hi guys
my lectures is useless he doesnt provide us with any answers to check if ours are right or wrong
can please help me with quation im stuck
diagram here
https://www.facebook.com/photo.php?fbid=10202425516857921&set=o.302689803084787&type=1&theater
(a)Name the two transistor connection configurations. (10 marks)
(b) Calculate a suitable value for resistor R1 so that the biasing is such that the quiescent
emitter voltage of Q2 (VEQ2) is at 7.5V. (60 marks)
(c) Determine the quiescent base current of Q1 (IBQ1) (10 marks)
(d) For an input signal of 1kHz, assuming that the capacitor impedances may be considered
to be very small ( 0), determine the input resistance of your amplifier. (30 marks)
(e) Without any output load connected, determine the overall signal voltage gain of the
amplifier, from I/P to O/P, in dB. (Assume that Ce is not connected.)
my lectures is useless he doesnt provide us with any answers to check if ours are right or wrong
can please help me with quation im stuck
diagram here
https://www.facebook.com/photo.php?fbid=10202425516857921&set=o.302689803084787&type=1&theater
(a)Name the two transistor connection configurations. (10 marks)
(b) Calculate a suitable value for resistor R1 so that the biasing is such that the quiescent
emitter voltage of Q2 (VEQ2) is at 7.5V. (60 marks)
(c) Determine the quiescent base current of Q1 (IBQ1) (10 marks)
(d) For an input signal of 1kHz, assuming that the capacitor impedances may be considered
to be very small ( 0), determine the input resistance of your amplifier. (30 marks)
(e) Without any output load connected, determine the overall signal voltage gain of the
amplifier, from I/P to O/P, in dB. (Assume that Ce is not connected.)
-
I sort of take issue with the implication that you'd be studying for a very similar exam one year later. It's difficult for me to imagine you one year later not being able to solve work that was required a year before. But I guess I don't care.
The 1st active stage in the circuit is a common-emitter amplifier. The 2nd active stage is an emitter-follower. That answers (a). I'm going to dismiss the Early effect in the following analysis.
Start out by calculating the quiescent emitter current of Q₂, based upon the assertion that the Q₂ quiescent emitter voltage is Ve₂ = 7.5V, as stated in (b):
Ie₂ = (7.5V) ⁄ (1kΩ) = 7.5mA
Noting that in the normal active region, the quiescent base and collector currents of Q₂ will then be:
Ib₂ = Ie₂ ⁄ (β+1) ≈ 37.3μA
Ic₂ = (β⋅Ie₂) ⁄ (β+1) ≈ 7.463mA
I have a mental benchmark that a 2N2222A (which has β=200) has a Vbe=0.656V at Ic=1mA. It's something I keep in the back of my mind, for some reason. Anyway, I can now estimate Vbe₂ as:
The 1st active stage in the circuit is a common-emitter amplifier. The 2nd active stage is an emitter-follower. That answers (a). I'm going to dismiss the Early effect in the following analysis.
Start out by calculating the quiescent emitter current of Q₂, based upon the assertion that the Q₂ quiescent emitter voltage is Ve₂ = 7.5V, as stated in (b):
Ie₂ = (7.5V) ⁄ (1kΩ) = 7.5mA
Noting that in the normal active region, the quiescent base and collector currents of Q₂ will then be:
Ib₂ = Ie₂ ⁄ (β+1) ≈ 37.3μA
Ic₂ = (β⋅Ie₂) ⁄ (β+1) ≈ 7.463mA
I have a mental benchmark that a 2N2222A (which has β=200) has a Vbe=0.656V at Ic=1mA. It's something I keep in the back of my mind, for some reason. Anyway, I can now estimate Vbe₂ as:
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