Question 12.7: Using the characteristic of UJT shown in Fig. 12.21 (a), cal......

Given,
\ f = 10 kHz,\ V_{s} = 10 V,\ V_{BB} = 15 V,\ V_{V} = 1 V,\ I_{P} = 14 μA.
\ I_{V} = 1 mA, η = 0.6
Therefore,
\ V_{P} = V_{s} + V_{V} = 10 + 1 = 11 V

\ R_{(max)} = \frac{V_{BB} − V_{P}}{I_{P}}= \frac{15 − 11}{14 × 10^{−6}} = 285 kΩ

\ R_{(min)} = \frac{V_{BB} − V_{V}}{I_{V}}= \frac{15 − 1}{1 × 10^{−3}} = 14 kΩ

R should lie between\ R_{(max)} and\ R_{(min)}.
Let R = 150 kΩ

Sweep time\ T_{s} = RC  \log_{e}\left( \frac{V_{BB} − V_{V}}{V_{BB} − V_{P}}\right)= RC  \log_{e} \left( \frac{15 − 1}{15 − 11}\right)

\ T_{s} = 1.252RC              \ T = \frac{1}{f} = \frac{1}{10 × 10^{3}} = 0.1 ms = 100 μs

As\ T_{r} = 0.5% of T,
\ T_{r} = 0.5μs
∴\ T_{s} = T − T_{r} = 100 − 0.5 = 99.5μs        \ RC = \frac{T_{s}}{1.252} = \frac{99.5 × 10^{−6}}{1.252} = 79.4728 × 10^{−6}s

Therefore,
\ C = \frac{79.4728 × 10^{−6}}{150 × 10^{3}} = 0.53 nF

Return time\ T_{r} = R_{1}C= 0.5μs
Therefore,

\ R_{1} = \frac{T_{r}}{C} = \frac{0.5 × 10^{−6}}{0.53 × 10^{−9}} = 943 Ω

The value of\ R_{2} is usually higher than\ R_{1}. So, choose\ R_{2} = 3 kΩ.