Question 12.1: Design a relaxation oscillator using a UJT, with VV = 3 V, η......
Design a relaxation oscillator using a UJT, with\ V_{V} = 3 V,\ η = 0.68 to 0.82,\ I_{P} = 2μA,\ I_{V} = 1 mA,\ V_{BB} = 20 V, the output frequency is to be 5 kHz. Calculate the typical peak-to-peak output voltage.
The given UJT has the following parameters:
\ V_{V} = 3 V,\ I_{P} = 2 μA,\ I_{V} = 1 mA,\ η = 0.68 to 0.82
\ η_{ave} = \frac{0.68 + 0.82}{2} = 0.75 \ V_{P} = V_{F} + ηV_{BB}
Therefore,
\ V_{P} = 0.7 + (0.75)(20) = 15.7 V
\ R_{max} = \frac{V_{BB} − V_{P}}{I_{P}} = \frac{ 20 − 15.7}{2 μA} = 2.15 MΩ \ R_{min} = \frac{V_{BB} − V_{V}}{I_{V}} = \frac{20 − 3}{1 mA} = 17 kΩ
Thus, R must be in the range 17k Ω to 2.15MΩ. If R is large, C must be very small. Therefore, choose R such that C is not very small.
Choosing R = 22 kΩ
\ T = \frac{1}{f} = \frac{1}{5 × 10^{3} HZ} = 200 μs \ T = RC \ln \frac{V_{BB} − V_{V}}{V_{BB} − V_{P}}
\ C = \frac{T}{R \ln\frac{20 − 3}{20 − 15.7}} = \frac{200 μs}{22 × 10^{3} \ln \frac{17}{4.3}} = 6600 pF
Choose C = 6800 pF (a standard value)
Peak-to-peak sweep amplitude = 15.7 − 3 = 12.7 V.