Inserting a resistor R_{E} in the emitter circuit as in Figure below causes *degeneration*, also known as negative feedback. This opposes a change in emitter current I_{E} due to temperature changes, resistor tolerances, beta variation, or power supply tolerance. Typical tolerances are as follows: resistor— 5%, beta— 100-300, power supply— 5%. Why might the emitter resistor stabilize a change in current? The polarity of the voltage drop across R_{E} is due to the collector battery V_{CC}. The end of the resistor closest to the (-) battery terminal is (-), the end closest to the (+) terminal it (+). Note that the (-) end of R_{E} is connected via V_{BB} battery and R_{B} to the base. Any increase in current flow through R_{E} will increase the magnitude of negative voltage applied to the base circuit, decreasing the base current, decreasing the emitter current. This decreasing emitter current partially compensates the original increase.

Note that base-bias battery V_{BB} is used instead of V_{CC} to bias the base in Figure above. Later we will show that the emitter-bias is more effective with a lower base bias battery. Meanwhile, we write the KVL equation for the loop through the base-emitter circuit, paying attention to the polarities on the components. We substitute I_{B}≅I_{E}/β and solve for emitter current I_{E}. This equation can be solved for R_{B} , equation: RB emitter-bias, Figure above.

Before applying the equations: RB emitter-bias and IE emitter-bias, Figure above, we need to choose values for R_{C} and R_{E} . R_{C} is related to the collector supply V_{CC} and the desired collector current I_{C} which we assume is approximately the emitter current I_{E}. Normally the bias point for V_{C} is set to half of V_{CC}. Though, it could be set higher to compensate for the voltage drop across the emitter resistor R_{E}. The collector current is whatever we require or choose. It could range from micro-Amps to Amps depending on the application and transistor rating. We choose I_{C} = 1mA, typical of a small-signal transistor circuit. We calculate a value for R_{C} and choose a close standard value. An emitter resistor which is 10-50% of the collector load resistor usually works well.

Our first example sets the base-bias supply to high at V_{BB} = V_{CC} = 10V to show why a lower voltage is desirable. Determine the required value of base-bias resistor R_{B}. Choose a standard value resistor. Calculate the emitter current for β=100 and β=300. Compare the stabilization of the current to prior bias circuits.

An 883k resistor was calculated for R_{B}, an 870k chosen. At β=100, I_{E} is 1.01mA.

For β=300 the emitter currents are shown in Table below.

*Emitter current comparison for β=100, β=300.*

Bias circuit | IC β=100 | IC β=300 |
---|---|---|

base-bias | 1.02mA | 3.07mA |

collector feedback bias | 0.989mA | 1.48mA |

emitter-bias, V_{BB}=10V | 1.01mA | 2.76mA |

Table above shows that for V_{BB} = 10V, emitter-bias does not do a very good job of stabilizing the emitter current. The emitter-bias example is better than the previous base-bias example, but, not by much. The key to effective emitter bias is lowering the base supply V_{BB} nearer to the amount of emitter bias.

How much emitter bias do we Have? Rounding, that is emitter current times emitter resistor: I_{E}R_{E} = (1mA)(470) = 0.47V. In addition, we need to overcome the V_{BE} = 0.7V. Thus, we need a V_{BB} >(0.47 + 0.7)V or >1.17V. If emitter current deviates, this number will change compared with the fixed base supply V_{BB},causing a correction to base current I_{B} and emitter current I_{E}. A good value for V_{B} >1.17V is 2V.

The calculated base resistor of 83k is much lower than the previous 883k. We choose 82k from the list of standard values. The emitter currents with the 82k R_{B} for β=100 and β=300 are:

Comparing the emitter currents for emitter-bias with V_{BB} = 2V at β=100 and β=300 to the previous bias circuit examples in Table below, we see considerable improvement at 1.75mA, though, not as good as the 1.48mA of collector feedback.

*Emitter current comparison for β=100, β=300.*

Bias circuit | IC β=100 | IC β=300 |
---|---|---|

base-bias | 1.02mA | 3.07mA |

collector feedback bias | 0.989mA | 1.48mA |

emitter-bias, V_{BB}=10V | 1.01mA | 2.76mA |

emitter-bias, V_{BB}=2V | 1.01mA | 1.75mA |

How can we improve the performance of emitter-bias? Either increase the emitter resistor R_{E} or decrease the base-bias supply V_{BB} or both. As an example, we double the emitter resistor to the nearest standard value of 910Ω.

