Question 17.1: Given the relaxation oscillator of Fig. 17.45 : a. Determine...
Given the relaxation oscillator of Fig. 17.45 :
a. DetermineR_{B_{1}} and R_{B_{2}} at I_E = 0 A.
b. Calculate V_P , the voltage necessary to turn on the UJT.
c. Determine whether R_1 is within the permissible range of values as determined by Eq. (17.8) to ensure firing of the UJT.
\frac{V – V_V}{I_V} \lt R_1 \lt \frac{V-V_P}{I_P} (17.8)
d. Determine the frequency of oscillation if R_{B_{1}}=100 \Omega during the discharge phase.
e. Sketch the waveform of v_{ C } for a full cycle.
f. Sketch the waveform of v_{R_{2}} for a full cycle.
a. \eta=\frac{R_{B_{1}}}{R_{B_{1}}+R_{B_{2}}}
0.6=\frac{R_{B_{1}}}{R_{B B}}
R_{B_{1}}=0.6 R_{B B}=0.6(5 k \Omega)= 3 k \Omega
R_{B_{2}}=R_{B B}-R_{B_{1}}=5 k \Omega-3 k \Omega=2 k \Omega
b. At the point wherev_{C}=V_{P}, if we continue with I_{E}=0 A, the network of Fig. 17.46 results, where
V_{P}=0.7 V +\frac{\left(R_{B_{1}}+R_{2}\right) 12 V }{\underbrace{R_{B_{1}}+R_{B_{2}}}_{R_{B B}}+R_{2}}=0.7 V +\frac{(3 k \Omega+0.1 k \Omega) 12 V }{5 k \Omega+0.1 k \Omega}=0.7 V +7.294 V
≅ 8 V
c. \frac{V-V_{V}}{I_{V}}<R_{1}<\frac{V-V_{P}}{I_{P}}
\frac{12V- 1V}{10 \ mA}<R_{1}<\frac{12V-8V}{10 \ \mu A}
1.1 k \Omega<R_{1}<400 k \Omega
The resistance R_{1}=50 k \Omega falls within this range.
d. t_{1}=R_{1} C \log _{e} \frac{V-V_{V}}{V-V_{P}}
=(50 k \Omega)(0.1 pF ) \log _{e} \frac{12 V -1 V }{12 V -8 V }
=5 \times 10^{-3} \log _{e} \frac{11}{4}=5 \times 10^{-3}(1.01)
= 5.05 ms
t_{2}=\left(R_{B_{1}}+R_{2}\right) C \log _{e} \frac{V_{P}}{V_{V}}=(0.1 k \Omega+0.1 k \Omega)(0.1 pF ) \log _{e} \frac{8}{1}
=\left(0.02 \times 10^{-6}\right)(2.08)
= 41.6 µs
and T=t_{1}+t_{2}=5.05 ms +0.0416 ms
= 5.092 ms
with f_{ osc }=\frac{1}{T}=\frac{1}{5.092 ms } \cong 196 H z
Using Eq. (17.17) gives
f \cong \frac{1}{R_{1} C \log _{e}[1 /(1-\eta)]}=\frac{1}{5 \times 10^{-3} \log _{e} 2.5}
= 218 Hz
e. See Fig. 17.47 .
f. During the charging phase, from (Eq. 17.9), we have
V_{R_{2}}=\frac{R_{2} V}{R_{2}+R_{B B}}=\frac{0.1 k \Omega(12 V )}{0.1 k \Omega+5 k \Omega}= 0 . 2 3 5 VWhen v_{C}=V_{P}, from (Eq. 17.12), we have
V_{R_{2}} \cong \frac{R_{2}\left(V_{P}-0.7 V \right)}{R_{2}+R_{B_{1}}}=\frac{0.1 k \Omega(8 V -0.7 V )}{0.1 k \Omega+0.1 k \Omega}= 3.65 V
The plot of v_{R_{2}}appears in Fig. 17.48 .
