# audio amplifier using common emitter and follower

#### kkrros

Joined Jan 21, 2024
2
i have this base circuit that we use for classes (pic in circuit example.jpg) ,
now, i want to deliver 1.5 w to the speaker (8ohms, max power of 2w).

in our input because of given impedance of 600,1V ptop, 1kHz, i have this clipper.

currently, i have this circuit (Vcc=12) with "assumed values" just to see the waveform. Now i need to solve for the real values of resistors and capacitors, but im really not sure on how to start. Also, as you can see i have the capacitors (beside R5 and before the speaker) removed because the waveforms are not aligned when i have it. or should i put it back?

#### Michal Podmanický

Joined May 11, 2019
27
For low frequencies like 1k you don’t need a cap across emitter resistor.

The ratio between bias resistors (R2 and R3) and load resistors (R4 and R5) of Q1 should be about 20:1 or 10:1. You have choosed about 100:1.

The ratio between R4 and R5 sets the gain. For 0.5V input it is too low. Increase the R4:R5 ratio a little.

Bias the collector voltage to about 7V.

#### kkrros

Joined Jan 21, 2024
2
For low frequencies like 1k you don’t need a cap across emitter resistor.

The ratio between bias resistors (R2 and R3) and load resistors (R4 and R5) of Q1 should be about 20:1 or 10:1. You have choosed about 100:1.

The ratio between R4 and R5 sets the gain. For 0.5V input it is too low. Increase the R4:R5 ratio a little.

Bias the collector voltage to about 7V.
oh! thanks for the feedback, that helps!
but also, the values of resistors and capacitors in my circuit are just "assumed" values that I have just to see the behavior of the output wave. I'm stuck though on how should I start to solve the actual values.

with these as given:
deliver 1.5 w to the speaker (8ohms, max power of 2w).
input: impedance of 600,1V ptop, 1kHz,
Vcc=12

#### crutschow

Joined Mar 14, 2008
34,386
deliver 1.5 w to the speaker (8ohms, max power of 2w).
That will require a pp voltage of 10V into the 8Ω load.
A single emitter-follower for this will dissipate a lot of power since it will require a very low emitter resistor with a high bias current to sink the load current for the negative half-cycle, so is not really a practical circuit.
Look into using a complementary transistor push-pull emitter follower output to drive the speaker load, which will minimize the power dissipation.

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#### MrAl

Joined Jun 17, 2014
11,448
i have this base circuit that we use for classes (pic in circuit example.jpg) ,
now, i want to deliver 1.5 w to the speaker (8ohms, max power of 2w).
View attachment 313301

in our input because of given impedance of 600,1V ptop, 1kHz, i have this clipper.
View attachment 313302

currently, i have this circuit (Vcc=12) with "assumed values" just to see the waveform. Now i need to solve for the real values of resistors and capacitors, but im really not sure on how to start. Also, as you can see i have the capacitors (beside R5 and before the speaker) removed because the waveforms are not aligned when i have it. or should i put it back?
View attachment 313303
Hi,

To get 1.5 watts into an 8 Ohm load you need a peak voltage of around 4.9 volts, and that means it has to go plus and minus 4.9 volts. That means that the input to the last transistor has to be about 0.7 volts higher than that, going 4.9+0.7 high and -4.9+0.7 low.
The output should be biased to about one-half of the supply voltage, which would be about 6 volts, which means the input (base) would be biased to about 6.7 volts.
The gain of the circuit with Q1 has to be enough to drive the collector voltage to at least the input voltage of Q2, which has a range of 9.8 volts.
To get 10 volts from 1 volt you need a gain of at least 10, and the collector has to be biased to about one-half of the supply voltage also.

The impedance presented by the 600 Ohm resistor and two diodes is 600 Ohms but only when the diodes are not conducting. That may or may not matter though. When the diodes conduct that impedance will be the parallel combination of the 600 Ohms and the dynamic resistance of either diode.

#### Audioguru again

Joined Oct 21, 2019
6,687
Your output transistor is a single-ended class A but almost ALL audio amplifiers have push-pull class AB output transistors.
Your 2N4401 output transistor can push the output positively 600mA into the 8 ohms speaker which is 4.8V peak (but but the transistor will become too hot) then there is no current in RB2 to pull the output down. So the output transistor will be a rectifier instead of a linear audio amplifier.

