Author Topic: amplifier circuit  (Read 2747 times)

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Offline yalectTopic starter

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amplifier circuit
« on: October 29, 2015, 10:54:43 am »
Hello,
I would like to ask you, how can I calculate the input impedance of the simple following amplifier and its bandwidth?
Thank you
 

Offline tron9000

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Re: amplifier circuit
« Reply #1 on: October 29, 2015, 11:16:47 am »
http://www.electronics-tutorials.ws/amplifier/input-impedance-of-an-amplifier.html

That is a Common Emitter AMp - I googled: input impedance of CE amp
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Offline tron9000

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Re: amplifier circuit
« Reply #2 on: October 29, 2015, 11:40:28 am »
http://www.electronics-tutorials.ws/amplifier/input-impedance-of-an-amplifier.html

That is a Common Emitter AMp - I googled: input impedance of CE amp

That looks like an emitter follower to me, since Rc is much smaller than Re.

Stand corrected, hmm, must have seen 100 and assumed 100K. Principle is still same for Rin.
Quote
I would use a SPICE simulator to solve it

+1. I would recommend SIMETRIX or LTSPICE
« Last Edit: October 29, 2015, 11:42:40 am by tron9000 »
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Online T3sl4co1l

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Re: amplifier circuit
« Reply #3 on: October 29, 2015, 04:27:23 pm »
It'll do both collector and emitter output, having little gain through either (about -0.09 out the collector and 0.99 out the emitter).  It's also "suicide biased" (hFE dependent).

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Offline LvW

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Re: amplifier circuit
« Reply #4 on: October 29, 2015, 05:37:46 pm »
  It's also "suicide biased" (hFE dependent).

Yes - it is a bad biasing scheme. Horowitz/Hill: Don`t use such a circuit.
The reason is as follows: The BJT is a voltage controlled device (VBE controls Ic according to Shockley`s famous formula.).
Now - the resistor RE introduces current-controlled voltage feedback, which only works if the base voltage is as stiff as possible.
That means: This feedback scheme is used in conjunction with a suitable voltage divider at the base (instead of a simple resistor as shown).

Nevertheless, one can compute the input resistance simply based on the feedback rules:
The input resistance (without feedback) is hie=VT/IB  with VT:Temperature voltage, IB:DC base current.
This resistance is increased due to feedback by the factor (1+RE*gm) with transconductance gm=hfe/hie

Therefore: r,in=[hie*(1+RE*gm)]||RB = (hie+hfe*RE)||RB
« Last Edit: October 29, 2015, 06:24:31 pm by LvW »
 


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