Transistor bias goes negative under input

Question

I am trying to make a simple class A audio amplifier with one transistor (TIP41, NPN) in common emitter configuration as shown in the pictures.

The problem I have is that when I add the input signal (more than 1 V), the transistor's base bias changes quickly from positive to negative in such a way that no usable amplification is possible. I can get 1 V/0.2 A output at best from the 10 V/0.57 A supply input. That's 3.5% efficiency.

Should the input signal stay under 1 V, is that the cause of bias change? Is the signal too strong for this base bias current? If so, how are signals amplified further (in large amplifiers,) if the final stages can't take more than 1 V or close? It can't be right.

Answer 1

This is happening because you're not using negative feedback.

For a common-emitter amplifier like this you need series-series feedback (AKA emitter degeneration). This will desensitize your transfer from beta (current gain) variations.

If you really need the gain, then you can also include a parallel capacitor with such resistor so that the emitter is grounded at AC (you still get the negative feedback at DC, so your transistor will be biased properly). However, this will make your AC transfer more sensitive to beta variations, so you'll suffer from more distortion.

Whatever you choose, you need feedback for linear amplification, there no other way.

If you must use stages like these for amplification, you can ditch the divider and make an inverting amplifier; connect the output to the 1st stage base node, and place another resistor in series so that you can define your gain as the classical inverting amplifier. This way you make good use of the large open-loop gain these common-emitter stages provide.

EDIT:

Perhaps I read your question too quick. If you're trying to have a predictable voltage gain and able to feed 3A into a load, there is no way you can do this with a single stage, much less with a textbook common-emitter amplifier.

The fundamental problem with a common-emitter amplifier is that, it's basically a transconductance amplifier with a predictable load (the collector resistor), i.e. it needs a predictable load for a predictable linear gain (and the quiescent collector current is fixed by the emitter voltage divided by the emitter resistor you added). You say you want to feed to a 4 ohm load that will be, effectively, in parallel with your collector resistor. Your gain will vary, and the current you have fixed with the series-series feedback (the emitter resistor) will now have to be shared among your load and collector resistor.

The only decent use for a common-emitter is as a preamplifier (or if your load is >2 orders of magnitude higher than your collector resistance), while subsequent buffer can provide the current capability you use.

I see you have agreed that you should change your approach for this design. Good luck!

Answer 2

Have a look at the behaviour of this transistor, as its base voltage is changed. Here's the circuit with just the transistor and a collector resistor:

^{simulate this circuit – Schematic created using CircuitLab}

This is what happens to the collector C potential, as the base B is swept from 0 to +1V (with respect to the emitter):

As you can see, the collector potential cannot ever rise above the positive supply of +10V, or below the emitter at 0V. This means that the only useful region of operation (from an audio perspective - digital is a different story) is in the part of this graph between the green markers.

Base potential must remain at all times between 0.67V and 0.75V, if you wish to avoid "clamping" the collector against either supply potential.

That's a range of less than 0.1V, so if you apply a signal with amplitude 1V, the base will spend most of its time well outside those limits, and the output will clamped to either +10V or 0V most of the time.

This common-emitter design is suitable only for very small amplitude inputs, significantly smaller than 0.1V.

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Now let's address your biasing of the base, with resistors 690Ω and 620Ω. These form a potential divider, which will attempt to provide a potential about half way between the supplies, at the transistor's base. That's only an estimate, based on the similarity of their values, a simulation will show us the real value, on the left voltmeter:

^{simulate this circuit}

That's already far above the 0.7V-ish point where you need the transistor's base to be "centered" around.

On the right is what happens to that potential with the transistor in place. The base and emitter of the transistor behave as if there was a diode between them (because, technically there is one), and this diode to "ground" prevents the base from rising much higher than 0.7V. Here the voltmeter shows 0.88V, which again is way outside the range of potentials you need to be providing at the base.

The transistor is clearly driven so hard into conduction, that its collector is super-glued to 0V. To get it out of this state would required a very strong input, able to pull very hard, downwards, the potential of the node at the resistors' junction and transistor base.

While what I am about to suggest is by no means a "good" solution, it might at least help you get somewhere with your experimentation. I suggest you change your resistances to try and achieve +0.7V at the base:

^{simulate this circuit}

Have a play with the simulation, to see what voltages appear in various places as you change the resistances. Pay close attention to collector voltage too.

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There are other issues, but these two must be fixed before I could begin to talk about the others. I hope you've got a better grasp why your circuit cannot hope to operate as an audio amplifier, in its current configuration.

Answer 3

The problem I have is that when I add the input signal (more than 1 V), the transistor's base bias changes quickly from positive to negative in such a way that no usable amplification is possible.

This phenomenon is not so much related to transistor amplifier circuits but rather to diode rectifier circuits. There it has been known and used for a long time (especially in TV receivers) in the so-called "DC restoration circuits". Here is my explanation.

Equivalent circuit

With a large input signal, the base-emitter junction can be considered as a diode; so the input part of the circuit can be simplified to the following form.

^{simulate this circuit – Schematic created using CircuitLab}

Simplified equivalent circuit

To see the basic idea clearly, we can simplify the circuit even more by removing the resistors and replacing the diode with an "ideal" one (VF = 0 V). Thus we obtain the following conceptual schematic.

^{simulate this circuit}

During the positive half-wave, the diode is "on" and the capacitor quickly charges through it; during the negative half-wave, the diode is "off" and the capacitor has nowhere to discharge.

As a result, the capacitor remains permanently charged to a voltage equal to the peak of the input voltage. This DC voltage is subtracted from the input AC voltage and drops it below zero.

Extremely simplified equivalent circuit

In fact, the diode is permanently "off". So we can remove it from the circuit diagram and replace the charged capacitor with a constant voltage source with the same voltage Voff = Vc. Thus we get an extremely simple conceptual circuit diagram.

^{simulate this circuit}

If we compare the two graphs, we see that they are the same except for the initial part where the capacitor must be charged.

Answer 4

Your transistor has way too much base current causing its bottom of the output waveform to be clipping. Since it is class-A without any negative feedback then it is normal for the top of its output waveform to be squashed.

I do not has a model for a TIP41 so I used a similar transistor. There is a high range of the hFE of a transistor part number anyway that needs selection or DC negative feedback to be used.

The distortion sounds awful and the heating is absurd for the small amount of output power. It will have much more trouble trying to drive a speaker.

Answer 5

The input \$1000\mu F\$ capacitor (I suspect is polarized) is correctly biased when the input is 1V. When 0V is applied the capacitor is essentially a short.

The input voltage must be less than 500 mV.

Edit: This also explains why an emitter resistor allows a higher input voltage.