Variations in bias due to temperature and beta may be reduced by moving the V_{BB} end of the base-bias resistor to the collector as in Figure below. If the emitter current were to increase, the voltage drop across R_{C} increases, decreasing V_{C}, decreasing I_{B} fed back to the base. This, in turn, decreases the emitter current, correcting the original increase.

Write a KVL equation about the loop containing the battery, R_{C} , R_{B} , and the V_{BE} drop. Substitute I_{C}≅I_{E} and I_{B}≅I_{E}/β. Solving for I_{E} yields the IE CFB-bias equation. Solving for I_{B} yields the IB CFB-bias equation.

Find the required collector feedback bias resistor for an emitter current of 1 mA, a 4.7K collector load resistor, and a transistor with β=100 . Find the collector voltage V_{C}. It should be approximately midway between V_{CC} and ground.

The closest standard value to the 460k collector feedback bias resistor is 470k. Find the emitter current I_{E} with the 470 K resistor. Recalculate the emitter current for a transistor with β=100 and β=300.

We see that as beta changes from 100 to 300, the emitter current increases from 0.989mA to 1.48mA. This is an improvement over the previous base-bias circuit which had an increase from 1.02mA to 3.07mA. Collector feedback bias is twice as stable as base-bias with respect to beta variation.

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