# A question about Transistor in forward-active biasing?

I have question about transistor biasing. Between picture 1 and 2, what's difference, why we add the resistor RE in pic 2 ? what's it effect to circuit ? Similarity for pics 3 and 4, instead of connecting RB to Vcc, why do they connect RB to the collector of BJT ? And which is the better way to bias a BJT among 4 pics ? thanks very much :)

• "Better way" depends very much on the specific application. You'd have to tell us something about what you're trying to accomplish before we could answer that. – Dave Tweed Mar 11 '15 at 13:07
• Circuit 2 and 4 has improved bias stability because of negative feedback. – nidhin Mar 11 '15 at 13:12
• In circuits 1 and 2 if the intention is to switch the transistor hard on or off I woildn't have the emitter resistor but if the intention is to set the collector at a particular voltage then the emmitter resistor makes it easier to tolerance as the emmitter will be about 0.6 volt below the base so we know the current in the emitter risistor and assuming hfe is large you will have almost the same current in the collector resistor. As for which is best - it depends what you are trying to achieve. – Warren Hill Mar 11 '15 at 13:15

As Olin said, circuit shown in #1 and #3 are completely open loop. So the bias stability of the circuit is less and it can even lead to thermal runaway. The bias stability can be improved by including a negative feedback mechanism in these circuits. Circuits #2 and #4 does that.

Negative feedback in circuit #2: Assume that the collector current increases. This leads to increase in voltage across $R_E$. So the voltage at emitter increases. Since $V_{BE}$ remains almost constant, the voltage at base also increases. This leads to decrease in base current as an effect the collector current also reduces. So there exists a negative feedback to stabilize the operating point.

$$I_C\uparrow I_ER_E\uparrow V_E\uparrow V_B\uparrow I_B\downarrow I_C\downarrow$$

Negative feedback in circuit #4: Assume that the collector current increases. This leads to increase in voltage across $R_C$. So the voltage at collector decreases. This leads to decrease in base current as an effect the collector current also reduces. So here also there is a negative feedback to stabilize the operating point.

$$I_C\uparrow (I_C+I_B)R_C\uparrow V_C\downarrow I_B\downarrow I_C\downarrow$$

PS: The negative feedback affects the ac signal also which will reduce the gain of the amplifier. To avoid that a bypass capacitor is usually connected in parallel to $R_E$ in circuit #2.

The big difference between #1 and #2 is that #1 is completely open loop. Transistor gain varies widely, so it's nearly impossible to come up with values for RB1 and RB2 so that Vce is near the middle of its range. The emitter resistor in #2 provides some feedback so that the operating point is less a function of the transistor gain. The downside is that the overall circuit gain will be lower.

#3 and #4 are again the same issue. #3 is open loop, so the operating point is directly a function of the transistor gain. The biasing method of #4 has feedback that stabalizes the operating point to make it less a function of gain. Consider #4 with the collector voltage being nicely in the middle of its range (roughly Vcc/2), but now a different transistor is used that has twice the gain. At the same base current, the collector current would be twice as much, which means twice as much voltage across Rc, bringing the collector operating point very low. However, as the collector goes lower, there is less voltage across Rb, which causes less base current, which causes less collector current, which raises the collector voltage.

Talking about "best" is pointless and bad engineering without a spec about what the circuit is supposed to do. Basically "best" is meaningless handwaving without a way to measure bestness.

Re in pic 2 provides a negative feedback against temperature induced current drift. The more current the junction starts to conduct the higher the drop and therefore less bias on the emitter junction.

Circuit 4 is also a feedback, don't have pencil at hand but it probably makes the amplification a factor of the resistor ratio rather than the beta which ends up appearing in a ratio to itself (and since beta can vary wildly, it's a good thing)