If you use a high current transistor and 1 ohm for RB2 then the output will produce 1.5W into 8 ohms but the output transistor current will be an average of about 5.4A all the time, even when the input signal is zero and it will heat with 12V x 5.4A= 27W.

#### Ian0

Joined Aug 7, 2020
9,765
For low frequencies like 1k you don’t need a cap across emitter resistor.
Actually you do. Otherwise the gain is limited to R4/R5. If you want more gain that that, you need a capacitor across R5.

#### MrAl

Joined Jun 17, 2014
11,448
Your output transistor is a single-ended class A but almost ALL audio amplifiers have push-pull class AB output transistors.
Your 2N4401 output transistor can push the output positively 600mA into the 8 ohms speaker which is 4.8V peak (but but the transistor will become too hot) then there is no current in RB2 to pull the output down. So the output transistor will be a rectifier instead of a linear audio amplifier.

If you use a high current transistor and 1 ohm for RB2 then the output will produce 1.5W into 8 ohms but the output transistor current will be an average of about 5.4A all the time, even when the input signal is zero and it will heat with 12V x 5.4A= 27W.
Hi there,

The output power problem with the transistor is a very good point, glad you brought this up, because that means the circuit will never work as planned if the output transistor is a 2N4401 or similar small transistor.
I do have to disagree with the current output though being 5 or more amps, because with no input and 6v output, an 8 Ohm resistor will draw about 750ma, but even that alone is too high for that tiny transistor. That means the transistor power dissipation at that point is 4.5 watts, which I am dang sure that tiny transistor can not handle.
This of course means a power transistor is required at the output.
Because there are other problems too, I don't think this particular design is going to work at the power level required for this project.
I like the idea of a voltage follower on the output, but that means the input voltage has to span the full range of the output voltage also, which is about plus and minus 5 volts. That means the base has to be driving by that first stage to plus and minus 5 volts, relative to the zero bias point of course, but that's still a span of 10 volts.
I dare not think about how much distortion we will see here, but that does not seem to be included in the spec's so we lucked out there I think

I would think the output load would have to be capacitor coupled also. Unfortunately, that means lowering the 120 Ohm resistor also which also increases DC power.

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#### Audioguru again

Joined Oct 21, 2019
6,687
Hi MrAl, I used a 1 ohm emitter resistor and an output bias of 5.5V. Then the output can produce the peak voltage of 5.5V peak in the total series output resistance of 9 ohms which makes exactly 1.5W into the 8 ohms speaker. The 1 ohm resistor has an idle current of 5.5A and heats with 30.3W.

If your emitter resistor is 8 ohms with 6V across it, then the max output is 3V peak which produces 2.12V RMS and only 0.56W in the 8 ohms speaker. Your 8 ohms emitter resistor has an idle current of 6V/8 ohms= 0.75A and heats with 4.5W.

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#### MrAl

Joined Jun 17, 2014
11,448
Hi MrAl, I used a 1 ohm emitter resistor and an output bias of 5.5V. Then the output can produce the peak voltage of 5.5V peak in the total series output resistance of 9 ohms which makes exactly 1.5W into the 8 ohms speaker. The 1 ohm resistor has an idle current of 5.5A and heats with 30.3W.

If your emitter resistor is 8 ohms with 6V across it, then the max output is 3V peak which produces 2.12V RMS and only 0.56W in the 8 ohms speaker. Your 8 ohms emitter resistor has an idle current of 6V/8 ohms= 0.75A and heats with 4.5W.
Hi,

Oh 1 Ohm wow, that's low. Yeah that would require a lot of power to run then.

I also see you added the output coupling capacitor which is a very good thing to do. The speaker would have a heck of a time running with 750ma of DC current

I have a feeling that the requirements of this circuit project will be hard to meet.
I did a 100 Ohms output load resistance version just to see how it would work, and with the reduced output current it seems to work given some modifications to the drive circuit. Of course 100 Ohms is a lot higher than the required 8 Ohms, but I just wanted to see what it would look like.

#### Audioguru again

Joined Oct 21, 2019
6,687
I have never heated my home with a class-A audio amplifier. My first amplifier was a Heathkit class-AB with tooobs. Its output tooobs needed to be replaced often to keep the power up and the distortion down.