  24 February 2009
Author: Giorgos Lazaridis
BJT Transistor theory

Transistor Circuit Essentials
Voltage Divider Biasing The most effective method to bias the base of a transistor amplifier is using a voltage divider. In the next chapter, we will analyze each transistor connection in detail and we will be using always this biasing method. Therefore, let's take some time to explain this method thoroughly.

The idea is that the voltage divider maintains a very stable voltage at the base of the transistor, and since the base current is many times smaller than the current through the divider, the base voltage remains practically unchanged. The resistor Re provides the negative feedback as explained before (Emitter Feedback Bias). Due to the fact that the base voltage remains unchanged, the negative feedback works very effectively and any unwanted increment of the current gain produces an -almost- equal negative feedback. The collector and emitter currents change just a few, and the Q point remains practically stable. Now, let's see in detail how this works...

Voltage Divider Bias Equations

We start with the assumption that the base current (IB) is many times smaller than the current through the voltage divider (IVD). A ratio of 20 is a good approach. This means that the base current must be at least 20 times smaller than the voltage divider current. This condition allows us to exclude IB from our calculations, with an error less than 5%. Now we can safely calculate the base voltage as follows:

VB = IVD x R2

Or using the classic voltage divider equation:

VB = (VCC x R2) / (R1 + R2)

The current that flows through the voltage divider is:

IVD = VCC / (R1 + R2)

From the base voltage we can calculate the emitter voltage and the collector-emitter voltage drop as follows:

VE = VB - VBE

VCE = VC - VE

The emitter current is calculated using the Ohm's law:

IE = VE / RE

And since the collector current is practically equal to the emitter current, we can calculate all the transistor currents and voltages:

VRC = IC x RC

VC = VCC - VRC = VCC - IC x RC

VCE = VCC - IC x RC - IE x RE = VCC - IC (RC + RE)

As you see, we can calculate everything we need without using any hybrid parameter. This is an amazing and unexpected result. Two transistors with different current gains can operate as amplifiers with exactly the same amplification, only because they are biased with a voltage divider.

Moreover, since VBE is many times smaller than VB and VB remains unchanged all the time, the emitter voltage VE remains also unchanged, hence maintaining a stable emitter current.

Firm and Stiff Voltage Divider

Previously, we made the assumption that the voltage divider current IVD is many times bigger than the base current IB, about 20 times as big. This is a good approach for an error less than 5%. This is not always possible thought. If the base current is high, the resistor values for the voltage divider must become very small, and this leads to numerous problems.

In such cases, we design the voltage divider with a ratio of 10 instead of 20. This approach has an error less than 10% when the IB is excluded from the calculations, which is still acceptable. The voltage divider that satisfies this condition is named Firm Voltage divider:

IVD > 10 x IB => RVD < 0.1 x Î²dc x RE (do not forget that Î² = hfe)

On the other hand, the application may require a very good Q stability with an error less than 1%. A ratio of 100 can be used to calculate the resistors, if this is possible:

IVD > 100 x IB => RVD < 0.01 x Î²dc x RE

The voltage divider that satisfies this condition is named Stiff Voltage divider, and has an error of less than 1%.

Condition Confirmation

Suppose that the designer wants to design a transistor amplifier with stiff voltage divider bias. He designed a circuit that has (along others) emitter current IE=1mA. The voltage divider he designed is calculated according to the stiff VDB condition, which means that the base current must be 100 times smaller than the voltage divider current. According to this calculation, the maximum base current cannot be greater than 40uA. The question now is: Does this circuit works efficiently for the whole hfe range?

The fact that IB and hfe are excluded from the calculations, does not mean that these values do not affect the operation. They still have a small affect, but this is very small (1 to 10%). What we have to confirm now is that this affect will always remain small, even at the worst case scenario.

But which is the "worst case scenario"? Well, simply: The worst case scenario is when the transistor operates with minimum current amplification. When this happens, the base current becomes maximum to supply the required emitter current. Suppose that the transistor that our designer used, has an hfe with range from 30 to 300. We have to confirm that the base current will remain under the calculated value (40uA) and still it will be able to provide full emitter current (1mA), even at the lowest hfe (30):

IE = Î² x IB => IB = IE / Î² = 1mA / 30 => IB = 33uA

So, the base current for the worst case scenario (33uA) is still less than the calculated base current (40uA), therefore we can say that this voltage divider remains stiff.

What each part does

Designing a transistor amplifier with VDB (Voltage Divider Bias) is not that hard, but sometimes it takes time to select the proper part values to begin with. And many times the designer has to change some parts to change the amplifier parameters. Here is a quick reference for the designer to know what each part controls:

• R1 - This resistor should be used to control the current through the voltage divider
• R2 - This resistor controls the base voltage VB
• RE - This resistor controls the emitter current IE
• RC - The collector resistor can control the VCE voltage

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