© Jonathan Bartlett 2020
J. BartlettElectronics for Beginnershttps://doi.org/10.1007/978-1-4842-5979-5_25

25. Transistor Voltage Amplifiers

Jonathan Bartlett1 
(1)
Tulsa, OK, USA
 

In Chapter 24, we started our study of the BJT NPN transistor. We noted that what the transistor actually amplified was current, so that the current coming into the collector was a multiple (known as β) of the current coming into the base.

Even though what a transistor does is provide current amplification, in this chapter, we will learn how to transform that into voltage amplification.

25.1 Converting Current into Voltage with Ohm’s Law

If the transistor provides us with current amplification, how might we translate an amplification in the amount of current into an amplification in the amount of voltage? The answer is simple—Ohm’s law describes the relationship between current and voltage: V = I ⋅ R. Therefore, a current amplification can be transformed into a voltage amplification if we use a resistor! The larger the resistor, the larger the change in voltage drop that a given change in current will induce for that resistor.

To see that happening, take a look at the circuit in Figure 25-1. Note that this circuit on its own is rather useless, but it is helpful for illustrating how the calculations work. In this circuit, the current at the base is controlled by the resistor RB. This current will be amplified into an increased current from the collector. However, the current at the collector is driven through a resistor, RC. Because this is through a resistor, that means that Ohm’s law will take effect, and the size of the voltage drop across RC will depend on the current running through it.

Remember Ohm’s law states that V = I ⋅ R, so any increase in current will increase the voltage drop across RC, at least until the voltage at the collector is equal to the base voltage (which, in this circuit, is 0.6 V). If that happens, there is nothing more the transistor can do—it will just treat the collector-emitter junction as a short circuit.

Let’s calculate what our circuit is actually doing. The voltage across the base is 5 V − 0.6 V = 4.4 V (remember we have to account for the diode-like voltage drop in the transistor from the base to the emitter). Therefore, using Ohm’s law, we can calculate the base current at I = V/R = 4.4/10000 = 0.0004 A.
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Figure 25-1

A Simple Current-to-Voltage Amplifier

Let’s assume the transistor beta is 100. Therefore, the current flowing at the collector will be 0.0004⋅100 = 0.040 A. So the voltage drop across the resistor can be calculated using Ohm’s law. V = IR = 0.040·50 = 2 V.

Now, let’s say that we change RB so that we have more current running in the transistor. Let’s decrease RB from 10 kΩ to 6 kΩ. Now the base current will be I = V/R = 4.4/6000 = 0.000733 A. Now the current flowing at the collector will be 100⋅0.000733 = 0.0733 A. So the voltage drop across the resistor is now V = IR = 0.0733⋅50 ≈ 3.67 V.

When we increase the current, we increase the voltage drop across the resistor. You may be wondering what happens to the extra voltage. That is, since the emitter of the resistor is at ground and the voltage across RC keeps changing, where is the remainder of the voltage? The transistor essentially swallows it up.

Remember, in our model of the transistor in Figure 24-2, the transistor acts as a variable resistor for the collector current. Therefore, the rest of the voltage drop happens within the transistor.

So, in effect, what we are doing is to translate changes in current at the base into changes in the voltage drop across RC (and likewise the VCE of the transistor).

As you might have noticed, when dealing with transistors, the place where the “action” occurs is not always right where you might expect it. In this circuit, the location where the voltage amplification actually occurs is at a resistor (RC) connected to the collector. So the transistor, acting as an amplifier, is not sending the amplified signal to the emitter. Instead, you can think of it as sending the amplified signal to the collector, where the resistor at the collector converts the amplified current into a corresponding voltage drop.
  • Example 25.28 In the circuit given in Figure 25-1, what is the voltage across the resistor RC if the base resistor RB goes up to 20 kΩ?

