Class A amp questions

[jackorocko]

I bought this tip31 transistor and it says on the package good for audio. I figured why not mess around with it in simulation to see what it is capable of doing. Here is my results.

I followed this guide for the design. http://www.electronics-tutorials.ws/amplifier/amp_2.html
Here is what I come up with:

What I don't really understand is the RC high-pass filter created by C1/R2 and C2/R5. I know this is what they call a decoupling capacitor, but I always thought the value of a decoupling cap was to be between 0.1uF and 10uF. Am I doing something wrong in that R2 should be a higher value? But according to the website above, I should make sure my biasing network has a current 10x larger then the base current, so the base current doesn't overload my biasing circuit?

Am I understanding this correctly? Is there anything wrong with using a 1000uF decoupling cap? Would this work better if I used a darlington pair made out of two tip31's so I have a higher gain which would mean less base current and smaller biasing current?

 

[(*steve*)]

This is a classic common emitter class A amplifier with a voltage divider for bias and an emitter resistor for negative feedback.

It is also not a very good amplifier But it's great for learning about amplifiers.

The first thing is that the bias network essentially sets the input impedance of an already low input impedance amplifier.

The second thing is that R4 severely reduces the gain of the stage. It's good, because the gain is no longer dependant on the varying characteristics of the transistor, but bad because it's so low.

The input capacitor in conjunction with the input impedance of the amplifier will act as a high pass filter, but the choice of a very large capacitor will reduce that to a much lower frequency.

The next step is to place a capacitor (say 10uF) across the emitter resistor to remove the negative feedback for AC, but retain it for the DC bias. Suddenly you'll have a lot more gain. Sometimes you see a capacitor and series resistor across the emitter resistor to increase the gain, but by a limited amount. This capacitor also affects gain with frequency, so you need to choose a value that will not reduce gain at frequencies that are of interest.

R1 and R2 seem very low in value. 22k and 4k would be more in the range I'd expect, but perhaps lower given that this is a higher current transistor.

With a circuit like this, you'll quickly see why class A amplifiers are not often used as power stages. I would imagine the TIP31 gets rather warm.

What happen in this circuit is that the voltage divider formed by R1 and R2 provides enough current to turn on the transistor. The combination of the base and collector current flow through R4, raising the emitter to 0.7V lower than the base. The R1/R2 voltage divider needs to be relatively stiff so that the current drawn from it doesn't cause its voltage to fall to 0.7V instead of the emitter voltage rising. Obviously the transistor's gain and the value of R3 and (to a greater extent) R4 are also involved. Surprisingly the transistor's gain is one of the least important.

A capacitor across R4 means that the emitter voltage does not rise and fall with an AC input voltage, which permits a greater voltage swing on the collector.

Was that helpful or confusing?

 

[jackorocko]

Was that helpful or confusing?

always helpful even if clouds still swirl around my head.

The first thing is that the bias network essentially sets the input impedance of an already low input impedance amplifier

we're talking impedance because of the AC signal input? Would the impedance of the biasing network be R1+R2 or just R2?

What happen in this circuit is that the voltage divider formed by R1 and R2 provides enough current to turn on the transistor. The combination of the base and collector current flow through R4, raising the emitter to 0.7V lower than the base. The R1/R2 voltage divider needs to be relatively stiff so that the current drawn from it doesn't cause its voltage to fall to 0.7V instead of the emitter voltage rising. Obviously the transistor's gain and the value of R3 and (to a greater extent) R4 are also involved. Surprisingly the transistor's gain is one of the least important.

This is where I lose you. What does "stiff" mean? The website I linked to said the biasing current should be 10x that of the base current. But when doing the math, I don't get anywhere near 22k/4k at 10x the current. Hell, even 220/40 would give me less then 5x the base current.

I did experiment around a little more with the bypass cap and found that @10u and a 10kHz 1V amplitude ac input signal, it drove the amp into saturation, reducing this value to 1u reduced the gain slightly and fixed the distortion. Since this is for audio, then I should be sizing my bypass cap for 20kHz?

 

[(*steve*)]

we're talking impedance because of the AC signal input? Would the impedance of the biasing network be R1+R2 or just R2?

I can probably save 1000 words by pointing you here.

What does "stiff" mean?

It means that it doesn't change much under the intended load. If you have 10x the current flowing in the voltage divider as opposed to the base current, then the voltage on the voltage divider will change only about 10% in response to the base current being drawn from it.

The website I linked to said the biasing current should be 10x that of the base current. But when doing the math, I don't get anywhere near 22k/4k at 10x the current. Hell, even 220/40 would give me less then 5x the base current.

You have a very low gain transistor used in a circuit not appropriate for high power use.

You should probably look at class B amplifiers for higher output, and consider a lower power, higher gain transistor for this circuit. The output of this circuit can drive a higher powered class B output stage.

Going back to your problem, if the gain of the transistor is 20 at the currents you're looking at, then the resistors making up your voltage divider will only be about twice what the collector and emitter resistors are. You will be making a heater.

I did experiment around a little more with the bypass cap and found that @10u and a 10kHz 1V amplitude ac input signal, it drove the amp into saturation, reducing this value to 1u reduced the gain slightly and fixed the distortion. Since this is for audio, then I should be sizing my bypass cap for 20kHz?

if you have a bypass cap on the emitter resistor, the better way of reducing the gain is to place a resistor in series with it.

The capacitor should be of an appropriate value for the lowest frequencies you want to amplify, not the highest.

 

[jackorocko]

I can probably save 1000 words by pointing you here.

Yep, that explained it.

It means that it doesn't change much under the intended load. If you have 10x the current flowing in the voltage divider as opposed to the base current, then the voltage on the voltage divider will change only about 10% in response to the base current being drawn from it.

This then effectively keeps the transistor close to it's Q-point which would help maintain stability of the circuit. I get that now

You will be making a heater.

I kinda assumed that when I started, but it always nice to see the math prove the theory. I am not sure how you measure power gain, as in rms, peak to peak, or amplitude. I did the math for amplitude of voltage * amp and got 750mW output power?

The capacitor should be of an appropriate value for the lowest frequencies you want to amplify, not the highest.

I am not sure I get you here, what is an appropriate value for low frequencies?

 

[(*steve*)]

Low frequencies for audio are maybe 20Hz. Maybe it's a little higher (say 100Hz) if your application isn't "hi-fi".

Power gain is (output power) / (input power)

You need to be able to measure the input and output power.

If you know the input impedance (and can measure the RMS voltage of the input signal) then you can calculate input power.

If you know the output device's impedance and the RMS voltage swing across it then you can calculate the output power.