Given the relaxation oscillator of Fig. 17.45 : a. Determine RB1 and RB2 at IE = 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. d. Determine the frequency

Question

Given the relaxation oscillator of Fig. 17.45 :

a. Determine[latex]R_{B_{1}}[/latex] and [latex]R_{B_{2}}[/latex] at [latex]I_E = 0 A[/latex].

b. Calculate [latex]V_P [/latex], the voltage necessary to turn on the UJT.

c. Determine whether [latex] R_1 [/latex] is within the permissible range of values as determined by Eq. (17.8) to ensure firing of the UJT.

[latex]\frac{V - V_V}{I_V} \lt R_1 \lt \frac{V-V_P}{I_P}[/latex] (17.8)

d. Determine the frequency of oscillation if [latex]R_{B_{1}}=100 \Omega[/latex] during the discharge phase.

e. Sketch the waveform of [latex]v_{ C }[/latex] for a full cycle.

f. Sketch the waveform of [latex]v_{R_{2}}[/latex] for a full cycle.

Accepted Answer

a. [latex]\eta=\frac{R_{B_{1}}}{R_{B_{1}}+R_{B_{2}}}[/latex]

[latex]0.6=\frac{R_{B_{1}}}{R_{B B}}[/latex]

[latex]R_{B_{1}}=0.6 R_{B B}=0.6(5 k \Omega)= 3 k \Omega[/latex]

[latex]R_{B_{2}}=R_{B B}-R_{B_{1}}=5 k \Omega-3 k \Omega=2 k \Omega[/latex]

b. At the point where[latex]v_{C}=V_{P}[/latex], if we continue with [latex]I_{E}=0 A[/latex], the network of Fig. 17.46 results, where

[latex]V_{P}=0.7 V +\frac{\left(R_{B_{1}}+R_{2} ight) 12 V }{\underbrace{R_{B_{1}}+R_{B_{2}}}_{R_{B B}}+R_{2}}[/latex]

[latex]=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[/latex]

≅ 8 V

c. [latex]\frac{V-V_{V}}{I_{V}}

[latex]\frac{12V- 1V}{10 \ mA}

[latex]1.1 k \Omega

The resistance [latex]R_{1}=50 k \Omega[/latex] falls within this range.

d. [latex]t_{1}=R_{1} C \log _{e} \frac{V-V_{V}}{V-V_{P}}[/latex]

[latex]=(50 k \Omega)(0.1 pF ) \log _{e} \frac{12 V -1 V }{12 V -8 V }[/latex]

[latex]=5 imes 10^{-3} \log _{e} \frac{11}{4}=5 imes 10^{-3}(1.01)[/latex]

= 5.05 ms

[latex]t_{2}=\left(R_{B_{1}}+R_{2} ight) C \log _{e} \frac{V_{P}}{V_{V}}[/latex]

[latex]=(0.1 k \Omega+0.1 k \Omega)(0.1 pF ) \log _{e} \frac{8}{1}[/latex]

[latex]=\left(0.02 imes 10^{-6} ight)(2.08)[/latex]

= 41.6 µs

and [latex]T=t_{1}+t_{2}=5.05 ms +0.0416 ms[/latex]

= 5.092 ms

with [latex]f_{ osc }=\frac{1}{T}=\frac{1}{5.092 ms } \cong 196 H z[/latex]

Using Eq. (17.17) gives

[latex]f \cong \frac{1}{R_{1} C \log _{e}[1 /(1-\eta)]}[/latex]

[latex]=\frac{1}{5 imes 10^{-3} \log _{e} 2.5}[/latex]

= 218 Hz

e. See Fig. 17.47 .

f. During the charging phase, from (Eq. 17.9), we have

[latex]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 V[/latex]

When [latex]v_{C}=V_{P}[/latex], from (Eq. 17.12), we have

[latex]V_{R_{2}} \cong \frac{R_{2}\left(V_{P}-0.7 V ight)}{R_{2}+R_{B_{1}}}=\frac{0.1 k \Omega(8 V -0.7 V )}{0.1 k \Omega+0.1 k \Omega}[/latex]

= 3.65 V

The plot of [latex]v_{R_{2}}[/latex]appears in Fig. 17.48 .

Question 17.1: Given the relaxation oscillator of Fig. 17.45 : a. Determine...