the half wave of the signal. This amplifier does not offer voltage gain, but a current gain. Exactly the thing we need in order to drive a speaker.

To the above schematic we can see a real Push-Pull amplifier. The “problem” of such an amplifier is the relatively low input impedance. Low as far as the driving of this stage by the common emitter amplifier we designed for the preamplifier stage. For this reason we will combine the this Push-Pull stage with a Common Collector amplifier.

pp-3

To the circuit above, we can see the “mixing” of the Push-Pull amplifier with a Common Collector.

We can see that the Common Collector topology is consisted of the transistors Q3, Q4. Those two transistor is combined as a Darlington pair. We can approach them as a single transistor with a β (beta) equal to the product of the β of each transistor. If for a example each transistor has a β of 100 the overall β will be equal to 100*100=10.000.

Resistor R4, R5 are used for the thermal equilibrium of the two power transistors Q1,Q2 of the Push-Pull stage. The internal resistor of each transistor changes with temperature. Without the resistors R4,R5, changes in temperature would significantly change the emitter current of the transistors.

The power transistors used is BD135 and BD136. The transistors for the Common Collector stage is 2N4403. Diodes D1,D2 is 1N4148. The resistor R4,R5 is 2 Ohms.

The selection of the parts is not optimal. It was based on the parts I had in hands at the time.

To the following circuit we can see the voltages and the currents of our amplifier section. We must calculate the resistors R1,R2 and R3.

As stated before, our amplifier must drive a 16 Ohm speaker with 5 volts peak-to-peak (or 2,5 volts peak). This means the current through the speaker will have a max value of:

This current will be the max current through the Q1 transistor (Q2 also, for the other half period). So:

We want our DC voltage to the output equals the half of the supply voltage (9 volts). So:

The voltage to the Emitter of Q1 is:

The current gain β (beta) of the power transistors is around 25. The base current, in order to the current we need to the output, must be:

The DC voltage to the base of Q1 is:

The current through R1 and the two diodes must be at least 10 times bigger of the base current of the power transistors, in order a stable DC base voltage. This current is the Emitter current of the Common Collector stage. So:

Now that we know the voltage across R1 and the current through it we can calculate it’s value:

The voltage to the Emitter of Q3, due to the forward voltage of the two diodes in series, will be about 1,4 volts less than the voltage in the Base of the Q1 transistor. So:

The voltage to the base of Q4 transistor will be 1,4 volts less than the voltage of the emitter of the Q3 transistor. The forward voltage of the two Base-Emitter junctions.

The overall gain of the Q3,Q4 transistors is 10.000. The base current of Q4 will be:

We need the current through R3 10 times bigger from the base current of Q4. So:

We now know the voltage across R3 (the voltage of the base of Q4) and the current through it, and we can calculate it’s value:

The current through R2 must be 11 times bigger from the base current of Q4. So:

We now know the voltage across R2 and the current through it, and we can calculate it’s value:

We were able to calculate R1,R2,R3.

  • R1=56 Ohm. Available from the E12 resistor series.
  • R2=92 kOhm. We are going to select 100k available from E12 series.
  • R3=43,2 kOhm. We are going to use a potentiometer of 47 kOhms to adjust precisely the bias of the transistors and compensate the change of the R2 value (29k to 100k).

The final design for the amplifier we designed to this two part article.

pp-final

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