Question 12.8: For the UJT relaxation oscillator shown in Fig.12.21(c), RBB......

Given\ R_{BB} = 3 kΩ,\ η = 0.7.
(a)
\ η = \frac{R_{B1}}{R_{B1} + R_{B2}} = \frac{ R_{B1}}{R_{BB}}

0.7 =\frac{R_{B1}}{3  kΩ}

\ R_{B1} = 0.7 × 3 kΩ = 2.1 kΩ
We have
\ R_{BB} = R_{B1} + R_{B2}

Therefore,
\ R_{B2} = R_{BB} − R_{B1} = 3 kΩ − 2.1 kΩ = 0.9 kΩ
(b) The circuit that enables the calculation of\ V_{P} is shown in Fig. 12.22(d). From Fig. 12.22(d):

\ V_{P} = 0.7  V + V_{BB} × \frac{R_{B1} + R_{1}}{R_{BB} + R_{1}} = 0.7 + 15 × \frac{(2.1 + 0.1)}{(3 + 0.1) }= 11.34 V

(c)
\ R_{(min)} = \frac{(V_{BB} − V_{V })}{I_{V}}= \frac{(15 − 2)}{10 × 10^{−3}} = 1.3 kΩ

\ R_{(max)} = \frac{(V_{BB} − V_{P})}{I_{P}}= \frac{(15 − 11.34)}{0.01 × 10^{−3}} = 366 kΩ

Choose
R = 100 kΩ

(d)\ T_{s} = RC \ln \frac{(V_{BB} − V_{V} )}{(V_{BB} − V_{P})} = 100 × 10^{3} × 50 × 10^{−9} × \ln \frac{(15 − 2)}{(15 − 11.34)} = 6.335 ms

\ f = \frac{1000}{6.335} = 157.85 Hz

\ T_{s} using the value of\ η is given as:
\ T_{s} = RC \ln\frac{1}{(1 − η)} = 100 × 10^{3} × 50 × 10^{−9} \ln \frac{1}{1 − 0.7} = 6.015 ms

\ f = \frac{1000}{6.015} = 166.25 Hz.

(e)\ T_{r} = (R_{B1} + R_{1})C × \ln \frac{V_{P}}{V_{V}}= (0.1 + 0.1)10^{3} × 50 × 10^{−9} × \ln \frac{11.34}{2} = 17.35 μs

\ T = T_{s} + T_{r} = 6.335 + 0.01735 = 6.352 ms
\ f = \frac{1000}{6.352} = 157.43 Hz

(f)\ V_{B1} during charging of C is given as:
\ V_{B1} = V_{BB} × \frac{R_{1}}{(R_{1} + R_{BB})} = \frac{15 × 0.1}{3 + 0.1} = 0.484 V

\ V_{B1} during discharge of C is given by:
\ V_{B1} = (V_{P} − V_{F} ) × \frac{0.1}{0.1 + 0.1} =\frac{(11.34 − 0.7)}{2}  = 5.32 V

\ V_{F} is the diode voltage when ON, Fig. 12.21(d).
(g) Waveforms of the output and\ V_{B1} are shown in Fig.12.21(e).