Improving the Amplifier through Feedback

Lesson 3 - Improving the Amplifier through Feedback

In the previous lesson, we designed a simple amplifier, simulated it and demonstrated its shortcomings using SPICE. In this lesson, we are going to improve on the simple design to make it into something useful.

Adding Feedback

Just like with most human endeavors, a common way to improve operation of a circuit is by providing feedback.
The basic problem with our previous circuit was that the bias circuit was fixed. The purpose of the bias circuit is to keep the collector voltage in the middle of the linear range by applying the right amount of base current, but our fixed bias circuit did not know if the collector voltage was actually in the middle of its voltage range, and was not able to detect and correct when a temperature change (or a part change) had moved the collector voltage outside its normal range and the transistor into saturation or cutoff.
Another problem is that we actually had to provide a bias voltage. In our initial model, the bias voltage came from the microphone, but it's impractical in real life. The circuit should provide its own bias voltage and do not depend on the source.
Ideally, we should design a bias circuit that adjusts the bias up and down as needed to keep the collector voltage in the middle of its range.
The parameters we are trying to compensate for are

1. variations in threshold voltage and
2. variations in current gain.

If we use a fixed bias circuit and the threshold voltage goes up (such as at low temperature), the transistor will turn off and the collector voltage will go up. To restore the operating point, we should increase the bias voltage to increase the base current and turn the transistor on harder.
If we use a fixed bias voltage and the current gain goes up, the transistor will turn on harder and the collector voltage will drop. In this case, we should reduce the base current.
In either case, when the collector voltage goes up, we need to turn the transistor on harder (increase the base current), and conversely when the collector voltage goes down, we need to reduce the base current.
Let's think about it for a minute: what circuit could we install to perform this operation?
.
.
.
Your time is up. A very simple solution is to install a resistor between collector and base. When the collector voltage goes up, the voltage across this resistor will go up, and therefore the current through it will go up as well.

You will notice two changes: 1) we added a resistor between collector and base (value R for now, we will calculate it next), and 2) we added a capacitor in series with the source. The purpose of a capacitor is to let AC signals go through, but block DC voltages. In this case, it will eliminate the consequences of a DC offset from the source. The offset could be zero or anything within the rating of the capacitor and it will not affect circuit operation. We will also compute the proper value of the capacitor later in another lesson.

Calculating the feedback resistor

The transistor controls the output voltage by drawing more or less current through the collector-emitter junction.
Therefore, the output voltage (the collector voltage) can possibly have any value between approximately zero (when the transistor is driven hard, we call this saturation) and the supply voltage V1 (when the transistor is turned off, we call this cut-off.)
The range of operation between saturation and cut-off is called the Linear Range.
The goal is to design the circuit so that without any input signal, the collector voltage will be approximately in the middle of the Linear Range (6 V in this case). In doing this, we will allow the maximum possible excursion of the collector voltage up and down before it is clipped.
To place the collector voltage at 6 V, the transistor needs to pull enough collector current to cause a 6 V drop across R1 (since the supply voltage is 12 V).
For the transistor to pull this much collector current (Ic), it should receive base current in the amount of Ic/Hfe where Ic is the collector current and Hfe is the current gain (ratio of collector current divided by the base current that caused it).

I_b = I_c / H_fe

So if we can calculate the amount of current we need through R1 to cause a 6 V drop, we will know the collector current. Then, knowing the current gain (from looking at the data sheet), we can calculate the necessary base current. Knowing the base current we need, we can compute the value of R3 since we know the voltage across R3.
Let's find the current gain of the transistor. For that, we should look up the part in the data sheet. The picture below is a part of the data sheet that deals with the current gain. For the complete document, click on the picture.

