Example –2
Determine R1 and extreme values of I1 in the circuit of Example-1, if Q 1 is replaced with the compound connection of fig. 2. Assume both transistors have a minimum hFE of 15 and negligible cutoff currents.
Solution:
The current through R1 should be selected about 50 percent higher than the maximum current required by the base of Q12. For worst-case design, the minimum hFE of 15 is used, yielding a composite hFE ≈ 15 (15) = 225 for both Q11 and Q12. This yields

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Fig. 2 |
We now select I2≈ 0.65 mA; R1 is calculated as follows:

where, 
The extreme values of I1 occurs when Vi is at its extremes. The minimum I1 is 0.65mA (as indicated above), while the maximum is

The variation in I1 is 1.5 - 0.65 = 0.85mA.
Example - 3
In the regulator shown in fig. 3 R1 = 50KΩ, R2 = 43.75KΩ and VZ = 6.3V. If the 15 V output drops 0.1 V, find the change in VBE2 that results.

Fig. 3
Solution:
When VO = 15 V,
Therefore, VBE2 = V2 - VZ = 7 V - 6.3 V = 0.7 V. When VO = 15 V - 0.1 V = 14.9 V, V2 becomes

Therefore, 
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