## User: stanzaa37

stanzaa37 •

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#### Posts by stanzaa37

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...
$R_{L}=800\Omega , R_{f}=40\Omega$
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
stanzaa37 •

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... **Subject:** Electronic Devices and Circuits 1
**Topic:** Unregulated Power Supplies
**Difficulty:** Medium
...

written 8 hours ago by
stanzaa37 •

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... 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
stanzaa37 •

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... **Subject:** Electronic Devices and Circuits 1
**Topic:** Voltage Amplifier
**Difficulty**: Medium
...

written 9 hours ago by
stanzaa37 •

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... 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
stanzaa37 •

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... ![enter image description here][1]
**Subject:** Electronic Devices and Circuits 1
**Topic:** Voltage Amplifier
**Difficulty:** Medium
[1]: https://i.imgur.com/do4pHkr.png ...

written 10 hours ago by
stanzaa37 •

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... 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
stanzaa37 •

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...
![enter image description here][1]
**Subject:** Electronic Devices and Circuits 1
**Topic**: Voltage Amplifier
**Difficulty:** Medium
[1]: https://i.imgur.com/zYU2WO5.png ...

written 11 hours ago by
stanzaa37 •

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... **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
stanzaa37 •

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... **Subject:** Electronic Devices and Circuits 1
**Topic:** Voltage Amplifier
**Difficulty**: Medium
...

written 12 hours ago by
stanzaa37 •

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