The data sheet looks probably more confusing than we would like at first. It does not give us a value of current gain, but seven!!! Please note that the current gain we denoted as ß in lesson 1, is here denoted h_fe. Both notations are equivalent and we will use either in the following. First of all, Note 1 indicates that the test is done in pulse mode, probably because the operating conditions would make the transistor hot if applied continuously, so by operating in pulse mode, we can control the heating. This information is not really relevant for our purpose.
Then the different values of gain are a function of the collector current (mostly) and one value is specified at 1.0V of collector voltage instead of 10 V. This information is useful because as the transistor nears saturation (when the collector voltage is very low), it's characteristics change from the more general "linear" mode.
Since we want to operate the transistor in linear mode, let's ignore the data at 1.0 V for now.
Finally, these are only minimum values. There is a maximum value given at 150 mA of collector current, and it is about 3 times the minimum value. There is no typical value specified. We will have to pick a value for our calculations.
In our circuit, with a 1 k ohm collector resistor, when the collector voltage is 6V, there will also be 6V across resistor R1, and therefore the current through it will be 6 mA. Therefore, the current gain will be somewhere between the values for 1.0 mA and 10 mA, we'll estimate the minimum gain to be around 60. Since the maximum value is about 3 times the minimum (180), we will also estimate the typical gain value at 6 mA to be somewhere in-between, around 120.
So, when the collector voltage is 6V, the collector current will be very close to the current in R1, and if the current gain is 120, the base current should be:

I_b = I_c / h_fe [Eq 1]

So, resistor R1 will have (6 - 0.7) Volt across it while carrying I_b therefore the value of R1 should be:

R1 = (6 - 0.7)/I_b [Eq 2]

We can consolidate and generalize these equations into:

R1 = (V_c - V_b) * H_fe / I_c [Eq 3]

With the values calculated above, R1 = 108 k ohm, which we will round to 100k ohm. Let's give C1 a value of 10 µF for now (we will learn how to compute it in another lesson) by right-clicking on it and entering a value of "10u", make sure temperature stepping is turned on (the .STEP statement is a directive, not a comment) and run the simulation, plotting the collector voltage.
Here is what I got:

What a difference with the circuit in Lesson 2!!!
Here are a few observations:

• The waveforms look the same at all temperatures
• They are slightly offset from each other (the value at the beginning of the simulation varies from about 3.8 to 4.4 V) as a result of temperature stepping.
• The peak-to-peak output voltage is around 4.0 V instead of the 6.0 V we designed the circuit for.

It looks like our circuit modification did what we wanted, the only significant departure from our calculations is the peak-to-peak output voltage, which at 4.0 V is about 2.0 V below what we calculated. The difference may be in the current gain. We used a specification from a certain vendor, Motorola in this case, where the 2N2222 model used by Spice may have a different nominal gain value. What is interesting, is that the waveforms look fine, and as long as we do not need the circuit to deliver more than 3 or 4 Volts peak-to-peak, it will work fine as it is.

Exercises

1. Find the effect, if any, of the DC Offset of Source V2 on the collector voltage or the waveforms.
2. Find a value of R3 which will set the average collector voltage close to 6.0 V when using the 2N2222 model.
3. Replace the 2N2222 transistor with a 2N3055 and run the simulation again. Observe the results and compare to the circuit in Lesson 2. Click here for the plots.
4. Modify Eq 3 as follows:
   • Vb (0.7 V) being negligible compared to Vc, replace (Vc - Vb) with Vc
   • Since we want Vc to be half the supply voltage Vs, we can replace Vc with Vs/2, or reciprocally Vs/2 with Vc.
   • The current gain is sufficient (> 50) to make Ib negligible compared to Ic. Therefore, replace Ic with an expression using the supply voltage Vs and R1.
   You should end up with an equation using only two parameters which is independent of the supply voltage.

Click here for the answers.

Conclusions of this lesson

• We learned about feedback and how it can stabilize a circuit over temperature and other parameter variations.
• We learned how to look for critical information in a part data sheet.
• We learned how to design a simple amplifier that would work over a broad range of conditions and part variations.

In the next lessons, we will learn how to compute other component values and how to optimize the circuit for various applications.