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User: stanzaa37

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stanzaa370
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Posts by stanzaa37

<prev • 320 results • page 1 of 32 • next >
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Answer: A: A diode having $R_{1}=40\Omega$ is used for half wave rectification. If V=20sin$
... $R_{L}=800\Omega , R_{f}=40\Omega$ i) Secondary voltage is given by $V_{secondary}=V_{ra} sin \omega t$ =20 sin $\omega t$ $v_{m}$=20V $I_{m}=\frac{V_{m}}{R_{L}+R_{F}}=\frac{20V}{800+40}$=23.8mA ........(i) ![enter image description here][1] $I_{dc}=\frac{I_{m}}{\pi}$=7.57mA.............(ii) ...
written 8 hours ago by stanzaa370
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A diode having $R_{1}=40\Omega$ is used for half wave rectification. If V=20sin$\omega t and R_{L}=800\Omega$(i) $I_{m},I_{dc},I_{rms}$ ii)efficiency and iii) DC output voltage
... **Subject:** Electronic Devices and Circuits 1 **Topic:** Unregulated Power Supplies **Difficulty:** Medium ...
edc1(33) written 8 hours ago by stanzaa370
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Answer: A: Design a single stage voltage amplifier so as to give the voltage gain of 150 st
... Let us select the transistor BC147B having $P_{D(max)}$=250mW$ \ \ \ \ $h_{fc(typ)}$=330 $h_{fe}$=4.5k$ \ \ \ \ \ $h_{oe}=30\mu-mho$ **i) Calculation of $R_{C}$** $A_{v}=\frac{-h_{fe}R_{C}}{h_{ic}}$ $\ \ \ \ \ $150=$\frac{330\times R_{C} }{4.5}k\Omega$ $R_{C}=2.045k\Omega$ Select $R_{C} ...
written 9 hours ago by stanzaa370
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Answer: A: The circuit shown in Fig Ex 4.22 uses transistor $h_{fe} h_{fe}=1k\Omega$ and $h
... a) $C_{2}$ determination $X_{C2}=h_{ib}=20\Omega$ $C_{2}=\frac{1}{2\pi f_{1}X_{C2}}=\frac{1}{2\times Hz\times 20\Omega}$ =79.6$\mu F$ (use 80$\mu F$ standard value) .......(i) b) Voltage gain $A_{v}$=$-\frac{-h_{fe}(R_{C}||R_{L})}{h_{ie}}$ $\frac{-50\times 3.3k\Omega}{1k\Omega}$=-165 ....... ...
written 9 hours ago by stanzaa370
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The circuit shown in Fig Ex 4.22 uses transistor $h_{fe} h_{fe}=1k\Omega$ and $h_{fb}=20\Omega$. Circuit lower cutoff frequency (f) is to be 100 Hz. Calculate the following quantities
... ![enter image description here][1] **Subject:** Electronic Devices and Circuits 1 **Topic:** Voltage Amplifier **Difficulty:** Medium [1]: https://i.imgur.com/do4pHkr.png ...
edc1(33) written 10 hours ago by stanzaa370
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Answer: A: Design a single stage CE amplifier to give voltage gain of 40 and the output vol
... The specifications of transistor 2N525 are $P_{D(max)}=225mW$ $h_{fe}=45(typ)$ $h_{c}=1.4 K\Omega \ \ \ \ h_{re}=3.2\times 10^{-4}, \ \ \ \ \ \ \ h_{oe}=25\mu mho$ **i)Calculation of $R_{C}$** The voltage gain is given as $A_{v}=\frac{h_{fe}R_{c}}{h_{fe}+\Delta hR_{c}}$ and $\Delta h= ...
written 10 hours ago by stanzaa370
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Design a single stage CE amplifier to give voltage gain of 40 and the output voltage of 4V(peak) and temperature stability factor S=10 Use transistor 2N525
... ![enter image description here][1] **Subject:** Electronic Devices and Circuits 1 **Topic**: Voltage Amplifier **Difficulty:** Medium [1]: https://i.imgur.com/zYU2WO5.png ...
edc1(33) written 11 hours ago by stanzaa370
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Answer: A: Design single stage CE voltage amplifier as to give the output voltage of $10V_{
... **i)Selection of operating point** The maximum voltage developed across the transistor collector emitter is $V_{CEQ}1+V_{0}$=1+10=11 V Also the quiescent voltage $V_{CEQ}\geq V_{p}+V_{CE}$ $\geq 5+0.3$ $V_{CEQ}\geq5.3 V=6V \ \ \ \ \ .......(i)$ $I_{Lp}=\frac{V_{P}}{R_{L}}=\frac{5}{2K}$= ...
written 11 hours ago by stanzaa370
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