# Question 12.3: For the Miller’s sweep shown in Fig. 12.12(a), VCC = 25 V, R......

For the Miller’s sweep shown in Fig. 12.12(a),$\ V_{CC}$ = 25 V,$\ R_{C2}$ = 5 kΩ,$\ R_{C1}$ = 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:$\ h_{fe}$ = 80,$\ h_{ie}$ = 1kΩ,$\ h_{oe}$ = 1/40 kΩ and$\ h_{re} = 2.5 × 10^{−4}$.

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(a)
$\ V_{s} = \frac{V_{CC}}{R_{C1}C_{s}} × T_{s}$

We have$\ V_{s} = V_{CC}$

Therefore,
$\ T_{s} = R_{C1}C_{s}$            $\ C_{s} = \frac{T_{s}}{R_{C1}}= \frac{5 × 10^{−3}}{10 × 10^{3}}$ = 0.5μF

(b)   Retrace time$\ T_{r} = R_{C2} × C_{s} = 5 × 10^{3} × 0.5 × 10^{−6}$ = 2.5 ms
(c)
$\ A_{I} = \frac{−h_{fe}}{1+ hoeR_{C2}} = \frac{−80}{1 + \frac{5}{40}}$= −71.11

$\ R_{i} = h_{ie} + h_{re}A_{I}R_{C2} = 1 + (2.5 × 10^{−4})(−71.11)(5) = (1 − 0.0889)$ = 0.91 kΩ
$\ A = A_{I}\frac{R_{C2}}{R_{i}}= −71.11 × \frac{5}{0.91}$ = −390.71

$\ e_{s(Miller)} = \frac{V_{s}}{V_{CC}} × \frac{1}{|A|} × \left(1 + \frac{R_{C1}}{R_{i}}\right)= \frac{25}{25} × \frac{1}{390.71} × \left(1 + \frac{10}{0.91}\right)$= 0.0307 = 3.07 %.

Question: 12.1

## Design a relaxation oscillator using a UJT, with VV = 3 V, η = 0.68 to 0.82, IP = 2μA, IV = 1 mA, VBB = 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: [latex...
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: 12.4

## 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...