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Figure 25-2

Reading the Amplified Signal from a Voltage Amplifier

$$ {displaystyle egin{array}{l}kern2.28em {I}_B=4.4/20000=0.00022;mathrm{A}\ {}{I}_C=100cdot {I}_B=100cdot 0.00022=0.022;mathrm{A}\ {}kern1.2em {V}_{RB}={I}_Ccdot {R}_B=0.022cdot 50approx 1.1;mathrm{V}end{array}} $$

Just to see where we are going, eventually we will use small voltage changes in the base to trigger current changes in the base which will then be amplified into a larger change in the voltage across RC.

25.2 Reading the Amplified Signal

So we have managed to create a voltage drop which changes in response to changes in current at the base. But how do we read this voltage drop? It is rather difficult to read it directly, but we can read its inverse directly.

Take a look at Figure 25-2. In this figure, we added some output signal line to show where we would read the output of the amplifier (i.e., where we would connect the rest of the circuit that receives the amplification). We put the output line between the collector resistor RC and the transistor. What this will do is give us the voltage of the source voltage (5 V) minus the voltage across RC. So, when we have a large voltage across RC, that will be reflected in a low voltage in our output. Likewise, when there is a low voltage across RC, that will be reflected in a high voltage in our output.

This sort of an output is known as an inverted output, because the output voltage is essentially reverse-amplified. That actually works just fine for audio signals, as it does not matter to the listener if the signal is inverted or not. However, if we needed to get it back to the non-inverted form, we could just add another amplification stage onto the end (we will see how to do this in Section 25.4, “Adding a Second Stage”).
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Figure 25-3

Components of a Transistor Biasing Circuit

Having said all that, I should point out that we still don’t know how to amplify an audio signal—yet. That is coming in the next section.

25.3 Amplifying an Audio Signal

What we really want to do is to amplify an audio signal. Imagine that someone is singing into a microphone, and we want to amplify the signal we get so that we can send it to a speaker. How would we do that?

There are a number of problems that you have to solve in order to get this done. You might imagine that you could just connect a microphone to the base of the transistor and just amplify directly. That’s a good idea, but sadly life is not always that easy. To understand why, remember that audio signals are basically alternating current. That means that the signal will swing both positive and negative. Also remember that the base voltage has to remain above the emitter voltage and the emitter is tied to ground. Therefore, if we tried to do this, we would lose the bottom (negative) half of the signal. In fact, if it was a small signal, we might lose the entire signal if it never reached the required 0.6 V above ground.

Therefore, the signal coming into the transistor needs to be biased. We basically need to bring the voltage of the incoming signal up so that it keeps the transistor on, but doesn’t overwhelm the signal coming in.

If you remember in Chapter 19 when we built the tone generator, we had to unbias the signal. We did that by taking the signal and coupling it through a capacitor. That allowed the two sides to exist at different base voltages, but transmit the changes in the voltage through the capacitor. This will be the same idea, but where the input is unbiased and the output is biased.

Figure 25-3 shows what this looks like in general terms. A simplistic way to think about this is to think of it as a voltage divider. The resistors are basically setting a bias voltage, and that voltage is varied based on the current that comes in through the capacitor.
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Figure 25-4

A Single-Stage Transistor Voltage Amplifier

While that is somewhat true, the details get more complicated as we add more components, especially the transistor. The emitter of the transistor will be connected to ground though through a resistor, RE. This means that, for the transistor to be on, the actual voltage coming in to the base of the transistor must always be 0.6 V (or above, based on RE). That means that the voltage divider itself must always be 0.6 V (or slightly above, due to RE) as well.

Figure 25-4 shows all of the parts of our simple transistor amplifier. The main question is, what do we set the values of each of these resistors to? There are many considerations, and they affect both the amount of gain you will get in your amplifier and whether or not the signal is clipped.