The calculated R_{B} = 39k is a standard value resistor. No need to recalculate I_{E} for β = 100. For β = 300, it is:

The performance of the emitter-bias circuit with a 910 emitter resistor is much improved. See Table below.

*Emitter current comparison for β=100, β=300.*

Bias circuit | IC β=100 | IC β=300 |
---|---|---|

base-bias | 1.02mA | 3.07mA |

collector feedback bias | 0.989mA | 1.48mA |

emitter-bias, V_{BB}=10V | 1.01mA | 2.76mA |

emitter-bias, V_{BB}=2V, R_{E}=470 | 1.01mA | 1.75mA |

emitter-bias, V_{BB}=2V, R_{E}=910 | 1.00mA | 1.25mA |

As an exercise, rework the emitter-bias example with the emitter resistor reverted back to 470Ω, and the base-bias supply reduced to 1.5V.

The 33k base resistor is a standard value, emitter current at β = 100 is OK. The emitter current at β = 300 is:

Table below below compares the exercise results 1mA and 1.38mA to the previous examples.

*Emitter current comparison for β=100, β=300.*

Bias circuit | IC β=100 | IC β=300 |
---|---|---|

base-bias | 1.02mA | 3.07mA |

collector feedback bias | 0.989mA | 1.48mA |

emitter-bias, V_{BB}=10V | 1.01mA | 2.76mA |

emitter-bias, V_{BB}=2V, R_{B}=470 | 1.01mA | 1.75mA |

emitter-bias, V_{BB}=2V, R_{B}=910 | 1.00mA | 1.25mA |

emitter-bias, V_{BB}=1.5V, R_{B}=470 | 1.00mA | 1.38mA |

The emitter-bias equations have been repeated in Figure below with the internal emitter resistance included for better accuracy. The *internal emitter resistance* is the resistance in the emitter circuit contained within the transistor package. This internal resistance r_{EE} is significant when the (external) emitter resistor R_{E} is small, or even zero. The value of internal resistance R_{EE} is a function of emitter current I_{E}, Table below.

*Derivation of r _{EE}*

r_{EE}= KT/I_{E}m where: K=1.38×10^{-23}watt-sec/^{o}C, Boltzman's constant T= temperature in Kelvins ≅300. I_{E}= emitter current m = varies from 1 to 2 for Silicon r_{EE}≅ 0.026V/I_{E}= 26mV/I_{E}

For reference the 26mV approximation is listed as equation rEE in Figure below.

*Emitter-bias equations with internal emitter resistance r _{EE} included..*

The more accurate emitter-bias equations in Figure above may be derived by writing a KVL equation. Alternatively, start with equations IE emitter-bias and R_{B} emitter-bias in Figure previous, substituting R_{E} with r_{EE}+R_{E}. The result is equations IE EB and RB EB, respectively in Figure above.

Redo the R_{B} calculation in the previous example emitter-bias with the inclusion of r_{EE} and compare the results.

The inclusion of r_{EE} in the calculation results in a lower value of the base resistor R_{B} a shown in Table below. It falls below the standard value 82k resistor instead of above it.

*Effect of inclusion of r _{EE} on calculated R_{B}*

r_{EE}? | r_{EE} Value |
---|---|

Without r_{EE} | 83k |

With r_{EE} | 80.4k |

**Bypass Capacitor for R _{E}**

One problem with emitter bias is that a considerable part of the output signal is dropped across the emitter resistor R_{E} (Figure below). This voltage drop across the emitter resistor is in series with the base and of opposite polarity compared with the input signal. (This is similar to a common collector configuration having <1 gain.) This degeneration severely reduces the gain from base to collector. The solution for AC signal amplifiers is to bypass the emitter resistor with a capacitor. This restores the AC gain since the capacitor is a short for AC signals. The DC emitter current still experiences degeneration in the emitter resistor, thus, stabilizing the DC current.

What value should the bypass capacitor be? That depends on the lowest frequency to be amplified. For radio frequencies Cbpass would be small. For an audio amplifier extending down to 20Hz it will be large. A “rule of thumb” for the bypass capacitor is that the reactance should be 1/10 of the emitter resistance or less. The capacitor should be designed to accommodate the lowest frequency being amplified. The capacitor for an audio amplifier covering 20Hz to 20kHz would be:

Note that the internal emitter resistance r_{EE} is not bypassed by the bypass capacitor.

Article extracted from Tony Kuphaldt’s Lesson in Electric circuits Volume III Semiconductors under the terms and conditions of Design Science License.