  24 February 2009
Author: Giorgos Lazaridis
BJT Transistor theory

Transistor Circuit Essentials
Choosing the right bias

After selecting the proper connection, the one that is most suitable for your application, you must select a biasing method. Biasing in general means to establish predetermined voltages and currents at specific points of a circuit, so that the circuit components will operate normally. For transistors, biasing means to set the proper voltage and current of the transistor base, thus setting the operating point, also known as quiescence point (Q). We will discuss in details the quiescence point within the next chapters. For now, you need to know that this point will determine how the transistor will operate (amplifier or switch). A correctly placed Q offers maximum amplification without signal distortion or clipping.

The most efficient and commonly used biasing method for transistor amplifiers, it the voltage divider bias (VDB). We will analyze this method in detail, but first we need to discuss the other biasing methods. In this chapter, we will use a Common Emitter NPN transistor amplifier to analyze the various biasing methods, but each method can be used for other connections as well.

Fixed bias This is the most rarely used biasing method with transistor amplifiers, but it is widely used when the transistor operates as a switch. The base current IB is controlled by the base resistor RB. From the second law of Kirchhoff, we have:

VCC = VB + VBE

VB is calculated using the Ohm's law:

VB = IB x RB

So, by selecting the proper base resistor RB, we can define the required base voltage VB and base current IB. Now we can calculate the collector current using the appropriate hybrid parameter. Since this is a common emitter circuit, we use the hfe:

IC = IB x hfe

The problem with this method is that the collector current is very sensitive is slight current gain changes. Suppose for example that this is a silicon transistor and operates as a B-class amplifier with current gain 300, RB=80 Kohms, RC=200 Ohms and VCC = 10 volts:

VCC = VB + VBE => VCC = IB x RB + VBE => IB = (VCC - VBE) / RB = (10-0.7) / 80000 = 112.25 uA

IC = IB x 300 = 33.67mA

The output of this circuit is taken from the collector resistor RC:

VRC = IC x RC = 6.7 Volts

Now suppose that the temperature rises. As we've discussed in earlier pages, this will increase the current gain. An increment by 15% is a realistic and rather small value. From 300 it will climb up to 345. This means that the collector current will become 38.7mA, and the output voltage will also become 7.7 Volts! A whole volt higher than before. That is why this biasing method is not used for transistor amplifiers.

On the other hand, due to the fact that this method is very simple and cost-effective, it is widely used when the transistor operates as a load switch, for example as a relay or LED driver. That is because the Q point operates from cut-off to hard saturation, and even large current gain changes have little or no effect at the output.

Emitter feedback bias (Fixed bias with emitter resistor) This is the first method that was historically used to fix the problem of the unstable current gain discussed previously. In a transistor circuit with fixed bias, a resistor was added at the emitter. This method never worked as it should, so it is not used anymore. This is how it was supposed to work. If the collector current is increased due to a temperature increment, the emitter current is also increased, thus the current through RE is also increased. The voltage drop across RE is increased (emitter voltage) which eventually increases the base voltage. Finally, this base voltage increment has as a result the decrement of the voltage across the base resistor RB, which eventually decreases the current of the base IB. The idea is that this base current decrement decreases also the collector current!

This sounds amazing since a change of the output of the circuit has an effect on the input. This effect is called "feedback" and more specifically it is a "negative feedback", since an output increment causes an input decrement. Here is how the new collector current is calculated:

IC = (VCC - VBE) / (RE + (RB / hfe))

Let's see how the previous circuit (Fixed bias) would react if we add a 100 Ohms RE feedback resistor.

IC = (10-0.7) / (100 + (80000 / 300)) = 9.3 / 366.6 = 25.3 mA

We assume again that the current gain is increased by 15%:

IC = (10 - 0.7) / (100 + (80000 / 345)) = 9,3 / 331,88 = 28mA

So, a 15% current gain increment caused a 15.1% output current increment. By adding a 100 Ohms feedback resistor at the emitter, a 15% current gain increment caused a 10.6% output current increment. The increment is 4.5% less which means that this method works somehow, but still the shifting of the Q-point is too large to be acceptable.

Collector feedback bias (Collector to base bias) The next method that the researchers used to stabilize the Q point is the collector feedback bias. According to this method, the base resistor is not connected at the power supply, instead it is connected at the collector of the transistor. If the current gain is increased due to temperature increment, the current through the collector is increased as well, and this decreases the voltage on the collector VC. But the base resistor is connected at this point, so less current will go through the resistor in the base. Less current through the base means less current through the collector.

Again, there is a negative feedback in this circuit. But how much is it? Lets do some math. The collector current is now calculated by the following formula:

IC = (VCC - VBE) / (RC + (RB / hfe))

To see the change, we will apply this formula in our first example (fixed bias):

IC = (10 - 0.7) / (100 + (80000 / 300)) = 9.3 / 366.6 = 25.3 mA

When the current gain is increased by 15%:

IC = (10 - 0.7) / (100 + (80000 / 345)) = 9.3 / 331.8 = 28 mA

The effectiveness of this method compared to the emitter resistor feedback bias shown before is exactly the same. The difference is that, RC is usually much larger than RE, which results in higher stability. Nevertheless, quiescence point Q cannot be considered stable.

Collector feedback bias (Collector to base bias) It did not take long before someone tried to utilize both the previous methods to work together to achieve better results. And indeed, the stabilization is much better than each one separately. The formula to calculate the collector current is the following:

IC = (VCC - VBE) / (RC + RE + (RB / hfe))

Let's apply this formula to our previous examples

IC = 9.3 / (100 + 100 + (80000 / 300)) => IC = 9.3 / 466.6 = 19.9 mA

With a 15% current gain increment:

IC = (VCC - VBE) / (RC + RE + (RB / hfe)) = 9.3 / (100 + 100 + (80000 / 345)) = 21.3 mA

So, a 15% current gain increment causes a 7% output current increment. Although it is better than the previous circuits, still the Q point is not stable enough. Add to this that hfe is extremely sensitive to temperature changes and the transistor generates a lot of heat when it operates as a power amplifier. So we need a much better stabilization technique.

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