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DC Biasing Bipolar Junction Transistors

concept
When we design amplifier circuits we need to set the voltages and currents on the transistor's terminals to the right values in order to get the amplifier to remain in the forward active mode and achieve the amplification we want. Our choices of battery values, resistors and how we connect everything together determines the characteristics of the amplifier circuit. There is no one "correct" way to bias a transistor, it depends on what kind of amplification we want and what our input signal is. It depends on how accurate we need the amplification to be, how free of noise, how large the input signal etc... In this topic we'll cover several options for biasing a BJT and the pros and cons of each as well as how to analyse them. When designing an amplifier circuit the DC bias is the most fundamental part and often the part that needs to be worked on iteratively to achieve the correct results.
fact
The aim of DC biasing a transistor is to have it operating at a particular current and voltage when no signal is applied. We call this point the 'Q' point for 'Quiescent' point. This Q point is selected so that the amplifier remains in the forward active region over the full swing of the input signal and creates the required amplification of the signal.
fact
In transistor amplifier circuits the AC signal input and output will be filtered by a capacitor to remove any DC bias. In DC analysis this means the capacitors act as open circuits and will often not be shown since they have no impact on the circuit analysis.

Fixed Bias Circuit

fact
The fixed bias circuit is shown below: KVL around the battery-base-emitter loop gives us: $$V_{CC} - I_BR_B - V_{BE} = 0$$ Which implies that: $$I_B = \frac{V_{CC} - V_{BE}}{R_B}$$ KVL around the battery-collector-emitter loop gives us: $$V_{CC} - I_CR_C - V_{CE} = 0$$ And therefore $$V_{CE} = V_{CC} - I_CR_C$$
example

Calculate the bias voltages and currents for the following circuit: First we assume $$V_{BE} = 0.7$$V as an initial guess so $$I_B = \frac{V_{CC} - V_{BE}}{R_B} = 47.08\mu$$A Then $$I_C = \beta I_B = 2.35$$mA Now we use our $$I_C$$ to check our $$V_{BE}$$ guess. If it's close we were correct, if not we use the new $$V_{BE}$$ to calculate a new $$I_B$$ then $$I_C$$ as above then find the new-new $$V_{BE}$$ and keep repeating until $$V_{BE}$$ converges. $$V_{BE} = V_T \log(\frac{I_C}{I_S}) = 0.026 \log(\frac{0.00235}{4.77\times 10^{-15}}) \approx 0.7$$ so our guess for $$V_{BE}$$ was good enough and our current calculations based off of it are accurate. Finally $$V_{CE} = V_{CC} - I_CR_C = 12 - 0.00235\times 2200 = 6.83$$V Now we need to check if the transistor is in the forward active mode. If it isn't we've been using the wrong equations. So we want to make sure that the collector current we calculated, $$I_C$$, isn't larger than the saturation current $$I_{C(sat)} = \frac{V_{CC}}{R_C}$$. For this circuit $$I_{C(sat)} = \frac{12}{2200} = 5.45m$$A which is more than our calculated $$I_C$$ so that checks out. Since $$V_{CE} \gt V_{BE}$$ our transistor is in forward active mode and it is therefore true that $$I_C = \beta I_B$$ (just as we wanted).

Emitter Resistance

fact
DC biasing with an emitter resistor provides better bias stability. This means that random variations in temperature, supply voltages and resistor's resistances will have a smaller effect on the transistor than in a circuit without this emitter resistor. We call this "degenerative resistance". A resistor on the emitter is very common in amplifier design for this very reason. fact
$$I_B = \frac{V_{CC} - V_{BE}}{R_B + (\beta + 1)R_E}$$
derivation
Looking at the base-emitter loop we do a KVL and see: $$V_{CC} - I_BR_B - V_{BE} - I_ER_E = 0$$ $$I_E = (\beta + 1)I_B \implies V_{CC} - I_BR_B = V_{BE} - (\beta + 1)R_E = 0$$ Rearranging we get $$I_B = \frac{V_{CC} - V_{BE}}{R_B + (\beta + 1)R_E}$$
example

Find $$V_{CE}$$ for the circuit below and determine whether it is in forward active mode. $$I_B = \frac{V_{CC} - V_{BE}}{R_B + (\beta + 1)R_E} = \frac{20V - 0.7V}{430k\Omega + 101\times 1k\Omega} = \frac{19.3V}{531k\Omega} = 36.35\mu$$A $$I_C = \beta I_B = 3.635$$mA $$\approx I_E$$ $$V_{CE} = V_{CC} - I_CR_C - I_ER_E = 20V - 3.635mA(2k\Omega ) - 3.635mA(1k\Omega ) \approx 9.1$$V Here we've assumed our guess for $$V_{BE}$$ is accurate enough to save some tedious calculations. You should not in general assume that $$V_{BE}$$ is DEFINITELY 0.7V but we do it here to save space and stay on point. To be in forward active we need $$V_{CE} \gt V_{BE}$$ which it is and we need $$I_C \lt I_{C(sat)} = \frac{20V}{2k\Omega + 1k\Omega} = 6.67$$mA so we're good there too. The transistor is in forward active mode.

DC Bias independent of $$\beta$$

concept
The bias topologies we've looked at so far depended on knowing the $$\beta$$ for the transistor. Unfortunately this is a pain since it varies from device to device by quite a lot, takes time to measure and changes as the temperature changes. What we want, ideally, is a way to bias the transistor that works more or less the same no matter which resistor we picked from our pile. This way we can all but guarantee that if we build 100 of these circuits they will all work the same way (a much harder challenge than many students realise). The next topology we look at aims to address this issue.
fact
The next topology is called the voltage divider bias circuit. It is used a lot with BJTs for reasons you will soon see. It's biasing currents and voltages don't change much with a changing $$\beta$$ value or different supply voltages / resistances
fact
The transistor's biasing values are given by: $$V_B = \frac{R_{B2}}{R_{B1} + R_{B2}}V_{CC}$$ $$V_E = V_B - V_{BE}$$ $$I_E = \frac{V_E}{R_E}$$ $$V_{CE} = V_{CC} - I_CR_C - I_ER_E$$ As you can see $$\beta$$ is never mentioned in these equations.
practice problems
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