Editor's note: The following tutorial is one of a series of six on transistor theory by Howard Skolnik, retired Burr-Brown designer.
In Parts 1 and 2 we explored the fundamental terminal impedances of a transistor and their relationship to small signal ac gain. In Part 3 we continue to refine the design topology towards practical, reproducible, solutions.
One way to desensitize the circuit to variations in VBE and temperature is to add an external RE that “swamps” the transistor's internal Re . In addition, this adds negative feedback to stabilize IE and greatly increases the input impedance to the base.
Let RL =10K and RE =1K with IE =1mA. The voltage gain is always RL / RE . But, now REtot = Re + RE . We might expect that RL and RE will dominate, so the voltage gain would be 10 (10K/1K). However, this is NOT exactly correct.
Let’s take a closer look. First, the actual REtot is 1K+26Ω = 1.026K. Second, from Figure 2 in part 1, we recall that Ro ≈100K. With RL =10K, the parallel combination is 9.09K. Therefore, the actual gain will be 9.09K/1.026K = 8.86. This might be good enough, but it is not 10! One needs to do the algebra to find the RL that when in parallel with 100K and divided by 1.026K = 10. The solution is found by solving for RL in [(100* RL )/(100+ RL )/1.026]=10. The answer, RL = 11.43K. Now, when RL (11.43K) is in parallel with Ro (100K), the result is 10.26K. And, 10.26K/1.026K = 10.
That’s the good news. However, with IC =1mA, the DC voltage drop on RL will be about 10V. This suggests other practical limitations. If, for example, the power supply (V1 ) is 15V, the maximum signal amplitude at the output will be limited to about 2Vpp . Further, if we wanted a voltage gain of 100, this would require a “total” RL of 102.6K, which cannot be achieved with the given Ro (100K). Remember, when putting resistors in parallel, the result is always less than any of the original elements. Even if we could yield a total RL of 102.6K, a minimum power supply of more than 110V would be required. A shocking thought!
These results do not take into account the likely loading effect on the signal source and the influence of whatever is connected to the output (external load). For “accurate” results, these factors must also be considered. However, doing so is not difficult. It is just a matter of parallel or series resistors or voltage divider effects that need to be accounted for.
The point of the above exercise is to demonstrate the need for taking real transistor characteristics into account to produce accurate analog performance.
In part 4 we consider the effects of a coupling capacitor and discover how to achieve higher gain.