Clipping will happen if the base voltage goes below 0.6 V because the transistor will simply stop conducting (and actually a little above that because of RE). Clipping will also happen if the RC resistor is too large, because if the current going through RC causes the voltage at the collector to get too low, the transistor will have reached its saturation point and therefore will stop amplifying. Since RC is converting the current into voltage, it also needs to be large enough to yield a wide voltage swing with the amount of current expected.

For the base resistors (RB1 and RB2), you need to keep in mind how much current you expect the input source to have. Remember that this current is going to be going to be alternating between the positive and negative directions. Therefore, when the input source swings negative, there has to be enough current coming through RB1 to keep the voltage positive and keep the transistor on. When the input source swings fully positive, RB2 can prevent clipping as well, as it provides an alternate path for incoming current that isn’t through the transistor.

You may be wondering what the resistor at RE is doing. RE is a feedback resistor. The gain (β) of a transistor varies both within manufacturing and based on temperature. A feedback resistor, by raising the voltage at the emitter of the transistor, will limit the amount of gain that the transistor can deliver based on its source input.

Feedback means that the output of a circuit is affecting the input in some way. Sometimes, feedback occurs because the output is wired back to the input. However, in this case, the resistor simply increases the voltage at the emitter, causing less current to flow in through the base and more current to flow through RB2.

Limiting the gain of the resistor stabilizes the gain over a variety of operating parameters, so that your circuit will continue to work as the transistor heats up. Additionally, were you to manufacture the circuit, limiting the gain would make it less susceptible to manufacturing differences between individual parts.

In any case, if the output of the circuit has a resistance RO, then the current gain from the input through the output can be calculated as
$$ mathrm{gain}=frac{R_{B2}{R}_C}{R_E{R}_O} $$
(25.1)

To understand why this formula works, see Appendix D, Section D.5, “Output Gain Calculations in BJT Common Emitter Applications.”1 The formula just gives you a starting point, though you normally have to play around with resistor values a little to get it to work.2 Also note that the formula only works when RO > RC.

When you think about your circuit, you will want to analyze it at two points—when the source current coming in at the input is at its highest and at its lowest (which in this case will be negative). If you have developed your circuit correctly, this should give a good swing in the voltage across RC, but without ever clipping (the transistor turning off or going into saturation).

The best place to start is to recognize that your voltage divider must always provide enough current to keep the transistor on even when the source current swings negative. Therefore, the current coming in to the base through the voltage divider should be a little larger than the signal’s current. This is known as the quiescent current .

In our case, we are going to develop our circuit to handle passive microphones, so the source signal current will only be on the order of about 10 microamps (0.01 mA). Therefore, we will set our quiescent current to be 0.1 mA. The means that RB1 should be a little less than 50 kΩ. We will use a standard 47 kΩ resistor here. RB2 can tolerate a wide range of values—basically anything that is at least a quarter the size of RB1 (we can’t let the voltage drop below the emitter voltage, or the transistor will stop conducting). We’ll make it 20 kΩ here.

We want a value for RC that will swing the voltage about half as far as we can go (we have to save headroom for both the 0.6 V drop on the transistor and the actual output to the rest of the system). If we want a 25x gain, then, if our max base current will be 0.11 mA (quiescent current plus source current), that will put our max collector current at 2.75 mA (25 0.11). We want RC to be dropping about 1.5 V at this level (we need to save some headroom for a number of purposes, including staying above the voltage drop at RE, the additional diode drop to stay above the base voltage, and some additional which will be used for the current going out of this stage of the amp). Therefore, the resistance for a 1.5 V voltage drop at 2.75 mA is $$ R=frac{V}{I}=frac{1}{0.00275}approx 545;Omega $$. Anything near that value should work fine, so we will go with a 510 Ω resistor.

