Question 12.2: For the circuit shown in Fig. 12.8, it is given that: VYY = ......
For the circuit shown in Fig. 12.8, it is given that:\ V_{YY} = 20 V,\ V_{Z1} = 6.8 V,\ V_{Z2} = 3.8 V,\ h_{rb} = 3 × 10^{−4},\ h_{ib} = 20Ω,\ h_{ob} = 0.5μmhos, α = 0.98 and\ R_{E} = 1kΩ. Find the slope error: (a) when\ R_{L} = ∞ (b) when\ R_{L} = 200 kΩ and (c) when\ R_{L} = 50 kΩ.
\ V_{EE} = V_{Z1} + V_{Z2} = 6.8 + 3.8 = 10.6 V
\ V_{CC} = V_{YY} − V_{EE} = 20 − 10.6 = 9.4 V
Taking the junction voltages into account, the sweep voltage is:
\ V_{s} = V_{CC} − V_{BE} = 9.4 − 0.6 = 8.8 V.
(a) The slope error of a transistor constant current sweep is:
\ V_{i} = V_{EE} − V_{γ} = 10.6 − 0.5 = 10.1 V and V_{s} = 8.8 V
\ e_{s} = \frac{8.8}{10.1}\left[3 × 10^{−4} + \frac{0.5 × 10^{−6}}{0.98} (1020)\right]× 100 = \frac{8.8}{10.1} (0.081) = 0.07 %
This error is small and hence, the sweep is linear.
(b) If\ R_{L} is connected as the load, then\ h_{ob} and\ R_{L} are in parallel. The effective admittance is :
If\ R_{L} = 200 kΩ, then the effective admittance:
\ = 0.5 × 10^{−6} + 0.5 × 10^{−5} = 5.5 × 10^{−6}mhos
\ e_{s} = \frac{8.8}{10.1}\left[3 × 10^{−4} + \frac{5.5 × 10^{−6}}{0.98} (1020)\right] × 100 = \frac{8.8}{10.1} (0.60) = 0.52%
In this case the slope error becomes large.
\ e_{s} ≈ 0.52 %
(c) \ R_{L} = 50 kΩ, then the effective admittance
\ = 0.5 × 10^{−6} + 0.2 × 10^{−4} = 20.5 × 10^{−6}mhos
\ = \frac{8.8}{10.1} (2.16) = 1.88%
In this case, the slope error becomes large.
\ e_{s} ≈ 1.88%
The slope error progressively becomes large as\ R_{L} decreases.