In this section we will consider a number of BJT circuits and perform the DC circuit analysis. For those circuits with an active mode BJT, we’ll assume that V_{BE} = 0.7V (npn) or V _{EB} = 0.7V (pnp). Example N12.1 (text example 5.4). Compute the node voltages and currents in the circuit below assuming ß = 100.
If the BJT is in the active mode, V_{be} = 0.7V then
With I_{c} Ia_{e} = then
Consequently, using KVL
V_{c}=10I_{e}R_{c}=
100.99*10^{3>}4.7*10^{3}=5.3V
Finally, using KCL I_{B}+I_{C}=I_{E},or I_{B}=I_{E}I_{C}=10.99=0.01mA
Now we’ll check to see if these values mean the BJT is in the active mode (as assumed).
• V_{CB} = 5.34 =1.3 V. This is greater than zero, which means the CBJ is reversed biased.
• V_{BE} =0.7 V. This is greater than zero, which means the EBJ is forward biased. Because the CBJ is reversed biased and the EBJ is forward biased, the BJT is operating in the active mode. Note that in the text, they show a technique for analyzing such circuits right on the circuit diagram in .
Example N12.2 (text example 5.5). Repeat the previous
example but with V_{B} =6 V. Assuming the BJT is operating in the active mode:
From the last calculation 2.57 C V_{c = 2.57VÞ VCB }= 3.43 V. Consequently, the BJT is not in the active mode because the CBJ is forward biased. A better assumption is the transistor is operating in the saturation mode. We’ll talk more about this later. For now, suffice it to say that in the saturation mode V_{CEsat} »0.2 V (see Section 5.3.4). Assuming this and reanalyzing the circuit:
Notice that
This ratio is often called forced ß . Observe that it’s not equal to 100, as this ratio would be if the transistor were operating in the active mode (see Section 5.3.4).
Example N12.3 (text example 5.7). Compute the node voltages and currents in the circuit below assuming ß = 100. To begin, we’ll assume the pnp transistor is operating in the active mode.
Now check if the BJT is in the active mode:
• EBJ? Forward biased.
• CBJ? Reversed biased.
So the BJT is in the active mode, as originally assumed.
Example N12.4 (text exercise D5.25). Determine the largest RC that can be used in the circuit below so that the BJT remains in the active mode. (This circuit is very similar to the one in the previous example.)
We’ll begin by assuming the BJT is operating in the active
mode. In the active mode, the CBJ needs to be reversed biased. The lowest voltage across this junction for operation in the active mode is V_{CB}=0 ÞV_{C}=V_{B}=0V
Therefore, by KVL
10+R _{C}I_{c}=0
or
This value of RC and smaller is required for the BJT to operate in the active mode. Example N12.5 (text example 5.10). Determine the node voltages and currents in the circuit shown below. Assume the BJT is operating in the active mode with ß=100 .
First, we’ll use Thévenin’s theorem to simplify the base circuit
The Thévenin equivalent resistance and voltage are then
Using this Thévenin equivalent for the base circuit, the overall circuit is then
To find the emitter current, we’ll apply KVL over the loop
shown giving 5=33.3*10^{3}.I_{B}+0.7+3,000.I_{E}The quantity of interest is I_{B}. With CB I_{C}= Iß _{B}=and I_{C} =Ia_{E} for a BJT in the active mode, we find
Using this in the KVL equation 50.7=[33.3*10^{3}+3,000(ß+1)]I_{b} With ß=100 then solving this equation we find
Next, by KCL
The node voltages are then
Lastly, let’s check if the BJT is operating in the active mode.
• BE B E V_{BE}= V_{B} V_{E} = 4.57 3.87 =0.7 V This is 0.7 V originally assumed for a forward biased EBJ.
• V_{BC}= V_{B} V _{C}=4.57 8.6=4.03 V. This is less than zero,
which means the CBJ is reversed biased. Therefore the BJT is operating in the active mode, as originally assumed.

