10. The input voltage of the potentiometer shown in Fig. 2 is 200 V. To what % should the wiper be adjusted so that the output voltage becomes

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1 mplifier. voltage divider consists of the resistors R kω and R 0 kω. a) What is the put voltage taken away from the R resistor if the put voltage is 30 V? b) What is the voltage ga? c) What is the power ga? (Neglect the difference between R and R.) d) What is the power ga level decibel units?. How many kiloohms should be the R resistor of the voltage divider, if we want to take away 5% of the put voltage from the R 00 Ω resistor? 3. The power ga level of a voltage divider is - 3. (Neglect the difference between R and R.) a) What is the power ga? b) What is the voltage ga? c) How many ohms should be the R resistor if the put voltage is taken away from R kω? 4. For the voltage divider shown Fig. 00 V, R 700 Ω, R 300 Ω. Fd and. 5. For the voltage divider shown Fig. 00 V, R 700 Ω, R 300 Ω. Fd and. 6. For the voltage divider shown Fig. 50 V, R 800 Ω, 75 V. Fd R. Fig. 7. For the voltage divider shown Fig. 50 V, R 800 Ω, 75 V. Fd R. 8. For the voltage divider shown Fig and R kω. Fd R. 9. For the voltage divider shown Fig and R 4 kω. Fd R. 0. The put voltage of the potentiometer shown Fig. is 00 V. To what % should the wiper be adjusted so that the put voltage becomes a) 40 V b) 00 V c) 0 V d) 00 V?. The put voltage of the potentiometer shown Fig. is 0 V. What is the put voltage if the wiper stands at 0%? Fig.

2 . n amplifier amplifies the amplitude of the put signal voltage by one thousand fold. Fd the power ga level. (Suppose R R.) 3. n amplifier amplifies the signal power by a factor of one thousand. Fd a) the voltage ga (suppose R R) and b) the power ga level. 4. By how many times does an amplifier of 43 amplify the signal power? (R R) 5. How many change does the halvg of the signal voltage correspond to? (R R) 6. We double the voltage of a signal with an amplifier. What is the change signal power expressed on the decibel scale? (R R) 7. How many decibels is the power ga level if the put power belongg to the W put power is a) 000 W b) 00 W c) 4 W d) W e) W f) 0. W g) 0 W? 8. To what power does an amplifier amplify the 5 W put signal power if the power ga level for this signal is a) 50 b) 3.7 B c) 0 d) e) 0 f) - g) -? 9. Two harmonic (i.e. susoidal) signals of equal power but different frequencies (f and f) are amplified with an amplifier the power ga level of which is 30 at the f and 7 at the f frequency. What is the power ratio of the amplified signals? 0. Two harmonic signals of equal power but different frequencies (f and f) are amplified with an amplifier the power ga level of which is 50 at the f and 33 at the f frequency. What is the power ratio of the amplified signals?. The power ga level of an amplifier with feedback is 50. What will be the power ga level if % of the put signal voltage is fed back opposite phase? The difference between R and R can be neglected.. The power ga level of an amplifier is 3. It decreases to 0 as a result of negative feedback. What part of the put voltage is fed back? The difference between R and R is to be neglected. 3. The power ga level of an amplifier is 6. It decreases to 0 as a result of negative feedback. What percentage of the put voltage is fed back? The difference between R and R is negligible.

3 4. The voltage ga of an amplifier is 00. 3% of the put voltage is fed back opposite phase with a feedback loop. R and R can be considered equal. a) What is the power ga level with feedback? b) To what value does the voltage ga change as a result of the feedback? c) What will be the power ga level as a result of the negative feedback?

4 Formulæ R I (Ohm's law) I R I R (electric power) R (voltage divider) R R + (defition of voltage ga)! # "! # " $ & % $ & % R R R R R! $ # & R R R " % R R!## " ## $ R R (defition of power ga) n 0nB 0 log 0 log $!!!#!!!" R R (defition of power ga level), NFB + β (voltage ga of amplifier with negative feedback)

5 Solutions R kω. a) 30V 0. 9V R + R kω + 0kΩ 0.9V b) % 30V c) % d) n 0 log 0 log B. R R 00Ω 00Ω Ω R R R R 3. a) 0 n 0 log % 3 b) % R R kω c) R R kω 3.4kΩ 34Ω R + R V; V; R 867 Ω 7. R 343 Ω 8. R 05.3 Ω 9. R 46 kω 0. a) 0% b) 50% c) 0% d) 00%. V. n 0 log 0 log 0 log a) b) n 0 log 0 log log log 0 log log 0log 6.0

6 7. 0 log 0log a) 30 b) 7 c) 3 d) 0 e) - 3 f) - 0 g) log 0log a) 500 kw b) 5 kw c) 50 W d) 6.3 W e) 5 W f) 3.97 W g) 0 W n 0 log 0 log 0 30, 0 0 7, 0 0,, , , n 0 log , + β NFB n 0 log 0 log n 0 log , NFB , NFB β + β, NVCS 0.094

7 3., NFB β β, NVCS a) 0log 0log b) 5, + β NFB c) n 0 log 0 log 5 8