Editor's Note: The following tutorial is one of a series of six on transistor theory by Howard Skolnik, retired Burr-Brown designer.
In Part 1 of this series we explored the terminal impedances of a bipolar transistor and identified key operating parameters. In Part 2 we define the ac gain equation and expose factors affecting accurate results.
Again, without getting into semiconductor theory, please accept that the impedance looking into the emitter is a constant at a given operating current (IE ) and temperature and changes inversely with Ie in a linear fashion. This is shown in Figure 3.
If temperature is considered constant at 25o C, Re = 26mV / IE .
To account for a different temperature (T):
As an example, let the temperature of interest be 25o C and the DC operating point (IE ) be 1mA. From the equations above, Re = 26Ω. At 100uA, Re is 260Ω and so forth.
If a small ac signal (Vac ) is applied to the input terminal (transistor base), Vac will be impressed on Re . This will result in a change in Ie (Δ Ie = Vac /Re ). Recall that IC is almost identical to IE (Δ Ie ≈ Δ IC ). Therefore, ΔVout can be considered to be Δ Ie * RL . Substituting (Vac /Re ) for Δ Ie yields:
The “minus” sign, above, accounts for the signal inversion between the base and collector.
Thus, to obtain a gain of 10, would require RL = 10 * Re . With the operating point at 1mA, Re = 26Ω and the required RL is 260Ω. Likewise, a collector load of 2600Ω would yield a gain of 100. These values of RL will yield accurate results because RL << RO (100K).
As far as we have gone, bipolar transistors really are this simple !
If we keep increasing RL (to increase the gain), we will run into limitations due to RO and the available power supply voltage. For example, if RL = 100K, we would expect a gain of about 3800 (100K/26). But wait, RL = RO ! Now we need to consider the parallel combination of RO and RL . Further, 1mA in 100K yields a voltage drop of 100V! Thus, the supply voltage would need to be greater than 100V, which is not likely to be practical. As we will see, there are better ways to achieve high gain.
Improving the design
The simple circuit in Figure 3 has a grounded emitter. Biasing (setting the operating point) used a voltage divider. An alternative could have been a single resistor connected from the transistor base to the supply voltage. However, this technique is also not reliable and not recommended for a real design. A typical silicon transistor has a VBE of about 0.65V. But, the VBE versus current relationship is highly non-linear and strongly temperature dependent. This makes these bias circuits unpredictable with temperature and transistor parameter variations.
One way to desensitize the circuit to variations in VBE and temperature is to add an external RE that “swamps” the transistors internal Re . In addition, this adds negative feedback to stabilize IE and greatly increases the input impedance to the base.
In part 3 we will extend the discussion to understand the effects of RE on overall amplifier performance.