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.

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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.

Question: 12.2

For the circuit shown in Fig. 12.8, it is given that: VYY = 20 V, VZ1 = 6.8 V, VZ2 = 3.8 V, hrb = 3 × 10^−4, hib = 20Ω, hob = 0.5μmhos, α = 0.98 and RE = 1kΩ. Find the slope error: (a) when RL = ∞ (b) when RL = 200 kΩ and (c) when RL = 50 kΩ. ...

$\ V_{EE} = V_{Z1} + V_{Z2}$ = 6.8 + ...
Question: 12.3

For the Miller’s sweep shown in Fig. 12.12(a), VCC = 25 V, RC2 = 5 kΩ, RC1 = 10 kΩ. The duration of the sweep is 5 ms. The sweep amplitude is 25 V. Calculate (a) the value of C; (b) the retrace time and (c) the slope error. The transistor has the following parameters: hfe = 80, hie = 1kΩ, ...

(a) \ V_{s} = \frac{V_{CC}}{R_{C1}C_{s}} × ...
Question: 12.4

The transistor bootstrap circuit in Fig. 12.16(a) has the following parameters, VCC = 15 V, VEE = −10 V, RB = 30 kΩ, R1 = 10 kΩ, RE = 5 kΩ, C1 = 0.005 μF, C3 = 1.0 μF. The input trigger is negative and has an amplitude of 2 V and a width of 60 μs. The transistor parameters are hFE = hfe = 50, ...

Referring to the circuit in Fig. 12.16(a): (a)   S...
Question: 12.5

Design a transistor bootstrap sweep generator to provide an output amplitude of 10 V over a time period of 1 ms. The ramp is to be triggered by a negative going pulse with an amplitude of 5 V, a pulse width of 1 ms and a time interval between the pulses is 0.1ms. The load resistance is 1 kΩ and ...

Refer to the bootstrap circuit shown in Fig. 12.16...
Question: 12.6

AUJT has characteristic as shown in Fig. 12.20(a) and the UJT relaxation oscillator is shown in Fig.12.20(b) Find the values of : (a) Sweep amplitude, (b) the slope and displacement errors, (c) the duration of the sweep. and (Assume η = 0.6 and VF = 0.7 V for silicon.) ...

The waveform of the sweep generator is shown in Fi...
Question: 12.7

Using the characteristic of UJT shown in Fig. 12.21 (a), calculate the values of R, C, R1 and R2 of the relaxation oscillator shown in Fig. 12.21; (b) to generate a sweep with a frequency of 10 kHZ and amplitude of 10V, Tr is 0.5 % of T. ...

Given, $\ f$ = 10 kHz,\ V_{s}[...
Question: 12.8

For the UJT relaxation oscillator shown in Fig.12.21(c), RBB = 3 kΩ, R1 = 0.1 kΩ, η = 0.7, VV = 2V, IV = 10 mA, IP = 0.01 mA. (a) Calculate RB1 and RB2 under quiescent condition (i.e., when IE = 0). (b) Calculate the peak voltage, VP. (c) Calculate the permissible value of R. (d) Calculate the ...

Given$\ R_{BB}$ = 3 kΩ,\ η[/la...
Question: 12.9

The bootstrap sweep circuit is shown in Fig.12.22. A square wave whose amplitude varies between 0 and −4 V and duration 0.5 ms is applied as a trigger. a) Calculate all the quiescent state currents and voltages. b) Determine the sweep amplitude, sweep time and sweep frequency. Assume hFE(min) = 30 ...

(a) Current through$\ R_{1}$ is: [lat...
To calculate$\ R_{1}$: Applying KVL t...