Then, we mentioned earlier that the total gain will be given by $$ frac{R_{B2} RC}{R_E{R}_O} $$. So far, we have $$ frac{20,000cdot 510}{R_E{R}_O} $$. We can estimate the next stage of the amplifier as having a resistance of 2,000 Ω (this is the base resistor of the next stage of the amplifier—see Appendix D, Section D.5, “Output Gain Calculations in BJT Common Emitter Applications,” for why this works as an estimate). Therefore, to get a 25x gain, we can do
$$ {displaystyle egin{array}{l}mathrm{gain}=frac{R_{B2}{R}_C}{R_E{R}_O}\ {}kern0.48em 25=frac{20,000cdot 510}{R_E2,000}\ {}kern0.48em {R}_E=frac{20,000cdot 510}{2,000cdot 25}\ {}kern0.48em {R}_E=204end{array}} $$

In testing, I found that a smaller resistor worked better for RE, so I used a 47 Ω resistor. Again, the equations are greatly simplified and only give a starting point for experimentation.

25.4 Adding a Second Stage

A single amplification stage is not always enough. If your signal source is weak enough, sometimes you need more power just to hear it. This can be accomplished in a number of ways. The simplest, conceptually, is to add another output stage to your amplifier.

In order to do this, we need to design the next stage to take into account the output of our first stage. We will need to use a coupling capacitor to handle the change in voltage from the output of the first stage into our bias circuit for the second stage. Without the coupling capacitor, the voltage characteristics of the transistor would severely alter the way that our previous stage works. Therefore, the coupling capacitor helps us to isolate the different parts of the circuit.

Now, the current coming in to this stage will be significantly larger than the previous stage. Our previous stage had about 0.01 mA coming in. While not exact, with an actual 25x gain, the current coming in will probably be about 0.25 mA. Therefore, we have to plan this stage differently (i.e., using a different quiescent current).

Using the same procedure as we used for the previous stage, we can use 0.75 mA as the quiescent current and utilize a 5.1 kΩ resistor (to choose a standard value) for RB1 and a 2 kΩ resistor for RB2. Our largest swing forward will now be 1 mA, and therefore, the amplified current will be 25 ∗ 1 = 25 mA. To get the 1.5 V drop, we will need a resistor for RC of $$ frac{1.5}{0.0025}=60;Omega $$. We will approximate to a 100 Ω resistor.

The output resistance will be our headphones, which we can estimate at 16 Ω. Doing the calculation gives
$$ {displaystyle egin{array}{l}mathrm{gain}=frac{R_{B2}{R}_C}{R_E{R}_O}\ {}kern0.48em 25=frac{2,000cdot 100}{16cdot {R}_E}\ {}kern0.48em {R}_E=frac{2,000cdot 100}{25cdot 16}\ {}kern0.48em {R}_E=100;Omega end{array}} $$

Again, I had to modify the RE to get better performance. This is unsurprising because the equation is only even moderately reliable when RO > RC. Lowering RE of this stage to 10 Ω seemed to work well.

The output of the second stage will again be coupled through a coupling capacitor to your headphone output and can be attenuated by another resistor (or a variable resistor for volume control) if needed.

Figure 25-5 shows the complete two-stage amplifier. With this circuit, you can use most microphones and most headphones to get enough output power to hear yourself in your headphones (in fact, if you don’t have a microphone, headphones can actually work as a microphone in a pinch). Figure 25-6 shows the two-stage amplifier built into a breadboard.

All of the capacitors can be pretty much any standard capacity above around 100 nF. The ones that couple the DC bias signal, if they are polarized, should have their negative side toward the unbiased signal. The coupling capacitor between the two stages should ideally be non-polarized, though with these power levels it doesn’t matter too much. If you do need to use a polarized capacitor here, it is best to have the negative side facing the second stage, since that should be right around 0.6 V, while the positive side should be swinging in the positive range.

One thing in this circuit we haven’t discussed is the bleed-off resistors connected right next to the coupling capacitors for the input and output circuit. These are very large (1,000,000 Ω) resistors. They are large so that they do not have any effect on the calculations in the circuit itself, but they do provide a place for the negative side of the capacitor to drain out when nothing is connected. Otherwise, any residual charge on the capacitor has no place to go and will remain even when the circuit is turned off and the devices are unplugged.

25.5 Using an Oscilloscope

Designing transistor amplifiers can be tricky because there are a lot of things that can go wrong. Although there is some leeway, using the wrong resistors can result in distortion or a loss of gain. Misconnecting a single component can render the entirety of the circuit inaudible.

Because of the variety of things that can go wrong when building audio circuits, it is best to use an oscilloscope. An oscilloscope can help you visually see what the voltages look like at each point in your circuit. Oscilloscopes can cost as little or as much as you want them to. There are pocket oscilloscopes that you can purchase for less than a nice dinner out, and there are bench oscilloscopes that you would need several months’ salary to purchase. Any of those are helpful in analyzing your circuit, so you can see what is happening to your voltages at every point in the circuit.
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Figure 25-5

The Complete Two-Stage Amplifier Circuit

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Figure 25-6

The Two-Stage Amplifier Built into a Breadboard

Review

In this chapter, we learned the following:
  1. 1.

    Although transistors provide current amplification, current changes can be transformed into voltage changes using Ohm’s law.

     
  2. 2.

    Using a resistor at the collector allows us to “read” a voltage change based on the changes in the currents going through the transistor.

     
  3. 3.

    Using a resistor in this way inverts the waveform—it will show low voltage when there is a lot of current and high voltage when there isn’t much current.

     
  4. 4.

    Adding in an emitter resistor limits the amount of voltage gain in the circuit in order to compensate for variable/drifting transistor betas.

     
  5. 5.

    In order to amplify an audio signal, we have to add a DC bias to the signal so that the transistor stays in its operating range (positive voltage).

     
  6. 6.

    The simplest way to use a transistor to amplify a signal is to add a voltage divider to the base of the transistor, with a coupling capacitor feeding the signal into the voltage divider.

     
  7. 7.

    The neutral, “no signal” design point is known as the quiescent point of a circuit. The quiescent point is the state of the circuit when the AC signal coming in is neutral (0 V).

     
  8. 8.

    The AC signal is coupled into the voltage divider through a coupling capacitor in order to manage the difference between the pure AC signal and the DC biased signal.

     
  9. 9.

    The circuit should be analyzed both at the highest and lowest swings around the quiescent point from the input signal.

     
  10. 10.

    The collector resistor should be chosen in order to maximize voltage swing while preventing overdrive.

     
  11. 11.

    Weak AC signals often need multiple amplification stages to provide sufficient output power for driving outputs.

     
  12. 12.

    The output of one amplifier can be coupled through a capacitor into the input of a second amplifier.

     
  13. 13.

    If the signal sources and outputs are not permanently connected, adding a bleed-off resistor will enable the coupling capacitors to drain out after the jacks are disconnected.

     
  14. 14.

    There are numerous things that can go wrong in a transistor amplifier, including clipping the audio signal, having a bad quiescent current, or accidentally losing your gain through bad resistor choices, not to mention just simply building the circuit wrong.

     
  15. 15.

    Because of the number of things that can go wrong, it is easiest to diagnose problems in an amplifier using an oscilloscope, which allows you to visualize what is happening at each point in the circuit.

     

Apply What You Have Learned

  1. 1.

    What is the purpose of the resistor in the collector of a transistor amplifier?

     
  2. 2.

    What is the purpose of the resistor in the emitter of a transistor amplifier?

     
  3. 3.

    Why is there a bias voltage on the base of the transistor? Why can’t the signal just be connected in directly to the base?

     
  4. 4.

    Why is the signal coupled in through a capacitor?

     
  5. 5.

    Why does the single stage voltage amplifier discussed in this chapter invert its output?

     
  6. 6.

    If the output of the two-stage amplifier is coupled into a third stage, the signal current would swing 1.85 mA in either direction. Design a third amplification stage which can handle this amount of current.

     
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